url
string | fetch_time
int64 | content_mime_type
string | warc_filename
string | warc_record_offset
int32 | warc_record_length
int32 | text
string | token_count
int32 | char_count
int32 | metadata
string | score
float64 | int_score
int64 | crawl
string | snapshot_type
string | language
string | language_score
float64 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
http://www.richtherrn.net/physics/lakecomp.htm
| 1,542,251,015,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-47/segments/1542039742338.13/warc/CC-MAIN-20181115013218-20181115034401-00024.warc.gz
| 477,274,092 | 33,738 |
Physics Fun: THE INVESTIGATIONS!
The Study of Mechanics, Energy, Force & Motion
Investigation #1: Zoomerang Coaster
Mass of each car = 1500 pounds or 680 kg
Number of cars = 7
Maximum Height = 36.91 meters Lift 1
35.5 meters Lift 2
Top of Loop = 19.325 meters
Total distance traveled = 286 meters (1 way)
Diameter of loop = __________________
Ride the Zoomerang or watch from the upper midway. Match the statement with the
letter from the photo above. Letters may be used more than once and there may be
more than one answer for each statement. Keep in mind that the Coaster zooms
both frontward and backwards.
_____ 1. Where do you have the greatest potential energy?
_____ 2. Where do you have the greatest kinetic energy?
_____ 3. This location is where you have the greatest velocity.
_____ 4. This location is where you have the slowest velocity.
_____ 5. This is where you feel almost weightless.
_____ 6. This is where you feel very heavy.
_____ 7. Where is work being done?
_____ 8. Where do you feel the greatest deceleration?
_____ 9. Where so you experience the greatest G-Force on you?
_____10. This is where the greatest friction is produced.
Observations and Calculations:
1. How many riders are on the Zoomerang? Find the average
number per ride from 3 runs of the coaster.
1st trip number
2nd trip number
3rd trip number
average number / trip
2. Use your watch or stopwatch to determine how long the ride
time = _________ seconds = _________ minutes = _______ hours.
3. Based on the time for one ride, calculate how many rides could
be run in an 8 hour day.
___________ rides
4. Using the average number of people per trip and the number
of rides per day, how many people could ride the Zoomerang
in:
One Day ? _______ One Season (85 days)? _________
5. Work is the force causing something to be displaced. W = Fd
a. Calculate the work needed to lift the Zoomerang to the
top of the first lift hill. Note: For vertical work here, the force
is the weight of the train and the riders and the distance is the
height Therefore, W = mgh. Use the max number of riders
at 63.5 kg each.
b. Power is the rate of doing work. Calculate the power needed to
lift the train to the top of the first hill.
Power = _________ Watts = __________ horsepower
6. Explain qualitatively the energy transformation in one complete trip.
first peak and then the change from this point to the end of the ride.
7. What measurements must be made to evaluate the maximum
potential energy of the Zoomerang?
a.
b. Carefully make these measurements and record the results here.
c. Calculate the maximum potential energy of the Zoomerang
(include riders). Where is the Zoomerang located when it has
potential energy? What is the speed at this point? (Label and
d. What is the Zoomerang’s maximum kinetic energy? Where is it
located at this time? What is its speed at this point? (Show work)
e. Compare the speed calculation obtained in (d) above with the speed
calculation obtained in the next problem. Comment on any
correlation.
It is said that the speed of a roller coaster as it travels through a
loop depends on the height of the hill from which the coaster has just
descended. The equation s = 8 √h – 2r gives the speed s in feet per
second, where h is the height of the hill and r is the radius of the loop.
Using the data assembled at the start of the lab, determine how fast the
Zoomerang travels through its initial “loop.”
8. a. What force or forces do you feel at the top of the loop?
b. Draw a free body diagram representing the forces acting on you at:
9. What is the minimum speed you can have when upside down and not fall
out? (assuming no restraints). Show your work!
Measurements
Your mass = __kg Time for first car to reach top of first hill
Angle of rise, first hill 0= ‑o Time for first car to travel down from B to C
Sensations (Normal, Heavier, Lighter):
At B,just before descending
At C, bottom of the curve
At D, top of the loop
Observations
1. What is the advantage of a long, shallow first incline?
2. Why is the first hill always the highest?
S. Why is the track of the roller coaster banked?
5. How do you feel at this point?
6. What does the near zero reading tell you about the track at that point?
7. Where does the meter give a maximum reading? Why is it a maximum here?
force meter =
force meter =
force meter =
force meter =
Investigation #2: The Rotor
1. Below are the measurements needed to determine the centripetal force on
you or one of your lab partners. Show forces & velocity on the diagram.
Fill in the values.
2. From the radius, calculate the circumference of the Rotor.
3. Calculate the centripetal acceleration on you, and explain how this relates to
centripetal force. ( ac = v2/r ; v = Circumference divided by the time for
one trip around.)
4. G force is a comparison of the normal force of gravity on you to the force
you experience in an accelerated frame of reference. A force of 1 g is equal
to your normal weight in pounds or newtons.
a. What force holds you to the wall of the Rotor?
b. Explain in detail what this force is and how it is produced.
c. Calculate this force on you. (Show your work.) Fc = mv2/r
d. Calculate the g force acting on you in the Rotor.
5. Draw a free body diagram of the forces acting on you while in the Rotor.
(Label all forces.)
6. a. If you do not slide down the wall, what does this tell you about
the force of friction?
b. What is the normal force in this diagram equal to?
c. Calculate the coefficient of friction (μ )
Investigation #3: The Pirate
Ride capacity (number of Riders)
Approximate weight (full)
Maximum height of the Pirate
Weight of Pirate Empty = 14300 lbs or 6490 kg
1. Explain the energy transformation which occurs when the Pirate is in
operational mode.
2. Calculate the maximum velocity of the Pirate and show where this
3. Measure the period of the Pirate (Use a stop watch and time several
oscillations).
Period =
Calculate the frequency of the Pirate. (Show all work).
4. Draw a free body diagram of the forces acting on you when (a) you
are at the bottom of the swing and (b) at the top of the swing.
5. Using the Pirate as a pendulum, and the information from above,
calculate the acceleration due to gravity at this park. (Show your work).
g =
Investigation #4: The Carousel
on the Carousel. Use two different distances from
the center.
Data:
Trial #1 Trial #2
4. Centripetal force acting on you
1. What effect on the centripetal force did changing your location produce?
2. If you are near the center of the Carousel, explain what strategy you would
use to throw a ball to a partner on the outside edge.
3. If the output of the engine is 25 hp. Calculate the work
required to turn the Carousel once.
(Note: 1 hp = 550 ft-lbs/sec or 746 watts).
Work _________________
Investigation #5: The Wildcat
Data:
Angle of the first hill: _______________ o
Time for train to travel up the first hill: __________________ sec.
Elevation of first hill = 73’7’’ from grade level (__________ meters)
1. Calculate your average speed going up the first hill.
2. What is your potential energy at the top of the first hill? (Show work)
3. How much work was needed to get you to the first hill?
(Show work)
4. What force was used to get you to the top of the hill?
(Show work)
5. The Wildcat track at the bottom of the first hill is at an elevation of
5’ 2’’ from grade level or _____________ meters. (Show all work)
a. How much potential energy is remaining at the bottom of the
first drop?
b. How much kinetic energy do you have at the bottom of the
first drop?
c. Calculate your velocity at the bottom of the first drop.
First Turn: Radius of curvature = _________ feet ( _______ meters)
Elevation = 57’ 9’’ or ( _________ meters)
7. Calculate the centripetal force on you in this turn. (Show your work)
8. Why is the track banked in the curve?
9. What is the g force on you in the turn? (Show your work)
End of Ride: Time to stop: _______________
Braking point elevation = 21’ 1’’ ( _________ meters)
Braking distance = 250 feet ( _________ meters)
10. What is your velocity at the braking point? (Show your work)
Investigation #6A: Wave Swinger
1. Estimate the radius of the circle traveled by a chair in the outer ring as the
ride operates
2. Using the above value, calculate the distance (circumference) traveled by the
chair in one revolution.
3. Calculate the linear speed of the moving chair.
4. Estimate the mass of the chair and the average rider.
5. What is the centripetal force needed to keep the chair with rider moving in a
circle? ( Assume the swing chair has a mass of 9 kg.)
6. Measure or estimate the angle between the chair chains and the
vertical.
7. What is the tension needed in the chain to supply the centripetal
force in Question #5?
8. Diagram the ride at the following times:
a. at rest
b. when it is moving, but not tilted
c. when it is moving and tilted.
9. Determine the length of the entire chain.
10. What causes the swings to move out as the wheel turns?
11. Where does “down” appear to the riders?
12. Describe the reasons for the different sensations on the ride at the
following points:
a. when moving, but not tilted.
b. going down when tilted.
c. going up when tilted.
13. Measure the period of a swing when:
a. moving and not tilted.
b. moving and tilted.
14. How does the angle of an empty swing compare with the angle of an
occupied one at the same radius? Does the mass of the rider seem to
make any difference?
15. Although the hub is rotating at a constant rate, it does not seem that
way when the ride is tilted. Indeed, your tangential velocity is NOT
constant. Why?
16. Determine the tangential speed at which the outer swing is moving
when the hub is moving and tilted. Give the answer for both the bottom
and top of the orbit.
17. Determine the tangential acceleration of the outer swing when the hub
is moving and tilted. Give the answer for both the bottom and top of
the orbit.
18. Find the centripetal force of an empty swing when the hub is moving and
tilted. Give the answer for both the bottom and top of the orbit.
19. The swing angle is the difference of the vector combination of the
gravitational and centripetal forces. Calculate the theoretical angle
the swing should have (when the ride is not tilted), and compare it
with the measured value.
20. Calculate the gravitational, centripetal, and tensional forces acting on
the swing when you are in it. Do this for the following four cases:
a. at rest
b. moving, but not tilted
c. moving, tilted, and at the top of the orbit.
d. moving, tilted, and at the bottom of the orbit.
How do these compare with the same quantities when the swing is
empty?
Investigation #6B: The American Flyers
The American Flyers
are similar to the Wave Swinger in that the cars
swing out away from the axis of rotation during
the ride. The American Flyers are different
because the rider controls the amount of the
swing by positioning the “sail” on the front of
the vehicle.
1. Estimate the mass of the car and passenger.
2. Estimate or find the period of rotation.
3. Estimate the radius of the circle traveled by the “flying car” when
the passengers do not touch or move the “sail.”
4. Estimate the amount of centripetal force needed to keep the vehicle flying
in the circle described in Question #3. Explain.
5. Estimate the amount of centripetal force needed to make the vehicle
fly as far as possible from the ride’s axis.
6. Estimate the amount of centripetal force needed to make the vehicle fly
as close as possible to the ride’s axis.
7. What is the approximate outward force that can be provided by the “sail.”
How did you get this answer?
8. What is the approximate inward force that can be provided by the “sail”?
How did you get this answer?
Investigation #7: Saw Mill Plunge
(A Water Roller Coaster)
Reminder: Use the information provided by your teacher concerning the bench marks for
hard-to-measure locations.
Data and Measurement:
Length of boat: 9 ft = ______ meters Mass of boat: 350 lb = ______ kg
Vertical drop of hill: ______ meters Angle of down hill: ______
Time for whole boat to pass any given
point before going up to top of hill: __________ seconds t1
Time for boat to come down hill: __________ seconds t2
Time duration of the splash at the
bottom of the hill: __________ seconds t3
Time for whole boat to pass any given
point after splashing at bottom of hill: __________ seconds t4
Observations:
1. Why is there water on the slide or hill and not just at the bottom of
the slide?
2. If there is a great deal of mass in the front of the boat, is the splash
larger or smaller than if there is a smaller mass in the front?
If so, explain.
3. Is there an observable splash-time difference with greater mass in the
front than if the greater mass is in the rear? If so, explain.
4. Is there any place on the ride where riders “lunge” forward involuntarily?
Where does this occur? Explain why.
Calculations:
1. Determine the average velocity of the log before going up the hill.
2. Calculate the length of the hill.
3. Determine the average speed of the log down the hill.
4. Assuming the speed of the log at the top of the hill is the same as the speed
before the hill, calculate the speed of the log at the bottom of the hill just prior
to splashing.
5. Calculate the average acceleration of the log going down the hill.
6. Calculate your momentum at the bottom of the hill before splashing.
7. Calculate your momentum after splashing is complete.
8. Using the time of splash, calculate the average force you experience during
the splash.
9. List several purposes of having water as part of this ride.
10. Compare this ride to a roller coaster. What are the similarities?
Investigation #8: Compounce Mt. Skyride
1. Determine the length of the Skyride in meters. Describe the method you used
to determine the total distance (round trip) that a single chair travels during its
circuit.
2. Observe the ride for one full circuit (or ride the ride yourself). How long does
is take for the ride to reach the top of the mountain?
3. How many chairs are on the Skyride? Each chair can carry up to 4 adults.
Using information from your answers to the above questions, what is the
average number of people that can ride the chairlift in an 8 hour day if all
chairs are used?
4. What is the distance between chairs? Explain how you arrived at your answer.
5. The Skyride is a continuously moving attraction. If you are on the chair at the
base of the mountain heading upward, is this potential energy or kinetic
energy?
Investigation #9: Ferris Wheel
1. If you were sitting on a bathroom scale, where on the above diagram would
you see a greater weight than normal?
2. At which position in the above diagram would you see a smaller weight?
3. Estimate the maximum speed of the ride in rpms (revolutions per minute).
4. Does the size of the Wheel affect your perception of its speed? Why or why not?
5. How many gondolas are there on the Ferris Wheel?
6. What are the maximum numbers of passengers that the wheel can carry with
a capacity of 6 adults or of 8 children per gondola?
7. Estimate / calculate the full height of the wheel from its base.
_______ meters
8. Estimate / calculate the radius of the wheel. _______ meters
9. Calculate the circumference of the wheel.
_______ meters
10. If light bulbs are to be placed 6 centimeters apart around the front edge of
the perimeter (circumference) of the wheel, give a close approximation of
how many bulbs would be needed.
_____________
11. Compute the mechanical advantage if the radius of the Ferris Wheel is
12.2 meters and the diameter of the axle is 12.0 inches.
_____________
12. Time 1 complete period (use a particular chair as your starting point). Find
the height for each angle (use triangulation). Make a data table of angles,
times, and heights from starting point. Label time in seconds and height in
meters. Plot 1 period of a time vs. height graph.
Angle time height
0
π/ 4
π /2
3 π /4
π
5π /4
6π /4
7π /4
2 π
13. Write a sine equation for your graph.
14. At what height will the chair used as the starting point be after
Investigation #10: Enterprise
A road is banked to differing
angles and curves based upon
the speed that cars and trucks
will use when traveling the road.
The suspended cars of the
Enterprise will swing out at
some angle when they travel
in a circle. The angle depends
upon the radius of the circular
path and the speed of the wheel.
1. Measure the radius of the wheel.
2. Measure the angle each car makes with the vertical as the wheel
approaches full speed while rotating horizontally. Is each car
uniformly the same angle, regardless of the position around the
wheel?
the ride:
a. at rest
b. at full speed, but while horizontally oriented.
c. at full speed, but at maximum vertical orientation
i. at the top
ii. halfway down
iii. at the bottom
iv. halfway up
4. Record your apparent weight changes (sensations) and compare
with the readings in Question #3.
5. Carefully observe the angle of each car relative to its suspension point
as it goes around when the arm is at its maximum vertical elevation.
Why is it different when approaching the very top from when it is
approaching the very bottom?
6. At what point do you feel the lightest?
The heaviest?
Why is there a difference?
7. Determine the period of motion when the car is rotating at its
maximum rate.
8. Calculate the accelerations and the number of g’s experienced
when:
a. at rest
b. at full speed, but while still horizontally oriented.
c. at full speed, but at maximum vertical orientation:
i. at the top.
ii. halfway down
iii. at the bottom
iv. halfway up
9. Calculate the force the seat exerts on you at the bottom when the ride
is vertical.
10. Draw a force diagram showing all of the forces acting on your body in
each of the following situations:
a. at rest
b. at full speed, but while tilt horizontally oriented
c. at full speed, but at maximum vertical orientation
i. at the top
ii. halfway down
iii. at the bottom
iv. halfway up
11. Knowing the force acting on a rider when the car is rotating at top
speed in a horizontal circle (Question 10b), derive an expression for
calculating the theoretical angle of tilt of the cars at this speed.
12. Using the results from Question 11, calculate the theoretical angle of
tilt of the cars at top speed. Compare and contrast this theoretical
value of tilt with the measured value.
13. Calculate the gravitational, centripetal, and tensional forces acting on
you while you are on this ride. Do this for the following four cases:
a. at rest
b. moving, but not tilted
c. moving, tilted , and at the top of the orbit
d. moving, tilted, and at the bottom of the orbit
Investigation #11: The Bumper Cars
Participate in this investigation with a partner.
1. What happens in a collision to each car when:
a. one bumper car is not moving?
b. a rear-end collision occurs?
c. a head-on collision occurs? (speculate)
d. there is a collision with a stationary object (the side rail)?
e. cars sideswipe?
2. Describe how you feel when any type of collision occurs. Are you a well-packaged passenger? Please explain your answer.
3. How is electrical energy supplied to the bumper cars? Describe the complete circuit for one of the cars.
4. Why do the cars have rubber bumpers?
5. Mass of the bumper car: 385 lbs ___________ kg
Mass of rider (you) ___________ lbs ___________ kg
Mass of car and rider ___________ lbs ___________ kg
Total mass of your partner and his/her car __________ kg
7. Is the mechanical energy (kinetic + potential) of the bumper cars conserved? Please explain your answer.
8. Estimate the average speed of a bumper car in motion.
9. Estimate the stopping distance of a bumper car in an average collision. Try to observe the approximate amount of “give” of a bumper car in a number of different collisions where the car comes close to stopping after the collision.
10. Find an average negative acceleration of a bumper car in an “average” collision. How many g’s is this? (Show your work)
11. Assume that you are traveling at 2 m/s. for momentum mv=mv = impulse (f*t)
a. Calculate the momentum of you and your car.
b. You collide with a wall and rebound at a speed of 1 m/s.
Calculate the momentum of you and your car after the
collision with the wall bumper. (Keep in mind that momentum
is a vector quantity!)
c. Calculate the change in momentum of you and your car.
d. Assume that you are moving at 2 m/s. You strike a wall
bumper and come to a rest in 0.5 seconds. Calculate the impulse
acting on you and your car during the collision.
e. Calculate the force that caused the change in momentum.
Investigation #12: Thunder Rapids Raft Ride
Data and Measurements:
Mass of raft: 681.8 kg
Vertical length of lift conveyor: _________ meters
Time for whole raft to pass any given
point before going up to top of hill: ________ seconds
Time for raft to cycle the route: ________ seconds
Time duration of the raft in the
Time for whole raft to pass any
given point after entering the station
until it drops off the conveyor ________ seconds
Observations:
1. Why is there water on the slide or hill and not just at the bottom of
the slide?
2. If there is a great deal of mass on one side of the raft, is the splash
larger or smaller than if there is a smaller mass on a side?
If so, explain.
3. Is there an observable splash-time difference with the greater mass of
a fully loaded raft than if the greater mass is on one side?
If so, explain.
4. Is there any place on the ride where the riders “lunge” forward
involuntarily? Where does this occur? Explain why.
Calculations:
1. Determine the average velocity of the raft before going up hill.
2. Calculate the length of the conveyor hill.
3. Determine the average speed of the raft up the hill.
4. Assuming the speed of the raft at the top of the hill is the same as the
speed before the hill, calculate the speed of the raft at the end of
5. Calculate the average acceleration of the raft as it leaves the conveyor.
6. Calculate your momentum at the bottom of the trough before
7. What happens to your momentum as water splashes down on you
at Lover’s Rock.
8. List several purposes of having water as part of this ride.
9. Compare this ride to a roller coaster. What are the similarities?
10. What would happen to the time length of the ride if the inflatable tubes
were to be over inflated? Under inflated?
13. _ _ _ ( ( ( ( . . . = = = Boulder Dash
This mountain coaster, new in 2000, is marvel of engineering. It is the longest
wooden roller coaster on the east coast and the only one of its kind, built on a
750 ft. mountain, which forms the western boundary of Lake Compounce Park.
The course is determined by the mountain topography and designed to disturb
as little of the natural setting as possible, including the trees, bushes, ledges,
and boulders.
The coaster ride begins in the north end of the Park near the Ferris Wheel
(located in Bristol) travels to the south end of the Park near the Skyride (located
in Southington) and back again over 4500 plus ft. of track. For a breathtaking
two minutes, guests race through dense woods, past rugged rock facing, and
between large boulders at up to 60 miles per hour.
The unusual design tries to keep the coaster a hill hugger and very fast. The
speed doesn’t change greatly during the ride as with most roller coasters. Heavily
dependent on gravity from the top of the first initial drop on, it maintains a high
range speed throughout the ride. For true coaster lovers (as well as everyone who
dares to ride), a deluxe assortment of other specialties complete the unparalleled
ride. In amusement park lingo, your experience might include sideways jogs,
bunny hops, ejector or floater airtime, laterals, and a feisty 180-degree
turnaround.
In short, Boulder Dash may be one of the coolest psychologically thrilling rides in
the world. Because you are actually riding on a real intact mountain, many
unexpected “blind” surprises may have your hair standing on end.
Now, that you know the scoop, give it a try!
Data:
One train: Mass of each car: 1134 kg
Number of cars: 6
Capacity of one train: _____________
Total distance traveled: 4672 ft. = _________meters
Total time of ride: _________ seconds
Estimate height of first hill: _______ ft = _______meters
Estimate angle of first hill: _______
Time to climb first hill: ___________ seconds
Estimate height of largest drop hill: ______ ft =_____meters
Time to descend largest drop hill: _________ seconds
1. Calculate the distance up the first hill.
2. Calculate your average speed going up the first hill.
3. What is your potential energy at the top of the first hill?
4. How much work was needed to get you to the first hill?
5. What force was used to get you to the top of the hill?
6. From observation: Does relatively the same speed appear to be maintained throughout
the ride ? How about as the rider?
7. Did the speed appear to be faster because of the boulders and trees?
8. Were there any backward leaning zones?
Any forward leaning zones?
9. What percentage of your ride appeared to be airtime?
10. Compare your adventure on Boulder Dash to that of the Wildcat and/or the
Zoomerang.
a. On which coaster did you experience less sideward g forces?
b. On which coaster did you have more airtime?
c. Did you experience differences in speed?
Congratulations! You are now officially a bold, brave, bona fide Boulder Dasher!
14. A heart pumping, adrenaline flowing, white knuckled, and literally hair raising experience, DownTime
is a vertical drop tower with attitude and turbo action.
After the guests are seated, the cart is raised slightly and weighed. Then it is steadily lifted to the top
of the tower where it is locked in brakes. Stationary for a few seconds, the cart is then abruptly
launched toward the ground with chilling negative g-force acceleration. The ride softens with a
bungee like bounce before reaching the bottom of the tower and rebounds for a few soft bounces
before descending slowly back to the ground. Air pressure, power cylinders, pistons, and air powered
brakes work in harmony with each other to provide guests with some exciting ups and downs.
Whether you’re a watcher or a rider, DownTime is an interesting phenomenon to investigate.
1. How many guests can the ride accommodate?
2. Why do you think the cart needs to be weighed?
3. Measure the overall cycle time of the ride from start to finish to gain perspective about
the ride. You’ll need a watch with a second hand or one with a stopwatch mode.
4. Measure the time it takes for the cart to be lifted to the top of the tower. Start measuring at the
end of the weigh sequence.
5. Measure the time of the cart’s turbo descent Start immediately after release of the braking
mechanism. Hint: Don’t look away or you’ll miss it! Stop just as the cart is ready to bounce.
6. Calculate the height of the DownTime tower (including the flagpole) using triangulation. The
distance from the center DownTime tower panel to the right front corner of the retail building
(facing building) is 83.79 feet = ______ meters. (Reminder: The height of the ride = height from
sighting + height of your eye.)
7. Calculate the height of the DownTime tower (excluding the flagpole).
DownTime is 185 feet tall (without the flagpole). The dynamic distance is 165 feet (the distance
through which all the action occurs.) The first bounce occurs about 40 feet up the tower (from
height without flagpole).The empty cart weighs approximately 2000 lbs.
185 ft = _______m 165 ft = _______m 40 ft = _______m 2000 lbs = _______kg
8. Calculate the average speed of the cart moving up the tower (begin end of weigh sequence).
9. Calculate the average speed of the cart moving back down toward the ground (begin release to
just as cart is entering first bounce).
10. Calculate the momentum of cart filled with riders (140 lb average), as it is entering first bounce.
11. Consider the following overview of the DownTime ride cycle, filling in the blanks as you proceed.
A. Load/unload passengers: Cart is at tower bottom. Air pressure in all components except the
air supply tank is at ambient.
B. With the cart lifted slightly and held at constant height, the weight is established. The ride
control system determines the ______________________ required to accomplish the desired
ride action.
C. The cart is dispatched and moves from bottom to top of tower. The “up-valve” admits air into
the power _________ to accomplish this. The air admitted into the cylinders acts on the top
side of the ________ and drives them downward. The passenger cable system lifts the cart in
proportion to ___________ movement. Air is exhausted from the “dump-valve and exhaust”
located on the bottom of each cylinder.
D. The cart is latched in the air-powered ______________ at the top of the tower. As the cart is
held, the air In the upper portion of the cylinders and main valve is vented to atmosphere
through the port valve filters, and calculated ________________ is introduced into the bottom
side of power cylinders and the turbo tank.
E. Cart launch: The brakes release the cart and air pressure accelerates the cart downward with
an initial acceleration of approximately ______ g. The power cylinders top side begins to build
pressure. The power cylinders bottom side begins to dissipate pressure.
F. The cart reaches the bottom of the first bounce about 40 ft up the tower as air is compressed
of approximately _____ g. The air pressure in the upper portion of the power cylinders and
the main valve reaches its ___________. This pressure depends on the _________________.
G. The cart bounces softly several times and descends slowly back to the ground.
14. If DownTime was a free fall drop tower rather than a turbo drop tower, what differences would
you expect to find in the ascent and descent? Consider time, speed, and g forces.
You’re done?! Had fun?! Then you’ve earned some Down Time!
To aid in triangulation measurements, 4 “bench marks” have been placed in the
Park at carefully measured distances from hard-to-measure locations, as follows.
1. A yellow marker can be found on the rail of the fence in front of the
Zoomerang, facing the loop. It is exactly 107’ 7” from a point directly under
the center of the Loop. It is also 185’ from a point under the starting end of
the Track (Lift 1).
2. A nail has been driven into the asphalt directly in front of, and 100’ from
the Pirate. It is at the intersection of perpendicular lines from the lamp post
in front of the Pirate and a nearby lamp post in front of the Twister.
3. To gauge the drop height at the Saw Mill Plunge, a stone marker is in the
grass lawn with trees in front of the water pumps at the lift slope, outside
the fence, and up the stone=walled terrace. The marker is 75 meters
horizontally from a point under the top of the drop hill. The ground at the
marker is 1 m below the water level at the bottom of the drop hill.
4. Another nail has been placed into the asphalt in front of the raised garden
with Carousel horse) in the entrance Plaza. It is aligned with the highest
point of the first hill of the Wildcat roller coaster and is 7’ 4” from the garden
wall when facing and aligned with the flag atop the Wildcat. The nail head is
186’ from a point below the highest elevation of the Wildcat.
| 7,849 | 33,458 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.96875 | 4 |
CC-MAIN-2018-47
|
latest
|
en
| 0.835661 |
https://kr.mathworks.com/matlabcentral/cody/problems/313-pythagorean-perfect-squares-find-the-square-of-the-hypotenuse-and-the-length-of-the-other-side/solutions/113492
| 1,582,682,230,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875146176.73/warc/CC-MAIN-20200225233214-20200226023214-00223.warc.gz
| 443,612,445 | 15,868 |
Cody
# Problem 313. Pythagorean perfect squares: find the square of the hypotenuse and the length of the other side
Solution 113492
Submitted on 16 Jul 2012 by Richard Zapor
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% seed=3;n=6; [Z,y] = findPerfectZ(seed,n); Z_correct = 25; y_correct = 4; assert(isequal(Z,Z_correct)); assert(isequal(y,y_correct));
2 Pass
%% seed=5;n=15; [Z,y] = findPerfectZ(seed,n); Z_correct = 169; y_correct = 12; assert(isequal(Z,Z_correct)); assert(isequal(y,y_correct));
3 Pass
%% seed=4;n=6; [Z,y] = findPerfectZ(seed,n); Z_correct = 25; y_correct = 3; assert(isequal(Z,Z_correct)); assert(isequal(y,y_correct));
4 Pass
%% seed=12;n=15; [Z,y] = findPerfectZ(seed,n); Z_correct = 169; y_correct = 5; assert(isequal(Z,Z_correct)); assert(isequal(y,y_correct));
5 Pass
%% seed=6;n=9; [Z,y] = findPerfectZ(seed,n); Z_correct = 100; y_correct = 8; assert(isequal(Z,Z_correct)); assert(isequal(y,y_correct));
6 Pass
%% seed=8;n=9; [Z,y] = findPerfectZ(seed,n); Z_correct = 100; y_correct = 6; assert(isequal(Z,Z_correct)); assert(isequal(y,y_correct));
| 389 | 1,218 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.671875 | 4 |
CC-MAIN-2020-10
|
latest
|
en
| 0.506932 |
https://www.vedantu.com/question-answer/how-do-you-graph-y54x-class-11-maths-cbse-600a539cf9efda579596f414
| 1,723,634,853,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-33/segments/1722641107917.88/warc/CC-MAIN-20240814092848-20240814122848-00247.warc.gz
| 789,310,486 | 33,564 |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
How do you graph $y=5-4x$?
Last updated date: 12th Aug 2024
Total views: 391.2k
Views today: 9.91k
Verified
391.2k+ views
Hint: The given equation $y=5-4x$ is linear with respect to both the variables $x$ and $y$. So it represents a straight line. Now, the equation is similar to the equation $y=x$. So by transforming the graph of the equation $y=x$, we can obtain the graph of the given equation $y=5-4x$. For this we first need to do the shifting of the graph of $y=x$ to obtain the graph of the equation $y=x+5$. Then we need to do the scaling of the graph of $y=x+5$ to obtain the graph of $y=4x+5$. Finally, on inverting the graph of $y=4x+5$ with respect to the y-axis, we will obtain the graph of $y=-4x+5$ or $y=5-4x$.
Complete step-by-step solution:
The equation to be graphed is given in the question as
$y=5-4x$
Since it is linear in $x$ and $y$, its graph is a straight line. Now, we can obtain its graph from the graph of the equation $y=x$ which is drawn as
Now, let us change the independent variable $x$ in $y=x$ is changed to $x+5$ to obtain $y=x+5$. Since the variable $x$ is changed to $x+5$, so the above graph must be shifted by $5$ units in the negative x-direction to obtain the graph of $y=x+5$ as
Now, we change the independent variable $x$ in $y=x+5$ to $4x$ to obtain $y=4x+5$. Since the variable $x$ is changed to $4x$, so the above graph must be contracted by $4$ units in the horizontal direction to get the graph of $y=4x+5$ as
Finally, to obtain the graph of $y=-4x+5$ from the previous graph of $y=4x+5$, we only need to change the variable $x$ to $-x$ so that the above graph must be inverted with respect to the y-axis to finally obtain the graph of $y=-4x+5$ or $y=5-4x$ as
Hence, we have graphed $y=5-4x$.
Note: We must note that while transforming one graph to obtain the other, the change is reflected in the direction of the variable changed. The method of transformation of graphs is used for obtaining the graphs of many complex equations from the basic equations. However, the simple graph of a straight line can easily be plotted by finding the coordinates of any two points satisfying the given equation of line and joining them together to get the final plot.
| 658 | 2,286 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.84375 | 5 |
CC-MAIN-2024-33
|
latest
|
en
| 0.922986 |
https://www.visanifah.com/2022/09/what-is-cube-of-6.html
| 1,669,453,848,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446706285.92/warc/CC-MAIN-20221126080725-20221126110725-00066.warc.gz
| 1,120,987,124 | 55,719 |
VqI14dIZgOPEqICDVdzsdHohm6R1qA6BYQ86dmeQ
### Search This Blog
Theme images by Igniel
Wall Visanifah
Visit profile
# What Is The Cube Of 6
Side length in feet in centimeters Meters are measured in centimeters, feet, and inches. Volume of a 6-sided cube 216 cubic in. 0.125 cu. ft. 0.0046296 square yards 0.0035396 square meters 3,596,6 cubic centimeters 3.5396 gallons 0.93506 liters (results may be rounded) Formula for Cube Volume A cube's volume equals the length of one side cubed. Volume = 3rd Side
Measuring the cube's side is simple. The calculation result, whether using our volume of a cube calculator or not, will always be in the length unit used, cubed. So, if it was measured in inches, the outcome would be in cubic inches. The answer will be in cubic feet if the length was in feet, and so on for cubic yards, cubic miles, cubic millimeters, cubic centimeters, and cubic meters. If you need to get the volume in a certain unit, you must first convert the length of the side to that unit before substituting it in the calculation. How do you find the volume of a cube?
Six three-centimeter-long wooden cubes are set in a row, face to face. Calculate the total surface area of the solid created in this way. There is no link to explain with a figure. - Surface Areas and Volumes in Math l'= 6 3 = 18 c m b r e a d t h (b) = 3 c m h e I g h t (h) = 3 c m h e I g h t (h) = 3 c m Software update for T S A.5g0035820a
The cube root's properties and practical applications
Cubic roots are helpful for dividing an angle in three, that is, when looking for an angle whose measure is one-third of a given angle. This is known as angle trisection. Cube roots are used to identify the edge of a cube whose volume is twice that of another cube with the same edge.
### What Is The Cube Of 64
A cube is a three-dimensional object with equal-length edges. The product of a cube's three dimensions gives its volume. For example, if the corners of a cube are "a" cm apart, the volume of the cube is given by the product "a an a," which is equal to the third power. The corners of the Rubik's cube depicted are equal to 3 units. As a result, its volume is:a3 = an an a = a3 = 3 3 3 = 27 units (27 cm3) (if we choose our volume unit as cubic centimeters).
Innovative Design - Combines ice cube trays, release pad, ice bucket, sealing cover, and scoop into a single product, allowing you to release all ice cubes with a single push! It increases efficiency and convenience. Food-Grade Material - The ice tray is made of PP + TPE, the release pad is made of PP, the ice bucket is made of PET, and the ice lid is made of ABS resin, making the mold reusable and safe for drinking.
### What Is The Cube Of 625
Find the volume of a cube, for example. The length of one of a cube's sides is the sole variable required to determine its volume. Because both sides are equal, it makes no difference which side is supplied precisely. If the length of a side is 5 inches, the volume equation yields 53 = 5 x 5 x 5 = 125 in3 (cubic inches).
Small prices are quite appealing, but when we combine them with high-quality ecological bikes, there are no businesses that can compete. The battery-powered bicycle has been around for a while and has been more popular in recent years. The biggest benefit is their environmental friendliness, since they emit no carbon dioxide, but they also provide a delightful flight without requiring you to pedal constantly. Cube e bike 625 watt: Get yours at a great price and online.
cube root definition
A cube root of a number an is a number x for which x3 = a, or a number x whose cube equals a. 2 is the cube root of 8 since 23 = 222 = 8, and -2 is the cube root of -8 because (-2)3 = (-2)(-2)(-2) = -8.
### What Is The Cube Of 60
Flexible Outputs Adapt to Any Playing Environment Aside from the Control function, the CUBE-60 has a variety of output possibilities for recording or live performance. Recording/Phones Out, Line Out, and Ext Speaker Out have their own output connectors. Furthermore, the Tuner Out jack allows you to connect an outboard tuner like as Roland's TU-Series.
This is one of the simplest bodies for which the volume can be calculated. You just need one measurement, then multiply it by itself and again, which is comparable to increasing it to the power of three. Finally, convert the unit to cubic equivalent. It's as simple as that. In practice, you seldom know ahead of time if anything is a cube, therefore you may need to measure at least three sides before you can be certain. If you have a blueprint or an engineering schematic with all of the measurements, your job will be much simpler.
Sweetwater has more to offer.
Sweetwater is a one-of-a-kind music gear utopia. One part outstanding customer service. One part zeal. Sweetwater has been dedicated to providing the best buying experience for musicians since 1979. Whether it's our customized advise, a free 2-year guarantee on almost everything we offer, our penchant for rapid, free delivery, our almost "obsessive" attention to detail, free tech support, or the charming way we say "thank you," we've got you covered. These are just a few of the many reasons why more musicians shop at Sweetwater than anyplace else. So, whether you're purchasing your first guitar or constructing your ideal studio, acquire your gear from a store you can rely on, from people who care. Sweetwater has further information.
To guarantee consistency, the seller must specify the condition of each bike sold on buycycle. He is in charge of his evaluation. If the bike comes in a condition that differs from what was advertised, the customer is entitled to a refund. You may choose from the following statuses on buycycle:
Related Posts
| 1,389 | 5,744 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.15625 | 4 |
CC-MAIN-2022-49
|
latest
|
en
| 0.897021 |
https://www.fmaths.com/interesting-about-math/readers-ask-what-is-a-slope-math.html
| 1,620,674,387,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243991759.1/warc/CC-MAIN-20210510174005-20210510204005-00598.warc.gz
| 796,673,049 | 11,293 |
# Readers ask: What Is A Slope Math?
## What does slope mean in math?
The slope of a line is a measure of its steepness. Mathematically, slope is calculated as “rise over run” (change in y divided by change in x).
## How do you find a slope?
The slope of a line characterizes the direction of a line. To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points.
## What does a slope of 0.5 mean?
Definition: The slope of a line is a number that measures its “steepness”, usually denoted by the letter m. Notice that for every increase of one unit to the right along the horizontal x-axis, the line moves down a half unit. It therefore has a slope of -0.5.
## What are 4 types of slopes?
Slopes come in 4 different types: negative, positive, zero, and undefined. as x increases.
## What are the 3 slope formulas?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form. We review all three in this article.
## What is a positive slope?
A positive slope means that two variables are positively related—that is, when x increases, so does y, and when x decreases, y decreases also. Graphically, a positive slope means that as a line on the line graph moves from left to right, the line rises.
You might be interested: What Is Math In Real Life?
## How do I find the slope-intercept form?
Slope – intercept form, y=mx+b, of linear equations, emphasizes the slope and the y- intercept of the line.
## What is a zero slope?
A zero slope is just the slope of a horizontal line! The y-coordinate never changes no matter what the x-coordinate is! In this tutorial, learn about the meaning of zero slope.
## How much is a 1% slope?
1 % as a decimal is 0.01 and hence the slope is 0.01. That means for a run of pipe of a certain length the rise must be 0.01 times the length. Thus for you example, since the length of the run is 80 feet which is 80 × 12 = 960 inches the rise must be 0.01 × 960 = 9.6 inches.
## What is a 1 to 2 slope?
Slope percent is important in figuring how to contour land. A 1 / 2 -percent slope is used for applications that involve drainage of land and concrete, such as patios and sidewalks. It also figures significantly in conservation farming irrigation, where the land most vulnerable to rain erosion gets a 1 / 2 -percent slope.
## How can you tell if a slope is greater than 1?
If you imagined these lines to be hills, you would say that line B is steeper than line A. Line B has a greater slope than line A. Next, notice that lines A and B slant up as you move from left to right. We say these two lines have a positive slope.
x y
− 1 −5
2 10
## What does negative slope look like?
A negative slope means that two variables are negatively related; that is, when x increases, y decreases, and when x decreases, y increases. Graphically, a negative slope means that as the line on the line graph moves from left to right, the line falls.
You might be interested: What Is Math Quiz Bee?
## How do you identify a slope type?
From the previous section, you have discovered that there are four types of slope.
1. postive slope (when lines go uphill from left to right)
2. negative slope (when lines go downhill from left to right)
3. zero slope (when lines are horizontal)
4. undefined slope (when lines are vertical)
## What does an infinite slope look like?
An infinite slope is simply a vertical line. When you plot it on a line graph, an infinite slope is any line which runs parallel to the y-axis. You can also describe this as any line that doesn’t move along the x-axis but stays fixed at one constant x-axis coordinate, making the change along the x-axis 0.
| 905 | 3,756 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.75 | 5 |
CC-MAIN-2021-21
|
latest
|
en
| 0.938181 |
https://artofproblemsolving.com/wiki/index.php?title=Power_set&diff=next&oldid=93024
| 1,604,012,664,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107905965.68/warc/CC-MAIN-20201029214439-20201030004439-00674.warc.gz
| 214,214,750 | 13,483 |
# Difference between revisions of "Power set"
The power set of a given set $S$ is the set $\mathcal{P}(S)$ of all subsets of that set. It is also sometimes denoted by $2^S$.
## Examples
The empty set has only one subset, itself. Thus $\mathcal{P}(\emptyset) = \{\emptyset\}$.
A set $\{a\}$ with a single element has two subsets, the empty set and the entire set. Thus $\mathcal{P}(\{a\}) = \{\emptyset, \{a\}\}$.
A set $\{a, b\}$ with two elements has four subsets, and $\mathcal{P}(\{a, b\}) = \{\emptyset, \{a\}, \{b\}, \{a, b\}\}$.
Similarly, for any finite set with $n$ elements, the power set has $2^n$ elements.
## Size Comparison
Note that for any nonnegative integer $n$, $2^n > n$ and so for any finite set $S$, $|\mathcal P (S)| > |S|$ (where absolute value signs here denote the cardinality of a set). The analogous result is also true for infinite sets (and thus for all sets): for any set $S$, the cardinality $|\mathcal P (S)|$ of the power set is strictly larger than the cardinality $|S|$ of the set itself.
### Proof
There is a natural injection $S \hookrightarrow \mathcal P (S)$ taking $x \mapsto \{x\}$, so $|S| \leq |\mathcal P(S)|$. Suppose for the sake of contradiction that $|S| = |\mathcal P(S)|$. Then there is a bijection $f: \mathcal P(S) \to S$. Let $T \subset S$ be defined by $T = \{x \in S \;|\; x \not\in f(x) \}$. Then $T \in \mathcal P(S)$ and since $f$ is a bijection, $\exists y\in S \;|\; T = f(y)$.
Now, note that $y \in T$ by definition if and only if $y \not\in f(y)$, so $y \in T$ if and only if $y \not \in T$. This is a clear contradiction. Thus the bijection $f$ cannot really exist and $|\mathcal P (S)| \neq |S|$ so $|\mathcal P(S)| > |S|$, as desired.
Note that this proof does not rely upon either the Continuum Hypothesis or the Axiom of Choice. It is a good example of a diagonal argument, a method pioneered by the mathematician Georg Cantor.
## Size for Finite Sets
The number of elements in a power set of a set with n elements is $2^n$ for all finite sets. This can be proven in a number of ways:
### Method 1
Either an element in the power set can have 0 elements, one element, ... , or n elements. There are $\binom{n}{0}$ ways to have no elements, $\binom{n}{1}$ ways to have one element, ... , and $\binom{n}{n}$ ways to have n elements. We add:
$\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n$
as desired.
### Method 2
We proceed with induction.
Let $S$ be the set with $n$ elements. If $n=0$, then $S$ is the empty set. Then
$P(S)=\{\emptyset \}$
and has $2^0=1$ element.
Now let's say that the theorem stated above is true or n=k. We shall prove it for k+1.
Let's say that Q has k+1 elements.
In set Q, if we leave element x out, there will be $2^k$ elements in the power set. Now we include the sets that do include x. But that's just $2^k$, since we are choosing either 0 1, ... or k elements to go with x. Therefore, if there are $2^k$ elements in the power set of a set that has k elements, then there are $2^{k+1}$ elements in the power set of a set that has k+1 elements.
Therefore, the number of elements in a power set of a set with n elements is $2^n$.
### Method 3
We demonstrated in Method 2 that if S is the empty set, it works.
Now let's say that S has at least one element.
For an element x, it can be either in or out of a subset. Since there are n elements, and each different choice of in/out leads to a different subset, there are $2^n$ elements in the power sum.
| 1,064 | 3,477 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 51, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375 | 4 |
CC-MAIN-2020-45
|
latest
|
en
| 0.829019 |
https://math.stackexchange.com/questions/1265285/solving-a-linear-system-in-function-of-a-parameter
| 1,558,593,556,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-22/segments/1558232257156.50/warc/CC-MAIN-20190523063645-20190523085645-00288.warc.gz
| 558,957,371 | 32,264 |
# Solving a linear system in function of a parameter
Problem: Solve the following system in function of the parameter $b$:
\begin{align*} \begin{cases} -bx + 2y - (2+b^2)z + bu &= -2 \\ x -2y + bz -u &= 0 \\ x + (2b-4)y + (2-b)z + (b-1)u &= 2 \\ x -2by -(3b+2)z + (4b-5)u &= 2b-4 \end{cases} \end{align*}
Attempt at solution: We write down the augmented matrix of this system, and then apply the operations: $R_1 \leftrightarrow R_2, R_2 \rightarrow R_2 + b R_1$. This gives us the matrix: \begin{align*} \left(\begin{array}{cccc|c} 1 & -2 & b & -1 & 0 \\ 0 & (2-2b) & (2+b^2) & 0 & -2b \\ 1 & (2b-4) & (2-b) & (b-1) & 2 \\ 1 & -2b & -(3b+2) & (4b+5) & (2b-4) \end{array}\right) \end{align*} After that, we do $R_3 \rightarrow R_3 - R_1$ and $R_4 \rightarrow R_4 - R_1$: \begin{align*}\left(\begin{array}{cccc|c} 1 & -2 & b & -1 & 0 \\ 0 & (2-2b) & (2+2b^2) & 0 & -2b \\ 0 & (2b-2) & (2-2b) & b & 2 \\ 0 & (-2b+2) & (-4b-2) & (4b-4) & (2b-4) \end{array}\right) \end{align*} Now I want a leading $1$ at the position $a_{22}$.
Case 1. Let $b = 1$. Then our matrix reduces to \begin{align*} \left(\begin{array}{cccc|c} 1 & -2 & 1 & -1 & 0 \\ 0 & 0 & 4 & 0 & -2 \\ 0 & 0 & 0 & 1 & 2 \\ 0 & 0 & -6 & 1 & -2 \end{array}\right) \end{align*} From the second row we see than that $u =2$. Substituting this in the last equation gives $z = - \frac{2}{3}$. But from the second row $z = - \frac{1}{2}$, which is a contradiction. Does this mean I can conclude the system has no solutions in this case?
Then if $b \neq 1$, should I just proceed with Gauss-elimination untill I hit another case distinction?
• Do you really need a leading $1$? If the only thing you need is solving the system, then this would be an innecesary complication. However, you're going to need those cases once you finish the reduction. – Daniel May 3 '15 at 21:09
• I don't know. How else could I solve it? – Kamil May 3 '15 at 21:13
• I mean, you can let the $(2-2b)$ as it is, and keep going with the other rows. However, if $2-2b=0$, then the matrix wouldn't be reduced. Don't get confused, you're doing it great, and yes, you must work all the cases as they appear. – Daniel May 3 '15 at 21:14
• Are the entries supposed to be real or complex numbers? If they are real numbers, you can divide row 2 by $a_{2,3}$: $2+2b^2$ because it can never be zero for real entries. By the way... it should probably be $2+b^2$ if you look closely on the equation system before. – mathreadler May 4 '15 at 5:53
| 938 | 2,467 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.6875 | 5 |
CC-MAIN-2019-22
|
latest
|
en
| 0.740545 |
https://fersch.de/beispiel?nr=geradegerade3&nrform=Geogeradegerade3&datei=1006geradegerade3Geogeradegerade3
| 1,618,837,729,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038879374.66/warc/CC-MAIN-20210419111510-20210419141510-00192.warc.gz
| 361,497,253 | 4,883 |
Beispiel Nr: 06
$\text{Gegeben:} \text{Gerade 1: } \vec{x} =\left( \begin{array}{c} a1 \\ a2 \\ a3 \\ \end{array} \right) + \lambda \left( \begin{array}{c} b1 \\ b2 \\ b3 \\ \end{array} \right) \\ \text{Gerade 2: } \vec{x} =\left( \begin{array}{c} c1 \\ c2 \\ c3 \\ \end{array} \right) + \sigma \left( \begin{array}{c} d1 \\ d2 \\ d3 \\ \end{array} \right) \\ \text{Gesucht:} \text{Die Lage der Geraden zueinander.} \\ \\ \textbf{Gegeben:} \\ \text{Gerade 1: } \vec{x} =\left( \begin{array}{c} 1 \\ -2 \\ 8 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 4 \\ -7 \\ -8 \\ \end{array} \right) \\ \text{Gerade 2: } \vec{x} =\left( \begin{array}{c} 9 \\ -5 \\ 3 \\ \end{array} \right) + \sigma \left( \begin{array}{c} -4 \\ -4 \\ -3 \\ \end{array} \right) \\ \\ \\ \textbf{Rechnung:} \\ \text{Gerade 1: } \vec{x} =\left( \begin{array}{c} 1 \\ -2 \\ 8 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 4 \\ -7 \\ -8 \\ \end{array} \right) \\ \text{Gerade 2: } \vec{x} =\left( \begin{array}{c} 9 \\ -5 \\ 3 \\ \end{array} \right) + \sigma \left( \begin{array}{c} -4 \\ -4 \\ -3 \\ \end{array} \right) \\ \text{Richtungsvektoren: } \\ \left( \begin{array}{c} 4 \\ -7 \\ -8 \\ \end{array} \right) =k \cdot \left( \begin{array}{c} -4 \\ -4 \\ -3 \\ \end{array} \right) \\ \begin{array}{cccc} 4&=&-4 k& \quad /:-4 \quad \Rightarrow k=-1 \\ -7&=&-4 k & \quad /:-4 \quad \Rightarrow k=1\frac{3}{4} \\ -8&=&-3 k & \quad /:-3 \quad \Rightarrow k=2\frac{2}{3} \\ \end{array} \\ \\ \Rightarrow \text{Geraden sind nicht parallel} \\ \left( \begin{array}{c} 1 \\ -2 \\ 8 \\ \end{array} \right) + \lambda \left( \begin{array}{c} 4 \\ -7 \\ -8 \\ \end{array} \right) = \left( \begin{array}{c} 9 \\ -5 \\ 3 \\ \end{array} \right) + \sigma \left( \begin{array}{c} -4 \\ -4 \\ -3 \\ \end{array} \right) \\ \begin{array}{cccccc} 1& +4\lambda &=& 9& -4\sigma& \quad /-1 \quad /+4 \sigma\\ -2& -7\lambda &=& -5& -4 \sigma& \quad /+2 \quad /+4 \sigma\\ 8& -8\lambda &=& 3& -3 \sigma& \quad /-8 \quad /+3 \sigma\\ \end{array} \\ \\I \qquad 4 \lambda +4 \sigma =8\\ II \qquad -7 \lambda +4 \sigma = -3 \\ III \qquad -8 \lambda -3 \sigma = -5 \\ \\ \text{Aus 2 Gleichungen }\lambda \text{ und } \sigma \text{ berechnen } \\ I \qquad 4 \lambda +4 \sigma =8 \qquad / \cdot\left(-7\right)\\ II \qquad -7 \lambda +4 \sigma = -3 \qquad / \cdot\left(-4\right)\\ I \qquad -28 \lambda -28 \sigma =-56\\ II \qquad 28 \lambda -16 \sigma = 12 \\ \text{I + II}\\ I \qquad -28 \lambda +28 \lambda-28 \sigma -16 \sigma =-56 +12\\ -44 \sigma = -44 \qquad /:\left(-44\right) \\ \sigma = \frac{-44}{-44} \\ \sigma=1 \\ \sigma \text{ in I}\\ I \qquad -28 \lambda -28 \cdot 1 =-56 \\ -28 \lambda -28 =-56 \qquad / +28 \\ -28 \lambda =-56 +28 \\ -28 \lambda =-28 \qquad / :\left(-28\right) \\ \lambda = \frac{-28}{-28} \\ \lambda=1 \\ \lambda \text{ und } \sigma \text{ in die verbleibende Gleichung einsetzen} \\ III \quad 8+1\cdot\left(-8\right)=3+1\cdot\left(-3\right) \\ 0=0 \\ \lambda \text{ oder } \sigma \text{ in die Geradengleichung einsetzen} \\ \\ \vec{x} = \left( \begin{array}{c} 1 \\ -2 \\ 8 \\ \end{array} \right) +1 \cdot \left( \begin{array}{c} 4 \\ -7 \\ -8 \\ \end{array} \right) \\ \text{Schnittpunkt: }S(5,-9,0) \\ \\$
| 1,377 | 3,206 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.5625 | 5 |
CC-MAIN-2021-17
|
latest
|
en
| 0.141036 |
https://math.stackexchange.com/questions/1784843/closed-form-for-sm-sum-n-1-infty-frac2n-cdot-nm-binom2nn-f
| 1,713,138,741,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816904.18/warc/CC-MAIN-20240414223349-20240415013349-00676.warc.gz
| 367,871,012 | 51,334 |
# Closed form for $S(m) = \sum_{n=1}^\infty \frac{2^n \cdot n^m}{\binom{2n}n}$ for integer $m$?
What is the (simple) closed form for $\large \displaystyle S(m) = \sum_{n=1}^\infty \dfrac{2^n \cdot n^m}{\binom{2n}n}$ for integer $m$?
Notation: $\dbinom{2n}n$ denotes the central binomial coefficient, $\dfrac{(2n)!}{(n!)^2}$.
We have the following examples (all verified by WolframAlpha) for $m\geq 0$:
$$\begin{eqnarray} S(0) &=&\sum_{n=1}^\infty \dfrac{2^n }{\binom{2n}n} = 2+ \dfrac{\pi}2 \\ S(1) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n }{\binom{2n}n} = 3+ \pi \\ S(2) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^2 }{\binom{2n}n} = 11+ \dfrac{7\pi}2 \\ S(3) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^3 }{\binom{2n}n} = 55+ \dfrac{35\pi}2 \\ S(4) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^4 }{\binom{2n}n} = 355 + 113\pi \approx \underline{709.9999}698 \ldots , \quad \text{So close to an integer? Coincidence?} \\ S(5) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^5 }{\binom{2n}n} = 2807+ \dfrac{1787\pi}2 \\ S(6) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^6 }{\binom{2n}n} = 26259+ \dfrac{16717\pi}2 \\ S(7) &=&\sum_{n=1}^\infty \dfrac{2^n \cdot n^7 }{\binom{2n}n} = 283623+ 90280\pi \\ \end{eqnarray}$$
And for $m<0$ as well (all verified by WolframAlpha as well):
$$\begin{eqnarray} S(-1) &=&\sum_{n=1}^\infty \dfrac{2^n \frac 1 n}{\binom{2n}n} = \dfrac{\pi}2 \\ S(-2) &=&\sum_{n=1}^\infty \dfrac{2^n \frac 1 {n^2} }{\binom{2n}n} = \dfrac{\pi^2 }8 \\ S(-3) &=&\sum_{n=1}^\infty \dfrac{2^n \frac 1 {n^3} }{\binom{2n}n} = \pi G - \dfrac{35\zeta(3)}{16} + \dfrac18 \pi^2 \ln2, \quad G \text{ denotes Catalan's constant} \\ \end{eqnarray}$$
So a natural question arise: Is there a closed form for $S(m)$ for all integers $m$?
For what it's worth, I couldn't get the (simple) closed form (using WolframAlpha alone) of $S(-4), S(-5), S(-6), \ldots$ and $S(8), S(9) , \ldots$ without expressing it in terms of hypergeometric functions.
My question is: Is there a closed form of $S(m)$ for all integers $m$ without using hypergeometric functions? And how do I compute all of these values?
Note that I don't consider hypergeometric functions to be a "legitimate" function because it defeats the purpose of this question.
My motivation: I was trying to solve this question and I decided to use the hint suggested by Mandrathrax, that is to use partial fractions to get
$$\dfrac{5n^5+5n^4+5n^3+5n^2-9n+9}{(2n+1)(2n+2)(2n+3)} = \dfrac{5n^2}8 - \dfrac{5n}8 - \dfrac9{n+1} + \dfrac{457}{64(2n+1)} + \dfrac{135}{64(2n+3)} + \dfrac{85}{32}$$
So if I can prove the values of $S(0), S(1), S(2)$ (which I failed to do so), then I'm pretty sure I'm halfway done with my solution.
Why do I still want to solve that question when it already has 21 upvotes? Because I think there's a simpler solution and I personally don't like to use polylogarithms.
My feeble attempt (with help from my friends Aareyan and Julian) to solve my own question: The Taylor series of $(\arcsin x)^2$ is $\displaystyle \dfrac12 \sum_{n=1}^\infty \dfrac1{n^2 \binom{2n}n} (2x)^n$. Differentiating with respect to $x$ then multiply by $x$ (repeatedly) gives some resemblance of $S(m)$, but these series only holds true for $|x| < 1$ and not $x=1$ itself. Now I'm stucked.
EDIT1 (14 May 2016, 1101 GMT): Twice the coefficients of $\pi$ for $S(m)$ with $m\geq0$ appears to follow this OEIS sequence, A014307.
EDIT2 (14 May 2016, 1109 GMT): The constant for $S(m)$ with $m\geq1$ appears to follow this OEIS sequence, A180875.
• It is odd, but I just noticed that the coefficient of $\pi$ for $m=3,5$, when you divide them, it is almost the constant added to the $\pi$ terms divided. Same for $m=5,6$ and $m=2,3$ May 14, 2016 at 10:58
• Why is it that $S(1) = S(-2)$ and $S(2) = S(-3)$ in sum notation? May 14, 2016 at 11:04
• Suppose we solved the formula for the partial sums of $$\sum_{n=1}^p\frac{2^ne^{nx}}{\binom{2n}n}$$If we differentiate this and the partial sum formula, we may be able to produce your problem. Integrate for negative values of course, and take the limit to infinity to solve May 14, 2016 at 11:09
• I do not think this question has anything to do with the riemann-zeta function. And I don't think pattern-recognition is the right tag either. Perhaps the sequences and series tag may be more useful. May 14, 2016 at 11:17
• @GohP.iHan: I updated my initial answer with rather simple generating functions that could interest you. Cheers, Oct 22, 2017 at 17:05
(this answer is about negative values of $$m$$: the power of $$n$$)
We want for $$m$$ any positive integer : $$\tag{1}S_{-m}(x) \equiv \sum_{n=1}^\infty \frac{(2x)^{2n}}{n^m\binom{2n}{n}}$$
Let's start again with : $$\tag{2}S_{-2}(x)=2\,\arcsin(x)^2=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2\binom{2n}{n}}$$ Differentiation and multiplication by $$\dfrac x2$$ returns : $$\tag{3}S_{-1}(x)=\frac x2\,S_2'(x)=2x\,\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n\binom{2n}{n}}$$ While multiplication of $$(1)$$ by $$\dfrac 2x$$ and integration gives us : $$\tag{4}S_{-3}(x)=\int_0^x \frac 4t\,\arcsin(t)^2\,dt=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^3\binom{2n}{n}}$$ This integral may be expressed using polylogarithms but it will be more convenient to write it using (real) Clausen functions : $$\tag{5}\operatorname{Cl}_{2m-1}(x):=\sum_{n=1}^{\infty }\frac{\cos(n\,x)}{n^{2m-1}},\;\operatorname{Cl}_{2m}(x):=\sum_{n=1}^{\infty }\frac{\sin(n\,x)}{n^{2m}}$$ These functions appear in the Fourier series from jump discontinuities (their variants are Bernoulli polynomials) and are obtained from successive integrations of $$\;\displaystyle\operatorname{Cl}_1(x)=-\log\left(2\sin\frac x2\right)\;$$ using $$\tag{6}\displaystyle\operatorname{Cl}_{2m}(x)=\int_0^x \operatorname{Cl}_{2m-1}(t)\,dt,\ \;\operatorname{Cl}_{2m+1}(x)=\zeta(2m+1)-\int_0^x \operatorname{Cl}_{2m}(t)\,dt$$
Let's set $$\;t=\sin(u/2)\,$$ and integrate by parts $$\,\log(2\sin(u/2))\,$$ to get : \begin{align} S_{-3}(x)&=\int_0^x \frac 4t\,\arcsin(t)^2\,dt\\ &=4\int_0^{2\arcsin(x)} (u/2)^2\,\frac{\cos(u/2)}{2\,\sin(u/2)}\,du\\ &=\left.u^2\log(2\sin(u/2))\right|_0^{2\arcsin(x)}-\int_0^{2\arcsin(x)} 2\,u\log(2\,\sin(u/2))\,du\\ \end{align}
This becomes $$\quad\displaystyle S_{-3}(x)=\,-2\log(2x)\,\operatorname{Ls}_2^{(1)}(2\arcsin(x))+2\,\operatorname{Ls}_3^{(1)}(2\arcsin(x))\;$$
using Leonard Lewin's notation $$(7.14)$$ for the generalized log-sine integral $$\;\operatorname{Ls}$$ :
(Lewin $$1981$$ "Polylogaritms and associated functions" and Kalmykov and Sheplyakov's $$2004$$)
$$\tag{7}\operatorname{Ls}_j^{(k)}(x)=-\int_0^x t^k\,\left(\log\left(2\sin\frac t2\right)\right)^{j-k-1}dt,\quad k\ge 0,\ j\ge k+1\\ \text{(or simply }\;\operatorname{Ls}_j(x)\;\text{for k=0)}$$
What makes the rewriting of $$S_{-3}$$ interesting is that it may be generalized to any $$\,S_{-m}\,$$ as shown by Borwein, Broadhurst and Kamnitzer $$2001$$ "Central binomial sums, multiple Clausen values, and zeta values". The derivation is rather clever (little typo : $$\Gamma(n)$$ should be $$\Gamma(k)$$) and will be reproduced in details and slightly generalized as in Kalmykov and Veretin $$2000$$ :
The gamma function is defined by $$\;\displaystyle\Gamma(m):=\int_0^\infty t^{m-1}e^{-t}\,dt=\int_0^\infty (nt)^{m-1}e^{-nt}\,d(nt),\;(n>0)$$
The substitution $$\;t=-\log u\;$$ gives $$\;\displaystyle\Gamma(m)=n^m\int_0^1 (-\log u)^{m-1}\,u^{n-1}\,du\;$$ so that for $$\;u:=y^2$$ :
\begin{align} S_{-m}(x)&=\sum_{n=1}^\infty \frac{(2x)^{2n}}{\binom{2n}{n}\,n^m}\\ &=\sum_{n=1}^\infty \frac{(2x)^{2n}}{\binom{2n}{n}\,\Gamma(m)}\int_0^1 (-2\log y)^{m-1}y^{2n-2}\,d(y^2)\\ &= \frac{(-2)^{m-1}}{(m-1)!}\int_0^1 (\log y)^{m-1}\,\sum_{n=1}^\infty (2x)^{2n}\frac{2\,y^{2n-1}}{\binom{2n}{n}}\,dy\\ &= -\frac{(-2)^{m-1}}{(m-2)!}\int_0^1 \frac{(\log y)^{m-2}}y\,\sum_{n=1}^\infty \frac{(2xy)^{2n}}{n\,\binom{2n}{n}}\,dy,\quad\text{(by parts)}\\ &=-\frac{(-2)^{m-1}}{(m-2)!}\int_0^1 (\log y)^{m-2}\frac{(2x)\,\arcsin(xy)}{\sqrt{1-(xy)^2}}\,dy,\quad\text{(using}\;S_{-1}(xy)\;\\ &=-\frac{(-2)^{m-1}}{(m-2)!}\int_0^{2\arcsin x} \left(\log\left(\frac 1x\sin\frac t2\right)\right)^{m-2}\frac t2\,dt,\quad\text{(setting}\;xy=\sin\frac t2\;\text{)}\\ \tag{8}S_{-m}(x)&= \frac 1{(m-2)!}\int_0^{2\arcsin x}\left[2\log(2x)-2\log\left(2\sin\frac t2\right)\right]^{m-2}\;t\;dt\\ \tag{9} S_{-m}(x)&= - \sum_{j=0}^{m-2} \frac{(-2)^j}{(m-2-j)!j!} \,\left(2 \log(2x)\right)^{m-2-j}\, \operatorname{Ls}_{j+2}^{(1)}\left(2\arcsin x\right)\\ \end{align}
Nan-Yue and Williams gave $$(8)$$ for $$x=\dfrac 12$$ in $$1995$$ "Values of the Riemann zeta function and integrals involving $$\log\left(2\sinh\frac{\theta}2\right)$$ and $$\log\left(2\sin\frac{\theta}2\right)$$".
The general identities $$(8)$$ and $$(9)$$ (with a link to BBK's paper) were given in :
The KV $$2000$$ paper contains too an additional intriguing formula using Nielsen's generalized polylogarithm $$\;\displaystyle \operatorname{S}_{n,p}(z):=\frac {(-1)^{n+p-1}}{(n-1)!p!}\int_0^1 \log^{n-1}(t)\,\log^p(1-zt)\,\frac{dt}t\;$$ that I'll rewrite using $$\,\operatorname{S}_{m-2,1}(z)=\operatorname{Li}_{m-1}(z)\,$$ as : $$\tag{10}S_{-m}(x)=\int_0^1\operatorname{Li}_{m-1}\left((2x)^2\,s(1-s)\right)\,\frac {ds}s$$ This formula is interesting too to evaluate $$\;S_{+m}(x)\;$$ (rational functions are integrated).
$$-$$ Now that $$(9)$$ allows us to express $$\,S_{-m}(x)\,$$ as "generalized log-sine integrals" $$\operatorname{Ls}_m^{(1)}\,$$ what can we do with that?
$$\;\operatorname{Ls}_2^{(0)}(x)\;$$ is simply the "Clausen integral" $$\,\operatorname{Cl}_{2}(x)\,$$ while (from Lewin's book $$1981$$ "Polylogaritms and associated functions" p. $$200$$) $$\,\operatorname{Ls}_{n+2}^{(n)}(x)\,$$ may be written as a sum of Clausen functions :
\begin{align} \frac{(-1)^m}{(2m-2)!}\int_0^x t^{2m-2}\log\left(2\sin\frac t2\right)dt&=\operatorname{Cl}_{2m}(x)+\sum_{k=1}^{m-1}(-1)^{k}\left(\frac{x^{2k-1}}{(2k-1)!}\operatorname{Cl}_{2m-2k+1}(x)+\frac{x^{2k}}{(2k)!}\operatorname{Cl}_{2m-2k}(x)\right)\\ \frac{(-1)^{m-1}}{(2m-1)!}\int_0^x t^{2m-1}\log\left(2\sin\frac t2\right)dt&=\zeta(2m+1)+\sum_{k=0}^{m-1}(-1)^{k}\left(\frac{x^{2k}}{(2k)!}\operatorname{Cl}_{2m-2k+1}(x)+\frac{x^{2k+1}}{(2k+1)!}\operatorname{Cl}_{2m-2k}(x)\right)\\ \end{align}
Concerning your specific choice of $$\;x=\dfrac 1{\sqrt{2}}\;$$ (so that $$\,\displaystyle 2\arcsin(x)=\frac{\pi}2\,$$ and $$\,2\log(2x)=\log(2)$$) some explicit results were given in DK $$2001$$ (appendix A) that I complete here :
\begin{align} \operatorname{Ls}_{2}\left(\frac{\pi}{2}\right) =&\;\beta(2)\\ \operatorname{Ls}_{3}\left(\frac{\pi}{2}\right) =&\;2\,\Im\,\operatorname{Li}_3\left(\tfrac {1+i}2\right)+\beta(2)\log(2)-\tfrac{23}{192}\pi^3-\tfrac 1{16}\pi\log^2(2)\\ \operatorname{Ls}_{4}\left(\frac{\pi}{2}\right) =&\;6\,\Im \,\operatorname{Li}_4\left(\tfrac {1+i}2\right) + 3\,\Im \,\operatorname{Li}_3\left(\tfrac {1+i}2\right)\log(2) - \tfrac 32\beta(4) + \tfrac 34\beta(2)\log^2(2) + \tfrac 34\pi\zeta(3) - \tfrac 1{16}\pi\log^3(2)\\ \hline \operatorname{Ls}_{2}^{(1)}\left(\frac{\pi}{2}\right) =&\;-\tfrac{1}8\pi^2\\ \operatorname{Ls}_{3}^{(1)}\left(\frac{\pi}{2}\right) =&\;\;\tfrac{1}2 \pi\,\beta(2)-\tfrac{35}{32}\zeta(3)\\ \operatorname{Ls}_{4}^{(1)}\left(\frac{\pi}{2}\right) =&\;-\tfrac{5}{96} \log^4(2)+\tfrac{5}{16} \zeta(2) \log^2(2) - \tfrac{35}{32} \zeta(3) \log(2) + \tfrac{125}{32} \zeta(4) + \tfrac{1}{2} \pi \operatorname{Ls}_{3}\left(\tfrac{\pi}{2}\right)- \tfrac{5}{4} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right)\\ \operatorname{Ls}_{5}^{(1)}\left(\frac{\pi}{2}\right) =&\;-\tfrac{1}{16} \log^5(2) + \tfrac{5}{16} \zeta(2) \log^3(2)- \tfrac{105}{128} \zeta(3) \log^2(2) - \tfrac{15}{8} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right) \log(2) - \tfrac{9}{8} \zeta(2) \zeta(3) \nonumber \\ &\quad + \tfrac{1}{2} \pi \operatorname{Ls}_{4}\left(\tfrac{\pi}{2}\right) - \tfrac{1209}{256} \zeta(5) - \tfrac{15}{8} \operatorname{Li}_{5}\left(\tfrac{1}{2}\right) \end{align} allowing us (thanks to Hypergeometric's powerful comments) to extend the OP's table using only the common special functions : \begin{align}S(-4)=&\;-2\pi\,\Im\,\operatorname{Li}_3\left(\tfrac {1+i}2\right) + \tfrac 52\operatorname{Li}_4\left(\tfrac 12\right) + \tfrac{19}{576}\pi^4 + \tfrac 1{48}\pi^2\log^2(2) + \tfrac 5{48}\log^4(2)\\ S(-5)=&\;4 \pi\,\Im\,\operatorname{Li}_4\left(\tfrac{1+i}{2}\right)-\tfrac{5}{2} \operatorname{Li}_5\left(\tfrac{1}{2}\right)+\tfrac 14{\pi ^2 \zeta (3)}-\tfrac{403}{64} \zeta (5)-\pi\,\beta\left(4\right)+\tfrac 1{48}{\log ^5(2)}+\tfrac{1}{144} \pi ^2 \log ^3(2)+\tfrac{19}{576} \pi ^4 \log (2)\\ \end{align} $$-$$ In Davydychev and Kalmykov $$2004$$ "Massive Feynman diagrams and inverse binomial sums" one finds some general results ($$\,z$$ is our $$(2x)^2\,$$) :
For $$\;\displaystyle\theta:=2\,\arcsin\left(\frac{\sqrt{z}}2\right),\ l_{\theta}:=\log\left(2\sin\frac{\theta}{2}\right)\;$$ we have : \begin{align} \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n} &= \theta\,\tan\frac{\theta}{2}&\\ \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n^2} &=\frac{1}{2}\,\theta^2 &\\ \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n^4} &= - 2 \operatorname{Ls}_{4}^{(1)}(\theta)+ 4 l_{\theta} \left[\operatorname{Cl}_3(\theta) + \theta \operatorname{Cl}_2(\theta) - \zeta(3) \right] +\theta^2 l_{\theta}^2 &\\ \end{align}
For the conformal variable $$\;\displaystyle y:=\frac{\sqrt{z-4}-\sqrt{z}}{\sqrt{z-4}+\sqrt{z}}\;$$ we have : \begin{align} \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n} & = \frac{1-y}{1+y}\;\log(y)\\ \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n^2} & = -\frac{1}{2}\;\log^2(y)\\ \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n^3} & = 2 \operatorname{Li}_3(y) - 2\;\log(y)\,\operatorname{Li}_{2}(y) - \log^2(y)\, \log(1-y) + \frac{1}{6}\,\log^3(y)- 2 \zeta(3) \\ \sum_{n=1}^\infty \frac{1}{\left( 2n \atop n\right) } \frac{z^n}{n^4} & = 4 \operatorname{S}_{2,2}(y) - 4 \operatorname{Li}_{4}(y) - 4 \operatorname{S}_{1,2}(y) \log(y) + 4 \operatorname{Li}_{3}(y) \log(1-y) \\ &+ 2 \operatorname{Li}_3(y) \log(y) - 4 \operatorname{Li}_{2}(y) \log(y) \log(1-y) - \log^2(y) \log^2(1-y) \\ &+ \frac{1}{3} \log^3(y)\log(1-y) - \frac{1}{24} \log^4(y) - 4 \log(1-y) \zeta(3) + 2 \log(y) \zeta(3)+ 3 \zeta(4) \end{align}
• THANKYOUUUUU!! It will take me decades to understand everything here. May 30, 2016 at 14:38
• The $z=-1$ case: $$\small \, _5F_4\left(1,1,1,1,1;\frac{3}{2},2,2,2;-\frac{1}{4}\right)=-\text{Li}_4\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right)+16 \text{Li}_4\left(\frac{1}{2} \left(\sqrt{5}-1\right)\right)-8 \text{Li}_3\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right) \text{csch}^{-1}(2)+8 \text{Li}_3\left(\frac{1}{2} \left(\sqrt{5}-1\right)\right) \text{csch}^{-1}(2)-\frac{7 \pi ^4}{45}+\frac{2}{3} \pi ^2 \text{csch}^{-1}(2)^2+\frac{1}{3} \text{csch}^{-1}(2)^3 \left(4 \cosh ^{-1}\left(\frac{3}{2}\right)-17 \text{csch}^{-1}(2)\right)$$ Oct 1, 2020 at 5:44
• A Logsine integral example: $$\small \, _6F_5\left(1,1,1,1,1,1;\frac{3}{2},2,2,2,2;\frac{1}{2}\right)=4 \pi \Im\left(\text{Li}_4\left(\frac{1}{2}+\frac{i}{2}\right)\right)-\frac{5 \text{Li}_5\left(\frac{1}{2}\right)}{2}+\frac{\pi ^2 \zeta (3)}{4}-\frac{403 \zeta (5)}{64}-\frac{1}{256} \pi \zeta \left(4,\frac{1}{4}\right)+\frac{1}{256} \pi \zeta \left(4,\frac{3}{4}\right)+\frac{\log ^5(2)}{48}+\frac{1}{144} \pi ^2 \log ^3(2)+\frac{19}{576} \pi ^4 \log (2)$$ See more in the link. Oct 1, 2020 at 6:01
• Two neat results indeed! The first one may be nicely rewritten using the golden ratio $o$ and its powers (and $\operatorname{csch}^{-1}(2)=\log(o)$) with the last term simply $-3\;\operatorname{csch}^{-1}(2)^4=-3\log(o)^4$. In your second result you may replace the two Hurwitz zeta terms simply by $-\pi\,\beta(4)$ (the Dirichlet beta function which could be useful for higher order generalizations!). Your summary of MZV identities is very interesting too! Oct 4, 2020 at 14:51
• Other arguments: $$\small \, _8F_7\left(1,1,1,1,1,1,1,1;\frac{3}{2},2,2,2,2,2,2;\frac{1}{4}\right)=\frac{17 \pi ^4 \zeta (3)}{810}+\frac{2 \pi ^2 \zeta (5)}{3}-\frac{493 \zeta (7)}{12}+\frac{\pi \left(\psi ^{(5)}\left(\frac{1}{3}\right)-\psi ^{(5)}\left(\frac{2}{3}\right)+\psi ^{(5)}\left(\frac{1}{6}\right)-\psi ^{(5)}\left(\frac{5}{6}\right)\right)}{311040 \sqrt{3}}$$ Oct 7, 2020 at 2:26
UPDATE 2017/10
We want the closed form for $\,\displaystyle S_m(x):=\sum_{n=1}^\infty \frac{n^m (2x)^{2n}}{\binom{2n}{n}}\,$ for nonnegative $m$ and will start with the well known expression for $m=-2$ :
$$\tag{1}F(x):=2\,\arcsin(x)^2=\sum_{n=1}^\infty \frac{(2x)^{2n}}{n^2\binom{2n}{n}}=S_{-2}(x)$$ From this we deduce : \begin{align} F'(x)&=4\,\frac{\arcsin(x)}{\sqrt{1-x^2}}&=4\sum_{n=1}^\infty \frac{(2x)^{2n-1}}{n\binom{2n}{n}}\\ (x\,F'(x))'&=4\,\frac{\frac{\arcsin(x)}{\sqrt{1-x^2}}(1)+x}{1-x^2}&=8\sum_{n=1}^\infty \frac{(2x)^{2n-1}}{\binom{2n}{n}}\\ (x\,(x\,F'(x))')'&=4\,\frac{\frac{\arcsin(x)}{\sqrt{1-x^2}}(1+2x^2)+3x}{(1-x^2)^2}&=16\sum_{n=1}^\infty \frac{n\;(2x)^{2n-1}}{\binom{2n}{n}}\\ &4\,\frac{\frac{\arcsin(x)}{\sqrt{1-x^2}}(1+10x^2+4x^4)+7x+8x^3}{(1-x^2)^3}&=32\sum_{n=1}^\infty \frac{n^2\;(2x)^{2n-1}}{\binom{2n}{n}}\\ &4\,\frac{\frac{\arcsin(x)}{\sqrt{1-x^2}}(1+36x^2+60x^4+8x^6)+15x+70x^3+20x^5}{(1-x^2)^4}&=64\sum_{n=1}^\infty \frac{n^3\;(2x)^{2n-1}}{\binom{2n}{n}}\\ &\cdots \end{align} Of course the idea is to take $\;x=\dfrac 1{\sqrt{2}}\,$ and obtain your results after multiplication by $\dfrac{\sqrt{2}}{2^{m+3}}\,$ if $m$ is the power of $n$ at the numerator.
A pattern seems to emerge from these laborious computations. Let's begin with : $$\tag{2}F_{m-1}(x):=\frac{a(x)P(x)+Q(x)}{(1-x^2)^m}$$ where $\,P(x),\;Q(x)\,$ are two polynomials while $\ a(x):=\dfrac {\arcsin(x)}{\sqrt{1-x^2}}\;$ so that $\ a(x)'=\dfrac{x\,a(x)+1}{1-x^2}$.
The derivative of $\,(x\;F_{m-1}(x))\,$ will then be given by : \begin{align} &=\frac{a(x)P(x)+Q(x)}{(1-x^2)^m}+x\left(\frac{a(x)P(x)+Q(x)}{(1-x^2)^m}\right)'\\ &=\frac{(a(x)P(x)+Q(x))\left(1+\dfrac{2mx^2}{1-x^2}\right)+x\left(a(x)'P(x)+a(x)P(x)'+Q(x)'\right)}{(1-x^2)^m}\\ &=\frac{(a(x)P(x)+Q(x))\left(1+\dfrac{2mx^2}{1-x^2}\right)+\dfrac{x^2\,a(x)+x}{1-x^2}P(x)+x(a(x)P(x)'+Q(x)')}{(1-x^2)^m}\\ &=\frac{(a(x)P(x)+Q(x))\left(1+(2m-1)x^2\right)+(x^2\,a(x)+x)P(x)+(x-x^3)(a(x)P(x)'+Q(x)')}{(1-x^2)^{m+1}}\\ &=\frac{a(x)\left[P(x)\left(1+2mx^2\right)+(x-x^3)P(x)'\right]+xP(x)+\left(1+(2m-1)x^2\right)Q(x)+(x-x^3)Q(x)'}{(1-x^2)^{m+1}}\\ \end{align}
which follows clearly our $(2)$ pattern with the recurrence for the polynomials (starting with $\;P_0(x)=1,\;Q_0(x)=x$) given by : \begin{align} \tag{3}P_m(x)&=P_{m-1}(x)\left(1+2mx^2\right)+(x-x^3)P_{m-1}(x)'\\ Q_m(x)&=x\,P_{m-1}(x)+\left(1+(2m-1)x^2\right)Q_{m-1}(x)+(x-x^3)Q_{m-1}(x)'\\ \end{align} From this recurrence we obtain : $$\tag{4}\boxed{\displaystyle\frac{a(x)P_m(x)+Q_m(x)}{(2\,(1-x^2))^{m+1}}=\sum_{n=1}^\infty \frac{n^m\;(2x)^{2n-1}}{\binom{2n}{n}}=\frac {S_m(x)}{2x}},\quad a(x)=\frac{\arcsin(x)}{\sqrt{1-x^2}}$$ that you may apply to your specific case $\;x=\dfrac 1{\sqrt{2}}\,$ for which $\;a(x)=\dfrac {\pi}{2\sqrt{2}}\,$ and $\,(2\,(1-x^2))=1\,$ (of course the $\sqrt{2}$ terms disappear after multiplication by $\,2x$ ). $$-$$ We didn't use the fact that $P(x),\;Q(x)$ were polynomials. Let's do that and suppose that $P_m(x)=\sum_{k=0}^m p_k\,x^{2k},\;Q_m(x)=\sum_{k=0}^m q_k\,x^{2k+1}$ then $(3)$ becomes :
\begin{align} P_m(x)&=\left(\sum_{k=0}^{m-1} p_k\, x^{2k}\right)\left(1+2mx^2\right)+(x-x^3)\left(\sum_{k=0}^{m-1} p_k\, x^{2k}\right)'\\ Q_m(x)&=x\,\left(\sum_{k=0}^{m-1} p_k\, x^{2k}\right)+\left(1+(2m-1)x^2\right)\left(\sum_{k=0}^{m-1} q_k\, x^{2k+1}\right)+(x-x^3)\left(\sum_{k=0}^{m-1} q_k\, x^{2k+1}\right)'\\ \end{align} Starting with $P_m(x)$ : \begin{align} P_m(x)&=\sum_{k=0}^{m-1} p_k\, x^{2k}+2m\sum_{k=1}^{m} p_{k-1}\, x^{2k} +\sum_{k=0}^{m-1} p_k\, 2k\,x^{2k}-\sum_{k=1}^{m} p_{k-1}\, 2(k-1)\,x^{2k}\\ &=p_0+(2m-2(m-1))\,p_{m-1}x^{2m}+\sum_{k=1}^{m-1} \left(p_k+2m\,p_{k-1}+p_k 2k-p_{k-1}\, 2(k-1)\right)\,x^{2k}\\ &=p_0+2\,p_{m-1}x^{2m}+\sum_{k=1}^{m-1} \left((2k+1)\,p_k+2(m-k+1)\,p_{k-1}\right)\,x^{2k}\\ \end{align}
The coefficients $p_k(m)$ of $P_m(x)$ may thus be obtained from the coefficients $p_k(m-1)$ of $P_{m-1}(x)$ (beginning with $\,p_0(m)=1\,$ since $p_0$ is the only $x^0$ term) : $$\tag{5}p_k(m)=\begin{cases} k=0 & 1 \\ 0<k<m& 2(m-k+1)\,p_{k-1}(m-1)+(2k+1)\,p_k(m-1)\\ k=m & 2\,p_k(m-1)\\ \text{else}& 0 \end{cases}$$ We may illustrate this by showing how the first coefficients $p_k(m)$ (in blue) were obtained
($k$ is indicated as $\,(k)\,$ in its appropriate diagonal) : $$\begin{array} {c|ccccccccccc} m&&&&&&&\color{blue}{p_k}\\ \hline &&&&&&&&(0)\\ 0&&&&&&&\color{blue}{1}\\ &&&&&&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 1}&&\rlap{\LARGE{\searrow}}\scriptsize {\times 2}&&(1)\\ 1&&&&&\color{blue}{1}&&&&\color{blue}{2}\\ &&&&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 1}&&\rlap{\LARGE{\searrow}}\scriptsize {\times 4}&&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 3}&& \rlap{\LARGE{\searrow}}\scriptsize {\times2}&&(2)\\ 2&&&\color{blue}{1}&&&&\color{blue}{10}&&&&\color{blue}{4}\\ &&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 1}&&\rlap{\LARGE{\searrow}}\scriptsize {\times6}&&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 3}&&\rlap{\LARGE{\searrow}}\scriptsize {\times 4}&&\rlap{\LARGE{\swarrow}}\scriptsize{\ \ \ \times 5}&&\rlap{\LARGE{\searrow}}\scriptsize {\times2}&&(3)\\ 3&\color{blue}{1}&&&&\color{blue}{36}&&&&\color{blue}{60}&&&&\color{blue}{8}\\ \end{array}$$ $\qquad\qquad\quad p_2(3)=\color{blue}{60}\,$ for example was obtained as $\;4\times \color{blue}{10}+5\times \color{blue}{4}$.
Concerning $Q_m(x)$ : \begin{align} Q_m(x)&=\sum_{k=0}^{m-1} p_k\, x^{2k+1}+\sum_{k=0}^{m-1} q_k\, x^{2k+1}+(2m-1)\sum_{k=1}^{m} q_{k-1}\, x^{2k+1}\\&\quad+\sum_{k=0}^{m-1} q_k\, (2k+1)\,x^{2k+1}-\sum_{k=1}^{m} q_{k-1}\, (2k-1)\,x^{2k+1}\\ &=(p_0+2q_0)x+\sum_{k=1}^{m-1}\left(p_k+q_k+(2m-1)\,q_{k-1}+q_k\, (2k+1)-q_{k-1}\,(2k-1)\right)\,x^{2k+1}\\ &=(1+2q_0)x+\sum_{k=1}^{m-1}\left(p_k+2(m-k)\,q_{k-1}+2(k+1)\,q_k\right)\,x^{2k+1}\\ \end{align} $$\tag{6}q_k(m)=\begin{cases} k=0 & \begin{cases} m=0 & 1 \\m>0&1+2\,q_0(m-1) \\ \end{cases}\\ 0<k<m& p_k(m-1)+2(m-k)\,q_{k-1}(m-1)+2(k+1)\,q_k(m-1)\\ \text{else}&0 \end{cases}$$
These results were obtained earlier by D.H. Lehmer in "Interesting Series Involving the Central Binomial" (pdf here "télécharger" Lehmer_binom.pdf).
The $p_k(m)$ triangle appears too in OEIS A156919 and in Savage and Viswanathan's paper "The $1/k$-Eulerian Polynomials" (for $k:=2$) and following generating function is provided (Alpha) : \begin{align} \tag{7}\sqrt{\frac {1-y}{\exp(2z(y-1))-y}}&=\sum_{n\ge 0}A_n^{(2)}(y)\frac {z^n}{n!}\\ &=1+z+(1+2y)\frac{z^2}{2!}+(1+10y+4y^2)\frac{z^3}{3!}+\cdots \end{align}
In your specific case $\,y=x^2=\dfrac 12$ this becomes this exponential generating function (e.g.f.) : $$\tag{8}\frac 1{\sqrt{2\exp(-z)-1}}=1+1z+2\frac{z^2}{2!}+7\frac{z^3}{3!}+35\frac{z^4}{4!}+226\frac{z^5}{5!}+\cdots$$ (OEIS A014307: exactly your numerators $\;1,2,7,35,226,\cdots\;$ for the $\dfrac {\pi}2$ terms!) $$-$$
Your integer sequence $\;1,3,11,55,355,\cdots$ is known too (OEIS A180875) but with no indicated generating function. An idea to get this one is to obtain the exponential generating function for the complete $S_m(x)$ terms first (subtracting $\frac {\pi}2$ times $(8)$ should then return the wished e.g.f. and sequence).
Let's multiply $(4)$ by $\;\displaystyle (2x)(2\,(1-x^2))^{m+1}\frac{z^m}{m!}\;$ and sum over $m$ to get : \begin{align} G_x(z)&:=(2x)\sum_{m=0}^\infty \left(a(x)P_m(x)+Q_m(x)\right)\frac{z^m}{m!}\tag{9}\\ &=\sum_{m=0}^\infty S_m(x) \left(2\left(1-x^2\right)\right)^{m+1}\frac{z^m}{m!}\\ &=\sum_{m=0}^\infty\sum_{n=1}^\infty \frac{n^m\;(2x)^{2n}}{\binom{2n}{n}} \left(2\left(1-x^2\right)\right)^{m+1}\frac{z^m}{m!}\\ &=2\left(1-x^2\right)\sum_{n=1}^\infty \frac{(2x)^{2n}}{\binom{2n}{n}} \sum_{m=0}^\infty\frac{\left(2n\left(1-x^2\right)z\right)^m}{m!}\\ &=2\left(1-x^2\right)\sum_{n=1}^\infty \frac{(2x)^{2n}\,\exp\left(2n\left(1-x^2\right)z\right)}{\binom{2n}{n}}\\ &=2\left(1-x^2\right)\sum_{n=1}^\infty \frac{\left(2x\exp\left(\left(1-x^2\right)z\right)\right)^{2n}}{\binom{2n}{n}}\\ &=2\left(1-x^2\right)S_0(u),\quad\text{for}\ \,u:=x\exp\left(\left(1-x^2\right)z\right)\\ G_x(z)&=2\left(1-x^2\right)u\,\frac{\large{\frac{\arcsin(u)}{\sqrt{1-u^2}}}+u}{1-u^2}\tag{10}\\ \end{align} since in our third equation $\,\displaystyle\frac x4(x\,F'(x))'\,$ gives $\;\displaystyle S_0(u)=\sum_{n=1}^\infty \frac{(2u)^{2n}}{\binom{2n}{n}}=u\,\frac{\large{\frac{\arcsin(u)}{\sqrt{1-u^2}}}+u}{1-u^2}$.
From the definition $(9)$ we may then use $(10)$ to evaluate for any $m\in \mathbb{N}$ the terms : $$\tag{11}(2x)\left(a(x)P_m(x)+Q_m(x)\right)=S_m(x)\left(2\left(1-x^2\right)\right)^{m+1}=\left.\left(\frac d{dz}\right)^m\right|_{z=0} G_x(z)$$ (the e.g.f. of the sole $\,S_m(x)$ terms is obtained by preferring $\;u:=x\,\exp(z/2)\;$)
In our specific case $\,x=\dfrac 1{\sqrt{2}}\,$ we have $\,u=\dfrac {\exp(z/2)}{\sqrt{2}}$ and $(10)$ and $(11)$ become : \begin{align} \tag{12}G_{\large{\frac 1{\sqrt{2}}}}(z)&=\frac{\large{u\frac{\arcsin(u)}{\sqrt{1-u^2}}}+u^2}{1-u^2}=\frac{{2\exp(z/2)}\frac{\arcsin(\exp(z/2)/\sqrt{2})}{\sqrt{2-\exp(z)}}+\exp(z)}{2-\exp(z)}\\ S_m\left(\frac{\pi}2\right)&=\frac{\pi}2 P_m\left(\dfrac 1{\sqrt{2}}\right)+\sqrt{2}\,Q_m\left(\dfrac 1{\sqrt{2}}\right)\\ \tag{13}&=\left.\left(\frac d{dz}\right)^m\right|_{z=0} \frac{{2\exp(z/2)}\frac{\arcsin(\exp(z/2)/\sqrt{2})}{\sqrt{2-\exp(z)}}+\exp(z)}{2-\exp(z)}\\ \tag{14}&=\left.\left(\frac d{dz}\right)^{m+1}\right|_{z=0} \frac{{2\exp(z/2)}\arcsin(\exp(z/2)/\sqrt{2})}{\sqrt{2-\exp(z)}}\\ \tag{15}&=\left.\left(\frac d{dz}\right)^{m+2}\right|_{z=0} 2\arcsin^2\left(\frac{\exp(z/2)}{\sqrt{2}}\right)\\ &\text{(integrating twice at the end)}\\ \end{align}
The last equation provides a very nice method to obtain all the $S(m)$ terms in the question
(btw $\,S(0)=1+\frac {\pi}2\,$ rather than $\,2+\frac {\pi}2\,$) :
• expand $\;\displaystyle 2\arcsin^2\left(\frac{\exp(z/2)}{\sqrt{2}}\right)$ in series : $$2\arcsin^2\left(\frac{\exp(z/2)}{\sqrt{2}}\right)=\frac{\pi^2}8+\frac{\pi}2z+\left(\frac {\pi}2+1\right)\frac{z^2}{2!}+\left(\frac{2\pi}2+3\right)\frac{z^3}{3!}+\left(\frac{7\pi}2+11\right)\frac{z^4}{4!}+\left(\frac{35\pi}2+55\right)\frac{z^5}{5!}+\cdots$$
• and compute the second derivative or simply ignore the two first terms and shift by $2$ the remaining ones!
Concerning an e.g.f. for $\;1,3,11,55,\cdots$ we will rewrite $(14)$ and combine it with $(8)$ to get : $$\frac{{2}\arcsin(\exp(z/2)/\sqrt{2})-\large{\frac {\pi}2}}{\sqrt{2\exp(-z)-1}}=1z+3\frac{z^2}{2!}+11\frac{z^3}{3!}+55\frac{z^4}{4!}+355\frac{z^5}{5!}+2807\frac{z^6}{6!}+\cdots$$
For an explicit general solution (instead of the recursive method provided) see this neat paper by the masters :
• Dyson, Frankel and Glasser "Lehmer's Interesting Series"
where they obtain following expression $(20)$ for $\;\displaystyle S_k(z) = \sum_{n=1}^{\infty} \frac{n^kz^n}{{2n \choose n}}\;$ :
$$\tag{16}S_k(z)=\sum_{n=1}^{k+1} n! \left({\frac{z}{4-z}}\right)^n\ E(k,n)\\ \left[\frac{1}{n} + \sum_{p=0}^{n-1} (-1)^p \frac{({\frac{1}{2}})_p}{(p+1)!}\ C^{n-1}_p \left({\frac 4z}\right)^{p+1} \left\{ \sqrt{\frac{z}{4-z}}\arcsin\left({\frac{\sqrt{z}}{2}}\right)- \frac{1}{2} \sum_{l=1}^p \frac{\Gamma(l)}{\left({\frac{1}{2}}\right)_l} \left(\frac{z}{4}\right)^l\right\}\right]$$
with $\;\displaystyle E(k,n) := \frac{(-1)^n}{n!} \sum_{m=1}^n (-1)^m\ C^{n}_m\ m^{k+1}$ the Stirling numbers of the second kind
and $\left({\dfrac{1}{2}}\right)_l$ the Pochhammer symbol.
Comparing this to $(1)$ we see that $z=(2x)^2$ while $k=m$ (of course you want $z=2$ here).
This paper includes too your interesting observation about the ratios approaching $\pi$ (page $12$).
A subsequent paper by Glasser ($2012$) is "A Generalized Apery Series".
• From the last edit of the OP it seems that he already had the link to this paper making all this much less interesting... May 14, 2016 at 13:03
• Way to make you sound less smart May 14, 2016 at 13:04
• @SimpleArt: well the answer was already provided (by Dyson, Frankel and Glasser ) making this de facto less challenging... May 14, 2016 at 13:08
• Extremely detailed. Now what about negative integers $m$? May 18, 2016 at 0:27
• @GohP.iHan: Well the pattern is rather different and increasingly difficult. Another answer should deserve these cases (I'll try...) May 19, 2016 at 22:18
| 12,786 | 28,502 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 73, "wp-katex-eq": 0, "align": 16, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.9375 | 4 |
CC-MAIN-2024-18
|
latest
|
en
| 0.554987 |
http://linksoftvn.com/caesar-cipher-in-python-the-crazy-programmer/
| 1,571,490,029,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986693979.65/warc/CC-MAIN-20191019114429-20191019141929-00471.warc.gz
| 113,417,148 | 30,449 |
Hello everyone, in this tutorial you’ll learn about in Python. If you have learned about cryptography then you should have known this term cipher. Well if you don’t know what is this then let me explain it to you.
What is Caesar Cipher?
In cryptography, Caesar cipher is one of the simplest and most widely known encryption techniques. It is also known with other names like Caesar’s cipher, the shift cipher, Caesar’s code or Caesar shift. This encryption technique is used to encrypt plain text, so only the person you want can read it. The method is named after Julius Caesar, who used it in his private correspondence.
In this encryption technique, to encrypt our , we have to replace each letter in the text by a some other letter at a fixed . Let’s say, there is a letter ‘T’ then with a right shift of 1 it will be ‘U’ and with a left shift of 1 it will become ‘S’. So here, the is 1 and the direction will also be same for a text. Either we can use left shift or right, not both in same text.
Let’s understand it with an example.
Example
Suppose we have text “the crazy programmer” to be encrypted. Then what we can do is replace each of letter present in the text by a another letter having fixed difference. Lets say we want right shift by 2 then each letter of the above text have to replaced by the letter, positioned second from the letter.
Plaintext: the crazy programmer
Ciphertext: vjg etcba rtqitcoogt
Now can’t read this text until he/she have the decrypt . Decrypt is nothing just the knowledge about how we shifted those letters while encrypting it. To decrypt this we have to left shift all the letters by 2.
That was the basic concept of Caesar cipher. If we see this encryption technique in mathematical way then the formula to get encrypted letter will be:
c = (x + n) mod 26
where, c is place value of encrypted letter,
x is place value of actual letter,
n is the number that shows us how many positions of letters we have to replace.
On other hand, to decrypt each letter we’ll use the formula given below:
c = (x – n) mod 26
## Program for Caesar Cipher in Python
Output:
enter string: the crazy programmer
enter shift number: 2
original string: the crazy programmer
after encryption: vjg etcba rtqitcoogt
So in above program we have used the same formula (with some modification) we mentioned above. But in computer science ‘A’ is different from ‘a’ thats why we have to write that formula twice, (for uppercase and lowercase letters).
As you can see in the program we have added and subtracted 65 (for Uppercase) and 97 (for lowercase) in that mathematical formula because the ascii value of ‘A’ is 65 and of ‘a’ is 97. The ord() method is used to get the ascii value of the letters.
Note 1: if you want left shift instead of right then please enter a negative number in ‘enter shift number: ’.
Note 2: the above program will work only for Python 3.x because input() method works different in both Python 2 and 3. To use the above program in Python 2, use raw_input() in place of input() method.
To decrypt this message, we will use the same above program but with a modification.
cipher = cipher + chr((ord(char) – shift – 65) % 26 + 65)
If you’ve any problem or suggestion related to caesar cipher in python then please let us know in comments.
SHARE
Next article10 best privacy apps for Android
| 790 | 3,359 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.640625 | 4 |
CC-MAIN-2019-43
|
latest
|
en
| 0.918361 |
http://math.colgate.edu/math102/Common/examples/ANOVA/anovanc.html
| 1,386,796,945,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2013-48/segments/1386164047228/warc/CC-MAIN-20131204133407-00016-ip-10-33-133-15.ec2.internal.warc.gz
| 138,796,586 | 2,058 |
A researcher would like to find out whether a man's nickname affects his cholesterol reading (though it is not clear why she believes it should). She records the cholesterol readings of 23 men nicknamed Sam, 24 men nicknamed Lou and 19 men nicknamed Mac; her data appears in the table to the right. She wants to know whether the differences in the average readings are significant; i.e., whether the average reading of all men nicknamed Sam is different from the average reading of all Lous or whether these averages differ from the average reading of all Macs.
SamLouMac
364260156
245204438
284221272
172285345
198308198
239262137
259196166
188299236
256316168
263216269
329155296
136212236
272201275
245175269
209241142
298233184
342279301
217368262
358413258
412240
382243
593325
261156
280
Using the ANOVA formulas, we have:
• I = 3;
• n1 = 23, n2 = 24, and n3 = 19, and x1,1 = 364, x1,2 = 260, x1,3 = 156, x2,1 = 245, etc.;
• N = 23 + 24 + 19 = 66;
• AV1 = 283.6, AV2 = 253.7, AV3 = 242.5 (at least approximately -- all figures except the degrees of freedom are approximate from here on);
• AV = [23(283.6) + 24(253.7) + 19(242.5)]/66 = 260.9;
• SSE = (364 - 283.6)2 + (260 - 253.7)2 + (156 - 242.5)2 + (245 - 283.6)2 + . . . (a total of 66 terms) = 406,259.7;
• SSG = 23(283.6 - 260.9)2 + 24(253.7 - 260.9)2 + 19(242.5 - 260.9)2 = 19,485.3;
• DFE = 66 - 3 = 63, and so MSE = 6448.6;
• DFG = 3 - 1 = 2, and so MSG = 9742.7; and hence, finally,
• F = 1.51.
And by consulting an F-distribution table with DFG = 2 and DFE = 63, we find that the probability of a F-value of at least is about 23%, not small enough to conclude statistical significance. We do not reject the null hypothesis and conclude that the average cholesterol readings of all Sams, all Lous and all Macs are not different.
Here is a screenshot of an Excel spreadsheet in which these computations are done.
| 634 | 1,880 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.859375 | 4 |
CC-MAIN-2013-48
|
longest
|
en
| 0.84738 |
https://cracku.in/57-a-plot-of-land-in-the-form-of-a-rectangle-has-a-di-x-mat-2006
| 1,718,864,961,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861883.41/warc/CC-MAIN-20240620043158-20240620073158-00332.warc.gz
| 154,549,948 | 25,013 |
Question 57
# A plot of land in the form of a rectangle has a dimension 240 m × 180 m. A drainlet 10m wide is dug all around it (outside) and the earth dug out is evenly spread over the plot, increasing its surface level by 25 cm. The depth of the drainlet is
Solution
Volume of the plot = 240 m x 180 m x 0.25 m = 10800 m$$^{3}$$
($$\because$$ The earth dug out is evenly spread over the plot, increasing its surface level by 25 cm)
Total volume of the drainlet is given by,
= (2 x 260 x 10 x h) + (2 x 180 x 10 x h) m$$^{3}$$
Volume of the plot is equal to the volume of the drainlet,
10800 m$$^{3}$$ = (2 x 260 x 10 x h) + (2 x 180 x 10 x h) m$$^{3}$$
10800 = h(5200 + 3600)
10800 = h(8800)
h = 1.227
Hence, option C is the correct answer.
• All Quant Formulas and shortcuts PDF
• 40+ previous papers with solutions PDF
| 280 | 836 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.03125 | 4 |
CC-MAIN-2024-26
|
latest
|
en
| 0.764539 |
https://www.jiskha.com/display.cgi?id=1389515012
| 1,500,779,853,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-30/segments/1500549424239.83/warc/CC-MAIN-20170723022719-20170723042719-00594.warc.gz
| 794,750,288 | 3,623 |
# ALGEBRA
posted by .
From a survey of 100 college students, a marketing research company found that 35 students owned iPods, 25 owned cars, and 5 owned both cars and iPods
How many students owned either a car or an iPod (but not both)?
How many students do not own either a car or an iPod?
236 Incorrect: Your answer is incorrect. .
• ALGEBRA -
Draw a Venn diagram
overlap between cars and Ipods is 5
so Ipod alone = 35-5 = 30
and car alone = 25-5 = 20
so Ipod or car but not both = 30+20 = 50
Ipod and car = 35
car alone = 20
sum = 55
100 - 55 = 45 own nothing
### Respond to this Question
First Name School Subject Your Answer
### Similar Questions
More Related Questions
Post a New Question
| 202 | 707 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.625 | 4 |
CC-MAIN-2017-30
|
longest
|
en
| 0.948371 |
https://web2.0calc.com/questions/please-help_72422
| 1,718,231,352,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861261.53/warc/CC-MAIN-20240612203157-20240612233157-00005.warc.gz
| 565,456,820 | 6,290 |
+0
# Please help!
0
32
5
+25
Rosencrantz and Guildenstern mix red, blue, and white paint in the following ratio:$$\text{red}:\text{blue}:\text{white}=3:2:1.$$They have $48$ cans of red paint, $40$ cans of blue paint, and $15$ cans of white paint. The cans are all the same size. What is the greatest number of cans of mixed paint they can make with the same ratio of colors, and why? How much paint of each color will be left over?
Jun 9, 2023
### 5+0 Answers
#1
+1
We can find the greatest number of cans of mixed paint they can make by finding the greatest common factor of the number of cans of each color. The greatest common factor of 48, 40, and 15 is 12. Therefore, they can make 12 cans of mixed paint. They will have 48−12=36 cans of red paint left over, 40−12=28 cans of blue paint left over, and 15−12=3 cans of white paint left over.
Jun 9, 2023
#2
+1
3 + 2 + 1 ==6 sum of all ratios
48 + 40 + 15 ==103 cans of all 3 colors.
103 / 6 ==17+1/6 cans of any one color.
But we have maximum of 15 cans of white paint, therefore:
15 x 6 ==90 cans be mixed up.
15 cans of red paint ==15 x 3 ==45 cans of red paint
15 cans of blue paint==15 x 2==30 cans of blue paint
15 cans of white paint ==15 x 1==15 cans of white paint.
So, we mix: 45 red + 30 blue + 15 white ==90 cans in total
48 - 45 ==3 cans of red paint left over
40 - 30 ==10 cans of blue paint left over
15 - 15 ==0 cans of white paint left over.
Jun 9, 2023
#3
+1
Rosencrantz and Guildenstern can make the greatest number of cans of mixed paint if they use all of the red paint, because it is the most. They can use 48 cans of red paint, which is 12 groups of 4 cans.
To make the same ratio of colors, they need to use 8 groups of 4 cans of blue paint and 3 groups of 4 cans of white paint. This will leave them with 20 cans of blue paint and 12 cans of white paint.
Therefore, the greatest number of cans of mixed paint they can make is 12.
Jun 9, 2023
#4
+1
Since the ratio of red paint to blue paint to white paint is 3:2:1, the greatest number of cans of mixed paint Rosencrantz and Guildenstern can make is the least common multiple of 3, 2, and 15, or 30 cans. To make the greatest number of cans, Rosencrantz and Guildenstern will use 30 cans of red paint, 20 cans of blue paint, and 10 cans of white paint. The amount of red, blue, and white paint left over is 48−30=18 cans, 40−20=20 cans, and 15−10=5 cans, respectively. Therefore, Rosencrantz and Guildenstern will have 18 cans of red paint, 20 cans of blue paint, and 5 cans of white paint left over.
To see why 30 is the greatest number of cans of mixed paint Rosencrantz and Guildenstern can make, notice that if they make 29 cans of mixed paint, they will have 1 can of red paint left over, and if they make 31 cans of mixed paint, they will have 2 cans of white paint left over.
Jun 10, 2023
#5
+25
0
thanks
Jun 14, 2023
| 887 | 2,892 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.5 | 4 |
CC-MAIN-2024-26
|
latest
|
en
| 0.871465 |
https://www.flightliteracy.com/aircraft-performance-charts-part-two/
| 1,585,440,359,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585370493121.36/warc/CC-MAIN-20200328225036-20200329015036-00375.warc.gz
| 942,877,098 | 14,823 |
# Aircraft Performance Charts (Part Two)
## Climb and Cruise Charts
Climb and cruise chart information is based on actual flight tests conducted in an aircraft of the same type. This information is extremely useful when planning a cross-country flight to predict the performance and fuel consumption of the aircraft. Manufacturers produce several different charts for climb and cruise performance. These charts include everything from fuel, time, and distance to climb to best power setting during cruise to cruise range performance.
The first chart to check for climb performance is a fuel, time, and distance-to-climb chart. This chart gives the fuel amount used during the climb, the time it takes to accomplish the climb, and the ground distance that is covered during the climb. To use this chart, obtain the information for the departing airport and for the cruise altitude. Using Figure 11-25, calculate the fuel, time, and distance to climb based on the information provided.
Figure 11-25. Fuel, time, and distance climb chart. [click image to enlarge]
Sample Problem 4
Departing Airport Pressure Altitude 6,000 feet Departing Airport OAT 25 °C Cruise Pressure Altitude 10,000 feet Cruise OAT 10 °C
First, find the information for the departing airport. Find the OAT for the departing airport along the bottom, left side of the graph. Follow the line from 25 °C straight up until it intersects the line corresponding to the pressure altitude of 6,000 feet. Continue this line straight across until it intersects all three lines for fuel, time, and distance. Draw a line straight down from the intersection of altitude and fuel, altitude and time, and a third line at altitude and distance. It should read three and one-half gallons of fuel, 6 minutes of time, and nine NM. Next, repeat the steps to find the information for the cruise altitude. It should read six gallons of fuel, 10.5 minutes of time, and 15 NM. Take each set of numbers for fuel, time, and distance and subtract them from one another (6.0 – 3.5 = 2.5 gallons of fuel). It takes two and one-half gallons of fuel and 4 minutes of time to climb to 10,000 feet. During that climb, the distance covered is six NM. Remember, according to the notes at the top of the chart, these numbers do not take into account wind, and it is assumed maximum continuous power is being used.
The next example is a fuel, time, and distance-to-climb table. For this table, use the same basic criteria as for the previous chart. However, it is necessary to figure the information in a different manner. Refer to Figure 11-26 to work the following sample problem.
Figure 11-26. Fuel time distance climb.
Sample Problem 5
Departing Airport Pressure Altitude Sea level Departing Airport OAT 22 °C Cruise Pressure Altitude 8,000 feet Takeoff Weight 3,400 pounds
To begin, find the given weight of 3,400 in the first column of the chart. Move across to the pressure altitude column to find the sea level altitude numbers. At sea level, the numbers read zero. Next, read the line that corresponds with the cruising altitude of 8,000 feet. Normally, a pilot would subtract these two sets of numbers from one another, but given the fact that the numbers read zero at sea level, it is known that the time to climb from sea level to 8,000 feet is 10 minutes. It is also known that 21 pounds of fuel is used and 20 NM is covered during the climb. However, the temperature is 22 °C, which is 7° above the standard temperature of 15 °C. The notes section of this chart indicate that the findings must be increased by ten percent for each 7° above standard. Multiply the findings by ten percent or .10 (10 × .10 = 1, 1 + 10 = 11 minutes). After accounting for the additional ten percent, the findings should read 11 minutes, 23.1 pounds of fuel, and 22 NM. Notice that the fuel is reported in pounds of fuel, not gallons. Aviation fuel weighs six pounds per gallon, so 23.1 pounds of fuel is equal to 3.85 gallons of fuel (23.1 ÷ 6 = 3.85).
The next example is a cruise and range performance chart. This type of table is designed to give TAS, fuel consumption, endurance in hours, and range in miles at specific cruise configurations. Use Figure 11-27 to determine the cruise and range performance under the given conditions.
Figure 11-27. Cruise and range performance.
Sample Problem 6
Pressure Altitude 5,000 feet RPM 2,400 rpm Fuel Carrying Capacity 38 gallons, no reserve
Find 5,000 feet pressure altitude in the first column on the left side of the table. Next, find the correct rpm of 2,400 in the second column. Follow that line straight across and read the TAS of 116 mph and a fuel burn rate of 6.9 gallons per hour. As per the example, the aircraft is equipped with a fuel carrying capacity of 38 gallons. Under this column, read that the endurance in hours is 5.5 hours and the range in miles is 635 miles.
Cruise power setting tables are useful when planning crosscountry flights. The table gives the correct cruise power settings, as well as the fuel flow and airspeed performance numbers at that altitude and airspeed.
Sample Problem 7
Pressure Altitude at Cruise 6,000 feet OAT 36 °F above standard
Refer to Figure 11-28 for this sample problem. First, locate the pressure altitude of 6,000 feet on the far left side of the table. Follow that line across to the far right side of the table under the 20 °C (or 36 °F) column. At 6,000 feet, the rpm setting of 2,450 will maintain 65 percent continuous power at 21.0 “Hg with a fuel flow rate of 11.5 gallons per hour and airspeed of 161 knots.
Figure 11-28. Cruise power setting. [click image to enlarge]
Another type of cruise chart is a best power mixture range graph. This graph gives the best range based on power setting and altitude. Using Figure 11-29, find the range at 65 percent power with and without a reserve based on the provided conditions.
Figure 11-29. Best power mixture range.
Sample Problem 8
OAT Standard Pressure Altitude 5,000 feet
First, move up the left side of the graph to 5,000 feet and standard temperature. Follow the line straight across the graph until it intersects the 65 percent line under both the reserve and no reserve categories. Draw a line straight down from both intersections to the bottom of the graph. At 65 percent power with a reserve, the range is approximately 522 miles. At 65 percent power with no reserve, the range should be 581 miles.
The last cruise chart referenced is a cruise performance graph. This graph is designed to tell the TAS performance of the airplane depending on the altitude, temperature, and power setting. Using Figure 11-30, find the TAS performance based on the given information.
Figure 11-30. Cruise performance graph. [click image to enlarge]
Sample Problem 9
OAT 16 °C Pressure Altitude 6,000 feet Power Setting 65 percent, best power Wheel Fairings Not installed
Begin by finding the correct OAT on the bottom left side of the graph. Move up that line until it intersects the pressure altitude of 6,000 feet. Draw a line straight across to the 65 percent, best power line. This is the solid line, that represents best economy. Draw a line straight down from this intersection to the bottom of the graph. The TAS at 65 percent best power is 140 knots. However, it is necessary to subtract 8 knots from the speed since there are no wheel fairings. This note is listed under the title and conditions. The TAS is 132 knots.
## Flight Literacy Recommends
Rod Machado's Private Pilot Handbook -Flight Literacy recommends Rod Machado's products because he takes what is normally dry and tedious and transforms it with his characteristic humor, helping to keep you engaged and to retain the information longer. (see all of Rod Machado's Products).
| 1,772 | 7,761 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.671875 | 4 |
CC-MAIN-2020-16
|
longest
|
en
| 0.914094 |
https://www.cracksat.net/sat/math-multiple-choice/question-no-400-answer-and-explanation.html
| 1,638,427,723,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-49/segments/1637964361169.72/warc/CC-MAIN-20211202054457-20211202084457-00423.warc.gz
| 779,063,389 | 3,434 |
# SAT Math Multiple Choice Question 400: Answer and Explanation
### Test Information
Question: 400
10. A mailing supply store sells small shipping boxes in packs of 8 or 20. If the store has 61 packs in stock totaling 800 small shipping boxes, how many packs have 20 boxes in them, assuming all the packs are full?
• A. 26
• B. 32
• C. 35
• D. 40
Explanation:
A
Difficulty: Medium
Category: Heart of Algebra / Systems of Linear Equations
Strategic Advice: Create a system of linear equations where e represents the number of packs with 8 boxes and t represents the number of packs with 20 boxes. Before selecting your final answer, make sure you answered the right question (the number of packs that have 20 boxes).
Getting to the Answer: The first equation should represent the total number of packs, e + t = 61. The second equation should represent the total number of boxes. Because e represents packs with 8 boxes and t represents packs with 20 boxes, the second equation should be 8e + 20t = 800. Now, solve the system using substitution. Solve the first equation to find that e = 61 - t. Then, substitute the result into the second equation:
We assigned the variable t to the number of packs with 20 boxes, so 26 packs have 20 boxes. This is what the question asks for, so you don't need to find the value of e.
| 319 | 1,328 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.28125 | 4 |
CC-MAIN-2021-49
|
latest
|
en
| 0.880642 |
http://gmatclub.com/forum/each-employ-on-a-certain-task-is-either-manager-or-director-29146.html
| 1,419,090,420,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2014-52/segments/1418802769990.68/warc/CC-MAIN-20141217075249-00073-ip-10-231-17-201.ec2.internal.warc.gz
| 119,009,239 | 42,840 |
Find all School-related info fast with the new School-Specific MBA Forum
It is currently 20 Dec 2014, 07:47
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# Each employ on a certain task is either manager or director.
Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Manager
Joined: 13 Mar 2006
Posts: 76
Followers: 1
Kudos [?]: 5 [0], given: 0
Each employ on a certain task is either manager or director. [#permalink] 04 May 2006, 21:34
00:00
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Each employ on a certain task is either manager or director. What % of the employees on the task force are directors?
(1) The average salary of managers on the TF is \$5,000 less that the av. sal. of all employees.
(2) The av. sal. of directors on the TF is \$15,000 greater that the av. sal. of all employees.
_________________
GMAT by 8th JUL
VP
Joined: 06 Jun 2004
Posts: 1061
Location: CA
Followers: 2
Kudos [?]: 36 [0], given: 0
[#permalink] 04 May 2006, 23:09
C
Let x be the average, m = manager, d = director. We are looking for d/(d+m)?
(1) x - 5000 ==> Not sufficient
(2) x = 15,000 ==> Not sufficient
(1) + (2)
m(x-5000) + d(x+15000)/(m+d) = x ==> After simplifying we get
3d = m and we can find out d/(d+m) ==> 1/4 ==> 25%
_________________
Don't be afraid to take a flying leap of faith.. If you risk nothing, than you gain nothing...
VP
Joined: 29 Apr 2003
Posts: 1405
Followers: 2
Kudos [?]: 17 [0], given: 0
[#permalink] 05 May 2006, 17:59
I also get E..
It says Average Salaray of ALL EMPLOYEEs. Not just the ones in the Task Force!
Manager
Joined: 27 Mar 2006
Posts: 136
Followers: 1
Kudos [?]: 1 [0], given: 0
[#permalink] 05 May 2006, 21:49
I say E
Director
Joined: 16 Aug 2005
Posts: 947
Location: France
Followers: 1
Kudos [?]: 10 [0], given: 0
[#permalink] 10 May 2006, 11:13
TeHCM your method of solving the problem I agree with, except those who are saying E has a point that (1) and (2) says "av. sal. of all employees". The question could be clear if it mentioned "av. sal. of all employees in the Task Force"
What is the source of this question?
Intern
Joined: 23 Feb 2006
Posts: 40
Followers: 0
Kudos [?]: 0 [0], given: 0
[#permalink] 11 May 2006, 05:05
Go for C.
here is the solution.
let m - manager salary, d-director salary.
n-number of manager,m-number of directors
From 1:
(m1+m2+...mn/n)+5000=(m1+...+mn+d1+...+dm)/(m+n) - nothing we can get since don't know relationship between m and n
From 2:
(d1+d2+...dm/m)-15000=(m1+...+mn+d1+...+dm)/(m+n) - nothing we can get again
Combying both we can solve equation.
[#permalink] 11 May 2006, 05:05
Similar topics Replies Last post
Similar
Topics:
Each employee on task force is either director or manager. 1 26 Jul 2008, 00:53
Each employee on a certain task force is either a manager or 1 15 May 2006, 06:53
Each employee on a certain task force is either a manager or 1 07 May 2006, 17:19
Each employee on a certain task force is either a manager or 7 20 Nov 2005, 00:23
Each employee on a certain task force is either a manager or 8 22 Jul 2005, 22:07
Display posts from previous: Sort by
# Each employ on a certain task is either manager or director.
Question banks Downloads My Bookmarks Reviews Important topics
Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.
| 1,224 | 4,039 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.90625 | 4 |
CC-MAIN-2014-52
|
latest
|
en
| 0.867609 |
https://www.math24.net/comparison-tests/
| 1,603,417,826,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107880519.12/warc/CC-MAIN-20201023014545-20201023044545-00451.warc.gz
| 804,034,457 | 12,722 |
# Comparison Tests
The $$N$$th term test, generally speaking, does not guarantee convergence of a series. Convergence or divergence of a series is proved using sufficient conditions. The comparison tests we consider below are just the sufficient conditions of convergence or divergence of series.
### The Comparison Tests
Let $$\sum\limits_{n = 1}^\infty {{a_n}}$$ and $$\sum\limits_{n = 1}^\infty {{b_n}}$$ be series such that $$0 \lt {a_n} \le {b_n}$$ for all $$n.$$ Then the following comparison tests hold:
• If $$\sum\limits_{n = 1}^\infty {{b_n}}$$ is convergent, then $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is also convergent;
• If $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is divergent, then $$\sum\limits_{n = 1}^\infty {{b_n}}$$ is also divergent.
### The Limit Comparison Tests
Let $$\sum\limits_{n = 1}^\infty {{a_n}}$$ and $$\sum\limits_{n = 1}^\infty {{b_n}}$$ be series such that $${a_n}$$ and $${b_n}$$ are positive for all $$n.$$ Then the following limit comparison tests are valid:
• If $$0 \lt \lim\limits_{n \to \infty } {\large\frac{{{a_n}}}{{{b_n}}}\normalsize} \lt \infty ,$$ then $$\sum\limits_{n = 1}^\infty {{a_n}}$$ and $$\sum\limits_{n = 1}^\infty {{b_n}}$$ are both convergent or both divergent;
• If $$\lim\limits_{n \to \infty } {\large\frac{{{a_n}}}{{{b_n}}}\normalsize} = 0,$$ then $$\sum\limits_{n = 1}^\infty {{b_n}}$$ convergent implies that the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is also convergent;
• If $$\lim\limits_{n \to \infty } {\large\frac{{{a_n}}}{{{b_n}}}\normalsize} = \infty,$$ then $$\sum\limits_{n = 1}^\infty {{b_n}}$$ divergent implies that the series $$\sum\limits_{n = 1}^\infty {{a_n}}$$ is also divergent.
The so-called $$p$$-series $$\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^p}}}\normalsize}$$ converges for $$p \gt 1$$ and diverges for $$0 \lt p \le 1.$$
## Solved Problems
Click or tap a problem to see the solution.
### Example 1
Determine whether $$\sum\limits_{n = 1}^\infty {\large\frac{{{e^{\frac{1}{n}}}}}{{{n^2}}}\normalsize}$$ converges or diverges.
### Example 2
Determine whether the series $$\sum\limits_{n = 1}^\infty {\large\frac{{{n^2} – 1}}{{{n^4}}}\normalsize}$$ converges or diverges.
### Example 3
Determine whether the series $$\sum\limits_{n = 1}^\infty {\large\frac{{{n^2}}}{{{n^3} – 3}}\normalsize}$$ converges or diverges.
### Example 4
Determine whether the series $$\sum\limits_{n = 1}^\infty {\large\frac{{3n – 1}}{{2{n^3} – 4n + 5}}\normalsize}$$ converges or diverges.
### Example 5
Determine whether $$\sum\limits_{n = 1}^\infty {\large\frac{{\sqrt n }}{{2{n^2} + n + 5}}\normalsize}$$ is convergent.
### Example 6
Determine whether the series $$\sum\limits_{n = 1}^\infty {\large\frac{n}{{{n^2} – 2n – 3}}\normalsize}$$ converges or diverges.
### Example 7
Determine whether the series
${\frac{1}{{\sqrt 2 }} + \frac{1}{{2\sqrt 3 }} }+{ \frac{1}{{3\sqrt 4 }} + \ldots }+{ \frac{1}{{n\sqrt {n + 1} }} + \ldots }$
converges or diverges.
### Example 1.
Determine whether $$\sum\limits_{n = 1}^\infty {\large\frac{{{e^{\frac{1}{n}}}}}{{{n^2}}}\normalsize}$$ converges or diverges.
Solution.
We easily can see that $${e^{\large\frac{1}{n}\normalsize}} \le e$$ for $$n > 1.$$ Then, by the comparison test,
${\sum\limits_{n = 1}^\infty {\frac{{{e^{\large\frac{1}{n}\normalsize}}}}{{{n^2}}}} } \le {\sum\limits_{n = 1}^\infty {\frac{e}{{{n^2}}}} } = {e\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} .}$
Since the series $$\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^2}}}\normalsize}$$ is convergent as a $$p$$-series with the power $$p = 2,$$ the original series also converges.
### Example 2.
Determine whether the series $$\sum\limits_{n = 1}^\infty {\large\frac{{{n^2} – 1}}{{{n^4}}}\normalsize}$$ converges or diverges.
Solution.
We use the comparison test. Note that $${\large\frac{{{n^2} – 1}}{{{n^4}}}\normalsize}$$ $${\large\frac{{{n^2}}}{{{n^4}}}\normalsize}$$ $$= {\large\frac{1}{{{n^2}}}\normalsize}$$ for all positive integers $$n.$$ As $$\sum\limits_{n = 1}^\infty {\large\frac{1}{{{n^2}}}\normalsize}$$ is a $$p$$-series with $$p = 2 \gt 1$$, it converges. Hence, the given series also converges by the comparison test.
Page 1
Problems 1-2
Page 2
Problems 3-7
| 1,557 | 4,207 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.375 | 4 |
CC-MAIN-2020-45
|
longest
|
en
| 0.585845 |
http://www.gradesaver.com/textbooks/math/algebra/algebra-2-common-core/chapter-6-radical-functions-and-rational-exponents-6-2-multiplying-and-dividing-radical-expressions-lesson-check-page-370/2
| 1,524,140,752,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125936914.5/warc/CC-MAIN-20180419110948-20180419130948-00680.warc.gz
| 337,126,848 | 13,801 |
## Algebra 2 Common Core
Published by Prentice Hall
# Chapter 6 - Radical Functions and Rational Exponents - 6-2 Multiplying and Dividing Radical Expressions - Lesson Check: 2
#### Answer
$-3\sqrt[3]{4}$
#### Work Step by Step
$=\sqrt[3]{(-27)(4)} \\=\sqrt[3]{(-3)^3(4)} \\=-3\sqrt[3]{4}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
| 134 | 455 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.84375 | 4 |
CC-MAIN-2018-17
|
latest
|
en
| 0.702259 |
http://scienceuniverse101.blogspot.com/2015/02/dijkstras-algorithm-to-find-single.html
| 1,529,836,991,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267866932.69/warc/CC-MAIN-20180624102433-20180624122433-00317.warc.gz
| 292,034,312 | 36,285 |
# Dijkstra’s Algorithm - to find Single Source Shortest Paths.
### Dijkstra’s Algorithm
#### - to find Single Source Shortest Paths
Some useful definitions:
Shortest Path Problem: Given a connected directed graph G with non-negative weights on the edges and a root vertex r, find for each vertex x, a directed path P (x) from r to x so that the sum of the weights on the edges in the path P (x) is as small as possible.
### Algorithm
• By Dutch computer scientist Edsger Dijkstra in 1959.
• Solves the single-source shortest path problem for a graph with nonnegative edge weights.
• This algorithm is often used in routing.
E.g : Dijkstra's algorithm is usually the working principle behind link-state routing protocols
#### ALGORITHM Dijkstra(G, s)
//Input: Weighted connected graph G and source vertex s
//Output: The length Dv of a shortest path from s to v and its penultimate vertex Pv for every vertex v in V
//initialize vertex priority in the priority queue Initialize (Q)
for every vertex v in V do
### The method
Dijkstra’s algorithm solves the single source shortest path problem in 2 stages.
Stage 1: A greedy algorithm computes the shortest distance from source to all other nodes in the graph and saves in a data structure.
Stage 2 : Uses the data structure for finding a shortest path from source to any vertex v.
• #### At each step, and for each vertex x, keep track of a “distance” D(x) and a directed path P(x) from root to vertex x of length D(x).
• Scan first from the root and take initial paths P( r, x ) = ( r, x ) with D(x) = w( rx ) when rx is an edge, D(x) = when rx is not an edge.
For each temporary vertex y distinct from x, set
#### D(y) = min{ D(y), D(x) + w(xy) }
Example:
Apply Dijkstra’s algorithm to find Single source shortest paths with vertex a as the source.
Solution:
Length Dv of shortest path from source (s) to other vertices v and Penultimate vertex Pv for every vertex v in V:
Conclusion:
• Doesn’t work with negative weights
• Applicable to both undirected and directed graphs
• Use unordered array to store the priority queue: Efficiency = Θ(n2)
• Use min-heap to store the priority queue: Efficiency = O(m log n)
| 521 | 2,193 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.875 | 4 |
CC-MAIN-2018-26
|
latest
|
en
| 0.892763 |
http://mathhelpforum.com/differential-equations/93641-quartic-oscillators-print.html
| 1,519,277,534,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-09/segments/1518891814002.69/warc/CC-MAIN-20180222041853-20180222061853-00307.warc.gz
| 221,743,631 | 3,685 |
Quartic Oscillators
• Jun 24th 2009, 10:53 AM
calculusfrk
Quartic Oscillators
Hi, here is a practice question. I'm not sure where the equation of motion comes in to play. Any help would be much appreciated. Thanks!
The equation of motion for a particle of mass 1 in a quartic oscillator V(x)=(1/4)x^4 is ẍ+x^3=0. Suppose that the maximum amplitude of the oscillator is x(max). Find an expression for the time T that it takes to go from x=0 to x=x(max) and show that this time is inversely proportional to x(max).
• Jun 25th 2009, 02:46 AM
the_doc
Using conservation of energy (let maximum of $x$ be $x_0$) you must have:
$V(x_0) = \frac{1}{2}m v^2 + V(x)$
but since $m=1$ we have
$v^2 =2 \left( V(x_0) - V(x)\right)$
$\iff v^2 = \frac{1}{2} \left(x_0^4 - x^4 \right)$
so
$\frac{dx}{dt} = \sqrt{\frac{1}{2} \left(x_0^4 - x^4 \right)}$
which is of separable variable form so
$\int_{0}^{T} \frac{1}{\sqrt{2}} \, \mathrm{d}t = \int_{0}^{x_0} \frac{1}{\sqrt{x_0^4 - x^4}} \, \mathrm{d}x$.
Let $x= x_0 u$ then $\mathrm{d}x = x_0 \, \mathrm{d}u$ so this becomes
$\frac{T}{\sqrt{2}} = \int_{0}^{1} \frac{1}{x_0^2 \sqrt{1 - u^4}} x_0 \, \mathrm{d}u$
$\iff T = \frac{\sqrt{2}}{x_0} \int_{0}^{1} \frac{1}{\sqrt{1 - u^4}} \, \mathrm{d}u$.
At this point we have answered the question as the integral has no dependence on $x_0$ so is simply a numerical constant and so we've shown that the time period is inversely proportional to the amplitude $x_0$.
If you must know what the integral evaluates to:
$\int_{0}^{1} \frac{1}{\sqrt{1 - u^4}} \, \mathrm{d}u = \frac{1}{4} \beta \left(\frac{1}{4},\frac{1}{2} \right) = \frac{4 \left[\left(\frac{1}{4}\right)! \right]^2}{\sqrt{2 \pi}}$,
so not anything nice in terms of elementary functions!
• Jun 25th 2009, 07:01 PM
zorop
then x'' + x^3 = 0 doesn't have any meaning here?
x'' =v*dv/dx = - x ^3 then v dv = - x^3 dx
so we can get the v= dx/dt
is this right?
• Jun 25th 2009, 09:46 PM
zorop
k, maybe it's the same in the end
• Jun 26th 2009, 04:53 AM
mr fantastic
Quote:
Originally Posted by zorop
then x'' + x^3 = 0 doesn't have any meaning here?
x'' =v*dv/dx = - x ^3 then v dv = - x^3 dx
so we can get the v= dx/dt
is this right?
The solution to $\ddot{x} + x^3 = 0$ gives the Conservation of Energy equation that the_doc used as his/her starting point:
$\ddot{x} + x^3 = 0$
$\Rightarrow \frac{d^2 x}{dt^2} + x^3 = 0$
$\Rightarrow \frac{d}{dx} \left[ \frac{v^2}{2}\right] + x^3 = 0$
$\Rightarrow \frac{v^2}{2} + \frac{x^4}{4} = C$.
| 967 | 2,488 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 20, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.09375 | 4 |
CC-MAIN-2018-09
|
longest
|
en
| 0.823798 |
https://www.instructables.com/topics/What-do-14-12-1-or-5-watts-values-mean-on-a-re/
| 1,532,091,168,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676591596.64/warc/CC-MAIN-20180720115631-20180720135631-00199.warc.gz
| 923,670,394 | 8,113 |
31407Views11Replies
# What do 1/4, 1/2, 1 or 5 watts values mean on a resistor? Answered
I know that this is such a basic question but what do those values mean when selecting a resistor?
Tags:
## Discussions
The value is the maximum wattage the part can dissipate without damage. The value specified is usually for room temp (25C), and the rating will need to be de-rated for higher temperatures.
Some types of resistors may need heatsinking to meet the stated power specs. The large 5, 10, 25, and 50 watt resistors with aluminum cases usually need a fairly large heat sink to achiew their ratings.
So, if P=I^2*R or P=V^2/R for the resistor in your application is less than the rated power (taking into consideration any derating at higher temps) then your part will be fine.
Suppose if I want to purchase the 15 milli Ohm resistor , when I am thinking about the Power Dissipation.
In my circuit,
Max expected current is 5A and maximum possible power dissipation,
Pd = ILoad (max)^2 * Rsh = 5A * 5A * 15 mOhm = 375 mW
Does it mean that the resistor I choose that should be more than 375mW ?
Because, Can I choose if the resistor 15mohm with 1W and 1% . In this case, should I consider this resistor or Can I use this resistor for my circuit with above given requirements?
you'll want to think in terms of your power supply, not the resistor you're using. For example, if you have 5V and 2A coming out of a charging cable then you have 5 x 2 = 10 watts coming into your circuit and you'll want resistors rated for 10w plus a factor of safety.
P=I x R means watts = Volts x Amps.
If you are not sure which value to pick, post your situation here and we may be able to recommend a value for you.
Let say a circuit where the voltage is 20 and 1 A current. So P=20, what should be the Watt value of the resistor?
Shouldn't I also take into consideration other components in the circuit?
The watt rating of a resistor is the MAX watts that it can handle. When selecting a proper resistor, you must select one that can handle more watts than you will be putting through it.
For example, if you know that .4 watts max will be passing through the resistor, select a 1/2 to 99 billion watt rated resistor. The wattage rating does not affect how the circuit operates, it is just used so that the resistor doesn't overheat.
Not correct: V = IR, means voltage = current x resistance (one volt = one amp-ohm). I think you meant P = I x V.
Yes, I did mean Watts = Volts x Amps, but it does NOT equal P=I x R. Thanks.
| 644 | 2,526 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.515625 | 4 |
CC-MAIN-2018-30
|
latest
|
en
| 0.928432 |
http://telliott99.blogspot.com/2010/11/normal-distribution-construction-in.html
| 1,531,966,130,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676590443.0/warc/CC-MAIN-20180719012155-20180719032155-00074.warc.gz
| 377,529,196 | 15,641 |
## Tuesday, November 9, 2010
### Normal distribution construction in Python
Here is an example with the normal distribution that will seem trivial after the t-distribution (here).
The basic form of the normal is exp {-x2/2}. We define that as a Python function f(x), vectorize it, and construct an array X of discrete points from -10 to +10 with interval dx = 0.001. We apply the vectorized function to the array to get the relative densities. So that we obtain the correct area under the curve, we multiply the height (value of f(x)) of each piece by its width, dx. When we sum up all the pieces, the total is equal to √2π as seen in the printout. We divide by this value to normalize the distribution so that its total area is equal to 1 and it becomes a pdf.
The form of the normal that includes a term for the standard deviation is normal(x). Everything is as before, except we substitute z for x, and at the end we find that the normalizing constant is 1/σ√2π. We plot it to have something pretty to look at.
[UPDATE:
In my discussions of probability distributions of late I've played a little fast and loose with terminology. The pdf (probability density function) has a value for any x. At x = μ the exponential term is equal to 1, and the value is 1/σ√2π. For the standard normal, this is about 0.4.
>>> 1.0/sqrt(2*pi) 0.3989422804014327
Of course, this is somewhat misleading, since the probability that x = any particular point approaches zero, because x is a real number with infinitely many values. The way we actually use the pdf is to ask what is the probability for a particular window, a range of values for a < x < b, and at least conceptually, this is done by integrating the function between these limits. That's where the discrete version that I developed comes in handy. We evaluate p(x) for a small enough interval and then multiply by the interval size to get a probability for that slice. To integrate between limits, just add the included slices. But for this to work, we have to "normalize" the pdf so that the total of all the little rectangles of width dx is equal to 1. It also helps to have the discrete version in developing the cdf (cumulative distribution function), since we can just accumulate the pdf as we move along the values of X generating the cdf as we go.
Technically, the pdf(x) is the slope of the cdf(x), which gets around the issue mentioned above for a continuous function. ]
output:
2.507 2.507 5.013 5.013
code listing:
from __future__ import division import numpy as np import math import matplotlib.pyplot as plt @np.vectorize def f(x): return math.e**(-0.5*(x**2)) @np.vectorize def normal(x,mu=0,sigma=1): z = (x-mu)/sigma return math.e**(-0.5*(z**2)) #================================== dx = 0.001 X = np.arange(-10,10+dx,dx) pdf = f(X) pdf *= dx print round(sum(pdf),3), S = math.sqrt(2*math.pi) print round(S,3), pdf /= S sigma = 2 pdfn = normal(X,0,sigma) pdfn *= dx print round(sum(pdfn),3), Sn = math.sqrt(2*math.pi) * sigma print round(Sn,3), pdfn /= Sn #================================== plt.plot(X,pdf,color='r',lw=4) plt.plot(X,pdfn,color='k',lw=4) ax = plt.axes() ax.set_xlim(-6,6) m = max(pdf) ymin,ymax = -m/100,m*1.1 ax.set_ylim(ymin,ymax) plt.text(3,0.75*ymax,s='$\sigma = 1$', color='r',fontsize=24) plt.text(3,0.65*ymax,s='$\sigma = 2$', color='k',fontsize=24) plt.savefig('example.png')
| 895 | 3,374 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.03125 | 4 |
CC-MAIN-2018-30
|
latest
|
en
| 0.936371 |
https://www.fatalerrors.org/a/0NV90j8.html
| 1,628,082,770,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046154805.72/warc/CC-MAIN-20210804111738-20210804141738-00587.warc.gz
| 742,214,970 | 7,443 |
# [problem solving report] advanced dynamic programming problem (Introduction to interval DP, tree DP and pressure DP)
Report on advanced dynamic programming practice of ACM team of grade 20 in Nanhua University
### 1. Stone merging (interval DP)
Title Link: Acwing stone merge
Problem analysis: the optimal solution of the last combined pile of stones can be decomposed into the optimal solution of the first two piles of stones, and the parent problem can be decomposed into subproblems to solve, so you can choose DP to solve
```//#pragma GCC optimize(2)
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#include<ctime>
#include<cstring>
#include<list>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef pair<int, int> PII;
const int N = 1e3 + 7;
int n;
int a[N];
int s[N]; //Prefixes and arrays
int dp[N][N]; //dp[i][j] represents the optimal solution between i~j regions
void solve()
{
cin >> n;
for (int i = 1; i <= n; i++)
cin >> a[i], s[i] = s[i - 1] + a[i];
for (int le = 2; le <= n; le++) //Enumeration interval length, the minimum interval length is 2
{
for (int i = 1; i + le - 1 <= n; i++) //Enumeration starting point
{
dp[i][i + le - 1] = 1e8;
for (int j = i; j <= i + le - 1; j++)
{ //Enumerate all the merging methods of the interval to find the optimal solution
dp[i][i + le - 1] = min(dp[i][i + le - 1], dp[i][j - 1] + dp[j][i + le - 1] + s[i + le - 1] - s[i - 1]);
}
}
}
//Output answer
cout << dp[1][n] << endl;
}
int main()
{
//std::ios::sync_with_stdio(false);
//cin.tie(0), cout.tie(0);
solve();
return 0;
}
```
### 2. Ring stone merging (ring interval DP)
Title Link: LibreOJ energy Necklace
Problem analysis: the difference between this problem and the previous one is that the interval is changed from a straight line to a ring, and the solution to the ring problem is to duplicate the original array so that the length of the array is doubled, and then the processing method is the same as the first one.
```//#pragma GCC optimize(2)
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#include<ctime>
#include<cstring>
#include<list>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef pair<int, int> PII;
const int N = 1e3 + 7;
int n;
int a[N]; //Original array
int s[N]; //Prefix and
int dp[N][N]; //minimum value
int dp2[N][N]; //Maximum
void solve()
{
cin >> n;
for (int i = 1; i <= n; i++)
cin >> a[i], s[i] = s[i - 1] + a[i];
for (int i = 1; i <= n; i++) //The array is doubled
s[i + n] = s[n + i - 1] + a[i];
int ma = -0x3f3f3f3f;
int mi = 0x3f3f3f3f;
mem(dp, 0x3f); //Initialize dp array
mem(dp2, -0x3f);
for (int le = 1; le <= n; le++) //Enumeration length
{
for (int i = 1; i + le - 1 <= 2 * n; i++) //Enumeration starting point, remember to enumerate to 2*n
{
if (i == i + le - 1)
dp[i][i + le - 1] = dp2[i][i + le - 1] = 0;
else
for (int j = i; j < i + le - 1; j++)
{ //Enumerates all possible merges
dp[i][i + le - 1] = min(dp[i][i + le - 1], dp[i][j] + dp[j + 1][i + le - 1] + s[i + le - 1] - s[i - 1]);
dp2[i][i + le - 1] = max(dp2[i][i + le - 1], dp2[i][j] + dp2[j + 1][i + le - 1] + s[i + le - 1] - s[i - 1]);
}
}
}
for (int i = 1; i + n - 1 <= 2 * n; i++) //Find the maximum
{
ma = max(ma, dp2[i][i + n - 1]);
mi = min(mi, dp[i][i + n - 1]);
}
cout << mi << endl;
cout << ma << endl;
}
int main()
{
//std::ios::sync_with_stdio(false);
//cin.tie(0), cout.tie(0);
solve();
return 0;
}
```
### 3. Energy Necklace (DP)
Title Link: LibreOJ energy Necklace
Topic analysis: the processing method of circular interval is similar to the previous one. Although this problem says that each necklace has two values, because the latter value of the former column must be equal to the latter bead, we can still directly deal with dp, but the enumeration length must start from 3.
```//#pragma GCC optimize(2)
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#include<ctime>
#include<cstring>
#include<list>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef pair<int, int> PII;
const int N = 210;
int n;
int dp[N][N],w[N];
void solve()
{
cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> w[i];
w[i + n]=w[i];
}
mem(dp, 0);
for (int len = 3; len <= n + 1; len++)
{
for (int l = 1; l - 1 + len <= n * 2; l++)
{
int r = l - 1 + len;
for (int k = l + 1; k < r; k++)
dp[l][r] = max(dp[l][r], dp[l][k] + dp[k][r] + w[l] * w[k] * w[r]); //After merging two intervals, add the product of three numbers
}
}
int res = 0;
for (int i = 1; i <=2* n; i++) //Find the maximum output
res = max(res, dp[i][i + n]);
cout << res << endl;
}
int main()
{
std::ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
solve();
return 0;
}
```
### 4. Binary apple tree (tree shape)
Title Link: LibreOJ binary apple tree
Topic analysis: a variant of knapsack problem, the selection of an item is related to the selection of another item, which also involves the storage and traversal of adjacency table, the first time to write or have a certain difficulty.
Adjacency table knowledge points: Chain forward star code analysis
```//#pragma GCC optimize(2)
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#include<ctime>
#include<cstring>
#include<list>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef pair<int, int> PII;
const int N = 110,M=N*2;
int n, m;
int h[N], e[M], w[M], ne[M], id; //Arrays and variables required by adjacency table
int dp[N][N];
void add(int a, int b, int c) //The addition of adjacency table will not be explained here
{
w[id] = c;
e[id] = b;
ne[id] = h[a];
h[a] = id;
id++;
}
void dfs(int u, int fa)
{
for (int i = h[u];~ i; i = ne[i])//i traverse all the children of u
{
if (e[i] == fa)continue;//This is the inverse that wants to recognize father as son. Since the graph is treated as an undirected graph when it is saved, it is necessary to skip the double edge
dfs(e[i], u);
for (int j = m; j >= 0; j--)//Volume, how many sides
{
for (int k = 0; k < j; k++)
{ //Traversing the sub nodes to determine whether to add the subtree with i as the root node in the final answer, knapsack model
dp[u][j] = max(dp[u][j], dp[u][j - k - 1] + dp[e[i]][k] + w[i]);
}
}
}
}
void solve()
{
cin >> n >> m;
mem(h, -1); //Initialize the h array, adjacent table knowledge points
for (int i = 0; i < n - 1; i++)
{
int a, b, c;
cin >> a >> b >> c;
//The undirected graph storage method corresponds to the above judgment of whether it is an inverse
add(a, b, c);
add(b, a, c);
}
dfs(1, -1); //1 is the parent node, h[1]=-1
cout << dp[1][m];
}
int main()
{
//Turn off synchronization flow
std::ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
solve();
return 0;
}
```
### 5. Dance without boss (tree shaped DP)
Title Link: LibreOJ anniversary party
Topic analysis: the code template is similar to the previous question, but the state transition equation needs to be changed. In the code, we need to realize that if a leader is selected, his immediate subordinate will not come to the dance; If the boss doesn't come, we have to judge whether his immediate subordinate will come or not, which is the best for the final result. We add a dimension with depth of 2 to the dp array, and judge whether the leader will come or not through 0 and 1.
```//#pragma GCC optimize(2)
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#include<ctime>
#include<cstring>
#include<list>
#define ll long long
#define ull unsigned long long
#define INF 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef pair<int, int> PII;
const int N = 6500;
int n;
int h[N], e[N], ne[N],id=1; //Adjacency table
int hasBoss[N]; //Used to find out who is the general manager without a boss
int happy[N]; //How happy everyone is
int dp[N][2]; //DP [i] [0] and DP [i] [1] respectively represent the maximum happiness of employee I who doesn't come and employee I who comes
int boss; //General manager's number, root node
void add(int a, int b)
{
e[id] = b;
ne[id] = h[a];
h[a] = id++;
hasBoss[b] = 1;
}
void dfs(int u) //u is the employee number
{
dp[u][1] = happy[u]; //When employee u comes, add his happiness value
for (int i = h[u]; i != -1; i = ne[i]) //Traversal of adjacency table
{
dfs(e[i]); //recursion
dp[u][1] += dp[e[i]][0]; //When u comes, none of his direct subordinates will come
dp[u][0] += max(dp[e[i]][0], dp[e[i]][1]); //If you don't come, it's better to judge whether the direct staff will come or not
}
}
void solve()
{
cin >> n;
for (int i = 1; i <= n; i++)
cin >> happy[i];
mem(h, -1);
for (int i = 1; i <= n-1; i++)
{
int x, y;
cin >> x >> y;
add(y, x);
}
for (int i = 1; i <= n; i++) //Looking for boss
if (!hasBoss[i])boss = i;
dfs(boss);
cout << max(dp[boss][0], dp[boss][1]);
}
int main()
{
//std::ios::sync_with_stdio(false);
//cin.tie(0), cout.tie(0);
solve();
return 0;
}
```
### 6. King (DP)
Title Link: King LibreOJ
Topic analysis: the so-called shape pressure is to save the status of chess pieces in each line of the graph into a dimension of the dp array with 01 binary number, so as to save the status of a line with a number. Greatly save time and space. We can also use binary operation to realize the restriction of playing chess pieces in the title. Then, if we can determine the case of line 1~i, the case of line i+1 is only related to the placement method of line i, so the derivation of our dp equation is derived from the previous layer. Therefore, in addition to Listing 1~n layers, we also need to list row 0 and row n+1. Because the situation of the first line will be introduced from line 0. And the final answer to the question exists in the case of n+1 line without putting anything.
For example, in the code:
1. You can't put two kings in a row --- (state > > I & 1) & (state > > I + 1 & 1)= 1;
2. You can't put the king's restriction on the next line around the eight points -- (A & B) & check (a | b)= 1;
Tip: the dp array needs to open long long. The code below uses "# define int long long" to see long long
```//#pragma gcc optimize(2)
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<string>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#include<ctime>
#include<cstring>
#include<list>
#define int long long
#define ull unsigned long long
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef pair<int, int> pii;
const int N = 12,M=1<<10,K=110;
int n, m;
vector<int> state;
vector<int> head[M];
int cnt[M];
int dp[N][K][M]; //DP [i] [J] [k] -- > there are j kings in the first row, and the number of schemes with K in the first row
bool check(int state) //Check if there are two consecutive functions on the same line
{
for (int i = 0; i < n; i++)
if ((state >> i & 1) && (state >> i + 1 & 1))
return false;
return true;
}
int count(int state) //A function that returns the number of Chinese kings in a row
{
int res = 0;
for (int i = 0; i < n; i++)
res += state >> i & 1;
return res;
}
void solve()
{
cin >> n >> m;
for (int i = 0; i < 1 << n; i++) //Save all binary numbers that meet condition 1
{
if (check(i))
{
state.push_back(i);
cnt[i] = count(i);
}
}
for(int i=0;i<state.size();i++) //Then save the number that meets condition 2 for each number that meets condition 1
for (int j = 0; j < state.size(); j++)
{
int a = state[i], b = state[j];
if (!(a & b) && check(a | b))
head[i].push_back(j);
}
dp[0][0][0] = 1; //Layer 0 can only have the possibility of not putting anything
for(int i=1;i<=n+1;i++) //Row i
for(int j=0;j<=m;j++) //A total of j, equivalent to the volume of the backpack
for(int a=0;a<state.size();a++) //Enumerate the legal placement of all lines in line i
for (int b=0;b<head[a].size();b++) //b represents the legal placement of i-1 line corresponding to a
{
int c = cnt[state[a]]; //c stands for the number of pieces in the legal situation a
if (j >= c) //If the volume of the backpack is larger than the item
dp[i][j][a] += dp[i - 1][j - c][b]; //Solution number of state transition
}
cout << dp[n + 1][m][0] << endl; //Output results, the final answer to the question exists in the case of n+1 line without putting anything.
}
signed main()
{
//std::ios::sync_with_stdio(false);
//cin.tie(0), cout.tie(0);
solve();
return 0;
}
```
Practice time: July 20, 2021
By Avalon Demerzel
Posted by fyg0072 at Jul 20, 2021 - 1:40 PM
| 4,121 | 13,399 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.515625 | 4 |
CC-MAIN-2021-31
|
latest
|
en
| 0.6325 |
http://mathhelpforum.com/differential-geometry/104136-triangulation-torus-print.html
| 1,529,874,948,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267867055.95/warc/CC-MAIN-20180624195735-20180624215735-00144.warc.gz
| 199,218,814 | 3,119 |
# Triangulation of a Torus
• Sep 24th 2009, 03:43 PM
Janu42
Triangulation of a Torus
I need help doing a triangulation of a torus. I don't need to use any formulas or anything, as we have just started doing this kind of thing. What I have to do is do the triangulation and then compute V-E+F and see if it equals 2.
My professor said a couple ways we can do this is use a square with opposite sides paired with one another for gluing.
Another way he described (and the way I think I want to go) is taking a cube with a square cut out through the middle. Since the top face can't have a hole, I would create 4 more edges from the corners of the object to the corners of the square hole, thus creating 4 faces on the top. The same process would happen on the bottom.
However, I don't understand how the triangulation is supposed to work... I know I can sorta just mold this figure into a torus, and I can figure out V,E, and F before I mold it, but how is this a "triangulation"? Do I need to turn the figure into smaller triangles or something, because I don't get how I would be able to compute V,E, and F for all of it.
I pretty much just need help figuring out what I need to do. My professor said using the square with opposite sides paired with one another would be the easiest way, so if someone can explain this to me that would be a big help as well.
• Sep 25th 2009, 05:35 AM
Opalg
I don't know if this will help. It's a diagram that I drew a few years ago (when teaching a geometry course) to illustrate how to triangulate a torus as a cube with a hole through the middle from top to bottom. It has 16 vertices, 32 edges and 16 quadrilateral faces. If you want triangular faces, you can divide each quadrilateral into two triangles by joining two opposite corners. That will double the number of faces, and also add 16 new edges.
| 458 | 1,843 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875 | 4 |
CC-MAIN-2018-26
|
latest
|
en
| 0.970091 |
http://www.ask.com/question/convert-mcg-to-milligrams
| 1,394,973,941,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2014-10/segments/1394678702690/warc/CC-MAIN-20140313024502-00021-ip-10-183-142-35.ec2.internal.warc.gz
| 187,305,051 | 17,289 |
# Convert MCG to Milligrams?
An Mcg is known as a microgram. It is one thousandth of a milligram. To convert Mcg to milligrams, you would divide the number of a thousand. An example is 5600 Mcg = 5.6 mg.
Q&A Related to "Convert MCG to Milligrams?"
An mcg (a microgram) is 10 -6 gm or one thousandth of a milligram. To convert micrograms to milligrams, divide by 1000 e.g. 5600 mcg = 5.6 mg Follow the link provided for a weight http://wiki.answers.com/Q/How+do+you+convert+500+m...
If mcg is microgram then: 400 mcg = 0.4 milligrams Why? 1 milligram is equal to 1000 micrograms http://answers.yahoo.com/question/index?qid=200709...
A gram is simply a unit of weight. The prefix milli represents 1000. So 1000 milligrams is equal to 1 gram. Take the number of milligrams and divide it by 1000 to find the number http://answers.ask.com/Reference/Other/how_to_conv...
1. Multiply the mass in milligrams by 1,000 to convert to micrograms. For example, if you have 59 mg, multiply 59 by 1,000 to get 59,000 mcg. 2. Divide the number of milligrams by http://www.ehow.com/how_8654930_convert-mg-mcg.htm...
Top Related Searches
Explore this Topic
Converting a weight needs to be done in various forms. The measurement that you need may need to either be in a small or large quantity. One Mcg equals 0.001 milligrams ...
1. Multiply the mass in milligrams by 1,000 to convert to micrograms. For example, if you have 59 mg, multiply 59 by 1,000 to get 59,000 mcg. 2. Divide the number ...
One can use a converter tool to find the conversion of a number that is in a mcg. form to find its mg form. For example, 1 mcg equals to .001 milligrams. One such ...
| 474 | 1,650 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.953125 | 4 |
CC-MAIN-2014-10
|
latest
|
en
| 0.822654 |
http://openstudy.com/updates/4eedee02e4b0367162f5ba04
| 1,448,975,393,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2015-48/segments/1448398467979.1/warc/CC-MAIN-20151124205427-00032-ip-10-71-132-137.ec2.internal.warc.gz
| 180,944,480 | 14,862 |
## A community for students. Sign up today
Here's the question you clicked on:
## abz_tech 3 years ago express in its simplest form: 5x-6/x²+x-6 - 3/x+3
• This Question is Closed
1. Tomas.A
write it in proper way
2. abz_tech
$\left(\begin{matrix}5x-6 \\ x²+x-6\end{matrix}\right)-\left(\begin{matrix}3 \\ x+3\end{matrix}\right)$
3. abz_tech
dont know how to get the division line soz, ignore the brackets aswell =/
4. Tomas.A
$\frac{5x-6}{x^2+x-6}-\frac{3}{x+3}$
5. abz_tech
ya thats it
6. Tomas.A
factor first fraction's denominator
7. abz_tech
(x-2)(x+3)
8. abz_tech
....what do i do now?
9. dj80
|dw:1324216715261:dw|
10. dj80
take lcm
11. abz_tech
can u show the steps plz?
12. abz_tech
i know thats the answer... im trying to find how my teacher got it
13. dj80
sure just a moment
14. abz_tech
tyvm
15. dj80
$(5x-6)\div(x-2)(x+3) - 3\div(x+3)$
16. abz_tech
lol its ok
17. abz_tech
i found it
18. dj80
then take x-2 as lcm a multiply it with 3
19. dj80
u got it??
20. abz_tech
forgot to divide my nominator by x
21. abz_tech
rofl
#### Ask your own question
Sign Up
Find more explanations on OpenStudy
Privacy Policy
| 431 | 1,165 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.53125 | 4 |
CC-MAIN-2015-48
|
longest
|
en
| 0.678523 |
https://www.teachoo.com/4894/724/Ex-7.9--10---Direct-Integrate-dx---root-1---x2--from-0-to-1/category/Ex-7.9/
| 1,686,342,107,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224656833.99/warc/CC-MAIN-20230609201549-20230609231549-00448.warc.gz
| 1,131,385,467 | 33,544 |
Ex 7.8
Chapter 7 Class 12 Integrals
Serial order wise
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Ex 7.8, 10 β«_0^1βππ₯/(1 + π₯2) Let F(π₯)=β«1βππ₯/(1 + π₯^2 ) =β«1β1/(1^2 + π₯^2 ) ππ₯ =1/1 .tan^(β1)β‘(π₯/1) =tan^(β1) π₯ Hence F(π₯)=tan^(β1) π₯ (Using β«1β1/(π₯^2 + π^2 ) ππ₯=1/π tan^(β1)β‘π₯) Now, β«_0^1βγππ₯/(1 + π₯^2 )=πΉ(1)βπΉ(0) γ =tan^(β1)β‘γ(1)βtan^(β1)β‘(0) γ =π/4β0 =π
/π
| 377 | 542 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.90625 | 4 |
CC-MAIN-2023-23
|
longest
|
en
| 0.420735 |
https://classroom.thenational.academy/lessons/to-recognise-identify-and-describe-unit-fractions-ccwpce
| 1,709,162,313,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474746.1/warc/CC-MAIN-20240228211701-20240229001701-00600.warc.gz
| 170,990,622 | 26,506 |
# To recognise, identify and describe unit fractions
In this lesson, we will build on equal parts by looking at sharing amounts equally. We will name the fractions, and using our understanding of the numerator, the vinculum and the denominator we will write the fraction.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.If I have 13 marbles, I cannot make 2 equal parts.
1/3
Q2.If I make 3 equal parts with 15 bean bags, how many would be in each part?
2/3
Q3.I have a certain number of bean bags. When I make 4 equal groups, there are 4 in each part. When I make 2 equal groups, there are 8 in each part. What is my whole number of bean bags?
3/3
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# Intro quiz - Recap from previous lesson
Before we start this lesson, let’s see what you can remember from this topic. Here’s a quick quiz!
Q1.If I have 13 marbles, I cannot make 2 equal parts.
1/3
Q2.If I make 3 equal parts with 15 bean bags, how many would be in each part?
2/3
Q3.I have a certain number of bean bags. When I make 4 equal groups, there are 4 in each part. When I make 2 equal groups, there are 8 in each part. What is my whole number of bean bags?
3/3
# Video
Click on the play button to start the video. If your teacher asks you to pause the video and look at the worksheet you should:
• Click "Close Video"
• Click "Next" to view the activity
Your video will re-appear on the next page, and will stay paused in the right place.
# Worksheet
These slides will take you through some tasks for the lesson. If you need to re-play the video, click the ‘Resume Video’ icon. If you are asked to add answers to the slides, first download or print out the worksheet. Once you have finished all the tasks, click ‘Next’ below.
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# To recognise, identify and describe unit fractions
In the quiz we seek to further embed our learning by challenging ourselves to remember what we have learnt in the lesson.
Q1.Which of these are NOT unit fractions?
1/3
Q2.The denominator shows us...
2/3
Q3.The numerator is...
3/3
#### Unit quizzes are being retired in August 2023
Why we're removing unit quizzes from the website >
Quiz:
# To recognise, identify and describe unit fractions
In the quiz we seek to further embed our learning by challenging ourselves to remember what we have learnt in the lesson.
Q1.Which of these are NOT unit fractions?
1/3
Q2.The denominator shows us...
2/3
Q3.The numerator is...
3/3
# Lesson summary: To recognise, identify and describe unit fractions
## Time to move!
Did you know that exercise helps your concentration and ability to learn?
For 5 mins...
Move around:
Climb stairs
On the spot:
Chair yoga
| 743 | 3,055 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.125 | 4 |
CC-MAIN-2024-10
|
latest
|
en
| 0.912215 |
https://www.studypool.com/discuss/1068672/find-the-maximum-or-minimum-value-of-the-function-round-your-answer-to-two-dec?free
| 1,511,416,484,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-47/segments/1510934806736.55/warc/CC-MAIN-20171123050243-20171123070243-00529.warc.gz
| 811,607,005 | 15,065 |
# Find the maximum or minimum value of the function. (Round your answer to two dec
label Algebra
account_circle Unassigned
schedule 1 Day
account_balance_wallet \$5
Find the maximum or minimum value of the function. (Round your answer to two decimal places.)
f(t) = 100 − 9t − 3t2
Jul 8th, 2015
---------------------------------------------------------------------------------------------------------------
$\\ f(t)=100-9t-3t^2\\ \\ f(t)=-3t^2-9t+100\\$
So, a = -3 , b = -9 and c = 100
Since a = -3 < 0, the parabola opens downwards and hence the vertex will be the highest point.
Vertex occurs where
$\\ t=\frac{-b}{2a}\\ \\ t=\frac{-(-9)}{2\times(-3)}\\ \\ t=\frac{9}{-6}\\ \\ t=-\frac{3}{2}$
For t = -3/2
$\\ f\bigg(\frac{-3}{2}\bigg)=-3\bigg(\frac{-3}{2}\bigg)^2-9\bigg(-\frac{3}{2}\bigg)+100\\ \\ f\bigg(\frac{-3}{2}\bigg)=-3\times\frac{9}{4}+\frac{27}{2}+100\\ \\ f\bigg(\frac{-3}{2}\bigg)=-\frac{27}{4}+\frac{27}{2}+100\\ \\ f\bigg(\frac{-3}{2}\bigg)=\frac{-27+27\times2+100\times4}{4}\\ \\ f\bigg(\frac{-3}{2}\bigg)=\frac{-27+54+400}{4}\\ \\ f\bigg(\frac{-3}{2}\bigg)=\frac{427}{4}\\ \\ f\bigg(\frac{-3}{2}\bigg)=106.75\\$
So, vertex is (-3/2, 106.75).
Hence, the maximum value of the function is 106.75 .
Maximum value = 106.75
---------------------------------------------------------------------------------------------------------------
Jul 8th, 2015
...
Jul 8th, 2015
...
Jul 8th, 2015
Nov 23rd, 2017
check_circle
| 520 | 1,443 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.46875 | 4 |
CC-MAIN-2017-47
|
latest
|
en
| 0.480816 |
albionexpertsonline.blogspot.com
| 1,516,220,804,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084886964.22/warc/CC-MAIN-20180117193009-20180117213009-00738.warc.gz
| 13,788,170 | 10,867 |
# Adhesive and Sealant Applications & Tools
Experts in the tools used for specific applications. The tool may cost you very little, but it accounts for over 50% of the success in the application.
## The Equation F=PA
Why do we have so many different types of tools, and how do we select the right tool for the right application? This is one of the hardest questions and one we don't give much thought to. Heck, a caulking gun is a caulking gun, right? Wrong, and here is why.
Force = Pressure x Area
This is a basic engineering formula that you can apply to dispensing tools. Simply put, it takes pressure in the cartridge, sausage, or bulk barrel (a material containment unit or MCU) to get material to flow out of the nozzle. Different amounts of pressure are required depending on:
a. How thick the material is.
b. The nozzle opening. Think of the pressure required to push material through a static mixer. A static mixer is a mixing nozzle which mixes two component materials before the material reaches the substrate.
PRESSURE makes things happen.
To determine the FORCE required of your dispensing tool you must calculate the area of the piston pushing on the material in the MCU. Lets compare the simple 1/10 gallon cartridge and the common quart cartridge. You don't generally see thick material in a quart cartridge. Why?
1/10 Gallon Cartridge Area = (Diameter Squared * Pi/4) = (1.8*1.8*3.1416)/4 = 2.55 square inches
Quart Cartridge Area = 5.19 square inches
5.19/2.56= 2.002
The Quart Cartridge Area is double that of a 1/10th Gallon Cartridge Area. This basic engineering formula reveals that it will require double the force to develop the same pressure in a Quart Cartridge versus a 1/10 Gallon Cartridge.
What can we draw from this?
1. Thin material requires less pressure and thus less force... and a lower-end dispensing tool.
2. Thicker material or very small or long nozzles require more pressure and thus more force... and thus a stronger or more forceful dispensing tool.
3. If you increase the diameter of the MCU you will need to develop more force from your dispensing tool. Most materials packaged in a Quart Cartridge are low pressure materials which require large nozzle openings and little accuracy in applying. A good example is subfloor adhesive. Your objective there is to get material down quickly.
In the next blog segment we will talk about how to measure force on a dispensing tool. Also, we will discuss what rod displacement means and how it needs to be factored in when selecting the right tool.
Labels: ,
Hi there! great stuff here, I'm glad that I drop by your page and found this very interesting. Thanks for posting. Hoping to read something like this in the future! Keep it up!
A wide variety of static mixers (also called motionless mixers) that are used with meter mix and dispensing machines and hand held cartridge dispensing systems. Included in the line of static mixers are the Statomix® and Quadro® brands.
| 670 | 2,973 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5625 | 4 |
CC-MAIN-2018-05
|
longest
|
en
| 0.871288 |
http://www.jiskha.com/display.cgi?id=1209220172
| 1,498,591,059,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-26/segments/1498128321536.20/warc/CC-MAIN-20170627185115-20170627205115-00430.warc.gz
| 570,357,502 | 3,882 |
# Algebra II
posted by .
Find the exact solution(s) of the system: (x^2/4)-y^2=1 and x=y^2+1
2)Write an equation for an ellipse if the endpoints of the major axis are at (-8,1) and (8,1) and the endpoints of the minor axis are at (0,-1) and (0,3).
Answer: (x^2/64) + (y-1)^2/4 = 1
Thanks a lot!!
• Algebra II -
(x^2/4)-y^2=1 and x=y^2+1
rewrite the second as y^2 = x-1 and sub into the first
x^2 /4 - (x-1) = 1
x^2 - 4x + 4 = 4 , I multiplied each term by 4
x(x-4) = 0
so x=0 or x=4
in y^2 = x-1
if x=0, y^2 = -1 ----> no solution
if x=4 , y^2 = 3 ---> y = ±√3
so they intersect at (4,√3) and (4,-√3)
2.
For the ellipse, the centre is (0,1), the midpoint of (-8,1) and (8,1)
(notice the (0,1) is also the midpoint of the minor axis)
a=8 and b=2, so....
(x^2)/64 + (y-1)^2 /4 = 1
### Answer This Question
First Name: School Subject: Answer:
### Related Questions
More Related Questions
Post a New Question
| 365 | 923 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.875 | 4 |
CC-MAIN-2017-26
|
latest
|
en
| 0.860111 |
http://alternativephysics.org/book/StaticElectricity.htm
| 1,519,110,593,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-09/segments/1518891812913.37/warc/CC-MAIN-20180220070423-20180220090423-00510.warc.gz
| 20,348,176 | 5,538 |
Home
# Static Electricity
Static electricity refers to an electrical charge that builds up due to contact between electrically dissimilar materials. Literally “electricity at rest”, it is understood to be caused by electrons being transferred from an object that attracts them weakly to an object that attracts them strongly. The extent to which an object attracts electrons is referred to as its ‘dielectric’ properties.
When two objects with different dielectrics come into contact the result is that one of the objects gains a net negative charge and the other a net positive charge. Once charged, the objects then display some unusual behaviour:
1. Objects with a net negative or net positive charge can attract other objects having a net neutral charge. This appears to violate Coulomb’s Law which states that only objects with opposite charge should attract. In other words, we should expect no attraction to occur.
2. Objects with a strong negative charge can attract objects with a weak negative charge. This appears to fully violate Coulomb’s Law which predicts that such objects should repel. If the like-charges are of a similar strength then they do indeed repel, as would be expected. But attraction appears to occur when the like-charges are largely different in magnitude.
The accepted explanation for this attractiveness is ‘polarization’ and is explained with this diagram:
On the left is an electrically neutral object. It has equal numbers of protons and electrons which are distributed uniformly throughout the object. The diagram on the right shows what happens when a negative charge is introduced nearby.
According to theory, the protons in the object will be attracted to the negative charge and the electrons repelled. While the protons cannot move, the electrons can, and they will migrate over to the side furthest from the negative charge, thus polarizing the object into oppositely charged areas. This is also known as ‘electric induction’.
Once the object is polarized, notice that the protons are now slightly closer to the negative charge than are the electrons. Because of this, the attractive force between the protons and the negative charge will be slightly greater than the repulsive force between the electrons and the negative charge. As a result the net force between the object and the negative charge is attractive, and this attractive force is what we observe as static electricity. If a positive charge was used in place of the negative charge, the electrons would migrate to the opposite end (near to the positive charge) and the end result would still be a net attractive force.
Another explanation involves rotating molecules. In some liquids such as water, the molecule is positive on one side and negative on the other. When exposed to a static charge, the molecules will freely rotate such that the oppositely charged side is closest to the charge. This polarizes the liquid and yields a net attraction to the static charge in a similar manner to the above [1].
#### Problems
While the polarizing concept appears to provide an explanation for the attractive force toward neutral objects, this theory has a number of problems:
1. If migration of electrons toward or away from the static charge was an explanation for an attractive force, then we should expect materials that conduct electricity like metals to be the most attractive because of their free floating electrons. Conversely we should expect materials that don’t conduct electricity to feel little attraction. In reality however we observe that static cling seems to apply to all substances almost equally. For example paper and rubber can be easily attracted despite being poor electrical conductors.
2. Polarization via rotating molecules can only serve as an explanation for certain molecules while in liquid form and possibly a few solids. It can’t serve as a general explanation for the static cling of all liquids and solids.
3. If polarization was the cause of static attraction then it can be shown that the net force function will vary in direct proportion with the depth of the object. E.g. an object with half the thickness will experience only half the force. This would mean that small pieces of paper should have very little attraction while facing a static charge, but much greater attraction when turned sideways. It would also mean that dust particles should have almost no attraction. We know that neither of these predictions prove true.
4. In addition to point 3, it can also be shown the force function will vary in inverse proportion to the cube of the distance from the static charge – not the inverse square as might be assumed. This would make the force function fall off very rapidly, e.g. doubling the distance would yield an eight-fold decrease and tripling would leave one 27th of the original force. We don’t observe this to be the case.
5. As for the observation that a strong negative static charge can attract a weak negatively charged object, polarization would have a very difficult time explaining this. Coulomb’s law suggests that the repulsive force between such objects should easily overpower a weak polarizing force in almost any situation.
This can be easily demonstrated as follows: Run a plastic comb through your hair or across some woollen fabric. The comb should now have a negative static charge due to the dielectric properties of the plastic. Next use it to pick up a small piece of paper via static cling. The paper is neutral originally but once it hits the comb some of the electrons should discharge onto the paper. At this point both the paper and comb have a negative charge. Yet the paper still clings! Now shake the paper off the comb and try picking it up again. The paper is still negative yet continues to be attracted. The attraction is probably weaker but still present. (You may need to ‘recharge’ the comb for better effect because its excess electrons tend to drain away.)
6. Positively charged objects do not attract neutral objects as well as negatively charged objects attract neutral objects. For example, when plastic is rubbed against fur, the dielectric properties will cause the fur to become positive and the plastic negative. The plastic will then attract neutral objects but the fur does not seem to, except weakly, despite that it must have an equal degree of charge. Logically we should expect the fur to exert an equal amount of force on the neutral object.
#### Another explanation for Static Cling
I would like to present another explanation to the phenomenon of static cling that overcomes these limitations of the polarization theory.
This explanation is based on an idea presented in an earlier chapter, in which I proposed that the electric force between charged particles depends not only on their charge and distance, but also velocity. I proposed a modification to Coulomb’s law, called Velocity Dependent Coulombs Law (VDCL) that incorporated velocity into the equation. Basically the force between charged particles is proportional to the cube of the difference between their velocity and the field velocity, i.e.
Where F is force, v is velocity and c is the field speed which is equal to light speed. Let’s see how this might apply to static electrical forces.
A neutral object contains equal number of protons and electrons. The protons are mostly standing still, while the electrons move in some sort of orbital motion around them. Let’s consider a single proton and a single electron in orbit. See below diagram:
On the right of the picture is a static charge shown by a single electron. This electron represents the excess negative charge of a static-charged object. This object also contains a large number of protons and electrons but we can ignore these since their charges negate each other.
The proton on the left is largely motionless, but the electron on the right is not. It is being buffeted back and forth due to thermal activity and interaction with the atoms in the charged object. This electron has an attractive force with the proton on the right. The VDCL predicts that this force should vary depending on its velocity and direction.
E.g. suppose the electron moved at 1% of light speed (0.01c) in each direction. The VDCL tells us that the force will be 0.970299 times normal force while moving toward and 1.030301 times normal while moving away from the static-proton. See the previous chapter on magnetism for details. The average attractive force will be 1.0003, i.e. slightly greater than if the electron was at rest.
Now what about the force between the two electrons? If they were both at rest their average repulsive force would be 1, and this would create a net attractive force of 0.0003 units. This weak attractive force is what might be perceived as static-electric cling. However it is not that simple.
Both these electrons are in motion and the VDCL tells us that this will increase their repulsive force. E.g. if both moved away from each other at 1% of light speed the force would decrease to 0.941192 (0.98 cubed); then as they moved toward each other, it would increase to 1.061208 (1.02 cubed). That’s an average repulsive force of 1.0012 times normal and greater than the attractive force. This would yield a net repulsive force of 0.0009 units. Obviously this is not going give us static cling.
And it gets worse because the electron orbiting the proton on the left would be moving much faster than the free-floating electron on the right. So the repulsion would be much stronger. How to resolve this quandary?
As it happens, the electrons won’t interact in this way. Keep in mind that electrons are light and very mobile. The electron orbiting the proton is doing so at high-speed and can easily adjust its orbit to absorb the much slower movements of the free electron on the right. Furthermore it will be shown in a later chapter that electrons in both objects will automatically synchronise themselves to absorb the motion of the other electrons around them. As a result the increased attraction of the free electron toward the proton is what dominates (because the proton is heavy and immobile) and this creates a net attractive force that we know as static electricity.
#### Anomalies explained
This explains the various anomalies above, such as why it is possible for a strongly negative object to attract a weakly negative one (point 5). The VDCL attractive force in this case is stronger than the ordinary repulsive Coulombs force.
It also explains why the positively charged fur (in point 6) doesn’t attract neutral objects like the plastic does. The fur has excess protons but these protons are not mobile like the free electrons in the plastic. So the protons can’t create an attractive VDCL force. Now it can be also observed that positively charged objects have a weak attraction toward neutral objects. This might be explained by polarization or by the fact that even neutral objects have a few free electrons floating about.
#### Conclusions
Many aspects of static electric attraction cannot be properly explained by polarization. The Velocity Dependent Coulombs Law is able to account for these anomalies and is more likely the proper explanation for how static electricity works.
Previous Contents Next chapter
| 2,231 | 11,307 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.515625 | 4 |
CC-MAIN-2018-09
|
latest
|
en
| 0.947813 |
http://mathhelpforum.com/differential-geometry/73011-solved-complex-numbers-cos-pi-5-a.html
| 1,481,150,896,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698542250.48/warc/CC-MAIN-20161202170902-00495-ip-10-31-129-80.ec2.internal.warc.gz
| 180,528,586 | 10,486 |
# Thread: [SOLVED] Complex Numbers- cos ( pi/5)
1. ## [SOLVED] Complex Numbers- cos ( pi/5)
I am trying to prove that cos(pi/5)= (1+sqrt(5))/4
I tried finding the 5th roots of unity and then just using the real part cis(pi/5) but then i dont know how to get it = (1+sqrt(5))/4.
I am stuck somewhere here.
2. i am still confused though. that website is for sin(pi/5) i let x=cos(theta) and now i have (x+y)^5 and i expanded it out. but now i am still stuck.
3. ## Cos pi/5
Hello kayla09
nzmathman has shown you how to find $\sin \pi/5$. Now use
$\cos^2(\pi/5) = 1 - \sin^2(\pi/5)$
$= 1 - \frac{5-\sqrt 5}{8}$
$= \frac{3+\sqrt 5}{8}$
$= \frac{6+2\sqrt 5}{16}$
$= \frac{1+2\sqrt 5+5}{16}$
$= \frac{(1+\sqrt 5)^2}{4^2}$
$\Rightarrow \cos(\pi/5) =\frac{1+\sqrt 5}{4}$, taking the positive square root.
| 314 | 810 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.859375 | 4 |
CC-MAIN-2016-50
|
longest
|
en
| 0.742867 |
https://nrich.maths.org/public/topic.php?code=-779&cl=3&cldcmpid=83
| 1,586,126,295,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585371611051.77/warc/CC-MAIN-20200405213008-20200406003508-00479.warc.gz
| 609,407,779 | 7,679 |
# Resources tagged with: Resilient
Filter by: Content type:
Age range:
Challenge level:
### There are 36 results
Broad Topics > Habits of Mind > Resilient
### Difference Sudoku
##### Age 14 to 16 Challenge Level:
Use the differences to find the solution to this Sudoku.
### Number Daisy
##### Age 11 to 14 Challenge Level:
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
### Reach 100
##### Age 7 to 14 Challenge Level:
Choose four different digits from 1-9 and put one in each box so that the resulting four two-digit numbers add to a total of 100.
### Product Sudoku
##### Age 11 to 16 Challenge Level:
The clues for this Sudoku are the product of the numbers in adjacent squares.
### Estimating Angles
##### Age 7 to 14 Challenge Level:
How good are you at estimating angles?
### Tower of Hanoi
##### Age 11 to 14 Challenge Level:
The Tower of Hanoi is an ancient mathematical challenge. Working on the building blocks may help you to explain the patterns you notice.
### M, M and M
##### Age 11 to 14 Challenge Level:
If you are given the mean, median and mode of five positive whole numbers, can you find the numbers?
### Cuboids
##### Age 11 to 14 Challenge Level:
Can you find a cuboid that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### Factors and Multiples Game
##### Age 7 to 16 Challenge Level:
A game in which players take it in turns to choose a number. Can you block your opponent?
### Triangles to Tetrahedra
##### Age 11 to 14 Challenge Level:
Imagine you have an unlimited number of four types of triangle. How many different tetrahedra can you make?
### Charlie's Delightful Machine
##### Age 11 to 16 Challenge Level:
Here is a machine with four coloured lights. Can you develop a strategy to work out the rules controlling each light?
### Isosceles Triangles
##### Age 11 to 14 Challenge Level:
Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?
### Nine Colours
##### Age 11 to 16 Challenge Level:
Can you use small coloured cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of each colour?
### Elevenses
##### Age 11 to 14 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Squares in Rectangles
##### Age 11 to 14 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### Frogs
##### Age 11 to 14 Challenge Level:
How many moves does it take to swap over some red and blue frogs? Do you have a method?
### Two and Two
##### Age 11 to 16 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
### Where Can We Visit?
##### Age 11 to 14 Challenge Level:
Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
##### Age 11 to 16 Challenge Level:
The items in the shopping basket add and multiply to give the same amount. What could their prices be?
### American Billions
##### Age 11 to 14 Challenge Level:
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
### Reflecting Squarely
##### Age 11 to 14 Challenge Level:
In how many ways can you fit all three pieces together to make shapes with line symmetry?
### Number Lines in Disguise
##### Age 7 to 14 Challenge Level:
Some of the numbers have fallen off Becky's number line. Can you figure out what they were?
### Substitution Cipher
##### Age 11 to 14 Challenge Level:
Find the frequency distribution for ordinary English, and use it to help you crack the code.
### Wipeout
##### Age 11 to 16 Challenge Level:
Can you do a little mathematical detective work to figure out which number has been wiped out?
### A Little Light Thinking
##### Age 14 to 16 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Funny Factorisation
##### Age 11 to 16 Challenge Level:
Using the digits 1 to 9, the number 4396 can be written as the product of two numbers. Can you find the factors?
### Factors and Multiples Puzzle
##### Age 11 to 14 Challenge Level:
Using your knowledge of the properties of numbers, can you fill all the squares on the board?
### Kite in a Square
##### Age 14 to 16 Challenge Level:
Can you make sense of the three methods to work out the area of the kite in the square?
### Odds and Evens
##### Age 11 to 14 Challenge Level:
Are these games fair? How can you tell?
### In a Box
##### Age 14 to 16 Challenge Level:
Chris and Jo put two red and four blue ribbons in a box. They each pick a ribbon from the box without looking. Jo wins if the two ribbons are the same colour. Is the game fair?
### Data Matching
##### Age 14 to 18 Challenge Level:
Use your skill and judgement to match the sets of random data.
### Same Length
##### Age 11 to 16 Challenge Level:
Construct two equilateral triangles on a straight line. There are two lengths that look the same - can you prove it?
##### Age 11 to 14 Challenge Level:
What happens when you add a three digit number to its reverse?
### Overlaps
##### Age 11 to 14 Challenge Level:
Can you find ways to put numbers in the overlaps so the rings have equal totals?
### Can You Make 100?
##### Age 11 to 14 Challenge Level:
How many ways can you find to put in operation signs (+ - x รท) to make 100?
| 1,421 | 5,935 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.875 | 4 |
CC-MAIN-2020-16
|
latest
|
en
| 0.845441 |
https://www.studypool.com/discuss/477501/somebody-please-explain-factoring-to-me?free
| 1,495,682,352,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463607963.70/warc/CC-MAIN-20170525025250-20170525045250-00285.warc.gz
| 950,200,218 | 14,219 |
##### somebody please explain factoring to me
Algebra Tutor: None Selected Time limit: 1 Day
studying for my g.e.d. and need factoring explained to me, please show me how to solve this in detail: 3c^2de^2+6cde^2+3de^2
Apr 15th, 2015
First compare the coefficients 3, 6, and 3. They have the common factor 3 and this is the greatest common factor (gcf) for them. Then look at the letters in the expressions 3c^2de^2, 6cde^2, and 3de^2. c is not present in in the third term, so it will not be present in the gcf. The smallest power of d for these three terms is d (or d^1) and the smallest power of e is the second one, e^2. We write the gcf as 3de^2.
Indeed 3c^2de^2 = 3de^2 * c^2; 6cde^2 =3de^2 * 2c; 3de^2 = 3de^2 * 1.
We can also factor the expression as 3c^2de^2 + 6cde^2 + 3de^2 = 3de^2 ( c^2 + 2c + 1).
Note that the expression in parentheses is a complete square c^2 + 2*c*1 + 1^2 = (c + 1)^2, so the complete factoring will look like 3c^2de^2 + 6cde^2 + 3de^2 = 3de^2 ( c + 1)^2.
Apr 15th, 2015
thank you
Apr 15th, 2015
You are welcome.
Apr 15th, 2015
...
Apr 15th, 2015
...
Apr 15th, 2015
May 25th, 2017
check_circle
| 449 | 1,142 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.125 | 4 |
CC-MAIN-2017-22
|
longest
|
en
| 0.85434 |
https://beautyfll.com/what-is-1l-in-grams/
| 1,685,887,237,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224649986.95/warc/CC-MAIN-20230604125132-20230604155132-00677.warc.gz
| 147,921,724 | 22,354 |
# What is 1l in grams?
How many grams is a liter?
Volume in Liters: Weight in Grams of:
Water Cooking Oil
1 l 1,000 g 880 g
2 l 2,000 g 1,760 g
3 l 3,000 g 2,640 g
03/12/2021
Regarding this, What is 0.05 ml on a syringe? Although it is labeled in “units” at the bottom of the syringe, each unit actually is one-hundredth of a milliliter (0.01 ml or 0.01 cc). Each small black mark equals 0.01 ml. A larger black mark and a number is found every 0.05 ml (i.e.,five-hundredths of a ml).
Is 1kg equal to 1 liter? One litre of water has a mass of almost exactly one kilogram when measured at its maximal density, which occurs at about 4 °C. It follows, therefore, that 1000th of a litre, known as one millilitre (1 mL), of water has a mass of about 1 g; 1000 litres of water has a mass of about 1000 kg (1 tonne or megagram).
Accordingly, How many ml is 5 grams?
Gram to Milliliter Conversion Table
Weight in Grams: Volume in Milliliters of:
Water Granulated Sugar
5 g 5 ml 7.1429 ml
6 g 6 ml 8.5714 ml
7 g 7 ml 10 ml
## What is 1 liter of milk in grams?
How much is 1 Litres in grams?
Volume in Liters: Weight in Grams of:
Water Cooking Oil
1 l 1,000 g 880 g
2 l 2,000 g 1,760 g
3 l 3,000 g 2,640 g
19/12/2021
How much is 0.5 on a syringe? For instance, your syringe may be marked with a number at every successive mL. In between you’ll see a mid-sized line that marks half mL units, like 0.5 milliliters (0.02 fl oz), 1.5 mL, 2.5 mL, and so on. The 4 smaller lines between every half mL and mL line each mark 0.1 mL.
How do you read a 0.5 mL syringe?
What is .5 mL on an insulin syringe? The number lines in an insulin syringe, measured in milliliters (mL), stand for the following: 0.3 mL syringes are for insulin doses under 30 units of insulin and are numbered at 1-unit intervals. 0.5 mL syringes are for 30 to 50 units of insulin and are numbered at 1-unit intervals.
## How many kilos is 5 Litres?
Liter to Kilogram Conversion Table
Volume in Liters: Weight in Kilograms of:
Water All Purpose Flour
5 l 5 kg 2.645 kg
6 l 6 kg 3.174 kg
7 l 7 kg 3.703 kg
What is 5 kg in Litres? Kilogram to Liter Conversion Table
Weight in Kilograms: Volume in Liters of:
Water Granulated Sugar
5 kg 5 l 7.1429 l
6 kg 6 l 8.5714 l
7 kg 7 l 10 l
Is 1l of milk 1kg? The density of milk is approximately 1.03 kilograms per litre so a litre of milk weighs very close to 1 kilogram.
Is 5g 5ml?
Contrary to what the other answer states, you can easily switch between grams and ml because they were meant to support each other so 1000 ml or 1 liter of water equates to 1000g or 1kg of water therefore in your case, 5ml would be synonymous to 5g.
## What is 5g in tablespoons?
Grams and tablespoons for sugar (granulated)
Grams to tablespoons Tablespoons to grams
50 grams = 4 tbsp 5 tbsp = 62.5g
60 grams = 4.8 tbsp 6 tbsp = 75g
70 grams = 5.6 tbsp 7 tbsp = 87.5g
80 grams = 6.4 tbsp 8 tbsp = 100g
Is 1g equal to 1 ml? The conversion from grams to ml for water is extremely easy. One gram of pure water is exactly one milliliter.
Is milk heavier than water?
A gallon is a measurement of volume and density is directly proportional to the mass of a fixed volume. Milk is about 87% water and contains other substances that are heavier than water, excluding fat. A gallon of milk is heavier than a gallon of water.
How many L are in a kg? Liter to Kilogram Conversion Table
Volume in Liters: Weight in Kilograms of:
Water Milk
1 l 1 kg 1.03 kg
2 l 2 kg 2.06 kg
3 l 3 kg 3.09 kg
## How many glasses of water is 1l?
Answer: One liter is equal to 4 glasses of water.
Let us understand this with the following explanation. Explanation: Although the capacity of a glass varies since it does not have a defined standard size. However, we consider the capacity of a glass of water to be equal to 8 ounces, and 1 liter is equal to 32 ounces.
How much is 0.6 mL in a syringe? measurement of medicines
0.2 ml 1/2 of the 0.4 mark
0.4 ml the first mark on the dropper
0.6 ml halfway between the 0.4 and 0.8 marks
0.8 ml the second mark on the dropper
1.0 ml the second mark on the dropper plus half of the first mark
What is 1 mL on an insulin syringe?
Syringe size and units
Syringe size Number of units the syringe holds
1/4 mL or 0.25 mL 25
1/3 mL or 0.33 mL 30
1/2 mL or 0.50 mL 50
1 mL 100
What are the 3 types of syringes? What are the types of Syringes?
• Insulin Syringe. One of the more common types of syringes, these are for single-use and are inexpensive.
• Tuberculin Syringe. Tuberculin syringes are small in size and hold up to 1ml of fluid.
• Multi-Shot Needle Syringe.
• Venom Extraction Syringe.
• Oral Syringe.
• Dental Syringe.
### Is 0.5 and 0.50 the same?
Equivalent decimals are decimal numbers that have the same value. For example, 0.5 and 0.50 are equivalent decimals. You can see in the models below that five tenths and fifty hundredths take up the same amount of space.
How much is 0.35 mL in a syringe? The number corresponds to the number of units per mL. So, if you are using a U-100 syringe, yes, 0.35 mL is the same as 35 Units. However if your syringe (or insulin) is U-40, then 35 Units would be equivalent to 0.88 mL.
We will be happy to hear your thoughts
Enable registration in settings - general
| 1,612 | 5,248 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.734375 | 4 |
CC-MAIN-2023-23
|
latest
|
en
| 0.894187 |
https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-006-introduction-to-algorithms-spring-2020/lecture-videos/lecture-10-depth-first-search/
| 1,643,371,610,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320305494.6/warc/CC-MAIN-20220128104113-20220128134113-00323.warc.gz
| 466,348,839 | 66,146 |
# Lecture 10: Depth-First Search
Flash and JavaScript are required for this feature.
This class builds on the previous lecture of breadth-first search (BFS) by introducing depth-first search (DFS) and full-BFS and full-DFS. The lecture continues with topological sorts and cycle detection.
Instructor: Justin Solomon
[CREAKING] [CLICKING] [SQUEAKING] [RUSTLING] [CLICKING]
JUSTIN SOLOMON: We just started a new unit on graph theory, which is going to be sort of our focus for the next couple of lectures in 6006. And so I thought we'd give it a little bit of review at the beginning the lecture because, as usual, I've muddled together a lot of notions in our previous lecture, and then start with some new ideas.
So basically, in our previous lecture, we talked about an algorithm called breadth-first search. And then almost always you see that paired with a second algorithm called depth-first search. And following tradition, and basically logic, we'll do the same thing in 006 today. But in any event, for today we'll stick to the technical material.
So as a little bit of review, I guess actually, the one thing I didn't do on this slide was actually draw a graph. So we should probably start with that. So if you recall, graph is a collection of nodes or vertices depending-- I don't know, is it like a European American thing or something-- and edges.
So here's an example, which as usual, I'm not managing to draw particularly clearly. So this graph is kind of like a cycle. So I have directed edges here, here, here, and here. And of course, there are many kind of variations on the theme, right?
So our basic sort of definition of a graph is that we have some set V, which is like the set of vertices. And then we have a set E, which is set of edges. And this was a subset of V cross V. And this is nothing more than fancy notation for saying that an edge is a pair of vertices, like a from and a to vertex.
Of course, there are many variations on this theme. You could have a directed versus an undirected graph. So this one is directed, meaning the edges look like arrows. If they didn't have arrowheads, they'd be undirected.
We define something called a simple graph where you have essentially no repeated edges. So for instance, you can't do something like this where you have the same edge twice.
And then there are a couple of different definitions that were kind of useful. So in particular-- I'm going to erase this, whoops-- useless edge here. Maybe make my graph slightly more interesting. So add another edge going in the reverse direction.
So maybe I have-- I'm going to give my vertices labels. x, y, z, and w. Then we talked about the neighbors of a given vertex, which are the vertices that you can reach by following edges in or out of your vertex. So in particular, the outgoing neighbors, which we sort of implicitly defined in our previous lecture but didn't call it out, we're going to notate with Adj+.
And these are all of the things that you can reach by going out of a vertex into the next one. So for example, Adj+ of w is going to be the set of vertices. We'll notice I can get from w to y and also from w to z. Yeah. So. Nope, nope. y comma z. OK.
So to continue just our tiny amount of review for the day, remember that a graph-- there are many different ways to represent a graph. The sort of brain dead one would be just like a big long list of edges. But of course, for our algorithms it's not a particularly efficient way to check things like, does this edge exist in my graph.
So the basic representation that I think we're mostly working from in this course is to think of a graph like a set of vertices, each of which maps to another set of vertices. So roughly every vertex maybe stores its outgoing set of edges. And so this is kind of nice because, of course, very quickly we can answer questions like, is this edge inside of our graph? Or we can iterate over the neighbors of a vertex and so on, which are the kind of typical things that we do in a lot of graph algorithms.
And then finally, in our previous lecture, we started talking about paths. So a path is like a chain of vertices that can get me from one vertex to the other only following edges of my graph. There is a term that I think I forgot to define last time because it didn't really matter a ton, which is a simple path, which is just a path that doesn't have the same vertex more than once.
And then, of course, there are many different questions you could ask about a graph that are basically different problems involving computing paths. So for instance, the shortest path between two vertices is sort of our canonical one in graph theory. Or you could ask questions about reachability and so on.
So there's our basic review from our previous lecture. Does our course staff have any questions about things so far? Excellent.
OK. And there's one additional piece of terminology that I fudged a little bit last time-- or rather, my co-instructor suggested a bit of an attitude adjustment. So I thought I'd better clarify really quick.
There's this interesting phrase, linear time, which we all know and love in computer science theory. And this sort of implicit thing, especially in this course, is that when we say linear time, we mean in the size of the input. Right? And so if we have a linear time graph algorithm, well, how much space does it take to store a graph? Well, we need a list of vertices and a list of edges, if nothing else.
So a reasonable way to interpret this phrase linear time is that it's an algorithm that looks like what we've shown on the screen. The times proportional to maybe the sum of the number of vertices and the number of edges. If that makes you uncomfortable like it does for me because one of these can kind of scale on the other, I think it's always fine to add more detail. Right? So if you want to say, linear in the sum of the number of vertices and edges, that's perfectly fine.
But if you see this phrase, that's how you should interpret it. Hopefully that's a fair way to put it. Excellent.
OK. So last time, we talked about an algorithm called breadth-first search-- BFS, for those in the know. Breadth-first search is an algorithm. And the reason we use the word breadth is because it's kind of, remember, we talked about level sets last time because we talked about breadth-first search in the context of computing shortest paths.
And in particular, we have our source node all the way on the left-hand side. And then breadth-first search constructed all the nodes that were distance 1 away. Right. That's the first level set, and then all the distance 2 away, and then all the distance 3 away, and so on. So in particular, the level set L3 isn't visited until we're completely done with level set L2.
Today, we're going to define another algorithm, which is called depth-first search, which doesn't do that, but rather, starts with its first vertex and just starts walking all the way out until it can't do that anymore. And then it kind of backtracks. That's one way to think about it.
And so somehow, in breadth-first search, we're like, drawing concentric circles. In depth-first search, we're doing the opposite. We're like, shooting outward until we reach the outer boundary, and then exploring the graph that way.
OK. And these are sort of the two extremes in terms of graph search kind of techniques that are typically used under the basic building blocks for algorithms in graph theory.
So in order to motivate and think about depth-first search, we're going to define a second problem, which is closely related to shortest path, but not exactly the same. And that's the reachability problem.
So here I have the world's simplest directed graph. So the black things are the edges. And the circles are the nodes or the vertices. And I've marked one special node in blue. And his name is the source node.
And now the question I want to ask is, what are all of the other nodes in my graph that I can reach by following edges-- directed edges-- starting with the source? So obviously, I can get to the node in the lower right, no problem. And of course once I get there, I can traverse and edge upward to get to that second green vertex.
Notice that I was really sneaky and evil, and I drew edges in this graph that might make you think that the red node is reachable. The red one being on the upper left. I'm realizing now that for colorblind people, this isn't a great slide.
But of course, because all the edges from the red vertex on the left here point out, I can't actually reach it from the blue source node. So the reachability problem is just asking, which nodes can I reach from a given source? Pretty straightforward, I think.
Of course, there are many ways to solve this. Right? In fact, one way we could do it would be to use our previous lecture. We could compute the shortest path distance from the source to all the other nodes. And then what would the length of the shortest path from the source to an unreachable node be? Any thoughts from our audience here?
Infinity. Thank you, Professor Demaine. Right.
So in addition to this, of course, a totally reasonable question, thinking back to our shortest path lecture, there are sort of two queries we might make. Right? One is just what is the length of the shortest path? The other is like, what is the actual shortest path from the source to a given vertex?
We can ask a very similar thing here, which is like, OK. You tell me that the green guy is reachable, but how? Give me a path as evidence or a certificate, if you want to be fancy about it.
So in order to do that, just like last time, remember, we defined a particular data structure that was the shortest path tree. We can do something very similar here. In particular, this is like the extent of my PowerPoint skills here. If I have a reachability problem, I can additionally store-- I can decorate every node in my graph with one other piece of information, which is the previous node along some path from my source to that thing. Right?
And just like last time, if I want to get an actual path from the source to w, what could I do? I can start with w and then just keep following those parent relationships until I get back to the source. Then if I flip the order of that list of vertices, I get a path from the source to the target that's valid.
So this object is called a path tree, just like we talked-- or a parent tree, rather. Just like we talked about in our last lecture, there's no reason why this thing should ever have a cycle in it. It's certainly a tree.
Right. So that's the basic reachability problem. And in addition to that, we can compute this object P, which is going to give me sort of information about how any given node was reachable. There's a slight difference between the parent tree that I've defined here and the shortest path tree, which I defined last time, which is, I'm not going to require that the shortest path I get-- oh, man-- the path I get when I backtrack along my tree P is the shortest path, it's just a path because for the reachability problem, I actually don't care.
Like, I could have a weird, circuitous, crazy long path. And it still tells me that a node is reachable. Right.
So that's our basic set up and our data structure. And now we can introduce a problem to solve reachability. Again, we already have an algorithm for doing that, which is to compute shortest paths. And remember that our shortest path algorithm from previous lecture took linear time and the size of the input. It took v plus e time.
Now the question is, can we do a little better? The answer, obviously, is yes, because I just asked it, and I gave you this problem. OK.
And here's a technique for doing that, which unsurprisingly, is a recursive algorithm. I'm going to swap my notes for my handwritten notes. And this algorithm is called depth-first search. And here's the basic strategy. I'm going to choose a source node and label that Node 1 here. I suppose it actually would have made sense for me to actually 0 index this. Maybe in the slides I'll fix it later.
But in any event, I'm going to mark my source node. And now I'm going to look at every node, every edge coming out of that node, and I'm going to visit it recursively. So that's our sort of for loop inside of this function visit.
And then for each neighboring node, if I haven't visited it before, in other words, I currently haven't given it a parent. That's our if statement here. Then I'm going to say, well, now they do have a parent. And that parent is me. And I'm going to recurse.
You guys see what this is doing? It's kind of crawling outward inside of our graph. So let's do the example on the screen.
And I purposefully designed this experiment-- or this example-- to look a little bit different from breadth-first search, at least if you choose to do the ordering that I did. So here's our graph. 1, 2, 5, 3, 4. OK.
And let's think about the traversal order that the depth-first search is going to do. Right. So here's our source. And now what does the source do? It rec-- so let's think about our recursion tree. So we have the source all the way up in here. And now he's going to start calling the visit function recursively.
So. And I'll go ahead and number these the same as on the screen. Well, he has one outgoing neighbor, and it hasn't been visited yet. So of course, the very first recursive call that I'll make is to that neighbor 2.
Now the neighbor 2 also recurses. Hopefully this kind of schematic picture makes some sense, what I'm trying to draw here. And well now, the 2 has two neighbors, a 3 and a 5. So let's say that we choose 3 first.
Well, the 3 now recurses and calls 4. And then the recursion tree is kind of done. So now it goes back out.
And then finally, well, now, the 3-- or, oh, boy. Yeah. The 2 looks at his next neighbor, which is the 5 and visits that recursively.
Notice that this is not following the level sets. Right? The depth-first search algorithm got all the way to the end of my tree in the recursive calls and then kind of backed its way out to the 2 before calling the 5. These are not the same technique. One goes all the way to the end and then kind of backtracks. When I say backtrack, what I mean is the recursion is kind of unraveling. Whereas in breadth-first search, I visit everything in one level set before I work my way out.
That distinction make sense? OK. So of course, we need to prove that this algorithm does something useful. So let's do that now.
So in particular, we need a correctness proof. So our claim is going to be that-- let's see here-- the depth-first search algorithm visits all, I guess reachable v, and that it correctly sets the parent in the process.
OK. So in order to prove this, of course, as with almost everything in this course, we're going to use induction. And in particular, what we're going to do is do induction on the distance from the source. So we're going to say that, like, for all vertices in distance k from the source, this statement is true. And then we're going to prove this inductively on k. OK?
So we want to do induction on k, which is the distance to the source vertex. So as with all of our inductive proofs, we have to do our base case and then our inductive step. So in the base case, k equals 0. This is a hella easy case because, of course, what is the thing that is distance 0 from the source? It's the source! Yeah.
And take a look at our strategy all the way at the top of the slide. We explicitly set the correct parent for the source, and in some sense, visit it because the very first thing we do is call visit of s. So there's kind of nothing to say here. Yeah? Or there's plenty to say if you write it on your homework. But your lazy instructor is going to write a check mark here. OK.
So now we have to do our inductive step. So what does that mean? We're going to assume that our statement is true for all nodes within a distance k. And then we're going to prove that the same thing is true for all nodes within a distance k plus 1. OK.
So let's do that. Let's consider a vertex v that's distance k plus 1 away. So in other words, the distance from the source to v is equal to k plus 1. And what's our goal? Our goal is to show that the parent of v is set correctly. Yeah?
What was that?
AUDIENCE: [INAUDIBLE].
JUSTIN SOLOMON: Oh, sorry. I forgot that the distances in this class are in order. Yeah. That's absolutely right. So it should be the distance from s to v. Yeah. Sorry. I'm really not used to thinking about directed graphs. But that's a good fix. OK.
So now what can we do? Well, there's this number is distance here. So in particular, the shortest path from s to v. So remember our argument last time that essentially, when we look at shortest path and we kind of truncate by 1, it's still shortest path? That property doesn't matter so much here. But at least we know that there's another vertex on the path, which is 1 distance from one less away.
So let's take u, which is also a vertex, to be the previous node on the shortest path from s to v. Right. And so in particular, we know that the distance from s to u is equal to k. And conveniently, of course, by our inductive hypothesis here, we know that our property is true for this guy. OK.
So now our algorithm, what do we know? Well, because our property is true, the visit function at some point in its life is called on this vertex u. That's sort of what our induction assumes.
So we have two cases. Right. So when we visit u, we know that when we call this visit function, well, remember that v kind of by definition is in Adj+ of u. Right. So in particular, DGS is going to consider v when it gets called. OK.
And now there's two cases. Right? So either when this happens, P of v does not equal None. Right.
Well, what does that mean? Well, it means that we already kind of found a suitable parent for v. And we're in good shape. Otherwise, p of v does equal None. Well, in this case, the very next line of code correctly sets the parent. And we're all set.
So in both of these two cases, we show that the parent of u was set correctly either by that line of code right here or just previously. And so in either case, our induction is done. All right. I guess given the feedback I received from our previous lecture, we now can end our LaTeX suitably.
OK. So what did we just show? We showed that the depth-first search algorithm can dig around in a graph and tell me all of the things that are searchable, or rather, are reachable from a given source, just basically by calling visit on that source and then expanding outward recursively. OK. So I think this is certainly straightforward from an intuitive perspective. It's easy to get lost when you write these kind of formal induction proofs because they always feel a tiny bit like tautology. So you should go home and kind of convince yourself that it's not.
OK. So of course, what do we do in this class? We always follow the same kind of boring pattern. The first thing we do, define an algorithm. Second thing we do, make sure that it's the right algorithm. What's the third thing we need to do?
AUDIENCE: Analyze it.
JUSTIN SOLOMON: Analyze it. That's right. In particular, make sure that it finishes before the heat death of the universe. And indeed, depth-first research doesn't really take all that long, which is a good thing.
So let's think about this a bit. So what's going to end up happening in depth-first search, well, we're going to visit every vertex at most once, kind of by definition here. And in each case, we're going to just visit its neighboring edges.
Can we ever traverse an edge more than one time? No. Right. Because the visit function only ever gets called one time per vertex. And our edges are directed. Right. So kind think about the from of every edge, the from vertex is only ever visited one time. And hence, every edge is only visited one time.
Do we ever visit-- ah, yes.
AUDIENCE: Does DFS work for an undirected graph?
JUSTIN SOLOMON: An undirected graph. Absolutely. So there's sort of different ways to think about it. One is to think of an undirected graph like a directed graph with two edges pointed either way, which I think is in this class how we actually kind of notated it in the previous lecture.
Yeah. Actually, that's probably a reasonable way to reduce it. So let's stick with that.
Right. Now, does DFS ever visit a vertex that is not reachable from the source? Well, the answer is no because all I ever do is recursively call on my neighbors. And so kind of by definition, if I'm not reachable, DFS will never see it. So if I think about my runtime carefully, it's not quite the same as breadth-first search.
Remember that breadth-first search took v plus e time. In depth-first search, it just takes order e time because I'm expanding outward from the source vertex, hitting every edge adjacent to every vertex that I've seen so far. But I never reach a vertex that I haven't-- that isn't reachable. Right?
And so because this only ever touches every edge one time, we're in good shape. And I see a question here. Yeah.
AUDIENCE: Does BFS reach vertices that are not reachable?
JUSTIN SOLOMON: Does BFS reach vertices that are not reachable? I guess not, now that you mention it. But at least in my boring proof of order v time last time, our very first step of BFS, reserve space proportional to v, which is enough to already make that runtime correct. Good question. Yeah.
So I guess the way that we've talked about it where you can stretch one little set after a time, if you think of that as reachability, then no. It doesn't reach it in the for loop. But just by construction, when we started we already took the time that we're talking about here.
So notice these run times aren't exactly the same. So for example, if my graph has no edges, BFS still is going to take time because it still has to take order v time, at least in the sort of brain-dead way that we've implemented it last time. Obviously, in that case, we could probably do something better. Whereas the way that we've defined the DFS algorithm, it only takes edge time.
I see confusion on my instructor's face. No? OK. Good.
The one thing to notice is that these are algorithms for slightly different tasks in some sense. The way that we wrote down breadth-first search last time, conveniently, it gives us the shortest path. There are breadth-first search algorithms that doesn't. I think in this class we kind of think of breadth-first search-- we motivate it in terms of the shortest path problem. But it's just kind of a strategy of working outwards from a vertex.
Whereas here, the way we've written down depth-first search, there's no reason why the path that we get should be the shortest. Right? So to think of a really extreme example, let's say that I have a cycle graph. So I get a big loop like this. Let's say that I do depth-first search starting from this vertex.
Well, what will happen? Well, this guy will call its neighbor recursively, who will then call its neighbor recursively, who will then call his neighbor recursively, and so on. So of course, when I do depth-first search, when I get to this vertex, there's a chain of 1, 2, 3, 4 vertices behind it.
Is that the shortest path from the source to the target here? Well, clearly not. Right? I could have traversed that edge. I just chose not to.
OK. So that's the depth-first search algorithm. It's just essentially a recursive strategy where I traverse all my neighbors, and each of my neighbors traverses their neighbors, and so on.
OK. So why might we want to use this algorithm? Well, we've already solved the reachability problem. So let's solve a few more things using the same basic strategy here.
So there's some notions that we've sort of-- actually, in some sense, already used in the lecture here. But we might as well call them out for what they are, which is this idea of connectivity. So a graph is connected if there's a path getting from every vertex to every other vertex. Right.
Now connectivity in a directed graph is kind of a weird object. Like, for instance, think of a directed graph with just two edges. And one edge goes from u to v. Then I can get from v to u, but not vise versa. That's kind of a weird notion.
So here in 6006 we'll mostly worry about connectivity only for undirected graphs because they're-- the vertices just basically come in like, big connected clumps. Or the more technical term for a big connected clump is a connected component. Yeah?
So let's see an example. So let's say that I have a graph, which has an edge and then a triangle. This is one graph. Do you see that? There's a collection of vertices, and there's a collection of edges.
But it has two connected components-- the guy on the right and the guy on the left, meaning that each vertex here is reachable from every other vertex here. Each vertex here is reachable from every vertex here. But there's no edge that goes from the triangle to the line segment. Yeah?
And so in the connected components problem, we're given a graph like this guy. And initially, we don't, you know-- OK. When I draw it like this, it's pretty clear that my graph has two connected components. Maybe my graph-embedding algorithm failed and it drew an edge like that. Well, then maybe-- I don't know-- it's still pretty obvious that there's two connected components.
But you can imagine a universe where you don't know that a priori. And the problem you're trying to solve is just to enumerate all these clumps of vertices that are reachable from one another in an undirected graph. And conveniently, we can use depth-first search to solve this problem pretty easily. Right?
So how could we do it? Well, in some sense how can we find one connected component? So let's say that I just choose a vertex in my graph. Well, what do I know about everything in its connected component?
Well, it's reachable from that vertex. Remember, we just solved the reachability problem, which says, if I have a vertex, I can now tell you all the other vertices that are reachable from this guy. So I could call DFS on, well, any vertex of this cycle here. Call the reachability thing.
And I know that for every vertex there's one of two things. Either the vertex has a parent in that object P, or it's the source. So I can very easily find the connected component corresponding to that vertex. Does that makes sense?
Have I found all the connected components? No. I found one. I found the one corresponding to the arbitrary vertex that I just chose. So how could I fix this?
Well, it's super simple. I could put a for loop on the outside, which just loops over all the vertices, maybe. And if that vertex is not part of a connected component yet, then I need to make a new one.
So then I call DFS on that vertex. I collect all the vertices that I got. And I iterate.
So this is the algorithm that in this class we're going to call full DFS. By the way, you could do the same thing with full breadth-first search. That's perfectly fine. Just kind of by analogy here.
Right. So what is full D-- oh, this chalk is easier. Well, I'm going to iterate over all my vertices. Where I stands for for loop. Of-- right. So if v is unvisited, then I'm going to do to DFS starting at v. I guess we used visit to refer to this in the previous slide.
And that's going to kind of flood fill that whole connected component. And then I can collect that connected component and continue. We have to be a little bit careful because, of course, we don't want like, checking things-- something to be visited to somehow take a bunch of time and make my algorithm slower than it needs to be. But of course, we have a set data structure that we know can do that and order one time at least in expectation.
OK. So this is the full DFS algorithm. It's really simple. Of DFS because I called DGS on every vertex. And it's full because I looped over all the vertices. Right.
And so if we think about it, how much time does this algorithm take? It's little bit sneaky because somehow I have a for loop over all the vertices. Then I could imagine a universe where I get, like, vertices times some other number because there's a for loop, and then there's something inside of it. I think that's how we're used to thinking about runtime of for loops.
But in this case, that actually doesn't happen because there's never a case where an edge gets traversed more than one time. Because if I'm in one connected component, then by definition, I can't be in another connected component. Right?
And so what happens is, in some sense, this innocent looking call to DFS-- I suppose if you were like a LISP or a programmer, you somehow wouldn't like this. It has a side effect, which is that I marked all the vertices in that connected component as "don't touch me again." Right. And so implicitly I kind of removed edges in this process.
So if you think through it carefully, the runtime of this full DFS algorithm is v plus e time because every edge is touched no more than one time. Kind of amortized over all the different calls to DGS here. And there's this for loop over vertices. So there's clearly an order v that you need here.
Does that argument make sense? So again, we call that linear in the size of the input. I'm going to say it as many times to get it in my own head correctly. OK.
Right. So this is the basic problem. This comes up all the time, by the way. Like, it seems like somehow a totally brain dead weird algorithm. Like, somehow, why would you want an algorithm that finds connected components. Like, why would you even have a graph that's disconnected or something?
But of course, that can happen a lot. So for instance, maybe you work at a social media company, and people have friends. But like, Eric and I are friends. And we're not friends with anybody else. We have a-- there's like, a blood oath kind of thing.
Then that might be not so easy to find in the graph because, of course, we're just two among a sea of students in this classroom, all of which have different interconnections that are just enumerated based on the list of edges. And so even though like, pictorially, it's kind of hard to draw a connecting component algorithm in a way that doesn't make it sound kind of like a useless technique from the start, because it's very clear there are two connected components there. Of course, we still have to be able to write code to solve this sort of thing.
OK. So for once, I think I'm almost on time in lecture today.
So we have one additional application of depth-first search in our class today, which is sort of on the opposite end of the spectrum. So we just talked about graphs that are undirected and thinking about cycles. Now, on the opposite end we might think of a DAG. So a DAG is a Directed Acyclic Graph.
Can anyone think of a special case of a DAG? I suppose I should define it first. And then we'll come back to that question, which means exactly what it sounds like. So it's a graph that has directed edges now and doesn't have any cycles in it.
So actually, the graph I gave you all the way at the beginning of lecture I think secretly was an example of one of these. So let's say that I have directed edges. Maybe if I make the head a triangle, it's a little easier to see. I'm not so sure. In any event, so I'm going to have an edge up and an edge to the right, and similarly, an edge down and an edge to the right.
This graph looks like a cycle. But it's not because the only direction that I can move is from the left-hand side to the right-hand side. So this is a directed graph. And it doesn't contain any cycles, meaning there's no path that it can take from a vertex that gets back to itself along the directed edges.
OK. And DAGs show up all the time. Now that I've defined what a DAG is, can somebody give me an example of a DAG that we've already seen in 6006?
AUDIENCE: A tree.
JUSTIN SOLOMON: A tree. At least if we orient all all the edges kind of pointing downward in the tree. Yeah. Otherwise, it gets kind of debatable over whether it's a DAG or not. If there's no direction to the edges, then somehow the definition just doesn't apply.
OK. So in processing directed acyclic graphs, there's a really useful thing that you can do that's going to show up in this class apparently quite a bit, which is kind of interesting to me, I'm curious to see what that looks like, which is to compute a topological order on the graph. We're at topologies here. So as a geometry professor in my day job, I get all excited.
But in this case, a topological order is a fairly straightforward thing. Actually, it's defined on the screen, and I have bad handwriting anyway. So let's just stick with that.
So topological ordering. So we think of f as a function that assigns maybe every node an index in array. I guess I shouldn't use the word array here. But just like an index, an ordering. So like, this is the first vertex. And this is the second vertex. And so on.
Then a topological order is one that has the properties shown here, which is that if I have a directed edge from u to v, then f of u is less than f of v. So in other words, if I look at the ordering that I get on my topological order, u has to appear before v. Yeah?
Let's look at our example again. So let's give our nodes names. So here's A, B, C, D. Well, what clearly has to be the first node in my topological order?
A. Right. It goes all the way to the left-hand side. Yeah.
Well, after that it's a bit of a toss-up. What do we know? We know that B and C have to appear before D. So maybe just to be annoying, I do A, C, B-- that's a B-- and then D. So it's a topological order. Notice that things that are on the left appear in my graph before things that are on the right, where the word "before" here means that there's an edge that points from one to the other. OK.
By the way, are topological orderings unique? No. So if we look at our graph example here, ABCD is also a topological order. And what that means is somehow very liberating. It means that when we design an algorithm for finding a topological order, so there's some design decisions that we can make. And we just have to find one among many.
But in any event, we're going to define a slightly different notion of order. And then we're going to show that they're closely linked to each other. And that is the finishing order.
So in the finishing order, we're going to call full DFS on our graph. Remember, that means we iterate over all our nodes. And if we haven't seen that node yet, we call DFS on it. And now we're going to make an order in which as soon as the call to a node in that visit function is complete, meaning I've already iterated over all my neighbors, then I add my node to the ordering.
That make sense? It's like a little bit backward from what we're used to thinking about. So it's the order in which full DFS finishes visiting each vertex. Yeah?
And now here's the claim, is that if we have a reverse finishing order, meaning that we take the finishing order and then we flip it backward. That's exactly going to give us a topological ordering of the vertices in our graph. Right.
So let's do that really quickly. So in other words, our claim here-- I think, yeah, let's see-- is that if I have a directed graph. So G is a DAG. Then let's see here. Then the-- oops. My notes are backwards. So I should switch to my-- Jason's notes, which of course, are correct.
Right. So if I have a graph that's a DAG, then the reverse of the finishing order is a topological order.
By the way, we're not going to prove the converse that if I have a topological order, that somehow that thing is the reverse of DFS, at least the way that maybe I coded it. There's a slightly different statement, which is, does there exist a DFS that has that ordering? But that's one that we'll worry about another time around piazza or whatever.
OK. So let's see here. So we need to prove this thing. So what are we going to do?
Well, what do we need to check is the topological order is that if I look at any edge of my graph, it obeys the relationship that I have on the screen here. So in particularly, we're going to take uv inside of my set of edges. And then what I need is that u is ordered before v using the reverse of the finishing order that we've defined here. OK.
So let's think back to our call to the DFS algorithm, where call this visit function. Right. So we have two cases. Either u is visited before v. Or it ain't. Yeah.
So let's do those two cases. So case Number 1 is, u is visited before v. OK.
All right. So what does that mean? Well, remember that there's an edge. Like, pictorially, what's going on?
Well, there's all kinds of graph stuff going on. And then there's u. And we know that there's a directed edge from u to v. That's our picture. Right? And maybe there's other stuff going on outside of us.
So in particular, well, just by the way that we've defined that visit function, what do we know? We know that when we call visit on u, well, v is one of its outgoing neighbors. So in particular, a visit on u is going to call visit v. And we know that because well, u is visited before v.
So currently, v's parent is l when I get to you That make sense?
Now, here's where reverse ordering, we're going have to keep it in our head because now, well, visit of u calls visit of v. So notice that visit of v has to complete before visit of u. Right?
V completes before visit of u. Well. So in reverse, sorting-- in reverse finishing order here, what does that mean? Well, if this completes before the other guy, then they get flipped backward in the list, which is exactly what I want because there's an edge from u to v.
OK. So Case 1 is done. Now we have Case 2, which is that v is visited before u.
OK. So now I'm going to make one additional observation. OK. So now I'm going to go back to my other notes because I like my schematic better.
Right. So what's our basic picture here? Oh, no. I-- Oh, you know what it was? I printed out another copy of this. That's OK. I can do it off the top of my head.
OK. So here's my source vertex. His name is S. Now, there's a bunch of edges and whatever. There's a long path. And now eventually, what happens?
Well. I have a node v. And somewhere out there in the universe is another node u. And what do I know?
I know that by assumption, I know that there's an edge from u to v. That make sense? So that's our sort of picture so far.
OK. So what do we know? We know that our graph is acyclic. Yeah? Kind of by definition, that's our assumption.
So can we reach u from v? In other words, does there exist a path from v to u? So that would look like this.
No because our graph is acyclic, and I just drew a cycle. So this is a big X. There's a frowny face here. Can't do it. He has hair, unlike your instructor.
OK. So right. So what does this mean? Well, OK. So by this picture, I suppose, we know that u cannot be reached from v.
Yeah. So what does that mean? Well, it means that the visit to v is going to complete and never see u because remember, the visit to v only ever call things that are kind of descendants of v. So in other words, visit of v completes without seeing u.
Well, that's exactly the same thing that we showed in our first case. Right? So by the same reasoning, what does that mean? In our reverse finishing order, the ordering from u to v is preserved.
OK. So that sort of completes our proof here that reverse finishing order gives me a topological order, which is kind of nice. And so this is a nice convenient way of taking all of the nodes in a directed acyclic graph and ordering them in a way that respects the topology or the connectivity of that graph.
So we're going to conclude with one final problem, which I don't have a slide on. But that's OK. And that's cycle detection.
So there's a bit of an exercise left to the reader here. So the problem that we're looking for now is that we're given a directed graph. There's a G in graph, in case you're wondering. But now, we don't know if it's a DAG or not.
And so the question that we're trying to ask is, does there exist a cycle in our directed acyclic graph? So we're just given our graph, and we want to know, can we do this?
Let's think through the logic of this a bit. So what do we know? We know that if our graph were a DAG, then I could call DGS, get the ordering out, and then I guess flip its ordering backwards. So I could compute the reverse finishing order. And it would give me a topological order of my graph.
So if I were a DAG, I would get a topological order when I call DFS. So let's say that I ran DFS. I got whatever ordering I got. And now I found an edge the points in the wrong direction.
I can just double check my list of edges, and I find one that does not respect the relationship that I see in the second bullet point here. Can my graph be a DAG?
No. Because if my graph were a DAG, the algorithm would work. I just proved it to you. Right?
So if my graph were a DAG, then I could do reverse finishing order. And what I would get back is a topological order. So if I found a certificate that my order wasn't topological, something went wrong, and the only thing that could go wrong is that my graph isn't a DAG. Yeah. Isn't a DAG.
In fact, sort of an exercise left to the reader and/or to your section-- do we still have section? I think we do, as of now-- is that this is an if and only if, meaning that the only time that you even have a topological ordering in your graph is if your graph is a DAG. This is a really easy fact to sanity check, by the way. This is not like, a particularly challenging problem.
But you should think through it because it's a good exercise to make sure you understand the definitions, which is to say that if you have a topological order, your graph is a DAG. If you don't have a topological order, your graph isn't a DAG.
So in other words, we secretly gave you an algorithm for checking if a graph is a DAG at all. Right? What could I do? I could compute reverse finishing order. Check if it obeys the relationship on the second bullet point here for every edge. And if it does, then we're in good shape. My graph is a DAG.
If it doesn't, something went wrong. And the only thing that could have gone wrong is not being a DAG.
OK. So in other words, secretly we just solved-- well, I guess the way that I've written it here, we've solved the cycle detection problem here, which is to say that, well, I have a cycle if and only if I'm not a DAG, which I can check using this technique. Of course, the word "detection" here probably means that I actually want to find that cycle, and I haven't told you how to do that yet.
All we know how to do so far is say, like, somewhere in this graph there's a cycle. And that's not so good. So we can do one additional piece of information in the two minutes we have remaining to sort of complete our story here, which is to modify our algorithm ever so slightly to not only say thumbs up, thumbs down, is there a cycle in this graph, but also to actually return the vertices as a cycle.
And here's the property that we're going to do that, which is following, which is that if G contains a cycle, right, then full DFS will traverse an edge from a vertex v to some ancestor of v. I guess we haven't carefully defined the term "ancestor" here. Essentially, if you think of the sort of the running of the DFS algorithm, then an ancestor is like something that appears in the recursive call tree before I got to v.
OK. So how could we prove that? Well, let's take a cycle. And we'll give it a name. In particular, we'll say that it's a cycle from v0 v1 to vk. And then it's a cycle, so it goes back to v0. OK.
And I can order this cycle any way I want. Notice that if I permute the vertices in this list in a cyclical way, meaning that I take the last few of them and stick them at the beginning of the list, it's still a cycle. That's the nice thing about cycles.
So in particular, without loss of generality, we're going to assume that v0 is the first vertex visited by DFS.
What does that mean? That means, like, when I do my DFS algorithm making all these recursive calls, the very first vertex to be touched by this technique is v0. OK.
Well, now what's going to end up happening? Well, think about the recursive call tree starting at v0. By the time that completes, anything that's reachable from v0 is also going to be complete. Do you see that?
So for instance, v0 somewhere in its call tree might call v2. And notice that v2 was not already visited. So in fact, it will. For v1 I got to call v2 and so on.
And in particular, we're going to get all the way to vertex k. Right? So in other words, we're going to visit a vertex vk. And notice what's going to happen. So remember our algorithm. In fact, we should probably just put it up on the screen would be easier than talking about it a bunch.
Well, vk is now going to iterate over every one of the neighbors of vk. And in particular, it's going to see vertex v0. Right? So we're going to see the edge from vk to v0, which is an edge kind of by definition because we took this to be a cycle here.
But notice that's exactly the thing we set out to prove, namely that full DFS traverses an edge from a vertex to one of its ancestors. Here's a vertex k. Here's the ancestor v0. Why do we know that it's an ancestor? Well, because v0 was called in our call tree before any of these other guys. Right?
So we wanted an algorithm that not only did cycle detection, but also actually gave me the cycle. What could I do? Well, it's essentially a small modification of what we already have. Right.
So-- whoops. Right. If I want to compute topological order or whatever, I can just do DFS. And that'll tell me like, yay or nay, does there exist a cycle.
If I want to actually find that cycle, all I have to do is check that topological order property at the same time that it traversed the graph during DFS. And the second that I find an edge that loops back, I'm done. And so that's our basic algorithm here. And this is a technique for actually just finding the cycle in a graph using the DFS algorithm.
| 10,756 | 46,365 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.578125 | 4 |
CC-MAIN-2022-05
|
latest
|
en
| 0.966018 |
https://mathstat.slu.edu/escher/index.php/Hyperbolic_Geometry_Exploration
| 1,603,531,595,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-45/segments/1603107882103.34/warc/CC-MAIN-20201024080855-20201024110855-00563.warc.gz
| 439,052,896 | 6,412 |
# Hyperbolic Geometry Exploration
Objective: Learn about distortion, geodesics, and triangles in the Poincaré disk model.
NonEuclid is a web applet, written by Joel Castellanos, for doing geometry in hyperbolic space. Go to the NonEuclid page and run the applet. It should display a large white circle - that's the Poincaré disk.
The program opens with a sample drawing. To clear the disk, choose "Clear All" under the orange "Measure/Modify" menu. Select "Draw Line Segment" from the blue "construction" menu. Draw some line segments.
1. What do lines that go through the center of the disk look like?
Clear your drawing, and draw a triangle. On the modify menu, choose "Move Point". Now you can drag the corners of your triangle and see how it changes.
On the measurements menu, select "Measure Triangle", and click the three corners of your triangle. You should see the side and angle measurements of the triangle in the box on the left. Go back to "Move Point" mode.
Move your triangle around to get a feel for the lengths of its sides and the sum of its angles.
1. Draw a triangle which appears large but really has sides of length under 4. Draw a triangle which appears small but really has sides of length over 10.
2. Try to make the angle sum 180°. What do you have to do?
3. Try to make all three sides of the triangle large. What happens to the angle sum?
Draw an infinite geodesic ("Draw Line" on the Constructions menu). Now use "Reflect" to make the reflection of your triangle across your infinite geodesic. Move the geodesic around, and notice the position and size of the triangle's reflection. The reflected triangle is the same size and shape as the original, it just appears different.
1. On your answer sheet, make a sketch showing the two reflected triangles and the reflection line.
If you have time, try these other web sites:
• Hyperbolic Applet by Paul Garrett. (Use "draw" to create a figure with click-and-drag; then use "move" with click-and-drag to move the entire Poincare disk, via hyperbolic translations, much like moving the sphere around in Spherical Easel.)
• Hyperbolic animations. by Jos Leys. (See animations of hyperbolic translations applied to various tessellations of the Poincare disk.)
Handin: A sheet with answers to all questions.
| 529 | 2,290 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.65625 | 4 |
CC-MAIN-2020-45
|
latest
|
en
| 0.89487 |
https://www.vistrails.org/index.php?title=User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates&curid=1679&diff=21096&oldid=21095
| 1,606,169,300,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-50/segments/1606141168074.3/warc/CC-MAIN-20201123211528-20201124001528-00193.warc.gz
| 906,269,120 | 15,523 |
# User:Tohline/Appendix/Ramblings/ConcentricEllipsodalCoordinates
(Difference between revisions)
Revision as of 16:01, 30 October 2020 (view source)Tohline (Talk | contribs) (→Speculation4)← Older edit Revision as of 16:21, 30 October 2020 (view source)Tohline (Talk | contribs) (→Speculation4)Newer edit → Line 1,231: Line 1,231:
- Note that, $~\hat{e}_3 \cdot \hat{e}_3 = 1$, which means that this also is a properly normalized ''unit'' vector. + Note that, $~\hat{e}_3 \cdot \hat{e}_3 = 1$, which means that this also is a properly normalized ''unit'' vector. Note, as well, that the dot product between our "first" and "third" unit vectors should be zero if they are indeed orthogonal to each other. Let's see … + + + + + + + +
+ $~\hat{e}_3 \cdot \hat{e}_1$ + + $~=$ + + $~ + (- q^2y \ell_q)x\ell_{3D} + (x\ell_q) q^2y\ell_{3D} = 0 \, . +$ +
+ Q.E.D.
+ + Now, even though we have not yet determined the proper expression for the "second" orthogonal coordinate, $~\lambda_2$, we should be able to obtain an expression for its associated unit vector from the cross product of the "third" and "first" unit vectors. Specifically we find, + + +
+ $~\hat{e}_3 \times \hat{e}_1$ + + $~=$ + + $~$ +
# Concentric Ellipsoidal (T6) Coordinates
## Background
Building on our general introduction to Direction Cosines in the context of orthogonal curvilinear coordinate systems, and on our previous development of T3 (concentric oblate-spheroidal) and T5 (concentric elliptic) coordinate systems, here we explore the creation of a concentric ellipsoidal (T6) coordinate system. This is motivated by our desire to construct a fully analytically prescribable model of a nonuniform-density ellipsoidal configuration that is an analog to Riemann S-Type ellipsoids.
## Orthogonal Coordinates
We start by defining a "radial" coordinate whose values identify various concentric ellipsoidal shells,
$~\lambda_1$ $~\equiv$ $~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, .$
When $~\lambda_1 = a$, we obtain the standard definition of an ellipsoidal surface, it being understood that, $~q^2 = a^2/b^2$ and $~p^2 = a^2/c^2$. (We will assume that $~a > b > c$, that is, $~p^2 > q^2 > 1$.)
A vector, $~\bold{\hat{n}}$, that is normal to the $~\lambda_1$ = constant surface is given by the gradient of the function,
$~F(x, y, z)$ $~\equiv$ $~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} - \lambda_1 \, .$
In Cartesian coordinates, this means,
$~\bold{\hat{n}}(x, y, z)$ $~=$ $~ \hat\imath \biggl( \frac{\partial F}{\partial x} \biggr) + \hat\jmath \biggl( \frac{\partial F}{\partial y} \biggr) + \hat{k} \biggl( \frac{\partial F}{\partial z} \biggr)$ $~=$ $~ \hat\imath \biggl[ x(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr] + \hat\jmath \biggl[ q^2y(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr] + \hat{k}\biggl[ p^2 z(x^2 + q^2 y^2 + p^2 z^2)^{- 1 / 2} \biggr]$ $~=$ $~ \hat\imath \biggl( \frac{x}{\lambda_1} \biggr) + \hat\jmath \biggl( \frac{q^2y}{\lambda_1} \biggr) + \hat{k}\biggl(\frac{p^2 z}{\lambda_1} \biggr) \, ,$
where it is understood that this expression is only to be evaluated at points, $~(x, y, z)$, that lie on the selected $~\lambda_1$ surface — that is, at points for which the function, $~F(x,y,z) = 0$. The length of this normal vector is given by the expression,
$~[ \bold{\hat{n}} \cdot \bold{\hat{n}} ]^{1 / 2}$ $~=$ $~ \biggl[ \biggl( \frac{\partial F}{\partial x} \biggr)^2 + \biggl( \frac{\partial F}{\partial y} \biggr)^2 + \biggl( \frac{\partial F}{\partial z} \biggr)^2 \biggr]^{1 / 2}$ $~=$ $~ \biggl[ \biggl( \frac{x}{\lambda_1} \biggr)^2 + \biggl( \frac{q^2y}{\lambda_1} \biggr)^2 + \biggl(\frac{p^2 z}{\lambda_1} \biggr)^2 \biggr]^{1 / 2}$ $~=$ $~ \frac{1}{\lambda_1 \ell_{3D}}$
where,
$~\ell_{3D}$ $~\equiv$ $~\biggl[ x^2 + q^4y^2 + p^4 z^2 \biggr]^{- 1 / 2} \, .$
It is therefore clear that the properly normalized normal unit vector that should be associated with any $~\lambda_1$ = constant ellipsoidal surface is,
$~\hat{e}_1$ $~\equiv$ $~ \frac{ \bold\hat{n} }{ [ \bold{\hat{n}} \cdot \bold{\hat{n}} ]^{1 / 2} } = \hat\imath (x \ell_{3D}) + \hat\jmath (q^2y \ell_{3D}) + \hat{k} (p^2 z \ell_{3D}) \, .$
From our accompanying discussion of direction cosines, it is clear, as well, that the scale factor associated with the $~\lambda_1$ coordinate is,
$~h_1^2$ $~=$ $~\lambda_1^2 \ell_{3D}^2 \, .$
We can also fill in the top line of our direction-cosines table, namely,
Direction Cosines for T6 Coordinates $~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)$ $~n$ $~i = x, y, z$ $~1$ $~x\ell_{3D}$ $~q^2 y \ell_{3D}$ $~p^2 z \ell_{3D}$ $~2$ --- --- --- $~3$ --- --- ---
### Other Coordinate Pair in the Tangent Plane
Let's focus on a particular point on the $~\lambda_1$ = constant surface, $~(x_0, y_0, z_0)$, that necessarily satisfies the function, $~F(x_0, y_0, z_0) = 0$. We have already derived the expression for the unit vector that is normal to the ellipsoidal surface at this point, namely,
$~\hat{e}_1$ $~\equiv$ $~ \hat\imath (x_0 \ell_{3D}) + \hat\jmath (q^2y_0 \ell_{3D}) + \hat\jmath (p^2 z_0 \ell_{3D}) \, ,$
where, for this specific point on the surface,
$~\ell_{3D}$ $~=$ $~\biggl[ x_0^2 + q^4y_0^2 + p^4 z_0^2 \biggr]^{- 1 / 2} \, .$
Tangent Plane
The two-dimensional plane that is tangent to the $~\lambda_1$ = constant surface at this point is given by the expression,
$~0$ $~=$ $~ (x - x_0) \biggl[ \frac{\partial \lambda_1}{\partial x} \biggr]_0 + (y - y_0) \biggl[\frac{\partial \lambda_1}{\partial y} \biggr]_0 + (z - z_0) \biggl[\frac{\partial \lambda_1}{\partial z} \biggr]_0$ $~=$ $~ (x - x_0) \biggl[ \frac{\partial F}{\partial x} \biggr]_0 + (y - y_0) \biggl[\frac{\partial F}{\partial y} \biggr]_0 + (z - z_0) \biggl[ \frac{\partial F}{\partial z} \biggr]_0$ $~=$ $~ (x - x_0) \biggl( \frac{x}{\lambda_1}\biggr)_0 + (y - y_0)\biggl( \frac{q^2 y }{ \lambda_1 } \biggr)_0 + (z - z_0)\biggl( \frac{ p^2z }{ \lambda_1 } \biggr)_0$ $~\Rightarrow~~~ x \biggl( \frac{x}{\lambda_1}\biggr)_0 + y \biggl( \frac{q^2 y }{ \lambda_1 } \biggr)_0 + z \biggl( \frac{ p^2z }{ \lambda_1 } \biggr)_0$ $~=$ $~ x_0 \biggl( \frac{x}{\lambda_1}\biggr)_0 + y_0 \biggl( \frac{q^2 y }{ \lambda_1 } \biggr)_0 + z_0 \biggl( \frac{ p^2z }{ \lambda_1 } \biggr)_0$ $~\Rightarrow~~~ x x_0 + q^2 y y_0 + p^2 z z_0$ $~=$ $~ x_0^2 + q^2 y_0^2 + p^2 z_0^2$ $~\Rightarrow~~~ x x_0 + q^2 y y_0 + p^2 z z_0$ $~=$ $~ (\lambda_1^2)_0 \, .$
Fix the value of $~\lambda_1$. This means that the relevant ellipsoidal surface is defined by the expression,
$~\lambda_1^2$ $~=$ $~x^2 + q^2y^2 + p^2z^2 \, .$
If $~z = 0$, the semi-major axis of the relevant x-y ellipse is $~\lambda_1$, and the square of the semi-minor axis is $~\lambda_1^2/q^2$. At any other value, $~z = z_0 < c$, the square of the semi-major axis of the relevant x-y ellipse is, $~(\lambda_1^2 - p^2z_0^2)$ and the square of the corresponding semi-minor axis is, $~(\lambda_1^2 - p^2z_0^2)/q^2$. Now, for any chosen $~x_0^2 \le (\lambda_1^2 - p^2z_0^2)$, the y-coordinate of the point on the $~\lambda_1$ surface is given by the expression,
$~y_0^2$ $~=$ $~\frac{1}{q^2}\biggl[ \lambda_1^2 - p^2 z_0 -x_0^2 \biggr] \, .$
The slope of the line that lies in the z = z0 plane and that is tangent to the ellipsoidal surface at $~(x_0, y_0)$ is,
$~m \equiv \frac{dy}{dx}\biggr|_{z_0}$ $~=$ $~- \frac{x_0}{q^2y_0}$
### Speculation1
Building on our experience developing T3 Coordinates and, more recently, T5 Coordinates, let's define the two "angles,"
$~\Zeta$ $~\equiv$ $~\sinh^{-1}\biggl(\frac{qy}{x} \biggr)$ and, $~\Upsilon$ $~\equiv$ $~\sinh^{-1}\biggl(\frac{pz}{x} \biggr) \, ,$
in which case we can write,
$~\lambda_1^2$ $~=$ $~x^2(\cosh^2\Zeta + \sinh^2\Upsilon)\, .$
We speculate that the other two orthogonal coordinates may be defined by the expressions,
$~\lambda_2$ $~\equiv$ $~x \biggl[ \sinh\Zeta \biggr]^{1/(1-q^2)} = x \biggl[ \frac{qy}{x}\biggr]^{1/(1-q^2)} = x \biggl[ \frac{x}{qy}\biggr]^{1/(q^2-1)} = \biggl[ \frac{x^{q^2}}{qy}\biggr]^{1/(q^2-1)} \, ,$ $~\lambda_3$ $~\equiv$ $~x \biggl[ \sinh\Upsilon \biggr]^{1/(1-p^2)} = x \biggl[ \frac{pz}{x}\biggr]^{1/(1-p^2)} = x \biggl[ \frac{x}{pz}\biggr]^{1/(p^2-1)} = \biggl[ \frac{x^{p^2}}{pz}\biggr]^{1/(p^2-1)} \, .$
Some relevant partial derivatives are,
$~\frac{\partial \lambda_2}{\partial x}$ $~=$ $~\biggl[ \frac{1}{qy}\biggr]^{1/(q^2-1)} \biggl[ \frac{q^2}{q^2-1} \biggr]x^{1/(q^2-1)} = \biggl[ \frac{q^2}{q^2-1} \biggr]\biggl[ \frac{x}{qy}\biggr]^{1/(q^2-1)} = \biggl[ \frac{q^2}{q^2-1} \biggr]\frac{\lambda_2}{x} \, ;$ $~\frac{\partial \lambda_2}{\partial y}$ $~=$ $~\biggl[ \frac{x^{q^2}}{q}\biggr]^{1/(q^2-1)} \biggl[ \frac{1}{1-q^2} \biggr] y^{q^2/(1-q^2)} = - \biggl[ \frac{1}{q^2-1} \biggr] \frac{\lambda_2}{y} \, ;$ $~\frac{\partial \lambda_3}{\partial x}$ $~=$ $~ \biggl[ \frac{p^2}{p^2-1} \biggr]\frac{\lambda_3}{x} \, ;$ $~\frac{\partial \lambda_3}{\partial z}$ $~=$ $~ - \biggl[ \frac{1}{p^2-1} \biggr] \frac{\lambda_3}{z} \, .$
And the associated scale factors are,
$~h_2^2$ $~=$ $~ \biggl\{ \biggl[ \biggl( \frac{q^2}{q^2-1} \biggr)\frac{\lambda_2}{x} \biggr]^2 + \biggl[ - \biggl( \frac{1}{q^2-1} \biggr) \frac{\lambda_2}{y} \biggr]^2 \biggr\}^{-1}$ $~=$ $~ \biggl\{ \biggl( \frac{q^2}{q^2-1} \biggr)^2 \frac{\lambda_2^2}{x^2} + \biggl( \frac{1}{q^2-1} \biggr)^2 \frac{\lambda_2^2}{y^2} \biggr\}^{-1}$ $~=$ $~ \biggl\{x^2 + q^4 y^2 \biggr\}^{-1} \biggl[ \frac{(q^2 - 1)^2x^2 y^2}{\lambda_2^2} \biggr] \, ;$ $~h_3^2$ $~=$ $~ \biggl\{x^2 + p^4 z^2 \biggr\}^{-1} \biggl[ \frac{(p^2 - 1)^2x^2 z^2}{\lambda_3^2} \biggr] \, .$
We can now fill in the rest of our direction-cosines table, namely,
Direction Cosines for T6 Coordinates $~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)$ $~n$ $~i = x, y, z$ $~1$ $~x\ell_{3D}$ $~q^2 y \ell_{3D}$ $~p^2 z \ell_{3D}$ $~2$ $~q^2 y \ell_q$ $~-x\ell_q$ $~0$ $~3$ $~p^2 z \ell_p$ $~0$ $~-x\ell_p$
Hence,
$~\hat{e}_2$ $~=$ $~ \hat\imath \gamma_{21} + \hat\jmath \gamma_{22} +\hat{k} \gamma_{23} = \hat\imath (q^2y\ell_q) - \hat\jmath (x\ell_q) \, ;$ $~\hat{e}_3$ $~=$ $~ \hat\imath \gamma_{31} + \hat\jmath \gamma_{32} +\hat{k} \gamma_{33} = \hat\imath (p^2z\ell_p) -\hat{k} (x\ell_p) \, .$
Check:
$~\hat{e}_2 \cdot \hat{e}_2$ $~=$ $~ (q^2y\ell_q)^2 + (x\ell_q)^2 = 1 \, ;$ $~\hat{e}_3 \cdot \hat{e}_3$ $~=$ $~ (p^2z\ell_p)^2 + (x\ell_p)^2 = 1 \, ;$ $~\hat{e}_2 \cdot \hat{e}_3$ $~=$ $~ (q^2y\ell_q)(p^2z\ell_p) \ne 0 \, .$
### Speculation2
Try,
$~\lambda_2$ $~=$ $~ \frac{x}{y^{1/q^2} z^{1/p^2}} \, ,$
in which case,
$~\frac{\partial \lambda_2}{\partial x}$ $~=$ $~ \frac{\lambda_2}{x} \, ,$ $~\frac{\partial \lambda_2}{\partial y}$ $~=$ $~ \frac{x}{z^{1/p^2}} \biggl(-\frac{1}{q^2}\biggr) y^{-1/q^2 - 1} = -\frac{\lambda_2}{q^2 y} \, ,$ $~\frac{\partial \lambda_2}{\partial z}$ $~=$ $~ -\frac{\lambda_2}{p^2 z} \, .$
The associated scale factor is, then,
$~h_2^2$ $~=$ $~\biggl[ \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2 + \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)^2 + \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2 \biggr]^{-1}$ $~=$ $~\biggl[ \biggl( \frac{ \lambda_2}{x} \biggr)^2 + \biggl( -\frac{\lambda_2}{q^2y} \biggr)^2 + \biggl( - \frac{\lambda_2}{p^2z} \biggr)^2 \biggr]^{-1}$
### Speculation3
Try,
$~\lambda_2$ $~=$ $~ \frac{(x+p^2 z)^{1 / 2}}{y^{1/q^2} } \, ,$
in which case,
$~\frac{\partial \lambda_2}{\partial x}$ $~=$ $~ \frac{1}{2y^{1/q^2}}\biggl(x + p^2z\biggr)^{- 1 / 2} = \frac{\lambda_2}{2(x + p^2z) } \, ,$ $~\frac{\partial \lambda_2}{\partial y}$ $~=$ $~ -\frac{\lambda_2}{q^2y} \, ,$ $~\frac{\partial \lambda_2}{\partial z}$ $~=$ $~ \, .$
### Speculation4
Here we stick with the primary (radial-like) coordinate as defined above; for example,
$~\lambda_1$ $~\equiv$ $~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} \, ,$ $~h_1$ $~=$ $~\lambda_1 \ell_{3D} \, ,$ $~\ell_{3D}$ $~\equiv$ $~[ x^2 + q^4y^2 + p^4 z^2 ]^{- 1 / 2} \, ,$ $~\hat{e}_1$ $~=$ $~ \hat\imath (x \ell_{3D}) + \hat\jmath (q^2y \ell_{3D}) + \hat{k} (p^2 z \ell_{3D}) \, .$
Note that, $~\hat{e}_1 \cdot \hat{e}_1 = 1$, which means that this is, indeed, a properly normalized unit vector.
Then, drawing from our earliest discussions of "T1 Coordinates", we'll try defining the second coordinate as,
$~\lambda_3$ $~\equiv$ $~ \tan^{-1} u \, ,$ where, $~u$ $~\equiv$ $\frac{y^{1/q^2}}{x} \, .$
The relevant partial derivatives are,
$~\frac{\partial \lambda_3}{\partial x}$ $~=$ $~ \frac{1}{1 + u^2} \biggl[ - \frac{y^{1/q^2}}{x^2} \biggr] = - \biggl[ \frac{u}{1 + u^2}\biggr]\frac{1}{x} \, ,$ $~\frac{\partial \lambda_3}{\partial y}$ $~=$ $~ \frac{1}{1 + u^2} \biggl[ \frac{y^{(1/q^2-1)}}{q^2x^2} \biggr] = \biggl[ \frac{u}{1 + u^2}\biggr]\frac{1}{q^2y} \, ,$
which means that,
$~h_3^2$ $~=$ $~ \biggl[ \biggl( \frac{\partial \lambda_3}{\partial x} \biggr)^2 + \biggl( \frac{\partial \lambda_3}{\partial y} \biggr)^2 \biggr]^{-1}$ $~=$ $~ \biggl[ \frac{u}{1 + u^2}\biggr]^{-2} \biggl[ \frac{1}{x^2} + \frac{1}{q^4y^2} \biggr]^{-1}$ $~=$ $~ \biggl[ \frac{1 + u^2}{u}\biggr]^{2} \biggl[ \frac{x^2 + q^4y^2}{x^2q^4y^2} \biggr]^{-1}$ $~\Rightarrow~~~h_3$ $~=$ $~ \biggl[ \frac{1 + u^2}{u}\biggr]xq^2 y \ell_q \, ,$ where, $~\ell_q$ $~\equiv$ $~[x^2 + q^4 y^2]^{-1 / 2} \, .$
The third row of direction cosines can now be filled in to give,
Direction Cosines for T6 Coordinates $~\gamma_{ni} = h_n \biggl( \frac{\partial \lambda_n}{\partial x_i}\biggr)$ $~n$ $~i = x, y, z$ $~1$ $~x\ell_{3D}$ $~q^2 y \ell_{3D}$ $~p^2 z \ell_{3D}$ $~2$ --- --- --- $~3$ $~-q^2 y \ell_q$ $~x \ell_q$ $~0$
which means that the associated unit vector is,
$~\hat{e}_3$ $~=$ $~ -\hat\imath (q^2 y \ell_{q}) + \hat\jmath (x \ell_{q}) \, .$
Note that, $~\hat{e}_3 \cdot \hat{e}_3 = 1$, which means that this also is a properly normalized unit vector. Note, as well, that the dot product between our "first" and "third" unit vectors should be zero if they are indeed orthogonal to each other. Let's see …
$~\hat{e}_3 \cdot \hat{e}_1$ $~=$ $~ (- q^2y \ell_q)x\ell_{3D} + (x\ell_q) q^2y\ell_{3D} = 0 \, .$
Q.E.D.
Now, even though we have not yet determined the proper expression for the "second" orthogonal coordinate, $~\lambda_2$, we should be able to obtain an expression for its associated unit vector from the cross product of the "third" and "first" unit vectors. Specifically we find,
$~\hat{e}_3 \times \hat{e}_1$ $~=$ $~$
| 6,185 | 14,353 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 252, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.671875 | 4 |
CC-MAIN-2020-50
|
latest
|
en
| 0.861207 |
www.northshoreanalytics.org
| 1,568,967,694,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-39/segments/1568514573908.70/warc/CC-MAIN-20190920071824-20190920093824-00144.warc.gz
| 290,735,954 | 7,131 |
# Survival modeling
A frequently asked question in healthcare analytics is: “What is the probability of survival for (at least) time $$t$$ from now ($$t_0=0$$)of an individual with specific conditions?” or, conversely, “What is the expected survival time of a given individual?”. Survival analysis is a form of regression that can help answer these questions. A standard procedure for evaluating survival probability, and, to some extent, expected survival time, is Cox survival analysis [12], [18],[26] . At the core of it is the semiparametric.
In the following analysis, we assume that an outcome of interest represents an irreversible state transition (e.g., alive to dead). The probability of an event of interest occurring before time $$t$$ is
$$P(t) = Pr(T \leq t) = \int_0^t p(x) dx\; ,$$ (3.26)
where $$T$$ is the time of the event and $$p(x)$$ is the probability density of the (possibly unknown) distribution of such an event. The probability of an individual surviving until (at least) time$$t$$ is termed the survival function and represents the complement of $$P(t)$$
$$S(t) = Pr(T > t) = \int_t^{\infty} p(x) dx\; .$$ (3.27)
The rate of arrival of outcomes of at time $$t$$ is equal to the instantaneous probability of an event at time $$t$$ conditional upon surviving until that time and can be calculated as[1]
\begin{eqnarray} h(t) = \lim_{\Delta t \to 0} \frac{P( t \leq T < t + \Delta t | T \geq t)}{\Delta t} \nonumber \\ = \frac{dP(t)}{dt} \frac{1}{S(t)} = \frac{p(t)}{S(t)} \; .[/itex] \end{eqnarray}
In the Cox model, hazard rate $$h(t)$$ is regressed against a set of predictors $$X_i$$ as
$$h(t) = h_0(t) e^{\sum_{i=1}^{N} b_i x_i } \; ,$$ (3.29)
where $$b_i$$ is the weighting of $$x_i$$, the $$i$$-th of $$N$$ explanatory variables. For the population of $$M$$ individuals, ([E.SurvMod.4]) can be rewritten as
$$h_i(t) = h_{i_0}(t) e^{\sum_{j=1}^{N} b_{j} x_{ij} } \; , i = \overline{1:N}, j = \overline{1:M} \; .$$ (3.30)
Taking the ()natural) logarithm of both sides of (3.30) we arrive at the equivalent of (3.18):
$$\ln \frac{h_i(t)}{h_{i_0}(t)} = \sum_{j=1}^{N} b_{j} x_{ij} \; , i = \overline{1:N}, j = \overline{1:M} \; .$$ (3.31)
The form of $$h_{i_0}(t)$$ is not formally specified; its shape is determined by empirical data in the training dataset giving rise to the unparametric portion of the model.[2] The solution of (3.26) is delivered by the maximum of the partial likelihood function defined in as
$$L_p = \prod_{i=1}^{N} \left [ \frac{e^{x_i \beta}}{\sum_{j=1}{N} Y_{ij} e^{x_i \beta}} \right ]^{\delta_i} \; , \\ Y_{ij} = \begin{cases} 0, \text{if } t_j < t_i \; , \\ 1, \text{otherwise} \;. \end{cases} \; , \\ \delta_{i} = \begin{cases} 0, \text{if event did not occur at time} t_i \; , \\ 1, \text{otherwise} \;. \end{cases} \; ,$$ (3.32)
A widely accepted standard for survival analysis in R is the survival package ; a noteworthy extension of it that takes into account relative survival probability is relsurv [27].
Two important variables to consider in survival analysis are time since first diagnosis and age. The former reflects the individual’s “lifetime” measured with respect to others with the same group of conditions, the latter relates his or her expected risk of experiencing a negative outcome to that of the general population.
It is important to distinguish survival analysis, which is characterized by an impenetrable boundary between the sets with null and eventful outcomes, from renewal analysis where such boundary can be crossed. Clearly, the transition from alive to deceased can occur only once whereas the transition between healthy and ill can occur multiple times. Renewal analysis is governed by a similar set of equations but is conceptually different from survival analysis.
An illustrative performance comparison between a regular logistic regression model and Cox proportional hazard model used for predicting one-year mortality among heart failure patients is presented in Fig. 3.6.
Figure 3.6. Example: Comparison of logistic regression and Cox proportional hazard model performance for predicting one-year mortality among heart failure patients.
As can be seen from Fig. 3.6, the AUC for model in question is approximately 0.81. The attained maxima are approximately 0.45 for $$F_1$$ score and 0.4 for Matthews' correlation coefficient, however, those maxima are attained at approximately 45% of the total population for the logistic regression model and 5% for the Cox proportional hazard model. This leads us to believe that, in this particular case, the latter achieves optimal accuracy for smaller population samples than the former, however, overall model accuracies is virtually identical.
The R code for performing predictive modeling in the example above can be found in Appendix D.
## Notes
1. conditional probability
2. Hence the term “semiparametric”.
| 1,318 | 4,879 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.765625 | 4 |
CC-MAIN-2019-39
|
longest
|
en
| 0.854716 |
https://countrybarnyard305.com/verify-746
| 1,669,707,209,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710690.85/warc/CC-MAIN-20221129064123-20221129094123-00276.warc.gz
| 218,920,335 | 6,721 |
## The Best Geometry app for ipad
Math can be a challenging subject for many students. But there is help available in the form of Geometry app for ipad. There's no need to be intimidated by equations with e in them - they're not as difficult to solve as they may first appear. Here's a step-by-step guide to solving equations with e. First, identify the term with e in it and isolate it on one side of the equation. Then, take the natural logarithm of both sides of the equation. This will result in an equation that only has numbers on one side, and e on the other. Next, use basic algebra to solve for the variable. Finally, take the exponential of both sides to undo the natural logarithm and arrive at the solution. With a little practice, you'll be solving equations with e like a pro!
A two equation solver can be a helpful tool for solving systems of linear equations. There are a variety of methods that can be used to solve systems of linear equations, and the two equation solver can help you to find the best method for your particular problem. In addition, the two equation solver can also help you to check your work for mistakes. This can be especially helpful if you are working with a large system of equations. Overall, the two equation solver can be a valuable tool for solving systems of linear equations.
How to solve perfect square trinomial? This is a algebraic equation that can be written in the form of ax2 + bx + c = 0 . If the coefficient of x2 is one then we can use the factoring method to solve it. We will take two factors of c such that their product is equal to b2 - 4ac and their sum is equal to b. How to find such numbers? We will use the quadratic formula for this. Now we can factorize the expression as (x - r1)(x - r2) = 0, where r1 and r2 are the roots of the equation. To find the value of x we will take one root at a time and then solve it. We will get two values of x, one corresponding to each root. These two values will be the solutions of the equation.
In addition, many of these websites also provide worked examples so that the student can see how the process works. With a little practice, using a math word problem solver online free can help students to become more confident and proficient in solving math word problems.
| 502 | 2,279 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.6875 | 5 |
CC-MAIN-2022-49
|
latest
|
en
| 0.96433 |
https://algebra-answer.com/tutorials-2/binomials/factoring-polynomials-2.html
| 1,631,843,902,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-39/segments/1631780053918.46/warc/CC-MAIN-20210916234514-20210917024514-00632.warc.gz
| 133,249,384 | 8,464 |
Factoring Polynomials
Some trinomials that can be factored do not look like the special trinomials from the
previous sections . Factor trinomials, written ax2 + bx + c, by doing the following rules:
1. Factor out the GCF of all three terms. Use the resulting trinomial for the rest of
the steps. If a is negative, also factor out –1 along with the GCF.
2. Check that the square root of b2 – 4ac is a whole number. If b2 – 4ac is negative,
then we cannot factor the trinomial. If the square root of b2 – 4ac is not a whole
number, then the factored form of ax2 + bx + c will have fractions or square root
signs in it . We will not be factoring these in this section.
3. Look at the sign of the constant term.
a. If the second sign (the one before the constant term) is a + sign, then both signs
in the factored form are whatever the first sign is.
ax2 + bx + c = ( __ + __ )( __ + __ )
or
ax2 – bx + c = ( __ – __ )( __ – __ )
b. If the second sign is a – sign, then the signs in the factored form are different .
ax2 + bx – c = ( __ + __ )( __ – __ )
or
ax2 + bx – c = ( __ + __ )( __ – __ )
4. Find two numbers that multiply together to give ac but add up to give b. Keep the
signs of a, b, and c with the numbers.
5. Rewrite in preliminary factored form:
a. Write the GCF on the outside of the factored form.
b. In both sets of the parentheses, write ax
c. Write the appropriate signs from step 3.
d. Write the two numbers found in step 4, one in each set of parentheses.
6. Factor out any GCFs from the sets of parentheses and throw them away.
7. Rewrite. This is the final factored form.
Examples:
1. Factor 3x2 + 5x + 4.
Step 1: Find GCF(3x2, 5x, 4). The GCF is 1 since none of the coefficients have a
factor in common .
Step 2: Find b2 – 4ac. In this trinomial, a = 3, b = 5, and c = 4, so
b2 – 4ac = (5)2 – (4)(3)(4) = 25 – 48 = -23. Since -23 is negative, we cannot factor
this polynomial.
2. Factor x2 + 5x + 6.
Step 1: Find GCF(x2, 5x, 6). The GCF of this trinomial is 1 since the only factor of
1 is 1.
Step 2: Find b2 – 4ac. Here, a = 1, b = 5, and c = 6, so b2 – 4ac = (5)2 – (4)(1)(6) =
25 – 24 = 1. Since , a whole number, then we can factor this trinomial and
have integer coefficients.
Step 3: Sign of the constant term. The constant term of this trinomial is +6, so both
the signs inside the factored form (a.k.a. the answer) will be the same. Since the
sign on the 5x is positive also , the factored form will look like ( __ + __ )( __ + __ ).
Step 4: Find 2 numbers. We need two numbers that multiply together to give (1)(6)
= 6 and add up together to give 5. Factors of 6 are 1 & 6 or 2 & 3. Since 2 + 3 = 5,
these are the numbers we will use.
Step 5: Preliminary answer. The preliminary factored form is 1(1x + 2)(1x + 3).
Step 6: Internal GCFs. There are no GCFs other than 1 inside each set of
parentheses.
Step 7: Rewrite. The factored form is (x + 2)(x + 3)
3. Factor the polynomial x 2 – 5x – 24.
Step 1: GCF(x2, -5x, -24) = 1.
Step 2: b2 – 4ac = (-5)2 – 4(1)(-24) = 25 + 96 = 121. , so we can factor
this polynomial.
Step 3: The sign of the constant term is negative, so the signs inside the sets of
parentheses will be different, ( __ + __ )( __ – __ )
Step 4: We want 2 numbers that multiply together to give (1)(-24) = -24 and add up
to -5. Factors of -24 and their sums are
The numbers 3 and -8 multiply together to give -24 and add up to -5.
Step 5: The preliminary answer is 1(1x + 3)(1x – 8).
Step 6: Neither set of parentheses have a GCF other than 1.
Step 7: The answer is (x + 3)(x – 8).
4. Factor x2 + 10x + 18.
Step 1: GCF(x2, 10x, 18) = 1.
Step 2: b2 – 4ac = (10)2 – (4)(1)(18) = 100 – 72 = 28. Since 28 is not a perfect
square (the square root of 28 is not a whole number), we cannot factor this trinomial
and have only integer coefficients in our answer . This trinomial is in its final
factored form.
5. Factor 2x2 + 18x – 72.
Step 1: GCF(2x2, 18x, 72) = 2. We must factor 2 out of our trinomial:
2x2 + 18x – 72 = 2(x2 + 9x – 36). Use the trinomial inside the parentheses for the
rest of the problem.
Step 2: b2 – 4ac = (9)2 – 4(1)(-36) = 81 + 144 = 225. Since , we can
continue with factoring.
Step 3: The constant term is negative, so the signs inside the sets of parentheses in
our answer will be opposite; that is, ( __ + __ )( __ – __ ).
Step 4: Find 2 numbers that add up to +9, but multiply to get (1)(-36) = -36:
The numbers -3 and 12 multiply together to get -36 and add together to get 9.
Step 5: The preliminary answer is 2(1x + 12)(1x – 3).
Step 6: Neither set of parentheses have a GCF other than 1.
Step 7: The factored form is 2(x + 12)(x – 3).
6. Factor the trinomial 6x2 – 27x + 12.
Step 1: GCF(6x2, -27x, 12) = 3, so factor 3 out of the trinomial: 6x2 – 27x + 12 =
3(2x2 – 9x + 4). The trinomial 2x2 – 9x + 4 will be used for the rest of the steps.
Step 2: b2 – 4ac = (-9)2 – 4(2)(4) = 81 – 32 = 49. , so we can factor.
Step 3: The constant term (12) is positive, so both sets of parentheses will have the
same sign. Since the sign of the middle term (-27x) is negative, the sets of
parentheses will look like ( __ – __ )( __ – __ ).
Step 4: Find two numbers that multiply together to give (2)(4) = 8 but add up to -9:
The numbers -1 and -8 are factors of 8 with a sum of –9.
Step 5: The preliminary answer is 3(2x – 1)(2x – 8).
Step 6: The second set of parentheses, the (2x – 8) part, has a GCF of 2, so divide
everything in this set of parentheses by 2 and discard the 2:
Step 7: The final factored form of 6x2 – 27x + 12 is 3(2x – 1)(x – 4).
7. Factor –20x3 – 80x2 – 75x.
Step 1: GCF(-20x3, -80x2, -75x) = -5x. Since the coefficient of the term with the
highest order is negative, we want to pull the negative out along with the GCF.
Factor out the -5: -20x3 – 80x2 – 75x = -5x(4x2 + 16x + 15).
Step 2: b2 – 4ac = (16)2 – 4(4)(15) = 256 – 240 = 16. Since 16 is a perfect square,
we can factor the trinomial 4x2 + 16x + 15.
Step 3: The constant and the middle terms are positive, so 4x2 + 16x + 15 will factor
into something that looks like ( __ + __ )( __ + __ ).
Step 4: Find two numbers that multiply to give (4)(15) = 60 but add up to 16:
The numbers 6 and 10 are factors 60 and have a sum of 16.
Step 5: The preliminary answer is -5x(4x + 10)(4x + 6).
Step 6: Both sets of parentheses have GCFs greater than one. GCF(4x, 10) = 2 and
GCF(4x, 6) = 2. Divide everything in the first set of parentheses by 2 and everything
in the second set of parentheses by 2, throwing away the GCFs afterward:
Step 7: Rewrite; the answer is -5x(2x + 5)(2x + 3).
Factor the following polynomials using any method . The answer should have only
integer coefficients.
1. x2 + 7x + 6
2. x2 + 2x + 1
3. 4x3 – 9x
4. 15x4 + 15x2
5. 2x2 + 8x + 4
6. -12x2 + 17x – 6
7. 7x2 – 14x – 56
8. –4x2 + 40x – 84
9. 10x2 – 13x – 3
10. 30x3 – 3x2 – 9x
11. x4 + 3x2 – 4
12. -18x3 + 54x2 – 8x + 24
Prev Next
Start solving your Algebra Problems in next 5 minutes!
2Checkout.com is an authorized reseller
of goods provided by Sofmath
Attention: We are currently running a special promotional offer for Algebra-Answer.com visitors -- if you order Algebra Helper by midnight of September 17th you will pay only \$39.99 instead of our regular price of \$74.99 -- this is \$35 in savings ! In order to take advantage of this offer, you need to order by clicking on one of the buttons on the left, not through our regular order page.
If you order now you will also receive 30 minute live session from tutor.com for a 1\$!
You Will Learn Algebra Better - Guaranteed!
Just take a look how incredibly simple Algebra Helper is:
Step 1 : Enter your homework problem in an easy WYSIWYG (What you see is what you get) algebra editor:
Step 2 : Let Algebra Helper solve it:
Step 3 : Ask for an explanation for the steps you don't understand:
Algebra Helper can solve problems in all the following areas:
• simplification of algebraic expressions (operations with polynomials (simplifying, degree, synthetic division...), exponential expressions, fractions and roots (radicals), absolute values)
• factoring and expanding expressions
• finding LCM and GCF
• (simplifying, rationalizing complex denominators...)
• solving linear, quadratic and many other equations and inequalities (including basic logarithmic and exponential equations)
• solving a system of two and three linear equations (including Cramer's rule)
• graphing curves (lines, parabolas, hyperbolas, circles, ellipses, equation and inequality solutions)
• graphing general functions
• operations with functions (composition, inverse, range, domain...)
• simplifying logarithms
• basic geometry and trigonometry (similarity, calculating trig functions, right triangle...)
• arithmetic and other pre-algebra topics (ratios, proportions, measurements...)
ORDER NOW!
| 2,891 | 8,787 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.875 | 5 |
CC-MAIN-2021-39
|
latest
|
en
| 0.875195 |
https://chelischili.com/arabesque-carpet
| 1,675,713,261,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500357.3/warc/CC-MAIN-20230206181343-20230206211343-00840.warc.gz
| 179,379,666 | 5,423 |
App that gives you any answer
One tool that can be used is App that gives you any answer. We can help me with math work.
The Best App that gives you any answer
In this blog post, we will show you how to work with App that gives you any answer. A system of equations is a set of two or more equations that share the same variables. In order to solve a system of equations, all of the equations must be satisfied simultaneously. This can be done by using a variety of methods, including substitution, elimination, and graphing. One useful tool for solving systems of equations is a compound inequality solver. A compound inequality solver is a tool that allows
To solve for all values of x by factoring, we need to factor the equation. To do this, we need to find all the factors of the equation and set them equal to zero. Once we have done this, we can solve for x.
Precalc can be a tough subject for many students. However, there are plenty of resources available to help you succeed. First, check with your teacher or professor. They can point you in the right direction for resources in the classroom or online. Additionally, there are numerous online resources available, including websites and forums dedicated to precalc help. With a little effort, you can find the help you need to succeed in precalc.
There's now an app for that! The Algebra Pro app can help you solve all kinds of algebra problems, from simple equations to complex calculus. Just enter your equation and let the app do the work. It's fast, accurate, and easy to use. So whether you're a student struggling with algebra or a professional who needs to brush up on their math skills, the Algebra Pro app is a great resource.
Use the exponent property to rewrite the equation as an exponential equation. 3. Solve the resulting exponential equation for x. 4. Check the answer by plugging it back into the original equation.
Help with math
Absolutely brilliant. If you are on any math related course or just have math questions in general life, USE THIS. It is perfect, and honestly, I use the calculator function far more than the camera function. Cannot rate this app any higher, it even shows you the solving steps so you can learn for yourself! It's honestly amazing.
Beatrix Collins
It is very helpful, as a student who is in Algebra 1 and is going into Algebra 2. It explains the steps better than some of my teachers. You don't even need internet to see it like you do for other apps. They don't cover everything but they do for most of Algebra 1 and 2. Though it would be nice to get some help with the trigonometry.
Nora Simmons
Plug in math equations for answers Pre algebra app for android Math problems online Take a picture of homework and get answers Factor math solver Take a picture of your homework and get answers
| 595 | 2,810 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.78125 | 4 |
CC-MAIN-2023-06
|
latest
|
en
| 0.948276 |
https://www.bartleby.com/questions-and-answers/scenario-141.-a-competitive-firm-sells-its-output-for-dollar20-per-unit.-when-the-firm-produces-200-/e10d5755-833a-47f4-a351-edbc60e0ad46
| 1,582,001,705,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875143505.60/warc/CC-MAIN-20200218025323-20200218055323-00312.warc.gz
| 684,230,690 | 31,970 |
# Scenario 14-1. A competitive firm sells its output for \$20 per unit. When the firm produces 200 units of output,average variable cost is \$16, marginal cost is \$18, and average total cost is \$23.15. Refer to Scenario 14-1. Calculate the firm’s total revenue, total cost, and profit at 200 units of output.Peter operates an ice cream shop in the center of Fairfield. He sells several unusual flavors of organic, homemade ice cream so he has a monopoly over his own ice cream, though he competes with many other firms selling ice cream in Fairfield for the same customers. Peter’s demand and cost values for sales per day are given in the table below. (Everyone who purchases Peter’s ice cream buys a double scoop cone because it’s so delicious.)QuantityPriceMRMCATC20\$5.60\$5.20\$2.20\$2.0540\$5.20\$4.40\$2.40\$2.1060\$4.80\$3.60\$2.60\$2.1580\$4.40\$2.80\$2.80\$2.20100\$4.00\$2.00\$3.00\$2.25120\$3.60\$1.20\$3.20\$2.30140\$3.20\$0.40\$3.40\$2.35160\$2.80-\$0.40\$3.60\$2.40180\$2.40-\$1.20\$3.80\$2.455. Refer to Scenario 16-3. When Peter maximizes his profits, what is his total cost per day?
Question
27 views
Scenario 14-1. A competitive firm sells its output for \$20 per unit. When the firm produces 200 units of output,
average variable cost is \$16, marginal cost is \$18, and average total cost is \$23.
15. Refer to Scenario 14-1. Calculate the firm’s total revenue, total cost, and profit at 200 units of output.
Peter operates an ice cream shop in the center of Fairfield. He sells several unusual flavors of organic, homemade ice cream so he has a monopoly over his own ice cream, though he competes with many other firms selling ice cream in Fairfield for the same customers. Peter’s demand and cost values for sales per day are given in the table below. (Everyone who purchases Peter’s ice cream buys a double scoop cone because it’s so delicious.)
Quantity Price MR MC ATC 20 \$5.60 \$5.20 \$2.20 \$2.05 40 \$5.20 \$4.40 \$2.40 \$2.10 60 \$4.80 \$3.60 \$2.60 \$2.15 80 \$4.40 \$2.80 \$2.80 \$2.20 100 \$4.00 \$2.00 \$3.00 \$2.25 120 \$3.60 \$1.20 \$3.20 \$2.30 140 \$3.20 \$0.40 \$3.40 \$2.35 160 \$2.80 -\$0.40 \$3.60 \$2.40 180 \$2.40 -\$1.20 \$3.80 \$2.45
5. Refer to Scenario 16-3. When Peter maximizes his profits, what is his total cost per day?
check_circle
Step 1
15.
The competitive firm sells 200 units of output and the average variable cost of the firm is given to be \$16. The marginal cost is given to be \$18 and the average total cost is given to be \$23. It is also given that the price that the firm receives from selling a unit of output is equal to \$20. The total revenue of the firm from selling 200 units of output can be calculated by multiplying the output with the per unit price as follows:
Step 2
The total cost of the firm can be calculated by multiplying the output with the average total cost of production. The average total cost of production is given to be \$23 per unit and thus, the total cost of the firm can be calculated as follows:
Step 3
The profit is the excess revenue made by the firm after deducting the cost of production from the total revenue. The total revenue i...
### Want to see the full answer?
See Solution
#### Want to see this answer and more?
Solutions are written by subject experts who are available 24/7. Questions are typically answered within 1 hour.*
See Solution
*Response times may vary by subject and question.
Tagged in
| 970 | 3,430 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5 | 4 |
CC-MAIN-2020-10
|
latest
|
en
| 0.87273 |
https://www.bearnaiserestaurant.com/blog/what-is-the-formula-for-carbon-14/
| 1,670,044,512,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446710924.83/warc/CC-MAIN-20221203043643-20221203073643-00040.warc.gz
| 691,350,987 | 14,058 |
# What is the formula for carbon-14?
## What is the formula for carbon-14?
Carbon-14 | CH4 – PubChem.
## What is the equation for carbon dating?
We can use our our general model for exponential decay to calculate the amount of carbon at any given time using the equation, N (t) = N0e kt . Modeling the decay of 14C. Returning to our example of carbon, knowing that the half-life of 14C is 5700 years, we can use this to find the constant, k.
What is the carbon-14 date method?
Radiocarbon dating (also referred to as carbon dating or carbon-14 dating) is a method for determining the age of an object containing organic material by using the properties of radiocarbon, a radioactive isotope of carbon.
### How do you calculate radiocarbon age?
Radiocarbon age is calculated from the δ13C-corrected Fraction Modern according to the following formula: Age = -8033 ln (Fm) Reporting of ages and/or activities follows the convention outlined by Stuiver and Polach (1977) and Stuiver (1980).
### What is half-life algebra?
Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value.
What is the easiest way to calculate half-life?
How to calculate half life? To find half-life: Find the substance’s decay constant. Divide ln 2 by the decay constant of the substance.
#### What is the half-life of 14 C?
5700 ± 30 yr
The half-life of radiocarbon (14C) is 5700 ± 30 yr, which makes it particularly useful for dating in archaeology. However, only an exceptional hindrance of the beta decay from 14C to 14N—a so-called Gamow-Teller ß-decay—makes this half-life so long.
#### What is the ratio of carbon-14 to its initial amount?
and the ratio of the current amount of Carbon-14 to its initial amount can be used to determine age. The last ratio is described by the function C (t) = 0.5^ (t/5730), where “t” is time (in years). Since 33.1% of Carbon-14 is lost, 66.9% is remained, or 0.669 of its initial amount.
What percentage of carbon 14 is lost when Carbon-14 is removed?
Since 33.1% of Carbon-14 is lost, 66.9% is remained, or 0.669 of its initial amount. Thus, you should solve an equation C (t) = 0.669, which is y = = , for unknown t.
## What is the half-life of carbon 14?
In the case of radiocarbon dating, the half-life of carbon 14 is 5,730 years. This half life is a relatively small number, which means that carbon 14 dating is not particularly helpful for very recent deaths and deaths more than 50,000 years ago. After 5,730 years, the amount of carbon 14 left in the body is half of the original amount.
## What happens to carbon-14 when a plant dies?
to take in atmospheric carbon, which is the beginning of absorption of carbon into the food chain. As long as the plant is alive, the ratio of C-14 to C-12 in its body remains constant. After the plant dies, carbon-14 continues to decay without being replaced.
• August 26, 2022
| 744 | 2,912 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875 | 4 |
CC-MAIN-2022-49
|
latest
|
en
| 0.90142 |
https://math.stackexchange.com/questions/3296233/about-partial-derivatives-of-implicit-functions
| 1,575,667,608,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540490972.13/warc/CC-MAIN-20191206200121-20191206224121-00442.warc.gz
| 468,013,188 | 32,739 |
# about partial derivatives of implicit functions
I want to explain the question I have in mind with an example.
$$F(x,y,z) = 0$$ system of equations : Let $$z$$ define as a implicit function of $$x$$ and $$y$$ . And $$F(x,y,z)=x^2-y+\ln(z)$$
$$(a)$$ $$\dfrac{\partial F}{\partial x}=F_x = 2x$$
$$(b)$$ $$\dfrac{\partial F}{\partial y}=F_y = -1$$
$$(c)$$ $$\dfrac{\partial F}{\partial z}=F_z = \dfrac{1}{z}$$
$$F(x,y,z)=0$$
Let's take the partial derivative of both sides of the equation with respect to $$y$$ .
then $$\dfrac{\partial}{\partial y}F(x,y,z)=\dfrac{\partial }{\partial y}(0)$$ $$=>$$ $$\dfrac{\partial F}{\partial y}=0$$
$$(b)$$ owing to $$\dfrac{\partial F}{\partial y}=-1$$
as a result : $$-1=0$$ is happening. Where am I doing wrong ? why did this happen ?
• z = f(x,y) , then $\partial_x z \ne 0$ and $\partial_y z\ne 0$ – Ajay Mishra Jul 18 at 0:19
What's going on is a very unfortunate abuse of notation. This is very common in differential calculus, and even more so in the context of implicit functions. It is only after you really understand what the function is (this includes knowing its domain and target space) vs where it is being evaluated that you'll understand what's going on here.
To start off, you're given a function $$F: U \to \Bbb{R}$$, where $$U = \Bbb{R}^2 \times \Bbb{R}^+$$ defined by \begin{align} F(x,y,z) = x^2 - y + \ln(z) \end{align} The statement "$$F(x,y,z) = 0$$ implicitly defines $$z$$ as a function of $$x$$ and $$y$$" is a convenient but somewhat imprecise way of saying the following more precise (and pedantic) statement:
There exists a function $$\zeta: \Bbb{R}^2 \to \Bbb{R}$$ such that for all $$(x,y) \in \Bbb{R} \times \Bbb{R}$$, we have that $$F(x,y,\zeta(x,y)) = 0$$.
I intentionally used $$\zeta$$ instead of $$z$$, because I think it is confusing (for a beginner atleast) to use the same letter $$z$$ in two different contexts, where they have different meanings. If you use $$z$$ to denote the "independent variable" in $$F(x,y,z)$$, and also as the "implicitly defined function" $$z: \Bbb{R}^2 \to \Bbb{R}$$, $$(x,y) \mapsto z(x,y)$$, then this is just a recipe for all sorts of confusion.
Now, the reason you are getting the apparent contradiction $$-1 = \dfrac{\partial F}{\partial y}$$ and $$\dfrac{\partial F}{\partial y} = 0$$ is because you're using the same letter $$F$$ to describe two different things. The correct and pedantic statement is:
For all $$(x,y,z) \in U$$, we have $$(\partial_2F)_{(x,y,z)} = -1$$.
(i.e we're taking the partial derivative of the function $$F$$ with respect to its second argument, and evaluating the function $$\partial_2F$$ at the particular point $$(x,y,z) \in U$$)
In slightly more common notation, we might say that for all $$(x,y,z) \in U$$, we have $$\dfrac{\partial F}{\partial y} \bigg|_{(x,y,z)} = -1$$.
Let us temporarily define a new function $$g: \Bbb{R}^2 \to \Bbb{R}^3$$ by the rule \begin{align} g(x,y) = (x,y,\zeta(x,y)) \end{align} Then, the first highlighted statement says that the composition $$F \circ g = 0$$ is the zero function; i.e for all $$(x,y) \in \Bbb{R}^2$$, $$(F \circ g)(x,y) = 0$$.
Later on, you make the statement:
\begin{align} \dfrac{\partial}{\partial y}F(x,y,z) = \dfrac{\partial}{\partial y}(0) = 0 \implies \dfrac{\partial F}{\partial y} = 0, \end{align}
which is riddled with all sorts of notational abuse (one of the several pitfalls of Leibniz notation). A more accurate statement is any of the following ones below:
• $$\partial_2(F \circ g) = 0 \, \,$$ (this is an equality of functions)
• For all $$(x,y) \in \Bbb{R}^2$$, $$\partial_2(F \circ g)_{(x,y)} = 0 \, \,$$ (this is pointwise version of the first statement, and is an equality of real numbers)
• For all $$(x,y) \in \Bbb{R}^2$$, $$\dfrac{\partial (F \circ g)}{\partial y}\bigg|_{(x,y)} = 0$$
The first two are the most notationally precise way of stating things. The last one is probably more common/convenient, but as always Leibniz's notation should always be used with caution.
I hope now you notice that there is no contradiction at all. We have two completely different statements; the first is that for all $$(x,y,z) \in U$$, $$(\partial_2F)_{(x,y,z)} = -1$$. The second is that for all $$(x,y) \in \Bbb{R}^2$$, $$\partial_2(F \circ g)_{(x,y)} = 0$$.
If you insist on Leibniz's notation, we could say that $$\dfrac{\partial F}{\partial y} = -1$$, whereas $$\dfrac{\partial (F \circ g)}{\partial y} = 0$$.
Of course, now you can use the chain rule to get some relationships between the partial derivatives: for any $$(x,y) \in \Bbb{R}^2$$, \begin{align} 0 &= \partial_2(F \circ g)_{(x,y)} \\ &= (\partial_1F)_{g(x,y)} \cdot (\partial_2g_1)_{(x,y)} + (\partial_2F)_{g(x,y)} \cdot (\partial_2g_2)_{(x,y)} + (\partial_3F)_{g(x,y)} \cdot (\partial_2g_3)_{(x,y)} \\ &= (2x) \cdot (0) + (-1) \cdot (1) + (\partial_3F)_{(x,y, \zeta(x,y))} \cdot (\partial_2 \zeta)_{(x,y)} \\ &= -1 + \dfrac{1}{\zeta(x,y)} \cdot (\partial_2 \zeta)_{(x,y)}. \end{align}
BTW, in the above computation $$g_i$$ is the $$i^{th}$$ component function of $$g$$. So, $$g_1(x,y) := x$$, and $$g_2(x,y) := y$$, and $$g_3(x,y) := \zeta(x,y)$$.
What I have done above is the notationally precise way of doing things, because I was very careful to distinguish between which function I'm differentiating and where I'm evaluating the derivatives. The more convenient, and also more misleading (if you're just starting out) way of performing the same computation is: \begin{align} 0 &= \dfrac{\partial }{\partial y} \left( F(x,y,z(z,y))\right) \\ &= \dfrac{\partial F}{\partial x} \cdot \dfrac{\partial x}{\partial y} + \dfrac{\partial F}{\partial y} \cdot \dfrac{\partial y}{\partial y} + \dfrac{\partial F}{\partial z} \cdot \dfrac{\partial z}{\partial y} \\ &= (2x) \cdot (0) + (-1) \cdot (1) + \dfrac{1}{z} \cdot \dfrac{\partial z}{\partial y} \\ &= -1 + \dfrac{1}{z} \dfrac{\partial z}{\partial y} \end{align}
• Thank you so much. You are a very good person . <3 – Tonyukuk Jul 18 at 20:51
As you said, $$z$$ is a imlpict function of $$x$$ and $$y$$, i.e. $$z=z(x,y)$$. So when you derivative both sides of $$F(x,y,z)=0$$ with respect to $$y$$, the left-hand side turns out to be $$\frac{\partial(x^2)}{\partial y}-\frac{\partial(y)}{\partial y}+\frac{\partial(\ln z)}{\partial y}=-1+\frac{\partial(\ln z)}{\partial z}\frac{\partial z}{\partial y}=-1+\frac1z\frac{\partial z}{\partial y}.$$
• $F_y$ equal to what , $(-1)$ or $(-1+\frac1z\frac{\partial z}{\partial y}).$ ? – Tonyukuk Jul 18 at 0:24
• @Tonyukuk If you regard $z$ as a function of $x,y$, then the latter one is right. – Feng Shao Jul 18 at 0:52
• Thank you so much. You've made me very happy. i am thankful to you . – Tonyukuk Jul 18 at 1:02
• @Tonyukuk You are welcome. – Feng Shao Jul 18 at 1:04
| 2,306 | 6,773 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 79, "wp-katex-eq": 0, "align": 5, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.25 | 4 |
CC-MAIN-2019-51
|
latest
|
en
| 0.880502 |
http://mathhelpforum.com/calculus/146892-finding-center-mass-print.html
| 1,500,987,975,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-30/segments/1500549425193.20/warc/CC-MAIN-20170725122451-20170725142451-00656.warc.gz
| 198,937,234 | 3,211 |
# Finding center of mass
• May 29th 2010, 12:14 PM
Em Yeu Anh
Finding center of mass
Q: Determine the mass and center of mass of the lamina for region D if D is bounded by $y=x^2$ and $x=y^2$; $p(x,y)=5\sqrt{x}$
Could someone assist me with setting up the limits on the integral for finding the mass?
• May 29th 2010, 01:03 PM
skeeter
Quote:
Originally Posted by Em Yeu Anh
Q: Determine the mass and center of mass of the lamina for region D if D is bounded by $y=x^2$ and $x=y^2$; $p(x,y)=5\sqrt{x}$
Could someone assist me with setting up the limits on the integral for finding the mass?
$dA = (\sqrt{x} - x^2) \, dx$
$dm = \rho \, dA = 5\sqrt{x}(\sqrt{x} - x^2) \, dx
$
$m = 5\int_0^1 x - x^{\frac{5}{2}} \, dx$
• May 29th 2010, 01:27 PM
AllanCuz
Quote:
Originally Posted by Em Yeu Anh
Q: Determine the mass and center of mass of the lamina for region D if D is bounded by $y=x^2$ and $x=y^2$; $p(x,y)=5\sqrt{x}$
Could someone assist me with setting up the limits on the integral for finding the mass?
Well,
$C.O.M = ( \bar x , \bar y )$
Where,
$\bar x = \frac{ \iint_D x P(x,y) dA }{ \iint_D P(x,y) dA}$
$= \frac{ 5 \int_0^1 x \sqrt{x} dx \int_{x^2}^{ \sqrt{x} }dy }{5 \int_0^1 \sqrt{x} dx \int_{x^2}^{ \sqrt{x} } dy }$
$= \frac{ 5 \int_0^1 x \sqrt{x} ( \sqrt{x} - x^2 ) dx }{ 5 \int_0^1 \sqrt{x} ( \sqrt{x} - x^2 ) dx }$
$= \frac{ 5 \int_0^1 (x^2 - x^{ \frac{7}{2} } ) dx }{ 5 \int_0^1 ( x - x^{ \frac{5}{2} } ) dx }$
And,
$\bar y = \frac{ \iint_D y P(x,y) dA }{ \iint_D P(x,y) dA}$
$= \frac{ 5 \int_0^1 \sqrt{x} dx \int_{x^2}^{ \sqrt{x} } y dy }{5 \int_0^1 \sqrt{x} dx \int_{x^2}^{ \sqrt{x} } dy }$
$= \frac{ 5 \int_0^1 \sqrt{x} ( \frac{ x^2 }{2} - \frac{x^4 }{2} ) dx }{ 5 \int_0^1 \sqrt{x} ( \sqrt{x} - x^2 ) dx }$
$= \frac{ \frac{5}{2} \int_0^1 ( x^{ \frac{5}{2} } - x^{ \frac{9}{2} } ) dx }{ 5 \int_0^1 ( x - x^{ \frac{5}{2} } ) dx }$
If you don't know double integrals yet that's fine, just compute them as the single integrals that I've reduced them to. This will serve as your C.O.M.
Your mass, as Skeeter pointed out is the denominator in the above calculations. So,
$Mass = 5 \int_0^1 ( x - x^{ \frac{5}{2} } ) dx$
| 893 | 2,153 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 22, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625 | 4 |
CC-MAIN-2017-30
|
longest
|
en
| 0.644164 |
http://hitchhikersgui.de/Orbit_equation
| 1,548,255,209,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-04/segments/1547584332824.92/warc/CC-MAIN-20190123130602-20190123152602-00569.warc.gz
| 105,209,068 | 12,456 |
# Orbit equation
In astrodynamics an orbit equation defines the path of orbiting body ${\displaystyle m_{2}\,\!}$ around central body ${\displaystyle m_{1}\,\!}$ relative to ${\displaystyle m_{1}\,\!}$, without specifying position as a function of time. Under standard assumptions, a body moving under the influence of a force, directed to a central body, with a magnitude inversely proportional to the square of the distance (such as gravity), has an orbit that is a conic section (i.e. circular orbit, elliptic orbit, parabolic trajectory, hyperbolic trajectory, or radial trajectory) with the central body located at one of the two foci, or the focus (Kepler's first law).
If the conic section intersects the central body, then the actual trajectory can only be the part above the surface, but for that part the orbit equation and many related formulas still apply, as long as it is a freefall (situation of weightlessness).
## Central, inverse-square law force
Consider a two-body system consisting of a central body of mass M and a much smaller, orbiting body of mass m, and suppose the two bodies interact via a central, inverse-square law force (such as gravitation). In polar coordinates, the orbit equation can be written as[1]
${\displaystyle r={\frac {\ell ^{2}}{m^{2}\gamma }}{\frac {1}{1+e\cos \theta }}}$
where ${\displaystyle r}$ is the separation distance between the two bodies and ${\displaystyle \theta }$ is the angle that ${\displaystyle \mathbf {r} }$ makes with the axis of periapsis (also called the true anomaly). The parameter ${\displaystyle \ell }$ is the angular momentum of the orbiting body about the central body, and is equal to ${\displaystyle mr^{2}{\dot {\theta }}}$.[note 1] The parameter ${\displaystyle \gamma }$ is the constant for which ${\displaystyle \gamma /r^{2}}$ equals the acceleration of the smaller body (for gravitation, ${\displaystyle \gamma }$ is the standard gravitational parameter, ${\displaystyle -GM}$). For a given orbit, the larger ${\displaystyle \gamma }$, the faster the orbiting body moves in it: twice as fast if the attraction is four times as strong. The parameter ${\displaystyle e}$ is the eccentricity of the orbit, and is given by[1]
${\displaystyle e={\sqrt {1+{\frac {2E\ell ^{2}}{m^{3}\gamma ^{2}}}}}}$
where ${\displaystyle E}$ is the energy of the orbit.
The above relation between ${\displaystyle r}$ and ${\displaystyle \theta }$ describes a conic section.[1] The value of ${\displaystyle e}$ controls what kind of conic section the orbit is. When ${\displaystyle e<1}$, the orbit is elliptic; when ${\displaystyle e=1}$, the orbit is parabolic; and when ${\displaystyle e>1}$, the orbit is hyperbolic.
The minimum value of r in the equation is
${\displaystyle r={{\ell ^{2}} \over {m^{2}\gamma }}{{1} \over {1+e}}}$
while, if ${\displaystyle e<1}$, the maximum value is
${\displaystyle r={{\ell ^{2}} \over {m^{2}\gamma }}{{1} \over {1-e}}}$
If the maximum is less than the radius of the central body, then the conic section is an ellipse which is fully inside the central body and no part of it is a possible trajectory. If the maximum is more, but the minimum is less than the radius, part of the trajectory is possible:
• if the energy is non-negative (parabolic or hyperbolic orbit): the motion is either away from the central body, or towards it.
• if the energy is negative: the motion can be first away from the central body, up to
${\displaystyle r={{\ell ^{2}} \over {m^{2}\gamma }}{{1} \over {1-e}}}$
after which the object falls back.
If ${\displaystyle r}$ becomes such that the orbiting body enters an atmosphere, then the standard assumptions no longer apply, as in atmospheric reentry.
## Low-energy trajectories
If the central body is the Earth, and the energy is only slightly larger than the potential energy at the surface of the Earth, then the orbit is elliptic with eccentricity close to 1 and one end of the ellipse just beyond the center of the Earth, and the other end just above the surface. Only a small part of the ellipse is applicable.
If the horizontal speed is ${\displaystyle v\,\!}$, then the periapsis distance is ${\displaystyle {\frac {v^{2}}{2g}}}$. The energy at the surface of the Earth corresponds to that of an elliptic orbit with ${\displaystyle a=R/2\,\!}$ (with ${\displaystyle R\,\!}$ the radius of the Earth), which can not actually exist because it is an ellipse fully below the surface. The energy increase with increase of a is at a rate ${\displaystyle 2g\,\!}$. The maximum height above the surface of the orbit is the length of the ellipse, minus ${\displaystyle R\,\!}$, minus the part "below" the center of the Earth, hence twice the increase of ${\displaystyle a\,\!}$ minus the periapsis distance. At the top the potential energy is ${\displaystyle g}$ times this height, and the kinetic energy is ${\displaystyle {\frac {v^{2}}{2}}}$. This adds up to the energy increase just mentioned. The width of the ellipse is 19 minutes times ${\displaystyle v\,\!}$.
The part of the ellipse above the surface can be approximated by a part of a parabola, which is obtained in a model where gravity is assumed constant. This should be distinguished from the parabolic orbit in the sense of astrodynamics, where the velocity is the escape velocity. See also trajectory.
## Categorization of orbits
Consider orbits which are at one point horizontal, near the surface of the Earth. For increasing speeds at this point the orbits are subsequently:
• part of an ellipse with vertical major axis, with the center of the Earth as the far focus (throwing a stone, sub-orbital spaceflight, ballistic missile)
• a circle just above the surface of the Earth (Low Earth orbit)
• an ellipse with vertical major axis, with the center of the Earth as the near focus
• a parabola
• a hyperbola
Note that in the sequence above, ${\displaystyle h}$, ${\displaystyle \epsilon }$ and ${\displaystyle a}$ increase monotonically, but ${\displaystyle e}$ first decreases from 1 to 0, then increases from 0 to infinity. The reversal is when the center of the Earth changes from being the far focus to being the near focus (the other focus starts near the surface and passes the center of the Earth). We have
${\displaystyle e=\left|{\frac {R}{a}}-1\right|}$
Extending this to orbits which are horizontal at another height, and orbits of which the extrapolation is horizontal below the surface of the Earth, we get a categorization of all orbits, except the radial trajectories, for which, by the way, the orbit equation can not be used. In this categorization ellipses are considered twice, so for ellipses with both sides above the surface one can restrict oneself to taking the side which is lower as the reference side, while for ellipses of which only one side is above the surface, taking that side.
1. ^ There is a related parameter, known as the specific relative angular momentum, ${\displaystyle h}$. It is related to ${\displaystyle \ell }$ by ${\displaystyle h=\ell /m}$.
| 1,735 | 7,011 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 46, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.625 | 4 |
CC-MAIN-2019-04
|
latest
|
en
| 0.844954 |
https://planetmath.org/concyclic
| 1,618,867,080,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-17/segments/1618038917413.71/warc/CC-MAIN-20210419204416-20210419234416-00210.warc.gz
| 549,635,280 | 3,623 |
# concyclic
In any geometry where a circle is defined, a collection of points are said to be concyclic if there is a circle that is incident with all the points.
Remarks. Suppose all points being considered below lie in a Euclidean plane.
• Any two points $P,Q$ are concyclic. In fact, there are infinitely many circles that are incident to both $P$ and $Q$. If $P\neq Q$, then the pencil $\mathfrak{P}$ of circles incident with $P$ and $Q$ share the property that their centers are collinear. It is easy to see that any point on the perpendicular bisector of $\overline{PQ}$ serves as the center of a unique circle in $\mathfrak{P}$.
• Any three non-collinear points $P,Q,R$ are concyclic to a unique circle $c$. From the three points, take any two perpendicular bisectors, say of $\overline{PQ}$ and $\overline{PR}$. Then their intersection $O$ is the center of $c$, whose radius is $|OP|$.
• Four distinct points $A,B,C,D$ are concyclic iff $\angle CAD=\angle CBD$.
Title concyclic Concyclic 2013-03-22 16:07:58 2013-03-22 16:07:58 CWoo (3771) CWoo (3771) 7 CWoo (3771) Definition msc 51-00
| 314 | 1,099 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 18, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.53125 | 4 |
CC-MAIN-2021-17
|
longest
|
en
| 0.856804 |
https://wiki.alioth.net/index.php/OXP_howto_dockable_stations
| 1,657,093,650,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-27/segments/1656104668059.88/warc/CC-MAIN-20220706060502-20220706090502-00645.warc.gz
| 663,573,614 | 10,888 |
# OXP howto dockable stations
### Introduction
A small tutorial by Svengali about making a Station dockable in Oolite.
I'll try to show you what Oolite expects from you to make a Station-Entity dockable and where the limits are.
A lot of users have asked this question in the past and sometimes it's a bit tricky to get it working. But it's really no magic.
Before we'll start, you should be familiar with modelling and texturing and a bit knowledge about setting up the structure of oxps won't hurt too.
For a deeper understanding you should take a look in the following pages - just in case .-)
### Chapter I
Point A - Absolute
The first step is to understand the basics how Oolite is calculating positions and distances, but don't worry, it's not so hard and we won't explore the Math behind it. It only needs some practice. So let's begin.
A Star-System in Oolite has a Middle-Point (0,0,0). You can imagine this as the BIGBANG-point.
This is the 'Absolute Middle-Point' (Point A) and is the Reference-point for every single Entity in Oolite. Planets, Moons, Stations and Ships are positioned in relation to this point! Everything!
You can see the relation to Point A in action by activating the FPS display (see Display Frame Rate). The displayed coordinates are translated to another Coordinates-System (here 'pwm' instead of 'abs'), but you'll get a feeling for it. For more info about the Coordinate-Systems see
Point E - Entity
Now the fun starts. Lets take a look at the pic "Point E".
Every Entity (Ship, Station, Asteroid, Cargo Cannister, etc.) has a own Middle-Point (Point E) that is used to determine the Entities position in space in relation to the Absolute Middle-Point (Point A) and also to calculate the distances and relative positions of Entities to each other. This point defines also (in conjunction with other parameters like the Ships mass) the Entity itself (e.g. Oolite places the models vertices in relation to that point and can check for collisions and laser hits then) and is used to place the Subentities if declared.
The Entities Middle-Point (Point E) is absolute (0,0,0) for the Entity itself AND relative (e.g. -12356.3, 20582,8, -238.76), because it is set in relation to the Star-Systems Middle-Point (Point A). This means positioning a Entity has to be seen in relation to Point A!
To talk about two middle-points sounds confusing, but it gives Oolite the possibility to calculate everything a lot faster and is easier to handle, because you only have ONE fixed point (Point A) and everything (Point E's) is set in relation to this point!
Let's take a look in one possible way to add a Entity that uses Point A as reference.
``` system.legacy_addShipsAt('trader', 1, 'abs', [-10,20,30]);
```
As you can see we're trying to create 1 trader. The Reference-Point is Point A and we are trying to position the trader at absolute X = -10, Y = 20 and Z = 30. Point A is somewhere nearby the Witchpoint Beacon, so we have to move to this point to see this ship. So a Ships middle-point (Point E) is related to the Star-Systems middle-point (Point A) and we'll get a absolute position for this Entity in the Star-System (e.g. [-10,20,30]).
In Javascript we can access this position, change it or calculate other things in relation to it.
You might think that's it - basics finished.
Oh, no - wait a moment...
Point D - Docking
The last point is now how Oolite is doing the docking. With all that knowledge packed it is fairly simple to see that Oolite checks for positions and distances of Entities (Point E's) to each other. This is probably a timeintensive task, but the development-team is doing a great job and the Math behind it is pretty spectacular for noobs like me. (Thanks Aegidian and Ahruman).
The docking procedure itself is simply triggered by another Point E, if a ship comes in range. This point is different to Ships middle-points, so lets call it Point D. It's more a virtual sphere than a point, but as I've said above - no magic .-)
This point is like the other points not visible, so no player will see this trigger. The docks you can see in Stations are only there to give your eyes and brain something to work with. But internally this sphere has some side effects. The collision detection is disabled inside this sphere and reduced in a corridor along the Z-Axis. That's also the reason why a dock has always to be oriented to show along the Z-Axis.
But this can also cause some problems for players trying to dock. The trigger and the corridor are invisible and if you don't have the right angle while approaching you can pass the point where the collision detection is disabled/lowered. The result is then probably a Finale Grande -Press Space-.
Another very important note is that the docking-sphere is attached to the Main-Entity, even if a Subentity is used for the visual part!
Basics done. Confused? Don't worry - it will become worse .-)
### Chapter II - The right models
Positioning
Svengali's Notes: A few words about orienting the models!!! (Main-Entity: 0,0,0 and Subent: 0,0,0 or relative to Main)
X/Y have to be centered (in relation to the Main-Entity) and port_radius defines the movement on Z.
A few words about naming the models and naming conventions in general.
And names of textures have to be exactly (case-sensitive) the same as in .dat defined.
Please try to avoid using too many Vertices and Faces for your models - try to find the balance between visual quality and used processing time. The same for the textures - try to reduce the size and help to keep Oolite smooth and fast.
The slits dimensions are fixed (69mx69mx250m), only the visible part can be adjusted with the standard way. Chapter IV will show you two other technics to do it in a more flexible way.
### Chapter III - Finalizing
shipdata examples
Svengali's Notes:
Some words about moving the Subentity via Quaternion for finetuning or in general.
Some words about how Oolite is finding the right Subent == dock
b. by name, so if 'dock' (LOWERCASE!!!) is used in the entityname it will be handled as docking-slit
c. if nothing is found and scan_class = Station Oolite will use a standard slit.
Remember: The slit itself has to be placed on the z-axis (X/Y = 0) and the port_radius defines Z.
Unfortunately all this means that you can't adjust the slits dimensions, only the visible part can be adjusted with the standard way. Chapter IV will show you two other technics to do it in a more flexible way.
### Chapter IV - Advanced options
Main Entity
Sub-Entity
Scriptable
Svengali's Notes:
Way 1)
port_dimensions
Some words about deprecated, but still working.
Way 2)
Scripted
Pro: Flexible positions
Contra: Slow and not failsafe
## Snippets from elsewhere
• Important: You need to include "station" or "carrier" as one of the "roles" in the Shipdata.plist, or docking will fail dismally, no matter the excellence of the dock itself!
• Aegidian (2003-6?): Quick and easy instructions for wings3D.
``` 1. Make a panel on your main model where the dock entrance will be. A rectangle 192x64 is ideal, but you can go bigger - and use different shapes.
2. Assign the special material hole to the rectangle.
3. While you have it selected make a note of the coordinates of the centre of the rectangle.
4. As a separate model, make a cube then scale it to the dimensions of your rectangle and a depth of 250. So a cube scaled to 192 x 64 x 250.
5. Select the face of the cube facing the same way as your dock entrance in (1.) and assign it the hole material.
6. While it's selected choose Tools - Center - All from the menu to move that face to the origin.
7. Select the object, and from the right-click menu choose Invert to turn the cube 'inside out'.
8. Texture your new dock. And convert it to an Oolite .dat model file.
9. Position the new dock as a subentity of the station, at the coordinates you noted down in (3.)
10. You should now be done. You can vary the shape of the dock entrance in stage 1. If you copy and paste the panel into a new model you can even use it to extrude the dock shape.
```
• Charlie on the above (2007): For those who prefer to use an alternate 3D package to Wings you don't have to use it for the hole material. Just chopping holes in your model & dock seems to work!
• Killer Wolf (2022): I actually did a dockable asteroid, the hacker outpost and salvage gang that was incorporated into the Anarchies OXP according to a search. You could fly inside it and then into the docking bay itself so you could probably reverse engineer that if needed. I wrote a post about docks (2009) on page 2 of the "things I wish I knew.." thread. Beyond that stations like the Hathor Trade Station also have docking tunnels that might help.
• Murgh (2022): For starters you just leave a 192m x 64m slit (material defined as a "hole" in the [station/dockable] model).. always facing the Z axis, with the hole's center at x=0 and y=0 if there will be rotation..
There are of course plenty of fine dock models you can directly borrow or modify, or build a 192x64x250 (or bigger) cube yourself, skinning its insides (In Wings3D, in "body mode", invert the dock's surfaces for UVmapping to look right) to your pleasure.. I'd recommend keeping the two models separate to make it easier (to keep distinct objects with regular and inverted texturing apart), since they will be two different .dat files and the unification of the two happens later in the shipdata.plist.
| 2,257 | 9,464 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.671875 | 4 |
CC-MAIN-2022-27
|
latest
|
en
| 0.919379 |
https://myschool.ng/questions/view/academic-questions/192077
| 1,590,921,620,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-24/segments/1590347413097.49/warc/CC-MAIN-20200531085047-20200531115047-00366.warc.gz
| 458,335,857 | 8,910 |
### The quantity y is partly constant and partly varies inversely as the square of z....
The quantity y is partly constant and partly varies inversely as the square of z. When y = 6, z = 2 and when y = 4.5, z = 4. Find y when z = 10?
To get notifications when anyone posts a new answer to this question,
Please don't post or ask to join a "Group" or "Whatsapp Group" as a comment. It will be deleted. To join or start a group, please click here
Gaby
1 week ago
Given the above statement,
QI P/x^2
Follow
Aliyu
1 week ago
y is partly constant
y varies inversely as the square of z
y=k/z^2
when y=6 , z=2
therefore k= 6×4 k=24
when y =4.5 and z=4 k=16x4.5=72
y=? when z=10
y=k/z^2=100y =1 y= 1/100
Follow
• Aliyu: for k = 24 y= 24/100 y =0.24
for k=72, y= 72 /100 y=0.72
1 week ago
• Charlotte: Thanks
1 week ago
| 285 | 814 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.65625 | 4 |
CC-MAIN-2020-24
|
latest
|
en
| 0.885008 |
https://engineeringinfohub.com/working-out-quantities-of-u-shaped-wall/
| 1,726,249,285,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651535.66/warc/CC-MAIN-20240913165920-20240913195920-00304.warc.gz
| 215,760,628 | 31,711 |
Costing & Estimation
# Working Out Quantities of U Shaped Wall
### what is u shaped wall?
Today I explain in this article the working out quantities of u wall, what is u shaped how to calculate the estimate of u shaped wall. and also explain the step on how to determine the length of the u wall. Working Out Quantities of U Shaped Wall
when two sides are formed in a wall when two sides of the third round wall it is called a U-shaped wall. the different names of the three walls are assigned to determine the fate. such as wall as like A B C etc on the surface of the structure, the total length is determined by adding the length of the three walls. for this, the length of straight walls is usually known while the diameter or radius of a round wall is known. this is done by extending the diameter or radius to the centerline of the wall to determine the length of the circular wall lengths of straight walls also goes in this length. In this way, the length of the wall on the sage on the superstructure is known. the following formula is then used to determine the length of a circular wall.
L=πR
Where L = length of the centerline of the circular wall
R = Radius of curve up to centerline.
we have given data in the sketch
### ESTIMATE OF U WALL
S.no description No L B H Q Remarks
1 excavation 1 13.15 0.7 0.5 4.60m3 13.15=3+3+(0.70-0.30)
2 concrete (1:4:8) 1 13.15 0.7 0.3 2.76me
3 bricks work in foundation and plinth
step 1 1 12.95 0.5 0.2 1.30m3
step 2 1 12.85 0.4 0.6 3.08m3
total 4.38m3
4 D.P.C 2.5″ thick (1:2:4) 1 12.75 0.3 3.83m3
5 bricks work in the superstructure 1 12.75 0.3 3 11.48
6 cement plaster(1:4)20mm thick
level 2 12.85 0.6 15.42m2
above to plinth level 2 12.75 3 72.50m2
total 91.92m2
7 whitewash as per item no 6 91.92m2
| 558 | 1,764 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.125 | 4 |
CC-MAIN-2024-38
|
latest
|
en
| 0.879851 |
schmahl.net
| 1,685,948,813,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-23/segments/1685224651325.38/warc/CC-MAIN-20230605053432-20230605083432-00780.warc.gz
| 575,995,204 | 7,588 |
# What Makes 100%?
What does it mean to give MORE than 100%?
Ever wonder about those people who say they are giving more than 100%? We have all been to those meetings where someone wants you to give over 100%.
What makes up 100% in life?
If:
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
Is represented as:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26.
Then:
K-N-O-W-L-E-D-G-E
11+14+15+23+12+5+4+7+5 = 96%
And
H-A-R-D-W-O-R-K
8+1+18+4+23+15+18+11 = 98%
But ,
A-T-T-I-T-U-D-E
1+20+20+9+20+21+4+5 = 100%
And,
B-U-L-L-S-H-I-T
2+21+12+12+19+8+9+20 = 103%
AND, look how far ass-kissing will take you.
A-S-S-K-I-S-S-I-N-G
1+19+19+11+9+19+19+9+14+7 = 118%
So, one can conclude with mathematical certainty, that while Knowledge and Hard Work will get you close, and Attitude will get you there, it’s the Bullshit and Ass Kissing that will put you over the top.
| 362 | 899 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.8125 | 4 |
CC-MAIN-2023-23
|
longest
|
en
| 0.876945 |
https://convertoctopus.com/373-6-fluid-ounces-to-pints
| 1,675,761,544,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764500392.45/warc/CC-MAIN-20230207071302-20230207101302-00078.warc.gz
| 211,708,489 | 7,421 |
## Conversion formula
The conversion factor from fluid ounces to pints is 0.0625, which means that 1 fluid ounce is equal to 0.0625 pints:
1 fl oz = 0.0625 pt
To convert 373.6 fluid ounces into pints we have to multiply 373.6 by the conversion factor in order to get the volume amount from fluid ounces to pints. We can also form a simple proportion to calculate the result:
1 fl oz → 0.0625 pt
373.6 fl oz → V(pt)
Solve the above proportion to obtain the volume V in pints:
V(pt) = 373.6 fl oz × 0.0625 pt
V(pt) = 23.35 pt
The final result is:
373.6 fl oz → 23.35 pt
We conclude that 373.6 fluid ounces is equivalent to 23.35 pints:
373.6 fluid ounces = 23.35 pints
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pint is equal to 0.042826552462527 × 373.6 fluid ounces.
Another way is saying that 373.6 fluid ounces is equal to 1 ÷ 0.042826552462527 pints.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that three hundred seventy-three point six fluid ounces is approximately twenty-three point three five pints:
373.6 fl oz ≅ 23.35 pt
An alternative is also that one pint is approximately zero point zero four three times three hundred seventy-three point six fluid ounces.
## Conversion table
### fluid ounces to pints chart
For quick reference purposes, below is the conversion table you can use to convert from fluid ounces to pints
fluid ounces (fl oz) pints (pt)
374.6 fluid ounces 23.413 pints
375.6 fluid ounces 23.475 pints
376.6 fluid ounces 23.538 pints
377.6 fluid ounces 23.6 pints
378.6 fluid ounces 23.663 pints
379.6 fluid ounces 23.725 pints
380.6 fluid ounces 23.788 pints
381.6 fluid ounces 23.85 pints
382.6 fluid ounces 23.913 pints
383.6 fluid ounces 23.975 pints
| 525 | 1,855 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.15625 | 4 |
CC-MAIN-2023-06
|
latest
|
en
| 0.766161 |
https://johnnyroccia.com/2022/07/26/five-coins/
| 1,679,966,740,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296948756.99/warc/CC-MAIN-20230328011555-20230328041555-00757.warc.gz
| 387,533,503 | 22,321 |
# Five Coins
The odds of a flipped coin coming up heads five times in a row is 1 in 32. If you’ve flipped a coin four times and it’s come up heads all four times, the chance of the fifth flip coming up heads is 1 in 2.
A lot of people struggle with that, conceptually. That’s the heart of the Gambler’s Fallacy: the belief that past events can influence raw in-the-moment probability. After all (some people think), if there’s only a 1 in 32 chance of getting five heads in a row, then surely this fifth flip coming up heads must be very unlikely!
Of course, you’re not evaluating the difference between getting five heads in a row versus the probability of getting one tails. You’re evaluating the probability of getting five heads in a row versus the probability of getting exactly four heads and one tails, in that order. And the chances of those two outcomes are equal – 50/50. So if you want the past four flips to “count,” then you have to count them both ways.
This all doesn’t really matter to the current flip, though. The thing that makes “five heads in a row” unlikely has already happened, so you’re only really asking “out of every time four heads get flipped in a row, how many times is the next flip also heads?” Which is again, 50/50.
Here’s what I notice about people.
If someone has 8 hours to complete a task, and they get distracted for 6 of them, they’ll stress. They’ll think the job can’t be done in 2 hours, that they’re a failure, that they’ve already lost. But if you just gave them 2 hours, they wouldn’t feel that way.
The past flips already happened; they don’t affect the present chances of success. They are what they are, good or bad. Call it.
| 396 | 1,682 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.828125 | 4 |
CC-MAIN-2023-14
|
latest
|
en
| 0.947136 |
https://www.sofolympiadtrainer.com/Synopsis/imo1_addition.jsp
| 1,716,880,043,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00202.warc.gz
| 858,487,834 | 19,623 |
1. Addition means to add. Suppose Ram eats two toffees on Sunday and six toffees on Monday, then total toffees he has eaten on Sunday and Monday is 2 + 6 = 8
2. These are 2 arrows
3. If one more arrow is added it is three arrows = 3 arrows
4. If two more arrows are added , there are four arrows 2 + 3 = 5
5. If six more arrows are added there are eight arrows, 2 + 6 = 8 arrows
6. Two marbles and three marbles make five marbles = 2 + 3 = 5 marbles = 4 balls + 3 balls = 7 balls
7. 5 + 4 or five plus four is equal to = 9
8. When we add 8 in 20 we get 28. It has 2 tens and 8 ones. 2 X 10 = 20 + 8 ones.
9. Observe these numbers - 35, 67, 24, 82, 76, 49.
10. When we arrange these from lowest to highest it is 24, 35, 49, 67, 76, 82. This arrangement is ascending order
11. When we arrange from highest to lowest the arrangement is 82, 76, 67, 49, 35, 24. This arrangement is descending order.
12. If we add zero to a number we get the same number. For instance 0 + 46 = 46.
13. In 75, there are seven tens which is equal to 7 X 10 = 70 and 5 ones which is equal to 5 X 1 = 5, hence it is 70 + 5 = 75
14. If you have 10 pencils and you purchase 6 more pencils, then you would have 16 pencils ( 10 + 6 )
15. You have 80 erasers and you have 9 more marbles than the number of erasers, then you have 80 + 9 = 89 marbles.
16. When we add 3 to 5 on the number line it would be at 8.
| 475 | 1,396 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.21875 | 4 |
CC-MAIN-2024-22
|
latest
|
en
| 0.921227 |
https://www.lsbin.com/3807.html
| 1,679,946,560,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-14/segments/1679296948684.19/warc/CC-MAIN-20230327185741-20230327215741-00042.warc.gz
| 953,753,885 | 20,349 |
算法题：如何计算从1到n的所有数字的数字总和？
2021年3月30日12:33:07 发表评论 425 次浏览
本文概述
``````Input: n = 5
Output: Sum of digits in numbers from 1 to 5 = 15
Input: n = 12
Output: Sum of digits in numbers from 1 to 12 = 51
Input: n = 328
Output: Sum of digits in numbers from 1 to 328 = 3241``````
C ++
``````// A Simple C++ program to compute sum of digits in numbers from 1 to n
#include<bits/stdc++.h>
using namespace std;
int sumOfDigits( int );
// Returns sum of all digits in numbers from 1 to n
int sumOfDigitsFrom1ToN( int n)
{
int result = 0; // initialize result
// One by one compute sum of digits in every number from
// 1 to n
for ( int x = 1; x <= n; x++)
result += sumOfDigits(x);
return result;
}
// A utility function to compute sum of digits in a
// given number x
int sumOfDigits( int x)
{
int sum = 0;
while (x != 0)
{
sum += x %10;
x = x /10;
}
return sum;
}
// Driver Program
int main()
{
int n = 328;
cout << "Sum of digits in numbers from 1 to " << n << " is "
<< sumOfDigitsFrom1ToN(n);
return 0;
}``````
Java
``````// A Simple JAVA program to compute sum of
// digits in numbers from 1 to n
import java.io.*;
class GFG {
// Returns sum of all digits in numbers
// from 1 to n
static int sumOfDigitsFrom1ToN( int n)
{
int result = 0 ; // initialize result
// One by one compute sum of digits
// in every number from 1 to n
for ( int x = 1 ; x <= n; x++)
result += sumOfDigits(x);
return result;
}
// A utility function to compute sum
// of digits in a given number x
static int sumOfDigits( int x)
{
int sum = 0 ;
while (x != 0 )
{
sum += x % 10 ;
x = x / 10 ;
}
return sum;
}
// Driver Program
public static void main(String args[])
{
int n = 328 ;
System.out.println( "Sum of digits in numbers"
+ " from 1 to " + n + " is "
+ sumOfDigitsFrom1ToN(n));
}
}
/*This code is contributed by Nikita Tiwari.*/``````
Python3
``````# A Simple Python program to compute sum
# of digits in numbers from 1 to n
# Returns sum of all digits in numbers
# from 1 to n
def sumOfDigitsFrom1ToN(n) :
result = 0 # initialize result
# One by one compute sum of digits
# in every number from 1 to n
for x in range ( 1 , n + 1 ) :
result = result + sumOfDigits(x)
return result
# A utility function to compute sum of
# digits in a given number x
def sumOfDigits(x) :
sum = 0
while (x ! = 0 ) :
sum = sum + x % 10
x = x / / 10
return sum
# Driver Program
n = 328
print ( "Sum of digits in numbers from 1 to" , n, "is" , sumOfDigitsFrom1ToN(n))
# This code is contributed by Nikita Tiwari.``````
C#
``````// A Simple C# program to compute sum of
// digits in numbers from 1 to n
using System;
public class GFG {
// Returns sum of all digits in numbers
// from 1 to n
static int sumOfDigitsFrom1ToN( int n)
{
// initialize result
int result = 0;
// One by one compute sum of digits
// in every number from 1 to n
for ( int x = 1; x <= n; x++)
result += sumOfDigits(x);
return result;
}
// A utility function to compute sum
// of digits in a given number x
static int sumOfDigits( int x)
{
int sum = 0;
while (x != 0)
{
sum += x % 10;
x = x / 10;
}
return sum;
}
// Driver Program
public static void Main()
{
int n = 328;
Console.WriteLine( "Sum of digits"
+ " in numbers from 1 to "
+ n + " is "
+ sumOfDigitsFrom1ToN(n));
}
}
// This code is contributed by shiv_bhakt.``````
的PHP
``````<?php
// A Simple php program to compute sum
//of digits in numbers from 1 to n
// Returns sum of all digits in
// numbers from 1 to n
function sumOfDigitsFrom1ToN( \$n )
{
\$result = 0; // initialize result
// One by one compute sum of digits
// in every number from 1 to n
for ( \$x = 1; \$x <= \$n ; \$x ++)
\$result += sumOfDigits( \$x );
return \$result ;
}
// A utility function to compute sum
// of digits in a given number x
function sumOfDigits( \$x )
{
\$sum = 0;
while ( \$x != 0)
{
\$sum += \$x %10;
\$x = \$x /10;
}
return \$sum ;
}
// Driver Program
\$n = 328;
echo "Sum of digits in numbers from"
. " 1 to " . \$n . " is "
. sumOfDigitsFrom1ToN( \$n );
// This code is contributed by ajit
?>``````
``Sum of digits in numbers from 1 to 328 is 3241``
``````sum(9) = 1 + 2 + 3 + 4 ........... + 9
= 9*10/2
= 45
sum(99) = 45 + (10 + 45) + (20 + 45) + ..... (90 + 45)
= 45*10 + (10 + 20 + 30 ... 90)
= 45*10 + 10(1 + 2 + ... 9)
= 45*10 + 45*10
= sum(9)*10 + 45*10
sum(999) = sum(99)*10 + 45*100``````
``sum(10d - 1) = sum(10d-1 - 1) * 10 + 45*(10d-1)``
``````1) Find number of digits minus one in n. Let this value be 'd'.
For 328, d is 2.
2) Compute some of digits in numbers from 1 to 10d - 1.
Let this sum be w. For 328, we compute sum of digits from 1 to
99 using above formula.
3) Find Most significant digit (msd) in n. For 328, msd is 3.
4) Overall sum is sum of following terms
a) Sum of digits in 1 to "msd * 10d - 1". For 328, sum of
digits in numbers from 1 to 299.
For 328, we compute 3*sum(99) + (1 + 2)*100. Note that sum of
sum(299) is sum(99) + sum of digits from 100 to 199 + sum of digits
from 200 to 299.
Sum of 100 to 199 is sum(99) + 1*100 and sum of 299 is sum(99) + 2*100.
In general, this sum can be computed as w*msd + (msd*(msd-1)/2)*10d
b) Sum of digits in msd * 10d to n. For 328, sum of digits in
300 to 328.
For 328, this sum is computed as 3*29 + recursive call "sum(28)"
In general, this sum can be computed as msd * (n % (msd*10d) + 1)
+ sum(n % (10d))``````
C ++
``````// C++ program to compute sum of digits in numbers from 1 to n
#include<bits/stdc++.h>
using namespace std;
// Function to computer sum of digits in numbers from 1 to n
// Comments use example of 328 to explain the code
int sumOfDigitsFrom1ToN( int n)
{
// base case: if n<10 return sum of
// first n natural numbers
if (n<10)
return n*(n+1)/2;
// d = number of digits minus one in n. For 328, d is 2
int d = log10 (n);
// computing sum of digits from 1 to 10^d-1, // d=1 a[0]=0;
// d=2 a[1]=sum of digit from 1 to 9 = 45
// d=3 a[2]=sum of digit from 1 to 99 = a[1]*10 + 45*10^1 = 900
// d=4 a[3]=sum of digit from 1 to 999 = a[2]*10 + 45*10^2 = 13500
int *a = new int [d+1];
a[0] = 0, a[1] = 45;
for ( int i=2; i<=d; i++)
a[i] = a[i-1]*10 + 45* ceil ( pow (10, i-1));
// computing 10^d
int p = ceil ( pow (10, d));
// Most significant digit (msd) of n, // For 328, msd is 3 which can be obtained using 328/100
int msd = n/p;
// EXPLANATION FOR FIRST and SECOND TERMS IN BELOW LINE OF CODE
// First two terms compute sum of digits from 1 to 299
// (sum of digits in range 1-99 stored in a[d]) +
// (sum of digits in range 100-199, can be calculated as 1*100 + a[d]
// (sum of digits in range 200-299, can be calculated as 2*100 + a[d]
// The above sum can be written as 3*a[d] + (1+2)*100
// EXPLANATION FOR THIRD AND FOURTH TERMS IN BELOW LINE OF CODE
// The last two terms compute sum of digits in number from 300 to 328
// The third term adds 3*29 to sum as digit 3 occurs in all numbers
// from 300 to 328
// The fourth term recursively calls for 28
return msd*a[d] + (msd*(msd-1)/2)*p +
msd*(1+n%p) + sumOfDigitsFrom1ToN(n%p);
}
// Driver Program
int main()
{
int n = 328;
cout << "Sum of digits in numbers from 1 to " << n << " is "
<< sumOfDigitsFrom1ToN(n);
return 0;
}``````
Java
``````// JAVA program to compute sum of digits
// in numbers from 1 to n
import java.io.*;
import java.math.*;
class GFG{
// Function to computer sum of digits in
// numbers from 1 to n. Comments use
// example of 328 to explain the code
static int sumOfDigitsFrom1ToN( int n)
{
// base case: if n<10 return sum of
// first n natural numbers
if (n < 10 )
return (n * (n + 1 ) / 2 );
// d = number of digits minus one in
// n. For 328, d is 2
int d = ( int )(Math.log10(n));
// computing sum of digits from 1 to 10^d-1, // d=1 a[0]=0;
// d=2 a[1]=sum of digit from 1 to 9 = 45
// d=3 a[2]=sum of digit from 1 to 99 =
// a[1]*10 + 45*10^1 = 900
// d=4 a[3]=sum of digit from 1 to 999 =
// a[2]*10 + 45*10^2 = 13500
int a[] = new int [d+ 1 ];
a[ 0 ] = 0 ; a[ 1 ] = 45 ;
for ( int i = 2 ; i <= d; i++)
a[i] = a[i- 1 ] * 10 + 45 *
( int )(Math.ceil(Math.pow( 10 , i- 1 )));
// computing 10^d
int p = ( int )(Math.ceil(Math.pow( 10 , d)));
// Most significant digit (msd) of n, // For 328, msd is 3 which can be obtained
// using 328/100
int msd = n / p;
// EXPLANATION FOR FIRST and SECOND TERMS IN
// BELOW LINE OF CODE
// First two terms compute sum of digits from
// 1 to 299
// (sum of digits in range 1-99 stored in a[d]) +
// (sum of digits in range 100-199, can be
// calculated as 1*100 + a[d]
// (sum of digits in range 200-299, can be
// calculated as 2*100 + a[d]
// The above sum can be written as 3*a[d] +
// (1+2)*100
// EXPLANATION FOR THIRD AND FOURTH TERMS IN
// BELOW LINE OF CODE
// The last two terms compute sum of digits in
// number from 300 to 328. The third term adds
// 3*29 to sum as digit 3 occurs in all numbers
// from 300 to 328. The fourth term recursively
// calls for 28
return (msd * a[d] + (msd * (msd - 1 ) / 2 ) * p +
msd * ( 1 + n % p) + sumOfDigitsFrom1ToN(n % p));
}
// Driver Program
public static void main(String args[])
{
int n = 328 ;
System.out.println( "Sum of digits in numbers " +
"from 1 to " +n + " is " +
sumOfDigitsFrom1ToN(n));
}
}
/*This code is contributed by Nikita Tiwari.*/``````
Python3
``````# PYTHON 3 program to compute sum of digits
# in numbers from 1 to n
import math
# Function to computer sum of digits in
# numbers from 1 to n. Comments use example
# of 328 to explain the code
def sumOfDigitsFrom1ToN( n) :
# base case: if n<10 return sum of
# first n natural numbers
if (n< 10 ) :
return (n * (n + 1 ) / 2 )
# d = number of digits minus one in n.
# For 328, d is 2
d = ( int )(math.log10(n))
"""computing sum of digits from 1 to 10^d-1, d=1 a[0]=0;
d=2 a[1]=sum of digit from 1 to 9 = 45
d=3 a[2]=sum of digit from 1 to 99 = a[1]*10
+ 45*10^1 = 900
d=4 a[3]=sum of digit from 1 to 999 = a[2]*10
+ 45*10^2 = 13500"""
a = [ 0 ] * (d + 1 )
a[ 0 ] = 0
a[ 1 ] = 45
for i in range ( 2 , d + 1 ) :
a[i] = a[i - 1 ] * 10 + 45 * ( int )(math.ceil(math. pow ( 10 , i - 1 )))
# computing 10^d
p = ( int )(math.ceil(math. pow ( 10 , d)))
# Most significant digit (msd) of n, # For 328, msd is 3 which can be obtained
# using 328/100
msd = n / / p
"""EXPLANATION FOR FIRST and SECOND TERMS IN
BELOW LINE OF CODE
First two terms compute sum of digits from 1 to 299
(sum of digits in range 1-99 stored in a[d]) +
(sum of digits in range 100-199, can be calculated
as 1*100 + a[d]. (sum of digits in range 200-299, can be calculated as 2*100 + a[d]
The above sum can be written as 3*a[d] + (1+2)*100
EXPLANATION FOR THIRD AND FOURTH TERMS IN BELOW
LINE OF CODE
The last two terms compute sum of digits in number
from 300 to 328. The third term adds 3*29 to sum
as digit 3 occurs in all numbers from 300 to 328.
The fourth term recursively calls for 28"""
return ( int )(msd * a[d] + (msd * (msd - 1 ) / / 2 ) * p +
msd * ( 1 + n % p) + sumOfDigitsFrom1ToN(n % p))
# Driver Program
n = 328
print ( "Sum of digits in numbers from 1 to" , n , "is" , sumOfDigitsFrom1ToN(n))
# This code is contributed by Nikita Tiwari.``````
C#
``````// C# program to compute sum of digits
// in numbers from 1 to n
using System;
public class GFG {
// Function to computer sum of digits in
// numbers from 1 to n. Comments use
// example of 328 to explain the code
static int sumOfDigitsFrom1ToN( int n)
{
// base case: if n<10 return sum of
// first n natural numbers
if (n < 10)
return (n * (n + 1) / 2);
// d = number of digits minus one in
// n. For 328, d is 2
int d = ( int )(Math.Log(n) / Math.Log(10));
// computing sum of digits from 1 to 10^d-1, // d=1 a[0]=0;
// d=2 a[1]=sum of digit from 1 to 9 = 45
// d=3 a[2]=sum of digit from 1 to 99 =
// a[1]*10 + 45*10^1 = 900
// d=4 a[3]=sum of digit from 1 to 999 =
// a[2]*10 + 45*10^2 = 13500
int [] a = new int [d+1];
a[0] = 0; a[1] = 45;
for ( int i = 2; i <= d; i++)
a[i] = a[i-1] * 10 + 45 *
( int )(Math.Ceiling(Math.Pow(10, i-1)));
// computing 10^d
int p = ( int )(Math.Ceiling(Math.Pow(10, d)));
// Most significant digit (msd) of n, // For 328, msd is 3 which can be obtained
// using 328/100
int msd = n / p;
// EXPLANATION FOR FIRST and SECOND TERMS IN
// BELOW LINE OF CODE
// First two terms compute sum of digits from
// 1 to 299
// (sum of digits in range 1-99 stored in a[d]) +
// (sum of digits in range 100-199, can be
// calculated as 1*100 + a[d]
// (sum of digits in range 200-299, can be
// calculated as 2*100 + a[d]
// The above sum can be written as 3*a[d] +
// (1+2)*100
// EXPLANATION FOR THIRD AND FOURTH TERMS IN
// BELOW LINE OF CODE
// The last two terms compute sum of digits in
// number from 300 to 328. The third term adds
// 3*29 to sum as digit 3 occurs in all numbers
// from 300 to 328. The fourth term recursively
// calls for 28
return (msd * a[d] + (msd * (msd - 1) / 2) * p +
msd * (1 + n % p) + sumOfDigitsFrom1ToN(n % p));
}
// Driver Program
public static void Main()
{
int n = 328;
Console.WriteLine( "Sum of digits in numbers " +
"from 1 to " +n + " is " +
sumOfDigitsFrom1ToN(n));
}
}
// This code is contributed by shiv_bhakt.``````
的PHP
``````<?php
// PHP program to compute sum of digits
// in numbers from 1 to n
// Function to computer sum of digits in
// numbers from 1 to n. Comments use
// example of 328 to explain the code
function sumOfDigitsFrom1ToN( \$n )
{
// base case: if n<10 return sum of
// first n natural numbers
if ( \$n < 10)
return ( \$n * ( \$n + 1) / 2);
// d = number of digits minus one in
// n. For 328, d is 2
\$d = (int)(log10( \$n ));
// computing sum of digits from 1
// to 10^d-1, d=1 a[0]=0;
// d=2 a[1]=sum of digit from 1 to 9 = 45
// d=3 a[2]=sum of digit from 1 to 99 =
// a[1]*10 + 45*10^1 = 900
// d=4 a[3]=sum of digit from 1 to 999 =
// a[2]*10 + 45*10^2 = 13500
\$a [ \$d + 1] = array ();
\$a [0] = 0;
\$a [1] = 45;
for ( \$i = 2; \$i <= \$d ; \$i ++)
\$a [ \$i ] = \$a [ \$i - 1] * 10 + 45 *
(int)( ceil (pow(10, \$i - 1)));
// computing 10^d
\$p = (int)( ceil (pow(10, \$d )));
// Most significant digit (msd) of n, // For 328, msd is 3 which can be obtained
// using 328/100
\$msd = (int)( \$n / \$p );
// EXPLANATION FOR FIRST and SECOND
// TERMS IN BELOW LINE OF CODE
// First two terms compute sum of
// digits from 1 to 299
// (sum of digits in range 1-99 stored
// in a[d]) + (sum of digits in range
// 100-199, can be calculated as 1*100 + a[d]
// (sum of digits in range 200-299, // can be calculated as 2*100 + a[d]
// The above sum can be written as
// 3*a[d] + (1+2)*100
// EXPLANATION FOR THIRD AND FOURTH
// TERMS IN BELOW LINE OF CODE
// The last two terms compute sum of digits in
// number from 300 to 328. The third term adds
// 3*29 to sum as digit 3 occurs in all numbers
// from 300 to 328. The fourth term recursively
// calls for 28
return ( \$msd * \$a [ \$d ] + ( \$msd * (int)( \$msd - 1) / 2) * \$p +
\$msd * (1 + \$n % \$p ) + sumOfDigitsFrom1ToN( \$n % \$p ));
}
// Driver Code
\$n = 328;
echo ( "Sum of digits in numbers " ), "from 1 to " , \$n , " is " , sumOfDigitsFrom1ToN( \$n );
// This code is contributed by Sachin
?>``````
``Sum of digits in numbers from 1 to 328 is 3241``
C ++
``````// C++ program to compute sum of digits
// in numbers from 1 to n
#include<bits/stdc++.h>
using namespace std;
int sumOfDigitsFrom1ToNUtil( int n, int a[])
{
if (n < 10)
return (n * (n + 1) / 2);
int d = ( int )( log10 (n));
int p = ( int )( ceil ( pow (10, d)));
int msd = n / p;
return (msd * a[d] + (msd * (msd - 1) / 2) * p +
msd * (1 + n % p) +
sumOfDigitsFrom1ToNUtil(n % p, a));
}
// Function to computer sum of digits in
// numbers from 1 to n
int sumOfDigitsFrom1ToN( int n)
{
int d = ( int )( log10 (n));
int a[d + 1];
a[0] = 0; a[1] = 45;
for ( int i = 2; i <= d; i++)
a[i] = a[i - 1] * 10 + 45 *
( int )( ceil ( pow (10, i - 1)));
return sumOfDigitsFrom1ToNUtil(n, a);
}
// Driver code
int main()
{
int n = 328;
cout << "Sum of digits in numbers from 1 to "
<< n << " is " << sumOfDigitsFrom1ToN(n);
}
// This code is contributed by ajaykr00kj``````
Java
``````// JAVA program to compute sum of digits
// in numbers from 1 to n
import java.io.*;
import java.math.*;
class GFG{
// Function to computer sum of digits in
// numbers from 1 to n
static int sumOfDigitsFrom1ToN( int n)
{
int d = ( int )(Math.log10(n));
int a[] = new int [d+ 1 ];
a[ 0 ] = 0 ; a[ 1 ] = 45 ;
for ( int i = 2 ; i <= d; i++)
a[i] = a[i- 1 ] * 10 + 45 *
( int )(Math.ceil(Math.pow( 10 , i- 1 )));
return sumOfDigitsFrom1ToNUtil(n, a);
}
static int sumOfDigitsFrom1ToNUtil( int n, int a[])
{
if (n < 10 )
return (n * (n + 1 ) / 2 );
int d = ( int )(Math.log10(n));
int p = ( int )(Math.ceil(Math.pow( 10 , d)));
int msd = n / p;
return (msd * a[d] + (msd * (msd - 1 ) / 2 ) * p +
msd * ( 1 + n % p) + sumOfDigitsFrom1ToNUtil(n % p, a));
}
// Driver Program
public static void main(String args[])
{
int n = 328 ;
System.out.println( "Sum of digits in numbers " +
"from 1 to " +n + " is " +
sumOfDigitsFrom1ToN(n));
}
}
/*This code is contributed by Narendra Jha.*/``````
Python3
``````# Python program to compute sum of digits
# in numbers from 1 to n
import math
# Function to computer sum of digits in
# numbers from 1 to n
def sumOfDigitsFrom1ToN(n):
d = int (math.log(n, 10 ))
a = [ 0 ] * (d + 1 )
a[ 0 ] = 0
a[ 1 ] = 45
for i in range ( 2 , d + 1 ):
a[i] = a[i - 1 ] * 10 + 45 * \
int (math.ceil( pow ( 10 , i - 1 )))
return sumOfDigitsFrom1ToNUtil(n, a)
def sumOfDigitsFrom1ToNUtil(n, a):
if (n < 10 ):
return (n * (n + 1 )) / / 2
d = int (math.log(n, 10 ))
p = int (math.ceil( pow ( 10 , d)))
msd = n / / p
return (msd * a[d] + (msd * (msd - 1 ) / / 2 ) * p +
msd * ( 1 + n % p) + sumOfDigitsFrom1ToNUtil(n % p, a))
# Driver code
n = 328
print ( "Sum of digits in numbers from 1 to" , n, "is" , sumOfDigitsFrom1ToN(n))
# This code is contributed by shubhamsingh10``````
C#
``````// C# program to compute sum of digits
// in numbers from 1 to n
using System;
class GFG
{
// Function to computer sum of digits in
// numbers from 1 to n
static int sumOfDigitsFrom1ToN( int n)
{
int d = ( int )(Math.Log10(n));
int []a = new int [d+1];
a[0] = 0; a[1] = 45;
for ( int i = 2; i <= d; i++)
a[i] = a[i-1] * 10 + 45 *
( int )(Math.Ceiling(Math.Pow(10, i-1)));
return sumOfDigitsFrom1ToNUtil(n, a);
}
static int sumOfDigitsFrom1ToNUtil( int n, int []a)
{
if (n < 10)
return (n * (n + 1) / 2);
int d = ( int )(Math.Log10(n));
int p = ( int )(Math.Ceiling(Math.Pow(10, d)));
int msd = n / p;
return (msd * a[d] + (msd * (msd - 1) / 2) * p +
msd * (1 + n % p) + sumOfDigitsFrom1ToNUtil(n % p, a));
}
// Driver code
public static void Main(String []args)
{
int n = 328;
Console.WriteLine( "Sum of digits in numbers " +
"from 1 to " +n + " is " +
sumOfDigitsFrom1ToN(n));
}
}
// This code contributed by Rajput-Ji``````
``Sum of digits in numbers from 1 to 328 is 3241``
| 6,844 | 18,809 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.9375 | 4 |
CC-MAIN-2023-14
|
latest
|
en
| 0.460429 |
http://matrix.skku.ac.kr/2018-DM-Sol/2018-DM-Sol.htm
| 1,569,168,996,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-39/segments/1568514575596.77/warc/CC-MAIN-20190922160018-20190922182018-00016.warc.gz
| 128,504,409 | 76,221 |
Action Learninghttp://matrix.skku.ac.kr/2018-DM-Sol/2018-DM-Sol.htm
Discrete Mathematics
Prof : Sang-Gu LEE (http://matrix.skku.ac.kr/sglee/ )
주경용 (조장) 생명과학과, 남영욱 시스템경영공학과, 박종호 시스템경영공학과, 김동준 융합생명공학과
Chapter 1
1. 1]Q&A
Question & Answer Chapter Ch1, Sets and Logics Date 2018.03.09 Type Writer Title Contents Question 조규상 Problems in solving Quiz 1,2 In Quiz1.Problem 3 (6)T/F The proposition "If all cats are white, then all dog is brown" is true Solution says it is True. I wonder if it is true because hypothesis is false. Or are there any other reason exists? 2. In Quiz2. Problem 1 Solution writes a truth table to determine the validity. Is it okay to solve in this way? p: “Socrates is a cat.” q : “Socrates is a mortal.” r: “Socrates is a man.” p → q r ∨ (¬q) r ∨ (¬q) ≡ (¬q) ∨ r ≡ q → r p → q q → r ∴p → r Answer 1 이상구 교수님 Final: Dear Mr. 조규상, Use truth table to determine^^ Final: Dear Mr. 조규상, Use truth table to determine^^ Dear Mr. 조규상, Here is my Comment^^ 1. In Quiz1. Problem 3 (6) T/F The proposition "If all cats are white, then all dog is brown" is true Because if it is true because the hypothesis (all cats are white) is false. [Yes you are right] 2. In Quiz2. Problem 1 Solution writes a truth table to determine the validity. p: “Socrates is a cat.” q : “Socrates is a mortal.” r: “Socrates is a man.” p → q r ∨ (¬q) r ∨ (¬q) ≡ (¬q) ∨ r ≡ q → r p → q q → r ________ ∴p → r Is it okay to solve in this way? 틀리지는 않았지만, truth table 을 이용하여 validity 를 determine 하라고 요구하였으니. 이 답을 부분 점수를 받을 수 있습니다. Use truth table to determine the validity. 로 다시 한번 해 보시는 것이 맞습니다. 2 박종호 final 박종호 : quiz2 problem1 In Quiz2.Problem 1 Solution writes a truth table to determine the validity. p: “Socrates is a cat.” q : “Socrates is a mortal.” r: “Socrates is a man.” p → q r ∨ (¬q) r ∨ (¬q) ≡ (¬q) ∨ r ≡ q → r p → q q → r ________ ∴p → r ->위 문제에 대한 진리표는 솔루션에 있으니 생략하고 r ∨ (¬q) ≡ (¬q) ∨ r ≡ q → r라고 작성하신 부분에서 동치가 나와 참고하시라고 작성합니다. 예제 1.2.13을 참고하시면 p->q의 부정과 p 또는 ㄱq와의 논리적 동치(logically equivalent)를 다뤘습니다. 같은 방식으로 동치 관계를 구한 것들을 올립니다. 1. ㄱ(p v q) ≡ ㄱp and ㄱq 2. ㄱ(p and q) ≡ ㄱp v ㄱq 3. p->q ≡ ㄱp and q 4. p<->q ≡ (p->q) and (q->p) p → q q → r ________ ∴p → r 작성자는 동치를 이용해 q->r을 이끌어내고 가설적 삼단논법(hypothetical syllogism)의 규칙으로 p->r이 타당함을 증명해냈습니다. 근데 문제에는 진리표를 작성하라는 말이 없음에도 불구하고 부분 점수를 받는군요~ 삼단논법은 그림 1.4.1과 예제 1.4.4를 참고하시면 이해가 편할 것 같습니다.
2. 2]Q&A Final Ok
Question & Answer Chapter Ch1, Sets and Logics Date 2018.03.10 Type Writer Title Contents Question 이세영 'Example 1.3.2' & 'Definition 1.3.3' I'm little bit confusing with 'Example 1.3.2'&'Definition 1.3.3' in the textbook. p→q p: The Mathematics Department gets an additional $60,000. q: The Mathematics Department will hire one new faculty member. the book says... we should define p→q to be true when p is false. But I think... If p is false, whatever q is, We cannot know whether p→q true or not. We should conclude 'it is true!' when we don't know exactly the result that p→q is true or not? Answer 1 이상구 교수님 Final: Answer: Dear 이세영, 'Example 1.3.2' & 'Definition 1.3.3' Dear 이세영, 설명을 읽어보고 이해가 되면 제목에 Final 을 주어 자신이 이해한 요약을 답으로 정리하여 달아주세요^^ p→q p: The Mathematics Department gets an additional$60,000. q: The Mathematics Department will hire one new faculty member. We did agree [p→q to be true when p is false] from the observation of the Truth table. 1. If p is false (The Mathematics Department 가 추가 예산을 확보 못하면), whatever q is (The Mathematics Department 가 새 교수요원을 고용하거나, 안 하거나), The Mathematics Department 는 false인 행동을 한 것이 전혀 없다 [예를 들어, "학장이 학과장들에게 만일 단과대학이 추가예산 5천만원을 확보하면 (p), 수학과에 교수요원을 한 두 명 고용하도록 도와 주겠다 (q)." 하고 얘기를 한 상황에서 생각해 보시오. 이 때 학장이 만을 재단으로부터 추가 예산을 확보 못하면, 학장이 추가 교수요원을 뽑아 주어도 OK 이고, 못 뽑아주어도 OK 입니다)] 따라서 2. (The Mathematics Department 가 새 교수요원을 고용한 것도 false 인 행동을 한 것이 전혀 없으니까 False 와 True 중에서 하나만 고른다면 True 로 정의하는 것이 더 합리적이고, 따라서 True 이고) 3. (The Mathematics Department 가 새 교수요원을 고용하지 않아도, false 인 행동을 한 것이 전혀 없으니까 False 와 True 중에서 하나만 고른다면 True 로 정의하는 것이 더 합리적이고, 따라서 True 입니다.) 결론 We know whether p→q is not False. So we agreed to conclude 'it is not False = It is true' when we know exactly the result that p→q is not false. Final 이세영 Final : summery (Tautology), Final OK by SGLee Final : summery (Tautology), Final OK by SGLee p→q p : The Mathematics Department gets an additional \$60,000. q : The Mathematics Department will hire one new faculty member. If p is false => whatever q is.... The Mathematics did not do "false" action. It is more reasonable to choose 'True' than 'False' in this 'Not false' action. 헷갈렸었는데 답을 듣고 확실히 알 수 있었습니다! 감사합니다: D ****** Final을 붙여 답을 다는 것이 이렇게 하는게 맞나요?
2. 3]Q&A
Chapter
Ch1, Sets and Logics
Date
2018.03.10
Type
Writer
Title
Contents
Question
이진용
Quiz 1 (4), (5)
Tautology contradiction 이 두 단어가 의미하는 것이 무엇인지 잘 모르겠습니다.
1
이상구
교수님
Final: Dear 이진용, contradiction (모순), tautology(항진(恒眞)명제) (4), (5)
In logicproof by contradiction is a form of proof, and more specifically a form of indirect proof, that establishes the truth or validity of a proposition. It starts by assuming that the opposite proposition is true, and then shows that such an assumption leads to a contradiction. Proof by contradiction is also known as indirect proofapagogical argumentproof by assuming the opposite, and reductio ad impossibilem. It is a particular kind of the more general form of argument known as reductio ad absurdum.
G. H. Hardy described proof by contradiction as "one of a mathematician's finest weapons", saying "It is a far finer gambit than any chess gambit: a chess player may offer the sacrifice of a pawn or even a piece, but a mathematician offers the game."[1]
Principle
Proof by contradiction is based on the law of noncontradiction as first formalized as a metaphysical principle by Aristotle. Noncontradiction is also a theorem in propositional logic. This states that an assertion or mathematical statement cannot be both true and false. That is, a proposition Q and its negation Q ("not-Q") cannot both be true. In a proof by contradiction, it is shown that the denial of the statement being proved results in such a contradiction. It has the form of a reductio ad absurdum argument. If P is the proposition to be proved:
2. P is assumed to be false, that is P is true.
3. It is shown that {\displaystyle \lnot } P implies two mutually contradictory assertions, Q and Q.
4. Since Q and Q cannot both be true, the assumption that P is false must be wrong, and P must be true.
An alternate form derives a contradiction with the statement to be proved itself:
1. P is assumed to be false.
2. It is shown that {\displaystyle \lnot } P implies P.
3. Since P and P cannot both be true, the assumption must be wrong and P must be true.
An existence proof by contradiction assumes that some object doesn't exist, and then proves that this would lead to a contradiction; thus, such an object must exist. Although it is quite freely used in mathematical proofs, not every school of mathematical thought accepts this kind of nonconstructive proof as universally valid.
Tautology- 항진(恒眞)명제(tautology)
논리식 혹은 합성명제에 있어 각 명제의 참·거짓의 모든 조합에 대하여 항상 참인 것을 항진(恒眞)명제(tautology) 라고 한다
에서 알 수 있는 바와 같이......
i) 가능한 모든 조건 아래에서 모든 진리치가 진이 되는 명제
이것은 원자명제 진리치에 상관없이 항상 진인 이런 명제를 항진명제 (tautology)라 한다
일반적으로어떤 진리함수적 명제의 가능한 진리치가 모두 진일 때, 오직 그 때에만, 그 명제를 tautology 라 부른다
항진명제는 그 논리적 형식상 결코 위가 될 수 없고 항상 진이어야만 한다.
2. 4]Q&A Final Ok
Question & Answer Chapter Ch1, Sets and Logics Date 2018.03.10 Type Writer Title Contents Question 함형빈 Quiz #1 Problem 1 number 3 Cardinality of P(T)를 물어봤는데 이중 P가 정확히 무엇을 의미하는지는 모르겠지만 solution에 답이 8로 나와있는데...... 그건 아닌 것 같다고 생각됩니다. P가 전체를 의미한다면 10이 나올 것이고, 중복되지 않는 것이나 T set에 있는 것과 같은 것의 총합이라 한다면 6이 답일텐데...... 왜 8이 나왔는지 모르겠습니다. Final 이상구 교수님 Final OK by SGLee: Answer : Quiz #1 Problem 1 number 3, Cardinality of P(T) = 8 Answer : Quiz #1 Problem 1 number 3, Cardinality of P(T) = 8 Answer Given set T=(2,4,8) , Cardinality of P(T), the power sot of T. empty set,(2), (4), (8), (2,4), (2,8), (4,8), (2,4,8) Total 8 Cardinality of P(T) = 8 그러므로 #Promblem1-(3)의 답이 8 입니다.
2. 5]Q&A Final Ok
Question & Answer Chapter Ch1, Sets and Logics Date 2018.03.10 Type Writer Title Contents Question 장호 Power set P( ) 질문 Quiz1의 problem1이나 problem2에서 P(T)와 P(S)가 나왔는데, 그 의미를 모르갰습니다. Answer 1 조규상 Final: Final OK by SGLee Quiz 1 Problem 1-3, Power set, 8 Final: Final OK by SGLee Quiz 1 Problem 1-3, Power set, 8 Power set으로 생각하시면 8으로 나옵니다 Q set T= {2, 4, 8} 의 Power set 안의 부분집합의 개수는? Answer Final: Answer Re: Quiz #1 Problem 1 number 3, Power set, 8 공집합, {2}, {4}, {8}, {2, 4}, {2, 8}, {4, 8}, {2, 4, 8} 총 8 개입니다. ******************* 그러므로 #Promblem1-(3)의 답이 8입니다. Final: Answer Re: Quiz #1 Problem 1 number 3, Power set, 8 위에도 답을 주셨습니다.
2. 6]Q&A
Question & Answer Chapter Ch1, Sets and Logics Date 2018.03.10 Type Writer Title Contents Question 홍석호 Quiz1 에 2번문제 질문드립니다. {0, {1, 2}, {3, 4, 5}} ⊆ Z. For any set S, S ⊆ P(S). 이 둘이 False인 이유를 모르겠습니다. 그리고 더불어 p ∨ (p → q) is a tautology. 진리표를 만들어보면 항진명제라는 것은 자명합니다. 그러나 진리표 없이 문맥상으로 항진, 모순 명제를 유추할수는 없는지요? 문맥상으로 보면 P와 Q의 합집합입니다. 명제는 집합으로는 접근하면 안되는지도 궁금합니다. Answer 1 이상구 교수님 Final: Dear 홍석호 군, 용도에 따라 다양한 증명이 있습니다. 이것을 2장에서 배웁니다. 홍석호 군, Dear 홍석호 군, quiz 1 problem 2 - (10) False 인 이유를 알려드립니다. For any set S, element S is in P(S) [True, # 9], but [element S⊆P(S) is False, # 10] 멱집합(冪集合, 영어: power set) 의 정의를 자세히 보시면, S 는 power set of S의 element 이기는 하지만 [True, # 9 ] ,, S 는 power set of S 의 subset 은 아닙니다. [ False, # 10] https://ko.wikipedia.org/wiki/%EB%A9%B1%EC%A7%91%ED%95%A9 p ∨ (p → q) is a tautology. 진리표를 만들어보면 항진명제라는 것은 자명합니다...... 진리표 없이 문맥상으로 항진, 모순 명제를 유추할수 있습니다. 문맥상으로 P와 Q의 합집합 임을 유추할 수 있습니다. 명제를 집합으로 접근하는 것이 벤 다이어그램을 이용하는 설명법입니다. 용도에 따라 다양한 증명이 있습니다. 이것을 2장에서 배웁니다.
2. 7]Q&A Final Ok
Question & Answer Chapter Ch1, Sets and Logics Date 2018.03.10 Type Writer Title Contents Question 강민석 quiz1 단어뜻질문드립니다 tautology가 무슨뜻인지 몰라서 인터넷에 검색해보니 동의어반복이라는데 정확히 무슨의미인지 궁급합니다 Answer 1 이동욱 항진명제입니다. 항진명제: 언제나 참인 명제라는 뜻입니다. Final 이상구 교수님 Final OK by SGLee to 이동욱, 강민석군, 항진(恒眞)명제(tautology) 는 ...... 강민석군, 항진(恒眞)명제(tautology) quiz1 단어 뜻 Tautology- 항진(恒眞)명제(tautology) 논리식 혹은 합성명제에 있어 각 명제의 참·거짓의 모든 조합에 대하여 항상 참인 것을 항진(恒眞)명제(tautology) 라고 한다. Tautology : 진리표 에서 알 수 있는 바와 같이,... i) 가능한 모든 조건 아래에서 모든 진리치가 진이 되는 명제 이것은 원자명제 진리치에 상관없이 항상 진인 이런 명제를 항진명제 (tautology) 라 한다. 일반적으로, 어떤 진리함수적 명제의 가능한 진리치가 모두 진일 때, 오직 그 때에만, 그 명제를 tautology 라 부른다. 항진명제는 그 논리적 형식상 결코 위가 될 수 없고 항상 진이어야만 한다.
2. 8]Q&A
Question & Answer Chapter Ch1, Sets and Logics Date 2018.03.10 Type Writer Title Contents Question 이동욱 quiz 1 problem 2 질문 드립니다. quiz 1 problem 2 - (10) 번 문제 질문드립니다. For any set S, S⊆P(S) I don't have any idea why this is False... Final 이상구 Final: Dear 이동욱 군, quiz 1 problem 2 - (10) False 인 이유를 알려드립니다. Dear 이동욱 군, quiz 1 problem 2 - (10) False 인 이유를 알려드립니다. For any set S, element S is in P(S) [True, # 9], but [element S⊆P(S) is False, # 10] 멱집합(冪集合, 영어: power set) 의 정의를 자세히 보시면, S 는 power set of S의 element 이기는 하지만 [True, # 9 ] ,, S 는 power set of S 의 subset 은 아닙니다. [ False, # 10] https://ko.wikipedia.org/wiki/%EB%A9%B1%EC%A7%91%ED%95%A9
2. 9]Q&A
Question & Answer Chapter Ch1, Sets and Logics Date 2018.03.11 Type Writer Title Contents Question 강민구 example 1.4.6 질문입니다 p → r and p ⇒ Using modus ponens to conclude r. modus ponen을 쓰면 p->r p ----- r 인데 r을 conclude 할 때는 p가 true라는 건 모르지 않나요? Final 이상구 Final: Answer: example 1.4.6 답입니다. (이 조교님, 학생들 질문에 답을 주어 보세요-TA 채점이랍니다) 답: p->r p ----- r 라는 의미는 p->r 이 true 이고 (1) p 도 true 이면 (2) ------------------------------------------------- r 이다 라는 의미인데, 이는 (2) p 가 True 이므로, (1) 에 의하여 p->r 이 true 가 됩니다. 따라서 r 이 Ture 임이 (2)와 (1) 에의하여 유도 되는 것입니다^^
2. 10]Q&A Final Ok
Question & Answer Chapter Ch1, Sets and Logics Date 2018.03.11 Type Writer Title Contents Question 김지만 Quiz1 - Problem 2 - (5) 질문드립니다. There are two partitions of a set with two element 두 개의 원소를 가진 집합을 partition(분할) 할 경우에는 오직 한가지의 경우, 2개의 원소가 두 개의 집합에 하나씩 나뉘어서 들어가는 경우 밖에 없기에 two partitions이 아니지 않나요? 답에는 True라고 되어 있었습니다. two Partitions이라는 단어가 분할이 된 두개의 집합이라는 뜻인가요? Answer & Final 이상구 교수님 Final OK by SGLee to 김지만 Ans. 1-Problem2-(5) 답: There are two partitions ... (T) 김지만 군, Ans. 1-Problem2-(5) 답: There are two partitions of a set with two element. (True) There are two partitions of a set with two element. 의 의미는 두 개의 원소를 가진 집합을 partition(분할) 할 경우에는 오직 한가지의 경우가 아니라 2가지 경우가 존재합니다. 1. 두 개의 집합에 각각 0개와 2개 씩 들어가게 나누는 방법과 2. 2개의 원소가 두 개의 집합에 하나씩 나뉘어서 들어가는 경우가 있습니다. 그래서 two partitions 가 존재하고 그래서 답에 True 라고 되어 있었습니다 Re-Final 박종호 final : dear 김지만 There are two partitions of a set with two elements. ->두개의 원소를 가진 집합은 2개의 분할을 갖느냐는 문제입니다. 교수님께서 잘 설명해주셨지만 좀 더 이해를 돕고자 예를 들어 표현해봤습니다. 집합 X={a, b} 라고 하면 집합 X의 분할은 {{a, b}}, {{a}, {b}}, 공집합 이렇게 3개가 나옵니다. 다음은 원소 3개를 가진 집합의 분할입니다. (참고) -> 공집합의 원소 포함 여부에 대한 부분에 대해서도 참고하시면 좋을 것 같습니다.
2. 11]Q&A Final Ok
Question & Answer Chapter Ch1, Sets and Logics Date 2018.03.11 Type Writer Title Contents Question 이상철 Quiz 1. Problem 2-(6) Quiz 1. Problem 2-(6) 문제가 True인데 정확히 어떻게 True 인지 알고 싶습니다. Answer 1 박종호 final : dear 이상철 quiz1 problem2-6 (6) If A and B are finite sets, then A X B is also finite 질문에서 집합 A,B가 유한집합이라면 그 데카르트 곱(Cartesian product)은 유한한가라고 물었습니다. 당연히 true입니다. 일반적으로 lA X Bl = ㅣAㅣXㅣBㅣ입니다. 예제 1.1.24 를 참고하시면 될 것 같습니다. Final 이상철 final OK by SGLee Dear 이상철 말씀하신 예제를 다시 보고 확인해보니ㅣAㅣ X ㅣBㅣ = ㅣA X Bㅣ 가 true일 수 밖에 없군요. 답변 감사합니다. 이해되었습니다.
2. 12]Q&A Final Ok
Question & Answer Chapter Ch1, Sets and Logics Date 2018.03.11 Type Writer Title Contents Question 황병재 English Q. 프린트내용질문 이해안되는 부분이있는데 1. We define the union of an arbitrary family S of be those element x belonging to at least one X in S 이거랑 2. We define the intersection of an abitrary family S of sets to be those elements x belonging to every set X in S 이부분이 해석이잘안되는데 정확히 무슨뜻인지 알려주시면 감사하겠습니다 Answer 1 이상구 교수님 Answer : 여러분이 그동안 배운 union 집합과 intersection 집합 의 집합을 유한개의 부분집합들 X_1, X_2, ...... , X_n 을 가진 S 또는 무수히 많은 부분집합들 X_1, X_2, X_3, ...... 에 대한 개념으로 확장한 (일반화한) 개념입니다. 1. We define the union of an arbitrary family S of be those element x belonging to at least one X in S 의 의미는 예를 들어 X_1, X_2, ...... , X_n 이 S 의 부분집합이라고 할 때 X_1, X_2, ...... , X_n 의 union 집합이란 S = {X_1, X_2, ......, X_n} 들 중 적어도 하나의 X_i 에 속하는 element x 들을 모은 집합 이라는 의미입니다. 2. We define the intersection of an abitrary family S of sets to be those elements x belonging to every set X in S 의 의미는 예를 들어 X_1, X_2, ...... , X_n 이 S 의 부분집합이라고 할 때 X_1, X_2, ...... , X_n 의 intersection 집합이란 모든 X_1, X_2, ...... , X_n 들에 속하는 element x 들을 모은 집합 이라는 의미입니다. 즉 여러분이 그동안 배운 union 집합 과 intersection 집합 의 집합을 유한개의 부분집합들 X_1, X_2, ...... , X_n 을 가진 S 또는 무수히 많은 부분집합들 X_1, X_2, X_3, ...... 에 대한 개념으로 확장한 (일반화한) 개념입니다. Final 박종호 final : dear 황병재 . We define the union of an arbitrary family S of be those element x belonging to at least one X in S 이거랑 2. We define the intersection of an abitrary family S of sets to be those elements x belonging to every set X in S 이부분이 해석이잘안되는데 정확히 무슨뜻인지 알려주시면 감사하겠습니다 교수님께서 예를 들어 자세히 설명해주셨지만 질문자가 정확한 뜻을 물었기에 번역해드립니다. 1.We define the union of an arbitrary family S of be those element x belonging to at least one X in S -> 임의의 집합족 S의 합집합은 S내의 적어도 하나의 집합 X에 속하는 원소 x들로 구성된 집합으로 정의한다. 2.We define the intersection of an abitrary family S of sets to be those elements x belonging to every set X in S -> 임의의 집합족 S의 교집합은 S 내의 모든 집합 X에 속하는 원소 x들로 구성된 집합으로 정의한다.
Chapter 2 PBL Report
Team6
1. 1] Summary Final OK
Summary Chapter 2 Proofs Date 2018.03.16 Type Writer Title Contents Summary 이세영 Final OK by SGLee summary : 2018.03.16 the Class-Lecture chapter 2 . Proof 2.1 Mathematical Systems, Direct Proofs, and Counterexamples 2.2 More Methods of Proof Problem-Solving Corner: Proving Some Properties of Real Numbers 2.3 Resolution Proofs (Skipped) 2.4 Mathematical Induction Problem-Solving Corner: Mathematical Induction 2.5 Strong Form of Induction and the Well-Ordering Property #수학적 체계, 직접 증명 및 반례(Mathematical Systems, Direct proofs and Countereexamples) *What is "axioms,definition,theorem,lemma,corollary" is? =>axioms(공리) : 참이라고 간주 definition(정의) theorem(정리) : 참이라고 증명된 명제 lemma(보조정리) : 다른정리를 증명하는데 유용하게 사용 corollary(따름증명) : 다른 정리로 부터 쉽게 얻어지는 정리 proof(증명) : 정리가 참임을 확립하는 논법 *Direct Proofs : A direct proof assumes that the hypotheses are true and then, using the hypotheses as well as other axioms, definitions, and previously derived theormes, show directly that the conclusion is true 2.2 More Methods of Proof Discuss several more methods of proof: Proof by contradiction, Proof by contrapositive, Proof by cases, Proofs of equivalence, existence Proofs #기타증명방법 : 실수 성질의 증명 *proof by contradiction(모순에 의한 증명) : A proof by contradiction assumes that the hypotheses are true and that the conclusion is false and then, using the hypotheses and the nagated couclusion as well as other axioms, definitions, and previously derived theorems, derives a contradiction. *indirect proof(간접증명) : is another name for proof by contradiction *proof by contrapositive(대우에 의한 증명) :To proof p->q, proof not p -> not q *proof by cases(경우에 의한 증명) *proof of Equivalence(동치 증명) : Proof of equivalence shows that two or more statements are all true or all false. *existence proof(존재 증명) *constructive existence proof(건설적 존재 증명) #분해 증명(Resolution proofs) 1시 30분 ~ #수학적 귀납법(mathmatical induction) *What is mathmatical induction? =>S(1)is true for all n >= 1 , if S(n) is true, then S(n+1)is true #강한 형식의 귀납법 및 정렬 순서 속성(Strong Form of Induction and the Well-Ordering Property) *Strong form of mathematical induction : allows us to assume the truth of all of the preceding statements. * Well-Ordering Property The Well-Ordering Property for nonnegative integers states that every nonempty set of nonnegative integers has a least element. This property is equivalent to the two forms of induction. We use the Well-Ordering Property to prove something familiar from long division: divided by is dividend, is divisor, is quotient, is remainder, Comment 이상구 summary : chapter 2 다른 학생이 조금 더 보충하고 이쁘게 하여 Final 선언 하세요^^ 다른 학생이 조금 더 보충하고 이쁘게 하여 Final 선언 하세요^^ Final 강민석 summary : chapter 2 .Mr Kang 강민석 and 홍석호, added some on it 1. 수학적 체계는 공리(항상 참이라고 간주함), 정의, 미정의 용어로 구성된다. 수학에서 중요한 것은 정리이다. 정리를 증명하기 위해서는 증명(proof)이라는 과정이 필요하다. 정리에서도 특이한 정리가 있는데, 그것은 보조정리(lemma), 따름정리(corollary)가 있다. 보조정리는 다른 정리를 증명하기 위해 필요한 도구이고 따름정리는 다른 정리로부터 쉽게 얻어지는 정리이다. 2. 증명의 종류 ①직접 증명(direct proof) 직접 증명은 명제p가 참이라 가정했을 때 공리, 정의 등을 이용하여 명제q가 참임을 직접적으로 증명하는 것이다. ②반증(disprove) 어떤 명제가 "모든 x에대해 P(x)가 참이다"라고 하자. 그 명제가 거짓인 것같으면 반례(counterexample)인 x하나만 찾으면 그 명제가 거짓임을 밝힐 수 있다. ③모순에 의한 증명(proof by contradiction) p→q를 증명하려 했는데 힘들다. 그러면 명제 r을 이용하여 모순임을 밝히면 된다. 논리적으로 따지자면 p→q와 (p∧￢q)→(r∧￢r), 이 둘이 동치이기 때문에 가능한 증명법이다. 모순에 의한 증명은 직접 증명보다 쉽다. 왜냐하면 도구가 하나 더 있기 때문이다. 직접증명은 도구 하나로 결론을 도출해야하는 반면 모순에 의한 증명은 두개로 결론을 도출할 수 있기때문이다. 이 증명에 대한 예시는 고등학교 때 배운 "√2가 무리수이다"에 대한 증명이다. ④대우에 의한 증명(proof by contrapositive) p→q와 ￢q→￢p가 동치이므로 가능한 증명법이다. 예를 들어, "모든 실수x에 대해 x^2가 무리수이면 x는 무리수이다"를 증명할 때 직접증명은 까다롭다. 대신 무리수의 반대인 유리수의 정의를 알고 있기 때문에 대우인 "모든 실수x에 대해 x가 유리수이면 x^2은 유리수이다"로 증명하면 매우 쉬워진다. ⑤경우에 의한 증명(proof by case) 고등학교 때 배운 확률과 통계의 경우의 수와 비슷하다. 경우의 수를 나눌 때 케이스분류를 하듯이, 증명에 필요한 변수를 범위에 따라 나누어 각각 증명하면 된다. 가끔 변수의 크기가 작으면 모두 검사하여 증명할 수 있는데 이를 전수 증명(exhaustive proof)라고 한다.
| 7,894 | 20,345 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.3125 | 4 |
CC-MAIN-2019-39
|
longest
|
en
| 0.456687 |
https://www.physicsforums.com/threads/extension-of-a-basis-exchange-theorem.281934/
| 1,508,673,765,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-43/segments/1508187825227.80/warc/CC-MAIN-20171022113105-20171022133105-00219.warc.gz
| 979,908,869 | 14,364 |
# Extension of a basis (exchange theorem)
1. Dec 29, 2008
### latentcorpse
hi. ok if i'm using the exchange theorem for extension to a basis. i have the standard basis of 4 dimensional real space is {e1,e2,e3,e4}. and v1=e1+e2
then i can say that the coefficient at e1 is 1 which is non zero therefore i can exchange and get {v1,e2,e3,e4} as a basis. however if v2 = v1-2e2 say then what would be the basis i could make by the exchange theorem.
i reckoned it would be "the coefficient at e2 is -2 therefore by exchange theorem {v1,-(1/2)v2,e3,e4} would be a basis."???
Unless all that matters is that the coefficient of e2 is non zero and then {v1,v2,e3,e4} would be the basis???
Help!!!
2. Dec 29, 2008
### HallsofIvy
Staff Emeritus
Yes, the only important point is that the coefficient is non-zero. Since the coefficient of e2 is non-zero, you can replace e2 by v2.
Suppose v= ae1+ be2+ ce3+ de4. Since v1= e1+ e2, e1= v1- e2 so v= a(v1- e2)+ be2+ ce3+ de4= av1+ (b-a)e2+ ce3+ de4. Since v2= v1- 2e2, e2= (1/2)(v2- v1) and v= av1+ (b-a)(1/2)(v2- v1)+ ce3+ de4= (1/2)(3a+ b)v1+(1/2)bv2+ ce3+ de4. The coefficients involve "1/2" but the basis doesn't have to. In fact, {v1,-(1/2)v2,e3,e4} is a basis if and only if {v1, v2, e3, e4} is a basis.
| 465 | 1,254 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625 | 4 |
CC-MAIN-2017-43
|
longest
|
en
| 0.83454 |
http://mymathforum.com/trigonometry/341752-trigonometry-homework-question.html
| 1,547,611,324,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-04/segments/1547583656665.34/warc/CC-MAIN-20190116031807-20190116053807-00075.warc.gz
| 154,473,768 | 9,497 |
My Math Forum Trigonometry Homework Question
User Name Remember Me? Password
Trigonometry Trigonometry Math Forum
September 10th, 2017, 11:35 AM #1 Newbie Joined: Sep 2017 From: Europe Posts: 1 Thanks: 0 Trigonometry Homework Question Hello, I'm given a right-angled triangle with an angle of 60 degrees, nothing else. The adjacent side is named x and the opposite side is named h. The question is 'write h in terms of x. I know that the answer will be h=(square root of 3)x, as I'm given the answers at the back of the book. However, I would like to know how to solve it. Any idea how? Last edited by skipjack; September 10th, 2017 at 03:44 PM.
September 10th, 2017, 02:51 PM #2 Global Moderator Joined: May 2007 Posts: 6,660 Thanks: 647 You need to use the cosine (1/2) of 60 deg.
September 10th, 2017, 06:06 PM #3 Global Moderator Joined: Dec 2006 Posts: 20,089 Thanks: 1902 Let A denote the vertex where the 60° angle is, B denote the vertex where the right angle is, and C denote the remaining vertex of the triangle. Let D be the point such that ABCD is a rectangle and O be the point of intersection of AC and BD. By symmetry, angle ABD = 60°, so triangle ABO is equilateral and AC = 2AO = 2x. By Pythagoras, h² = AC² - AB² = 4x² - x² = 3x², so h = √3x.
September 10th, 2017, 06:20 PM #4 Newbie Joined: Sep 2017 From: San Diego Posts: 8 Thanks: 0 tan(a) = h/x solve for h, h = tan(a)*x But tan(a)=sin(a)/cos(a) so h = (sin(a)/cos(a))*x If you plug in 60 for a and fill in the steps you get h = sqrt(3)*x
September 10th, 2017, 07:51 PM #5 Global Moderator Joined: Dec 2006 Posts: 20,089 Thanks: 1902 How would you know the values of sin(60°) and cos(60°) (yet not tan(60°))?
September 11th, 2017, 05:26 AM #6 Global Moderator Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,898 Thanks: 1093 Math Focus: Elementary mathematics and beyond Construct an equilateral triangle with sides equaling 1 unit. Construct an altitude that divides the triangle into to congruent right-angled triangles. Then √(1 - (1/2)$^2$) = √3/2, so cos(60) = 1/2, sin(60) = √3/2 and tan(60) = √3. tan(60) = h/x = √3, h = √3x. Last edited by greg1313; September 11th, 2017 at 05:36 AM.
Tags homework, question, trigonometry
Thread Tools Display Modes Linear Mode
Similar Threads Thread Thread Starter Forum Replies Last Post Jaket1 Probability and Statistics 3 January 30th, 2017 12:19 PM flight90 Probability and Statistics 1 November 9th, 2015 04:08 PM bosha Linear Algebra 1 April 19th, 2015 06:53 AM orionankh Linear Algebra 6 April 16th, 2015 06:04 PM TrimHonduras Trigonometry 1 July 21st, 2013 07:40 PM
Contact - Home - Forums - Cryptocurrency Forum - Top
| 857 | 2,698 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.03125 | 4 |
CC-MAIN-2019-04
|
latest
|
en
| 0.904498 |
https://nrich.maths.org/public/leg.php?code=31&cl=1&cldcmpid=6119
| 1,432,398,839,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2015-22/segments/1432207927824.26/warc/CC-MAIN-20150521113207-00303-ip-10-180-206-219.ec2.internal.warc.gz
| 884,507,008 | 10,216 |
# Search by Topic
#### Resources tagged with Addition & subtraction similar to Cubes Here and There:
Filter by: Content type:
Stage:
Challenge level:
### How Old?
##### Stage: 2 Challenge Level:
Cherri, Saxon, Mel and Paul are friends. They are all different ages. Can you find out the age of each friend using the information?
### X Is 5 Squares
##### Stage: 2 Challenge Level:
Can you arrange 5 different digits (from 0 - 9) in the cross in the way described?
### Hubble, Bubble
##### Stage: 2 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
### Two Egg Timers
##### Stage: 2 Challenge Level:
You have two egg timers. One takes 4 minutes exactly to empty and the other takes 7 minutes. What times in whole minutes can you measure and how?
### Prison Cells
##### Stage: 2 Challenge Level:
There are 78 prisoners in a square cell block of twelve cells. The clever prison warder arranged them so there were 25 along each wall of the prison block. How did he do it?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Polo Square
##### Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Rabbits in the Pen
##### Stage: 2 Challenge Level:
Using the statements, can you work out how many of each type of rabbit there are in these pens?
### Zargon Glasses
##### Stage: 2 Challenge Level:
Zumf makes spectacles for the residents of the planet Zargon, who have either 3 eyes or 4 eyes. How many lenses will Zumf need to make all the different orders for 9 families?
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### On Target
##### Stage: 2 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Pouring the Punch Drink
##### Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### Train Carriages
##### Stage: 2 Challenge Level:
Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
##### Stage: 2 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Here to There 1 2 3
##### Stage: 1 Challenge Level:
Move from the START to the FINISH by moving across or down to the next square. Can you find a route to make these totals?
### All Seated
##### Stage: 2 Challenge Level:
Look carefully at the numbers. What do you notice? Can you make another square using the numbers 1 to 16, that displays the same properties?
### Neighbours
##### Stage: 2 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
##### Stage: 2 Challenge Level:
Lolla bought a balloon at the circus. She gave the clown six coins to pay for it. What could Lolla have paid for the balloon?
### Seven Square Numbers
##### Stage: 2 Challenge Level:
Add the sum of the squares of four numbers between 10 and 20 to the sum of the squares of three numbers less than 6 to make the square of another, larger, number.
### Finding Fifteen
##### Stage: 2 Challenge Level:
Tim had nine cards each with a different number from 1 to 9 on it. How could he have put them into three piles so that the total in each pile was 15?
### Sealed Solution
##### Stage: 2 Challenge Level:
Ten cards are put into five envelopes so that there are two cards in each envelope. The sum of the numbers inside it is written on each envelope. What numbers could be inside the envelopes?
### How Much Did it Cost?
##### Stage: 2 Challenge Level:
Use your logical-thinking skills to deduce how much Dan's crisps and ice-cream cost altogether.
### A Square of Numbers
##### Stage: 2 Challenge Level:
Can you put the numbers 1 to 8 into the circles so that the four calculations are correct?
### Page Numbers
##### Stage: 2 Short Challenge Level:
Exactly 195 digits have been used to number the pages in a book. How many pages does the book have?
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Mrs Beeswax
##### Stage: 1 Challenge Level:
In how many ways could Mrs Beeswax put ten coins into her three puddings so that each pudding ended up with at least two coins?
### The Puzzling Sweet Shop
##### Stage: 2 Challenge Level:
There were chews for 2p, mini eggs for 3p, Chocko bars for 5p and lollypops for 7p in the sweet shop. What could each of the children buy with their money?
### Twenty Divided Into Six
##### Stage: 2 Challenge Level:
Katie had a pack of 20 cards numbered from 1 to 20. She arranged the cards into 6 unequal piles where each pile added to the same total. What was the total and how could this be done?
### Routes 1 and 5
##### Stage: 1 Challenge Level:
Find your way through the grid starting at 2 and following these operations. What number do you end on?
### Calendar Calculations
##### Stage: 2 Challenge Level:
Try adding together the dates of all the days in one week. Now multiply the first date by 7 and add 21. Can you explain what happens?
### Dodecamagic
##### Stage: 2 Challenge Level:
Here you see the front and back views of a dodecahedron. Each vertex has been numbered so that the numbers around each pentagonal face add up to 65. Can you find all the missing numbers?
### Prompt Cards
##### Stage: 2 Challenge Level:
These two group activities use mathematical reasoning - one is numerical, one geometric.
### Six Is the Sum
##### Stage: 2 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### The Dice Train
##### Stage: 2 Challenge Level:
This dice train has been made using specific rules. How many different trains can you make?
### Shapes in a Grid
##### Stage: 2 Challenge Level:
Can you find which shapes you need to put into the grid to make the totals at the end of each row and the bottom of each column?
### Find the Difference
##### Stage: 1 Challenge Level:
Place the numbers 1 to 6 in the circles so that each number is the difference between the two numbers just below it.
### Today's Date - 01/06/2009
##### Stage: 1 and 2 Challenge Level:
What do you notice about the date 03.06.09? Or 08.01.09? This challenge invites you to investigate some interesting dates yourself.
### Homes
##### Stage: 1 Challenge Level:
There are to be 6 homes built on a new development site. They could be semi-detached, detached or terraced houses. How many different combinations of these can you find?
### Sums and Differences 1
##### Stage: 2 Challenge Level:
This challenge focuses on finding the sum and difference of pairs of two-digit numbers.
### Sums and Differences 2
##### Stage: 2 Challenge Level:
Find the sum and difference between a pair of two-digit numbers. Now find the sum and difference between the sum and difference! What happens?
##### Stage: 1 Challenge Level:
In Sam and Jill's garden there are two sorts of ladybirds with 7 spots or 4 spots. What numbers of total spots can you make?
### Oh! Harry!
##### Stage: 2 Challenge Level:
A group of children are using measuring cylinders but they lose the labels. Can you help relabel them?
### Magic Constants
##### Stage: 2 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
### Fifteen Cards
##### Stage: 2 Challenge Level:
Can you use the information to find out which cards I have used?
### Journeys in Numberland
##### Stage: 2 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
### Super Shapes
##### Stage: 2 Short Challenge Level:
The value of the circle changes in each of the following problems. Can you discover its value in each problem?
### The Add and Take-away Path
##### Stage: 1 Challenge Level:
Two children made up a game as they walked along the garden paths. Can you find out their scores? Can you find some paths of your own?
### Big Dog, Little Dog
##### Stage: 1 Challenge Level:
Woof is a big dog. Yap is a little dog. Emma has 16 dog biscuits to give to the two dogs. She gave Woof 4 more biscuits than Yap. How many biscuits did each dog get?
### Consecutive Numbers
##### Stage: 2 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
| 2,294 | 9,631 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.859375 | 4 |
CC-MAIN-2015-22
|
longest
|
en
| 0.898464 |
http://web2.0calc.com/questions/an-electronics-store-has-received-a-shipment-of-20-table-radios-that-have-connections-for-an-ipod-or-iphone-twelve-of-these-have-two-slots
| 1,516,440,293,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-05/segments/1516084889542.47/warc/CC-MAIN-20180120083038-20180120103038-00778.warc.gz
| 369,737,040 | 5,663 |
+0
# An electronics store has received a shipment of 20 table radios that have connections for an iPod or iPhone. Twelve of these have two slots
0
1086
1
An electronics store has received a shipment of 20 table radios that have connections for an iPod or iPhone. Twelve of these have two slots (so they can accommodate both devices), and the other eight have a single slot. Suppose that eight of the 20 radios are randomly selected to be stored under a shelf where the radios are displayed, and the remaining ones are placed in a storeroom. Let X = the number among the radios stored under the display shelf that have two slots.Compute P(X = 2), P(X ≤ 2), and P(X ≥ 2)
Guest Mar 2, 2015
### Best Answer
#1
+81022
+5
The probability that X = 2 is given by
C(12,2)* C(8,6) / C(20,8) = about 1.47%
The probability that X ≤ 2 = probability that none, one or two of the two slot models are chosen =
C(12,0)* C(8,8) / C(20,8) + C(12,1)* C(8,7) / C(20,8) + C(12,2)* C(8,6) / C(20,8) = about 1.54%
The probabliity that X ≥ 2 = 1 minus the probability that none or one of the two slot models are chosen =
1 - [C(12,0)* C(8,8) / C(20,8) + C(12,1)* C(8,7) / C(20,8)] = about 99%
CPhill Mar 2, 2015
Sort:
### 1+0 Answers
#1
+81022
+5
Best Answer
The probability that X = 2 is given by
C(12,2)* C(8,6) / C(20,8) = about 1.47%
The probability that X ≤ 2 = probability that none, one or two of the two slot models are chosen =
C(12,0)* C(8,8) / C(20,8) + C(12,1)* C(8,7) / C(20,8) + C(12,2)* C(8,6) / C(20,8) = about 1.54%
The probabliity that X ≥ 2 = 1 minus the probability that none or one of the two slot models are chosen =
1 - [C(12,0)* C(8,8) / C(20,8) + C(12,1)* C(8,7) / C(20,8)] = about 99%
CPhill Mar 2, 2015
### 3 Online Users
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners. See details
| 681 | 1,984 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.6875 | 4 |
CC-MAIN-2018-05
|
latest
|
en
| 0.899565 |
https://metanumbers.com/37199
| 1,627,070,458,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-31/segments/1627046150000.59/warc/CC-MAIN-20210723175111-20210723205111-00566.warc.gz
| 402,814,922 | 10,838 |
## 37199
37,199 (thirty-seven thousand one hundred ninety-nine) is an odd five-digits prime number following 37198 and preceding 37200. In scientific notation, it is written as 3.7199 × 104. The sum of its digits is 29. It has a total of 1 prime factor and 2 positive divisors. There are 37,198 positive integers (up to 37199) that are relatively prime to 37199.
## Basic properties
• Is Prime? Yes
• Number parity Odd
• Number length 5
• Sum of Digits 29
• Digital Root 2
## Name
Short name 37 thousand 199 thirty-seven thousand one hundred ninety-nine
## Notation
Scientific notation 3.7199 × 104 37.199 × 103
## Prime Factorization of 37199
Prime Factorization 37199
Prime number
Distinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 37199 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 10.524 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 37,199 is 37199. Since it has a total of 1 prime factor, 37,199 is a prime number.
## Divisors of 37199
2 divisors
Even divisors 0 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 37200 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 18600 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 192.87 Returns the nth root of the product of n divisors H(n) 1.99995 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 37,199 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 37,199) is 37,200, the average is 18,600.
## Other Arithmetic Functions (n = 37199)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 37198 Total number of positive integers not greater than n that are coprime to n λ(n) 37198 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3937 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 37,198 positive integers (less than 37,199) that are coprime with 37,199. And there are approximately 3,937 prime numbers less than or equal to 37,199.
## Divisibility of 37199
m n mod m 2 3 4 5 6 7 8 9 1 2 3 4 5 1 7 2
37,199 is not divisible by any number less than or equal to 9.
## Classification of 37199
• Arithmetic
• Prime
• Deficient
### Expressible via specific sums
• Polite
• Non-hypotenuse
• Prime Power
• Square Free
## Base conversion (37199)
Base System Value
2 Binary 1001000101001111
3 Ternary 1220000202
4 Quaternary 21011033
5 Quinary 2142244
6 Senary 444115
8 Octal 110517
10 Decimal 37199
12 Duodecimal 1963b
20 Vigesimal 4cjj
36 Base36 spb
## Basic calculations (n = 37199)
### Multiplication
n×i
n×2 74398 111597 148796 185995
### Division
ni
n⁄2 18599.5 12399.7 9299.75 7439.8
### Exponentiation
ni
n2 1383765601 51474696591599 1914807238510891201 71228914465366641785999
### Nth Root
i√n
2√n 192.87 33.3819 13.8878 8.20552
## 37199 as geometric shapes
### Circle
Diameter 74398 233728 4.34723e+09
### Sphere
Volume 2.15617e+14 1.73889e+10 233728
### Square
Length = n
Perimeter 148796 1.38377e+09 52607.3
### Cube
Length = n
Surface area 8.30259e+09 5.14747e+13 64430.6
### Equilateral Triangle
Length = n
Perimeter 111597 5.99188e+08 32215.3
### Triangular Pyramid
Length = n
Surface area 2.39675e+09 6.06635e+12 30372.9
| 1,259 | 3,804 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.734375 | 4 |
CC-MAIN-2021-31
|
latest
|
en
| 0.807151 |
https://classbasic.com/2019/10/04/smase-asei-pdsi-method-on-equivalent-fractions-and-ordering-of-fractions/
| 1,624,367,668,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-25/segments/1623488517820.68/warc/CC-MAIN-20210622124548-20210622154548-00064.warc.gz
| 169,596,271 | 32,373 |
# Equivalent Fractions and Ordering of Fractions (Primary 4)
Last Updated on August 10, 2020 by Alabi M. S.
MATHEMATICS
SMASE – ASEI PDSI METHOD
FIRST TERM
WEEK 5
PRIMARY 4
TOPIC – ORDERING OF EQUIVALENT FRACTIONS
LEARNING AREA
• Lesson One – Introduction to Equivalent Fractions
• Lesson Two – Equivalent Fractions
• Lesson Three – Ordering of Fractions
• Revision and Weekly Assessment
TIME – 40 Minutes each.
LEARNING OUTCOME
By the end of the lesson, pupils should be able to:
1. obtain equivalent fractions of a given fraction.
2. ordering pairs of fractions.
3. solve quantitative reasoning on equivalent fractions.
RATIONALE
Fractions really are everywhere! Every time we cut an apple, an orange, or any kind of fruit, we are taking a piece of the whole. We can represent those pieces as fractions. Fractions are part of a whole. That’s part of something bigger. Fractions have two numbers, a numerator (the part) and a denominator (the whole). Equivalent Fractions are equal in value. They represent the same amount of the whole.
The concept of ordering equivalent fractions will enable the pupils understand fraction of the same value.
LEARNING MATERIALS
The teacher will teach the lesson with the aid of:
1. Orange
2. Paper cuttings of different shapes.
3. Fraction charts
4. Squares
5. Cardboards
6. Flow chart of quantitative reasoning.
7. Flash cards
PREVIOUS KNOWLEDGE
Sharing among members of the family is a part of daily activities. Sometimes, they equally or given part of their thing to friends or family.
REFERENCE MATERIALS
1. 9 – Years Basic Education Curriculum
2. Lagos State Scheme of work
3. New Method Mathematics Book
4. All Relevant Materials
5. Online Materials
LESSON PLAN/DEVELOPMENT
ASEI PDSI LESSON NOTES
#### LESSON DEVELOPMENT 3 – Ordering of Fractions – Ascending and Descending docx
LESSON DEVELOPMENT 4 – REVISION AND WEEKLY ASSESSMENT
Smart Teachers Plan Lesson Notes - ClassRoomNotes support teachers with hands-on lesson plans/notes, printable and thoughtful teaching resources. @ClassRoomNotes - We always love to hear from you always. Stay connected with your classroom.
error:
| 506 | 2,151 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.3125 | 4 |
CC-MAIN-2021-25
|
latest
|
en
| 0.833187 |
https://studylib.net/doc/11693713/unit-title---subtraction-strategies-length-of-unit--9-day.
| 1,582,120,167,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-10/segments/1581875144150.61/warc/CC-MAIN-20200219122958-20200219152958-00249.warc.gz
| 567,724,044 | 13,164 |
# Unit Title: Subtraction Strategies Length of Unit: 9 Days Grade/Teacher: 2 grade
```Unit Title: Subtraction Strategies
Unit Assessment Date:
Unit Standards: What standards will I explicitly teach and intentionally
assess?
2.OA.1 Use addition and subtraction within 100 to solve one- and two step word
problems involving situations of adding to, taking from, putting together, taking apart,
and comparing, with unknowns in all positions, e.g., by using drawings and equations
with a symbol for the unknown number to represent the problem.
2.NBT.5 Fluently add and subtract within 100 using strategies based on place value,
properties of operations, and/or the relationship between addition and subtraction.
2.NBT.9 Explain why addition and subtraction strategies work, using place value and
the properties of operations
2.OA.2 Fluently add and subtract within 20 using mental strategies. By end of Grade 2,
know from memory all sums of two one-digit numbers.
21st Century Skills/ Mathematical Practices —How can I provide opportunities
for my students to collaborate, communicate, think critically, and be creative?
1. Make sense of problems and persevere in solving them.
2. Reason abstractly and quantitatively.
3. Construct viable arguments.
4. Model with mathematics.
5. Use appropriate tools strategically.
6. Attend to precision.
7. Look for and make use of structure.
8. Look for and express regularity in repeated reasoning.
Day(s)
1
Length of Unit: 9 Days
Materials and Resources Needed:—Other than Envisions, what other
materials and resources will be utilized during unit instruction?
Reteaching, enrichment and practice pages based on students’ needs (should
be used in conjunction with manipulatives)
Number cubes, connecting cubes, number cards (o-20), subtraction fact cards,
two-color counters, Double ten-frame
Essential Question(s) — What BIG Questions will guide student learning?
What do good mathematicians do?
How can different strategies be helpful when solving problems?
Program Review Connection:
Writing/Communication: Writing to explain
Practical Living/Career Studies:
Arts & Humanities:
Primary:
Student-Friendly Targets—What do I want
students to know and be able to do?
to mastery of the targets?
Formative/Summative Assessment—How will
students demonstrate progress toward or
mastery of learning targets?
I can learn and use a strategy for subtracting 0, 1,
and 2 from greater numbers.
Whole Group: Discuss target. Problem Based
Interactive Learning (page 71). Pose a problem, model
and use vocabulary; group activity
Small Group: Guided Practice
Centers: Independent Practice page, game from
differentiated instruction, technology, fluency center
(part-part whole cards, dice, etc.), review center
Observation of student strategies during small group
Look for Ideas: Mastery of subtracting 0, 1, and 2 from
a number.
Quick Check 3-1
McBrayer Elementary
Math Unit Planning Guide
2
I can use addition doubles facts to subtract.
3
I can use addition facts to subtract.
4
I can use addition facts to 18 to help me subtract.
5
I can use the “make-10” strategy to subtract.
6
I can solve two-question story problems by using
the answer to one question to solve another
equation.
Whole Group: Review target, Quick Check, Self-Assess
Whole Group: Discuss target. Problem Based
Interactive Learning (page 75). Pose a problem, model
and use vocabulary; group activity
Small Group: Guided Practice
Centers: Independent Practice page, game from
differentiated instruction, technology, fluency center
(part-part whole cards, dice, etc.), review center
Whole Group: Review target, Quick Check, Self-Assess
Whole Group: Discuss target. Problem Based
Interactive Learning (page 79). Pose a problem, model
and use vocabulary; group activity
Small Group: Guided Practice
Centers: Independent Practice page, game from
differentiated instruction, technology, fluency center
(part-part whole cards, dice, etc.), review center
Whole Group: Review target, Quick Check, Self-Assess
Whole Group: Discuss target. Problem Based
Interactive Learning (page 83). Pose a problem, model
and use vocabulary; group activity
Small Group: Guided Practice
Centers: Independent Practice page, game from
differentiated instruction, technology, fluency center
(part-part whole cards, dice, etc.), review center
Whole Group: Review target, Quick Check, Self-Assess
Whole Group: Discuss target. Problem Based
Interactive Learning (page 87). Pose a problem, model
and use vocabulary; group activity
Small Group: Guided Practice
Centers: Independent Practice page, game from
differentiated instruction, technology, fluency center
(part-part whole cards, dice, etc.), review center
Whole Group: Review target, Quick Check, Self-Assess
Whole Group: Discuss target. Problem Based
Interactive Learning (page 91). Pose a problem, model
and use vocabulary; group activity
Small Group: Guided Practice
Centers: Independent Practice page, game from
differentiated instruction, technology, fluency center
(part-part whole cards, dice, etc.), review center
Whole Group: Review target, Quick Check, Self-Assess
Observation of student strategies during small group
Look for Ideas: Mastery of using addition doubltes
facts to subtract.
Quick Check 3-2
Observation of student strategies during small group
Look for Ideas: Mastery subtracting using related
Quick Check 3-3
Observation of student strategies during small group
Look for Ideas: Mastery subtracting using related
Quick Check 3-4
Observation of student strategies during small group
Look for Ideas: Mastery of using the make -10 strategy
to subtract.
Quick Check 3-5
Observation of student strategies during small group
Look for Ideas: Mastery solving two question problems.
Quick Check 3-6
McBrayer Elementary
Math Unit Planning Guide
7
8
9
Review, Reteach and Enrich
Review, Reteach and Enrich
(All Targets from Topic 3)
(All Targets from Topic 3)
(All Targets from Topic 3)
Learning
Target
Assessed
Day 1
Day 2
Day 3
Day 4
Day 5
Day 6
Day 7
Review Games; Can do unit assessment in small
groups
Reteaching & Extensions
Clear up Misconceptions
Unit Assessment
Summative/Common Assessment Overview
Multiple Choice Question Assessing Target
| 1,426 | 6,197 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.578125 | 4 |
CC-MAIN-2020-10
|
latest
|
en
| 0.885385 |
https://www.scootle.edu.au/ec/search?topic=%22Irregular%20polygons%22&sort=relevance
| 1,726,213,078,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-38/segments/1725700651510.65/warc/CC-MAIN-20240913070112-20240913100112-00414.warc.gz
| 917,489,457 | 13,718 |
Search results
Refine by topic
Main topic Specific topic
Related topic
Listed under: Irregular polygons
Shapes and objects: Year 4 – planning tool
This planning resource for Year 4 is for the topic of Shapes and objects. Students build on their knowledge of shape by combining and cutting 2D shapes.
Volume and surface area: Year 10 – planning tool
This planning resource for Year 10 is for the topic of Volume and surface area. Students extend their application of volume and surface area to solve problems on composite solids. Students will need to be able to visualise the individual elements of the composite solids and identify the areas where these elements touch.
Volume and mathematical modelling video
Use this video as a springboard to explore volume of composite shapes, adjusting numbers to make calculations friendlier and draw on reasoning and mathematical modelling.
reSolve: Geometry: Trapezium Pieces
This lesson explores different shapes that can be formed by cutting a trapezium in two with one straight line. Students are challenged to classify and name the shapes that are made, and justify their classifications based on the definitions and properties of shapes. The lesson is outlined in detail including curriculum ...
Fun with fractals
Do you know how to recognise a fractal? Watch this video to find out! What are the examples given of fractals found in nature? Can you think of any others? Why not have a go at doing your own drawing of the Sierpinski Triangle?
Area explorer
This is an interactive resource about calculating the area of a shape presented on a grid. Students are presented with a random figure and are required to calculate its area, in square units. Immediate feedback is provided and the student can make repeated attempts if a question is not answered correctly. The additional ...
What is a fractal?
Do you know what a fractal is? Basically, fractals are never-ending patterns created by repeated mathematical equations. In this clip, Yuliya, a student at MIT (in the USA) describes the properties of fractals and shows you where they can be found in technology and nature. Have a good look at the world around you and see ...
reSolve: Rectilinear Polygons
In this sequence of three lessons, students use geometric reasoning to establish relationships between angles in polygons and go on to make generalisations using algebraic expressions. Students explore and enumerate right angles in a series of rectilinear polygons and generalise their findings. They then explore the number ...
reSolve: Authentic Problems: Expanded Square
This sequence of four lessons explores concepts around informal area and symmetry. Students design an 'expanded square' where approximately half the area of the original square is flipped to the outside. The lessons provide opportunities for students to devise and use methods to informally measure area, record their mathematical ...
Sizing Them Up
This resource is a web page containing a short task to explore area of irregular shapes by informal means. Arrange irregular shapes in size order smallest to largest. This resource is an activity from the NRICH website.
Mixed Up Maths, Ep 13: Shapes glorious shapes
Did you know that not all pyramids have a square base? Investigate the bases and faces of some pyramids. Travel around the world as we view some famous structures. First stop, we're in search of a building that is a rectangular prism. Find out which world famous building is a pentagonal prism. See what type of 3 dimensional ...
MathXplosion, Ep 45: How to make an origami frog
Origami folds have associated geometric patterns or "paper trails" in which we are able to visualise different types of triangles, angles, polygons, lines and symmetry. Use these patterns to turn a two-dimensional flat sheet of paper into a three-dimensional hopping frog!
MathXplosion, Ep 4: Tessellation tricks
Learn how two shapes from a repeating tile cause a pattern to undergo a metamorphosis. Create the illusion of one animal slowly transforming into another, line by line. Is it a bird? Is it a fish?
| 824 | 4,112 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.609375 | 4 |
CC-MAIN-2024-38
|
latest
|
en
| 0.934172 |
http://www.ck12.org/book/Algebra-Explorations-Concepts-Pre-K-through-Grade-7/r1/section/9.7/
| 1,481,257,277,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698542680.76/warc/CC-MAIN-20161202170902-00285-ip-10-31-129-80.ec2.internal.warc.gz
| 378,172,861 | 34,623 |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
9.7: Yummy Yogurt
Difficulty Level: At Grade Created by: CK-12
Yummy’s sells yogurt desserts. Can you use the information in the sign to complete the table? Can you write a rule to represent the relationship between number of scoops and total cost? In this concept, we will practice making tables and writing rules to match given information.
Guidance
In order to make a table and write a rule for situations like the one above, we can use the problem solving steps to help.
• First, describe what you know. What does each scoop cost? What does the other part of the yogurt dessert cost?
• Second, identify what your job is. In these problems, your job will be to make a table and write a rule.
• Third, make a plan. In these problems, your plan should be to use the facts on the sign to fill in the table. Then, look for a pattern to help you write the rule.
• Fourth, solve. Implement your plan.
• Fifth, check. Make sure your rule works with the facts on the sign.
Example A
Celeste’s sells yogurt sundaes.
a. Use the information in the sign. Complete the table to show how total cost is related to number of scoops.
b. Write a rule to represent the relationship between number of scoops and total cost. Use \begin{align*}\textbf{\textit{s}}\end{align*} for number of scoops and \begin{align*}\textbf{\textit{T}}\end{align*} for total cost.
Solution:
We can use problem solving steps to help us to fill in the table and write the rule.
\begin{align*}& \mathbf{Describe:} && \text{The sign shows the cost of a scoop and the cost of fruit topping}. \\ & && \text{The table has two columns, one for number of scoops and the other for total cost}. \\ \\ & \mathbf{My \ Job:} && \text{Complete the table for}\ 1\ \text{through}\ 4\ \text{scoops of yogurt. Write a function rule} \\ & && \text{to show how total cost is related to number of scoops and the cost of the cone}. \\ \\ & \mathbf{Plan:} && \text{Start with the table. Remember to include the cost of the fruit topping in the total} \\ & && \text{cost. Then write the rule}. \\ \\ & \mathbf{Solve:} \end{align*}
Number of Scoops Total Cost
1 \begin{align*}\2.50\end{align*}
2 \begin{align*}\4.00\end{align*}
3 \begin{align*}\5.50\end{align*}
4 \begin{align*}\7.00\end{align*}
\begin{align*}\text{Rule}: \textbf{\textit{T}} \textbf{ \ = \1.50}\textbf{\textit{s}} \textbf{\ + \1.00} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\end{align*}
\begin{align*}& \mathbf{Check:} && \text{Use the rule to verify the table.} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \\ \\ & && \1.50 \times 1 + \1.00 = \2.50 \\ & && \1.50 \times 2 + \1.00 = \4.00 \\ & && \1.50 \times 3 + \1.00 = \5.50 \\ & && \1.50 \times 4 + \1.00 = \7.00\end{align*}
Example B
Delicioso’s sells yogurt banana splits.
a. Use the information in the sign. Complete the table to show how total cost is related to number of scoops.
b. Write a rule to represent the relationship between number of scoops and total cost. Use \begin{align*}\textbf{\textit{p}}\end{align*} for number of scoops and \begin{align*}\textbf{\textit{M}}\end{align*} for total cost.
Solution:
We can use problem solving steps to help us to fill in the table and write the rule.
\begin{align*}& \mathbf{Describe:} && \text{The sign shows the cost of a scoop and the cost of a banana and syrup.} \\ & && \text{The table has two columns, one for number of scoops and the other for total cost}. \\ \\ & \mathbf{My \ Job:} && \text{Complete the table for}\ 1\ \text{through}\ 4\ \text{scoops of yogurt. Write a function rule} \\ & && \text{to show how total cost is related to number of scoops and the cost of the cone}. \\ \\ & \mathbf{Plan:} && \text{Start with the table. Remember to include the cost of the banana and syrup in the total} \\ & && \text{cost. Then write the rule}. \\ \\ & \mathbf{Solve:} \end{align*}
Number of Scoops Total Cost
1 \begin{align*}\5.75\end{align*}
2 \begin{align*}\8.75\end{align*}
3 \begin{align*}\11.75\end{align*}
4 \begin{align*}\14.75\end{align*}
\begin{align*}\text{Rule}: \textbf{\textit{M}} \textbf{ \ = \3.00}\textbf{\textit{p}} \textbf{\ + \2.75} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\end{align*}
\begin{align*}& \mathbf{Check:} && \text{Use the rule to verify the table.} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \\ \\ & && \3.00 \times 1 + \2.75 = \5.75 \\ & && \3.00 \times 2 + \2.75 = \8.75 \\ & && \3.00 \times 3 + \2.75 = \11.75 \\ & && \3.00 \times 4 + \2.75 = \14.75\end{align*}
Example C
Flavor’s sells yogurt floats.
a. Use the information in the sign. Complete the table to show how total cost is related to number of scoops.
b. Write a rule to represent the relationship between number of scoops and total cost. Use \begin{align*}\textbf{\textit{n}}\end{align*} for number of scoops and \begin{align*}\textbf{\textit{L}}\end{align*} for total cost.
Solution:
We can use problem solving steps to help us to fill in the table and write the rule.
\begin{align*}& \mathbf{Describe:} && \text{The sign shows the cost of a scoop and the cost of root beer}. \\ & && \text{The table has two columns, one for number of scoops and the other for total cost}. \\ \\ & \mathbf{My \ Job:} && \text{Complete the table for}\ 1\ \text{through}\ 4\ \text{scoops of yogurt. Write a function rule} \\ & && \text{to show how total cost is related to number of scoops and the cost of the cone}. \\ \\ & \mathbf{Plan:} && \text{Start with the table. Remember to include the cost of the root beer in the total} \\ & && \text{cost. Then write the rule}. \\ \\ & \mathbf{Solve:} \end{align*}
Number of Scoops Total Cost
1 \begin{align*}\3.50\end{align*}
2 \begin{align*}\5.50\end{align*}
3 \begin{align*}\7.50\end{align*}
4 \begin{align*}\9.50\end{align*}
\begin{align*}\text{Rule}: \textbf{\textit{L}} \textbf{ \ = \2.00}\textbf{\textit{n}} \textbf{\ + \1.50} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\end{align*}
\begin{align*}& \mathbf{Check:} && \text{Use the rule to verify the table.} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \\ \\ & && \2.00 \times 1 + \1.50 = \3.50 \\ & && \2.00 \times 2 + \1.50 = \5.50 \\ & && \2.00 \times 3 + \1.50 = \7.50 \\ & && \2.00 \times 4 + \1.50 = \9.50\end{align*}
Concept Problem Revisited
a. Use the information in the sign. Complete the table to show how total cost is related to number of scoops.
b. Write a rule to represent the relationship between number of scoops and total cost. Use \begin{align*}\textbf{\textit{n}}\end{align*} for number of scoops and \begin{align*}\textbf{\textit{T}}\end{align*} for total cost.
We can use problem solving steps to help us to fill in the table and write the rule.
\begin{align*}& \mathbf{Describe:} && \text{The sign shows the cost of a cone and the cost of one scoop of yogurt}. \\ & && \text{The table has two columns, one for number of scoops and the other for total cost}. \\ \\ & \mathbf{My \ Job:} && \text{Complete the table for}\ 1\ \text{through}\ 4\ \text{scoops of yogurt. Write a function rule} \\ & && \text{to show how total cost is related to number of scoops and the cost of the cone}. \\ \\ & \mathbf{Plan:} && \text{Start with the table. Remember to include the cost of the cone in the total} \\ & && \text{cost. Then write the rule}. \\ \\ & \mathbf{Solve:} \end{align*}
Number of Scoops Total Cost
1 \begin{align*}\3.00\end{align*}
2 \begin{align*}\5.00\end{align*}
3 \begin{align*}\7.00\end{align*}
4 \begin{align*}\9.00\end{align*}
\begin{align*}\text{Rule}: \textbf{\textit{T}} \textbf{ \ = \2.00}\textbf{\textit{n}} \textbf{\ + \1.00} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\end{align*}
\begin{align*}& \mathbf{Check:} && \text{Use the rule to verify the table.} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \\ \\ & && \2.00 \times 1 + \1.00 = \3.00 \\ & && \2.00 \times 2 + \1.00 = \5.00 \\ & && \2.00 \times 3 + \1.00 = \7.00 \\ & && \2.00 \times 4 + \1.00 = \9.00\end{align*}
Vocabulary
One type of table shows a relationship between an input and an output. In this concept, the inputs of our tables were number of scoops and the outputs of our tables were total cost. A rule is an equation that can describe the relationship between the inputs and the outputs of a table. In this concept, we wrote rules that showed the relationship between the number of scoops and the total cost.
Guided Practice
1. Yaley's sells yogurt chocolate cups.
a. Use the information in the sign. Complete the table to show how total cost is related to number of scoops.
b. Write a rule to represent the relationship between number of scoops and total cost. Use \begin{align*}\textbf{\textit{b}}\end{align*} for number of scoops and \begin{align*}\textbf{\textit{C}}\end{align*} for total cost.
2. Mixer’s sells yogurt slushes
a. Use the information in the sign. Complete the table to show how total cost is related to number of scoops.
b. Write a rule to represent the relationship between number of scoops and total cost. Use \begin{align*}\textbf{\textit{t}}\end{align*} for number of scoops and \begin{align*}\textbf{\textit{R}}\end{align*} for total cost.
3. Fabulous’s sells yogurt sandwiches.
a. Use the information in the sign. Complete the table to show how total cost is related to number of scoops.
b. Write a rule to represent the relationship between number of scoops and total cost. Use \begin{align*}\textbf{\textit{q}}\end{align*} for number of scoops and \begin{align*}\textbf{\textit{V}}\end{align*} for total cost.
1. a.
Number of Scoops Total Cost
1 \begin{align*}\3.25\end{align*}
2 \begin{align*}\4.50\end{align*}
3 \begin{align*}\5.75\end{align*}
4 \begin{align*}\7.00\end{align*}
b. Rule: \begin{align*}\mathbf{C} = \1.25\textbf{\textit{b}} + \2.00\end{align*}
2. a.
Number of Scoops Total Cost
1 \begin{align*}\4.50\end{align*}
2 \begin{align*}\7.50\end{align*}
3 \begin{align*}\10.50\end{align*}
4 \begin{align*}\13.50\end{align*}
b. Rule: \begin{align*}\textbf{\textit{R}} = \3.00\textbf{\textit{t}} + \1.50\end{align*}
3. a.
Number of Scoops Total Cost
1 \begin{align*}\4.00\end{align*}
2 \begin{align*}\5.25\end{align*}
3 \begin{align*}\6.50\end{align*}
4 \begin{align*}\7.75\end{align*}
b. Rule: \begin{align*}\textbf{\textit{V}} = \1.25\textbf{\textit{q}} + \2.75\end{align*}
Practice
1. Use the information in the sign. Complete the table to show how total cost is related to number of scoops.
2. Write a rule to represent the relationship between number of scoops and total cost. Use \begin{align*}\textbf{\textit{n}}\end{align*} for number of scoops and \begin{align*}\textbf{\textit{C}}\end{align*} for total cost.
Yummy's sells apple pie yogurt.
1. Use the information in the sign. Complete the table to show how total cost is related to number of scoops.
2. Write a rule to represent the relationship between number of scoops and total cost. Use \begin{align*}\textbf{\textit{n}}\end{align*} for number of scoops and \begin{align*}\textbf{\textit{C}}\end{align*} for total cost.
Yummy's sells cheesecake yogurt.
1. Use the information in the sign. Complete the table to show how total cost is related to number of scoops.
2. Write a rule to represent the relationship between number of scoops and total cost. Use \begin{align*}\textbf{\textit{n}}\end{align*} for number of scoops and \begin{align*}\textbf{\textit{C}}\end{align*} for total cost.
Yummy's sells s'mores yogurt.
1. Use the information in the sign. Complete the table to show how total cost is related to number of scoops.
2. Write a rule to represent the relationship between number of scoops and total cost. Use \begin{align*}\textbf{\textit{n}}\end{align*} for number of scoops and \begin{align*}\textbf{\textit{C}}\end{align*} for total cost.
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Show Hide Details
Description
Difficulty Level:
Tags:
Date Created:
Jan 18, 2013
| 4,296 | 12,817 |
{"found_math": true, "script_math_tex": 65, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.875 | 5 |
CC-MAIN-2016-50
|
latest
|
en
| 0.817821 |
https://artofproblemsolving.com/wiki/index.php?title=2002_AIME_I_Problems/Problem_6&oldid=78224
| 1,713,174,330,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296816954.20/warc/CC-MAIN-20240415080257-20240415110257-00137.warc.gz
| 95,049,644 | 12,193 |
# 2002 AIME I Problems/Problem 6
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
The solutions to the system of equations
$\log_{225}x+\log_{64}y=4$
$\log_{x}225-\log_{y}64=1$
are $(x_1,y_1)$ and $(x_2,y_2)$. Find $\log_{30}\left(x_1y_1x_2y_2\right)$.
## Solution
Let $A=\log_{225}x$ and let $B=\log_{64}y$.
From the first equation: $A+B=4 \Rightarrow B = 4-A$.
Plugging this into the second equation yields $\frac{1}{A}-\frac{1}{B}=\frac{1}{A}-\frac{1}{4-A}=1 \Rightarrow A = 3\pm\sqrt{5}$ and thus, $B=1\pm\sqrt{5}$.
So, $\log_{225}(x_1x_2)=\log_{225}(x_1)+\log_{225}(x_2)=(3+\sqrt{5})+(3-\sqrt{5})=6$ $\Rightarrow x_1x_2=225^6=15^{12}$.
And $\log_{64}(y_1y_2)=\log_{64}(y_1)+\log_{64}(y_2)=(1+\sqrt{5})+(1-\sqrt{5})=2$ $\Rightarrow y_1y_2=64^2=2^{12}$.
Thus, $\log_{30}\left(x_1y_1x_2y_2\right) = \log_{30}\left(15^{12}\cdot2^{12} \right) = \log_{30}\left(30^{12} \right) = \boxed{012}$.
One may simplify the work by applying Vieta's formulas to directly find that $\log x_1 + \log x_2 = 6 \log 225, \log y_1 + \log y_2 = 2 \log 64$.
| 475 | 1,087 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 16, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.5625 | 5 |
CC-MAIN-2024-18
|
latest
|
en
| 0.520145 |
https://socratic.org/questions/596c7dbe7c01493774b41c8f
| 1,721,678,810,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763517915.15/warc/CC-MAIN-20240722190551-20240722220551-00205.warc.gz
| 454,169,030 | 7,185 |
# Evaluate the integral? : int \ x^2/(sqrt(x^2-25))^5 \ dx
Jul 17, 2017
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = - \frac{1}{75} {x}^{3} / {\left(\sqrt{{x}^{2} - 25}\right)}^{3} + C$
#### Explanation:
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \int {x}^{2} / {\left(5 \sqrt{{\left(\frac{x}{5}\right)}^{2} - 1}\right)}^{5} \mathrm{dx}$
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{5} ^ 5 \int {x}^{2} / {\left(\sqrt{{\left(\frac{x}{5}\right)}^{2} - 1}\right)}^{5} \mathrm{dx}$
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int {\left(\frac{x}{5}\right)}^{2} / {\left(\sqrt{{\left(\frac{x}{5}\right)}^{2} - 1}\right)}^{5} d \left(\frac{x}{5}\right)$
Substitute:
$\frac{x}{5} = \sec t$
$d \left(\frac{x}{5}\right) = \sec t \tan t$
to have:
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int {\sec}^{2} \frac{t}{\sqrt{{\sec}^{2} t - 1}} ^ 5 \sec t \tan t \mathrm{dt}$
use now the trigonometric identity:
${\sec}^{2} t - 1 = {\tan}^{2} t$
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int {\sec}^{2} \frac{t}{\sqrt{{\tan}^{2} t}} ^ 5 \sec t \tan t \mathrm{dt}$
If we restrict ourselves to the interval $x \in \left(5 , + \infty\right)$ so that $t \in \left(0 , \frac{\pi}{2}\right)$ we have that $\tan t > 0$ so $\sqrt{{\tan}^{2} t} = \tan t$:
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int {\sec}^{2} \frac{t}{\tan} ^ 5 t \sec t \tan t \mathrm{dt}$
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int {\sec}^{3} \frac{t}{\tan} ^ 4 t \mathrm{dt}$
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int \frac{1}{\cos} ^ 3 t {\cos}^{4} \frac{t}{\sin} ^ 4 t \mathrm{dt}$
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int \cos \frac{t}{\sin} ^ 4 t \mathrm{dt}$
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{25} \int \frac{d \left(\sin t\right)}{\sin} ^ 4 t$
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = - \frac{1}{75} \frac{1}{\sin} ^ 3 t + C$
To undo the substitution note that:
$\sin t = \sqrt{1 - {\cos}^{2} t} = \sqrt{1 - \frac{1}{\sec} ^ 2 t} = \sqrt{1 - \frac{25}{x} ^ 2} = \frac{\sqrt{{x}^{2} - 25}}{x}$
Then:
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = - \frac{1}{75} {x}^{3} / {\left(\sqrt{{x}^{2} - 25}\right)}^{3} + C$
Now consider $x \in \left(- \infty , - 5\right)$ and substitute $t = - x$:
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \int {\left(- t\right)}^{2} / {\left(\sqrt{{\left(- t\right)}^{2} - 25}\right)}^{5} d \left(- t\right)$
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = - \int {t}^{2} / {\left(\sqrt{{t}^{2} - 25}\right)}^{5} \mathrm{dt}$
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = \frac{1}{75} {t}^{3} / {\left(\sqrt{{t}^{2} - 25}\right)}^{3} + C$
and undoing the substitution:
$\int {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \mathrm{dx} = - \frac{1}{75} {x}^{3} / {\left(\sqrt{{x}^{2} - 25}\right)}^{3} + C$
so the expression is the same for $x \in \left(- \infty , 5\right) \cup \left(5 , + \infty\right)$
Jul 17, 2017
$\int \setminus {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \setminus \mathrm{dx} = - \frac{1}{75} \setminus {x}^{3} / {\left(\sqrt{{x}^{2} - 25}\right)}^{3} + C$
#### Explanation:
Compare the denominator to the trig identity:
${\tan}^{2} A \equiv {\sec}^{2} A - 1$
In an attempt to reduce the denominator to something simper.
Why this identity in particular? Well firstly it is a trig identity of the form $f n {1}^{2} = f n {2}^{2} - 1$ so with an appropriate factor it will reduce to being dependant upon $f n 1$ alone, and secondly, $\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x$ and $\frac{d}{\mathrm{dx}} \sec = \sec x \tan x$ so either derivative also appears in the identity, which should help with a substitution in the integral,
So we want to find:
$I = \int \setminus {x}^{2} / {\left(\sqrt{{x}^{2} - 25}\right)}^{5} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus {x}^{2} / {\left(\sqrt{25 \left({x}^{2} / 25 - 1\right)}\right)}^{5} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus {x}^{2} / {\left(5 \sqrt{{\left(\frac{x}{5}\right)}^{2} - 1}\right)}^{5} \setminus \mathrm{dx}$
So let us try a substitution $\sec \theta = \frac{x}{5}$ such that
$x = 5 \sec \theta$, and $\frac{\mathrm{dx}}{d \theta} = 5 \sec \theta \tan \theta$
Thus:
$I = \int \setminus {\left(5 \sec \theta\right)}^{2} / {\left(5 \left(\sqrt{{\sec}^{2} \theta - 1}\right)\right)}^{5} 5 \sec \theta \tan \theta \setminus d \theta$
$\setminus \setminus = \int \setminus \frac{{5}^{3} {\sec}^{3} \theta \tan \theta}{5 \left(\sqrt{{\tan}^{2} \theta}\right)} ^ 5 \setminus d \theta$
$\setminus \setminus = \int \setminus \frac{{5}^{3} {\sec}^{3} \theta \tan \theta}{{5}^{5} {\tan}^{5} \theta} \setminus d \theta$
$\setminus \setminus = \frac{1}{25} \setminus \int \setminus \frac{{\sec}^{3} \theta}{{\tan}^{4} \theta} \setminus d \theta$
$\setminus \setminus = \frac{1}{25} \setminus \int \setminus \frac{\frac{1}{\cos} ^ 3 \theta}{{\sin}^{4} \frac{\theta}{\cos} ^ 4 \theta} \setminus d \theta$
$\setminus \setminus = \frac{1}{25} \setminus \int \setminus \frac{\cos \theta}{{\sin}^{4} \theta} \setminus d \theta$
Which we can integrate by observation as:
$\frac{d}{\mathrm{du}} \frac{1}{\sin} ^ 3 u = - 3 \cos \frac{u}{\sin} ^ 4 u$
And so we conclude that:
$I = \frac{\frac{1}{25} \setminus \frac{1}{\sin} ^ 3 \theta}{- 3} + C$
$\setminus \setminus = - \frac{1}{75} \setminus \frac{1}{\sin} ^ 3 \theta + C$
If we refer back to our earlier substitution, we note:
$x = 5 \sec \theta \implies \sec \theta = \frac{x}{5}$
$\cos \theta = \frac{5}{x}$
And using the trig identity ${\cos}^{2} A + {\sin}^{2} A \equiv 1$ this gives:
${\sin}^{2} \theta + {\left(\frac{5}{x}\right)}^{2} = 1 \implies \sin \theta = \sqrt{1 - \frac{25}{x} ^ 2}$
Using this to reverse the earlier substitution we get:
$I = - \frac{1}{75} \setminus \frac{1}{\sqrt{1 - \frac{25}{x} ^ 2}} ^ 3 + C$
$\setminus \setminus = - \frac{1}{75} \setminus \frac{1}{\sqrt{1 - \frac{25}{x} ^ 2}} ^ 3 + C$
$\setminus \setminus = - \frac{1}{75} \setminus \frac{1}{\sqrt{\frac{1}{x} ^ 2 \left({x}^{2} - 25\right)}} ^ 3 + C$
$\setminus \setminus = - \frac{1}{75} \setminus \frac{1}{\frac{1}{x} \sqrt{{x}^{2} - 25}} ^ 3 + C$
$\setminus \setminus = - \frac{1}{75} \setminus {x}^{3} / {\left(\sqrt{{x}^{2} - 25}\right)}^{3} + C$
| 2,899 | 6,685 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 58, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.34375 | 4 |
CC-MAIN-2024-30
|
latest
|
en
| 0.205559 |
www.knowonlineadvertising.com
| 1,721,349,864,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763514860.36/warc/CC-MAIN-20240718222251-20240719012251-00354.warc.gz
| 727,806,010 | 57,612 |
It’s true mathematics never leave you and so not even in an online industry it has spared you, so let’s have some clear picture with examples of the formulas used in the online world so that you know well how things are working in online advertising industries ..(formulas in internet advertising)
• CPM CALCULATION :
cost per mille (1000) – This is one of the most used metrics on the web. It is the cost that has to be paid by an advertiser for serving 1000 impressions.
Example: In a campaign, say an Ad of 728×90 is running and the CPM set is \$5 and the impressions to be served is 2,00,000, what will be the actual cost to the advertiser?
The formula for CPM goes this way :
Cost to an Advertiser = CPM x (Impressions / 1000)
using the above metrics :
Cost to an Advertiser= 5 x (2,00,000/1000) = 1000
So, \$ 1000 is what the advertiser has to pay!
When CPM needs to be calculated: use the final cost.
CPM = Cost to an Advertiser x 1000 / Impressions .
• CPC CALCULATION :
As we know we have different rate models in online advertising, CPC – cost per click is one of the popular and well known used metrics. Here the advertiser has to pay as per clicks and not on impressions. The page views can be any number, as the advertiser is concerned about clicks here.
Example: Suppose a campaign having 300×250 size banner running at CPC of \$2 and the number of clicks the Ad has got is 1000, what is the amount that the advertiser has to pay actually?
The formula for CPC goes as below :
Cost to an Advertiser: CPC x number of clicks
using the above metrics,
Cost to an Advertiser = 2 x 1000 = 2000
So, \$ 2000 is what the advertiser has to pay!
CPC = Cost to an advertiser / number of clicks
Also , Cost = Impressions * CTR * CPC.
• CTR CALCULATION :
CTR is click through rate, it measures the effectiveness of any advertisement. It is calculated by using a simple formula as below :
Example: A campaign having 728×90 Ad has served 10,000 impressions and has generated 100 clicks so what will be CTR of that Ad?
CTR = (number of clicks / number of impressions) x 100
using above metrics ,
CTR = (100/10000) x 100 = 1 %
1% CTR means on every 100 impression there is one click.
• CR CALCULATION :
CR is conversion rate, to calculate the conversion rate a simple formula is used :
so if the number of conversions made is 20 in 1000 impressions, the conversion rate will be (20/1000)*100 = 2 %
• CPA CALCULATION :
Another metric that we have and is generally used is cost per acquisition which is irrespective of impressions and clicks both. This metric deals with any action or basically when something is acquired like user sign up or when any sale is made.
Cost to an advertiser = CPA x ( Impression x CTR X CR )
Suppose CPA is \$ 5, the number of impressions is 10,000, CTR is 3% and CR is 2%.
using the above metrics,
Cost to an advertiser = 5 x ( 10000 x 0.03 x 0.02 ) = \$30
Similarly, if you know the actual cost, we can easily calculate the CPA for the Ad using the below formula :
CPA = cost to an advertiser / ( Impressions x CTR x CR )
Also, Average Cost Per Acquisition (CPA) = Average Cost per Click / Conversion Rate
Some more formulas that are used :
CPL / CR = CPA or
Cost Per Lead divided by Conversion Rate = Cost Per Acquisition
and,
VPA – CPA = NP or
Value Per Acquisition minus Cost Per Acquisition equals Net Profit.
• CPV CALCULATION :
CPV is the cost per visit which is new with respect to CPM, CPC, and CPA but is now commonly used by companies. The costing is done as per the actual visits.
CPV = Total cost of the campaign/ Number of total visit or incremental visit after the campaign started
Suppose, the total cost is \$1000 and the visit increased from 50 to 250 visits so the incremental visit is equal to 200.
Applying the above numbers in the CPV formula, we get
CPV = 1000/200 = \$5
i.e. Each visit cost is \$5 to the advertiser
• eCPM CALCULATION : (Effective cost per mile)
Effective CPM is the actual CPM that is being applied, If the CPM set is \$2 and the eCPM is \$ 1.5, the net profit is \$ 0.5. eCPM helps you measure how well your ads are performing. It is calculated by dividing total earnings/total spend by total number of impressions in thousands.
eCPM = ( Total spent / Impressions delivered ) x 1000
Example: An ad size of 728 has delivered 213456 impressions and has also spent some \$300 with CPM set at \$1.5, what will be the eCPM?
Using the formula,
eCPM = ( 300 / 213456 ) x 1000 = \$1.40
We can also calculate eCPM using eCPC, but for that, we need to know the conversion rate. and then using the formula as below :
eCPM = eCPC x conversion rate x 1000
• eCPC CALCULATION :
eCPC is a metric used by Internet marketers to calculate the effectiveness of their online campaigns when the rate model used is CPC. eCPC can also be termed as ” profitable per click” so if the actual CPC is \$2 and the eCPC is coming as \$1 per click, then \$1 is the profit on each click.
Net profit = CPC – eCPC
eCPC = (Total spent or revenue / clicks)
Example : If eCPC needs to be calculated for a single Ad size say 300×250, we just need to figure out how much the Ad size has spent and how much clicks it has generated, so if it is like \$200 has been spent and the clicks generated are 100 with actual CPC set to \$3 .
using the formula,
eCPC = ( 200 / 100 ) = \$ 2
So , net profit = ( 3 – 2 ) = \$1.
• eCPA CALCULATION :
eCPA is the effective cost per Action which is calculated the same as eCPC and eCPM, It is the total spend by the total number of actions(i.e conversions) acquired.
Example: The total spent on the campaign is \$2000 and the total number of conversions made is 200, what will be the eCPA?
Formula of eCPA ,
eCPA = Total spent / number of conversions or actions
Using the above metrics ,
eCPA = 2000 / 200 = \$ 10
• Estimated LTV (LifeTime value) of a user:
eLTV = Avg. Purchase per Year * Avg. Purchase Size * Loyalty Duration in Years
eCAC = Time & Money Cost of Acquition / Number of Customers Acquired
## 62 thoughts on “Basic Formulas in Internet Advertising”
1. Sridhar says:
Hi,
I found that the CPM, CPC, CPA etc… missed abbreviations.
Cost per thousand impressions- CPM.
Cost per click- CPC.
Cost per action-CPA.
2. prathap says:
could you please provide formulae for eCPM,eCPC,eCPA .
1. Jayaram says:
How much revenue spent until now/impressions they got
3. Hi Prathap , I have updated the page with the formulas and example of eCPC , eCPC, and eCPA .. Please let me know if you need any help and clarification .
Enjoy surfing 🙂
4. Prathap says:
Why we using Iframe tag in creatives?
And how to identify the flash and video creatives.
5. Subhransu Sekhar says:
6. Prathap says:
Could you please provide formulae for CPV,and how to calc CPV to eCPM.
1. Hi Prathap,
I have updated the formula for CPV, v as visit and not views. Let me know if you need the formula for cost per views.
Regards,
Avinash
1. Mehdi says:
Yes please share for CPV too and also would like to ask how to calculate CPM from CPV (cost per view)
7. Sridhar S. says:
Hi,
The CR rate has been wrongly updated for that example mentioned above.
For 20 conversion in 1000 impression, it will be 0.02 but the conversion rate is 2℅. I think you missed to multiple by 100.
Please let me if my understanding is wrong.
Thanks,
Sridhar S.
9972482838
1. Hi Sridhar,
Thanks for the catch, a simple miss changing the result, it should be 2% and not 0.02%.
You are perfectly correct.
Regards,
Avinash
8. Subhomoy Goswmai says:
Hello i have registered with an ad network and they have shown me the statistics which i felt to understand…
Can any1 plz help me to explain it to me what does it mean.
The url is : http://picturespk.pk/images/vi
I know how to calculate CPM , CTR, eCPM
but as per ecpm we have to know sites previous income but from the above pictures how they have calculated my precious income…
I have created my account this 16 may…
plz explain how the earnings and ecpm is calculated from my picture…
1. Hey,
Can’t access the image link 🙁
1. Subhomoy Goswmai says:
9. I just want to know the mean of conversion and found online answer here:
Getting someone to open an email is a conversion. Having them click on the call-to-action link inside that email is another conversion. Going to the landing page and filling out a registration form to read your content is a conversion. And, of course, buying your product is the ultimate conversion.
10. John says:
Hi there,
I have couple of questions and doubts regarding these metrics. Can I have a short chat with you to resolve them?
Thanks
11. Awesome, a very simple calculation formulas. Thanks for sharing.
12. Ankita says:
Hello Friend!
CPM is not cost per mile. It is cost per thousand impressions.
A lot of people read this and get confused, so I would request you to change this in your article.
1. Hi Ankita,
Thanks for your correction, but any idea What’s mille basically?
1. Thanks for adding that Don, i hope Ankita now that’s clear to you.
13. Hi,
I have the Impression’s & CPM . Kindly help me to find out the Number of users and Page view’s
For Example : Impression’s : 24000 & CPM unit cost is 120
1. Hi Deepu,
There is no hard and fast rule or formula for page views and visits but more of a appropriate estimation, below is the working:
Impressions 24000
CPM 120
Cost 2880
CTR 0.20% (Banners)
Clicks 48
Estimate (Click to visit) 50%
Visits 24
Page Views (Generally 45%+ of Visits) 53
Let us know in case of any query
Regards,
kOA
14. yogesh says:
Great Keep it up 🙂
1. yogesh says:
welcome 🙂
Actually i am php web developer and i want to learn digital marketing, How i start it will please give me some advises 🙂
15. thanks for sharing basic formulaes for internet marketing with live examples
16. Mariam says:
17. Stacey Hay says:
Can you please tell me formula to work out cost per unique CPU?
18. Laura Buragaitė says:
Hi, can someone help to immediatelly?
I have clicks – 100
impressions – 750
pay for the click – 0,2 eur
pay for the impression 0,1 eur
I need to find –
CPC – i need answer in amount of euros and units;
CTR %
CPI (cost per impression) – i need answer in amount of euros and units
19. Eugene says:
In your formula for eCPM you wrongly put ‘Conversion Rate’ instead of CTR.
So the formula should be:
eCPM = CPC * CTR * 1000.
1. Hey Eugene,
We can also calculate using the above formula. Try out!
20. Anthony says:
In the formula for CR, are the conversions the total of people that clicked on the ad?
For example out of 1000 impressions there are 100 conversions.
Does that mean that 100 people clicked on the ad? If so, how do you determine the number of conversions?
1. Not really, it could be view through conversions also which accounts basis the last for first view i.e. impressions and not clicks. Also, it is impossible that all clicks can convert to conversions, the average percentage is around 2-5% which also differs depending on what the final conversion is.
Regards,
kOA
21. ruzgrid says:
How could I calculate the visits per million contacts?
1. Hey,
Can you little elaborate what is your ask and what are your trying to figure out from this calculation, we would be in a better position to help you out.
Regards,
kOA
22. amol mahajan says:
Can you please provide the formula of eCPL …????
1. Hey,
The simplest way to calculate is divide the cost spent on lead placements by total numbers of leads collected. You need to funnel the complete media plan and then look at the effective cost spent and also look at the lead quality to exactly relate it to ROI.
Regards,
kOA
23. thanks for the giving us you valuable infomation. It is very helpfull for me, specially for ctr. thanks again.
24. sagar says:
Hi,I just want to know that if suppose i want to advertise my education institute in india ,then what amount i have to pay to the advertisement agency and what are the standard rate of advertising an add in india
1. Hi Sagar,
If you look to get it done by an ad agency, then they work on commission percentage which completely varies, but you can take an average of 5% of total net spends. If the institute is not very big, please manage at your end as the learning plus transparency would be high and you would know the next steps to make the business grow. If you are not much aware of advertising, take consultancy from known media friends and go ahead yourself.
Regards,
kOA
25. Mr Mo says:
Hi kOA, I urgently need help calculating CAC from only CTR, CPC, LTV and CPM values. Any formula please?
1. Hey, CAC has a simple formula of dividing the total cost by total acquired customers. The best way to calculate the same is always weekly so that you can get to know the fluctuation of CAC. All the metrics you provided can also be well used to calculate the same, but it has to be one strategy. Was the buying done on CPC or CPM?
26. Mr Mo says:
Hi KoA, it was done on CPC
1. What is your acquisition here?
1. We need to derive the formula using CPC, CTR and CPM. So baiscally caluclate the cost using CPC and CTR, using which we need to put the same in CAC formula which is cost/no. of acq. and we need to assume here that when we run CPC 0.2% is our conversion rate. Using the same we can calculate the CAC.
27. Raju Naik says:
Good Info, Thank U
28. liana says:
Hi, I urgently need help calculating the conversion ratio. Any formula please?
Find the conversion ratio for the following:
Impressions: 847,381
Clicks: 435
Total Spend: \$152.53
Conversions: 2
Conversion Ratio: 1:1 !!! not Conversion rate
1. Hi Liana,
You need to know the price of the product to figure out conversion ratio. The easiest formula to get conversion ratio is Price of the product/Per conversion cost i.e. \$76.26 here.
Regards,
kOA
29. Lidia says:
I need a help with this:
| 3,555 | 13,885 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.40625 | 4 |
CC-MAIN-2024-30
|
latest
|
en
| 0.937274 |
https://strathmaths.wordpress.com/2011/09/02/when-it-all-goes-horribly-wrong-the-mathematics-of-natural-disasters/
| 1,529,328,026,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267860557.7/warc/CC-MAIN-20180618125242-20180618145242-00034.warc.gz
| 724,429,804 | 18,984 |
## When it all goes horribly wrong: the mathematics of natural disasters
Here’s a scene you never see in the movies. A volcano has erupted in the middle of the city; lava is pouring down the street; tall buildings are shuddering as the ground turns to quicksand; and the mayor is yelling down the phone “Get me a mathematician, now!”
In fact, mathematicians, and the tools we design, will already be at work. For example, when an earthquake strikes anywhere in the world, computers use the latest inverse methods to process seismic data and pinpoint the source (see this simplified account from the US Geological Survey). If the quake is underwater, fluid dynamical models are used to predict whether a tsunami will result. Thanks to modern numerical analysis, these models are fast enough to allow beaches to be evacuated before the wave strikes. Meanwhile, the emergency services are deploying on the basis of statistical hazard assessments, and using insights from graph theory and queueing theory to plan how to restore power and communications. Mathematicians may not be on the front line, but without us the aftermath of disasters would be far worse.
Although these topics lie at the frontiers of research, during your degree you will meet many of the ideas behind them.
One theme you may meet is wave theory: how oscillations travel through media such as the ocean or the solid earth. Often, energy travels away from the source at a particular speed c: calculating what c depends on is hard, but once we know it we can draw simple conclusions. For example, the speed of “primary” seismic waves through the earth’s crust is given by
$c = \sqrt{\displaystyle\frac{K+\frac{4}{3}\mu}{\rho}}$,
where $\rho$ is the density of the rock and where K and $\mu$ measure the stiffness of the crust. Crucially, c does not depend on the shape or the size of the waves. In places like California, where the crust is young and warm, K and $\mu$ are small, so even huge earthquakes do not travel very far or fast. In places like the eastern US, where the crust is colder and stiffer, waves can travel much further and faster — so the recent Virginia earthquake was felt as far away as Canada.
We can also understand why some tsunamis do so much damage. Suppose that an earthquake releases a given amount of energy E at a single point, causing a tsunami. The wave then spreads out in an expanding circle at speed c (now given by a different formula from that for seismic waves). After a time t this circle has radius r = ct: the energy is shared around the entire circumference, so it must be spread around a distance $2\pi r$. The energy density, which is related to the wave size, is proportional to $E/(2\pi r)$ — so the further away we are, the safer we are.
But now suppose that the energy is released along a line rather than at a point, as in the 2004 “Boxing Day” tsunami. Instead of spreading out as a circle, such waves travel perpendicularly to the original line. The lengths of these waves do not increase with distance from the source, so their size does not decrease: this is why Sri Lanka and India were so badly affected in 2004.
Another theme, from a completely different area of maths, is risk assessment. For example, given the record of eruptions in Iceland over the last two centuries, how likely is it that another eruption will affect the British Isles in the next twenty years? The problem is that the events we are interested in are much larger than “average” and also very rare. You are probably familiar with the normal distribution, with probability density
$f(x) = \displaystyle\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\displaystyle\frac{(x-\mu)^2}{2\sigma^2}\right)$
where $\mu$ and $\sigma^2$ are the mean and variance. If probability is normally distributed, the chance of a very large event (x much greater than $\mu$) decreases rapidly as x increases: the chance that $x > \mu+3\sigma$ is roughly 0.13%, while the chance that $x > \mu + 4\sigma$ is roughly 0.0032%.
In practice, the statistics of many events, including natural disasters, follow “heavy-tailed” distributions where the chance of a big event drops off much more slowly. For example, in a Pareto distribution the probability density is
$f(x) = \alpha\displaystyle\frac{x_m^{\alpha}}{x^{\alpha+1}}$,
where $\alpha$ and $x_m$ are related to the mean and variance. If, for example, we take $\alpha = 3$, we find that the chance that $x > \mu+3\sigma$ is roughly 1.4% (ten times larger than before); the chance that $x > \mu + 4\sigma$ is roughly 0.82% (over 250 times larger than before). Choosing the correct probability distribution for hazard forecasting is a difficult task for statisticians; persuading governments that these “technical” details are crucial can be harder still!
So, next time someone asks you what use maths is, tell them you can use it to understand earthquakes and decide whether volcanoes are worth planning for. You may get some sceptical looks, but admit it… doesn’t it sound more interesting than accountancy?
(DP)
| 1,154 | 5,039 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 18, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.6875 | 4 |
CC-MAIN-2018-26
|
latest
|
en
| 0.925096 |
http://www.drumtom.com/q/using-laws-of-log-how-come-1-4-log3-81-equal-1
| 1,481,047,359,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2016-50/segments/1480698541950.13/warc/CC-MAIN-20161202170901-00227-ip-10-31-129-80.ec2.internal.warc.gz
| 438,043,524 | 9,210 |
# Using laws of log. How come 1/4 log3 81 equal 1?
• Using laws of log. How come 1/4 log3 81 equal 1?
(81) = log3 (3 4). Now we use that ... 1 8) 2 2 = 1 4 2=log2 (1 4) 2 1 = 1 2 1=log2 (1 2) 20 =1 0=log 2 (1) ... logarithms. The second equal sign uses that ...
Positive: 60 %
The laws of logarithms, ... Where Did Logs Come From? ... Therefore log 125 5 = 1/3 and log 5 125 = 3, and 1/3 does indeed equal 1/3.
Positive: 57 %
### More resources
How to Solve Logarithmic Equations. ... equation 3 –4 = x, or 1/81 = x. ... combine the two logs that are adding into one log by using the ...
Positive: 60 %
Deriving Approximate Logarithms. ... We thus have the logarithm of a number between 1 and 10, which turns to equal ... To get log 3.73 we add to log3 ...
Positive: 55 %
1 4 1 4 4 4 x=6 x=6 x =1296 ... 5xlog6=(2x -1)log3 5xlog6=2xlog3-log3 ... Part V – Solving Using Log Rules solving using log rules: ...
| 339 | 911 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.25 | 4 |
CC-MAIN-2016-50
|
longest
|
en
| 0.800138 |
https://www.thedonutwhole.com/how-many-grams-of-hydrogen-are-in-1-gram-of-water/
| 1,716,523,361,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058677.90/warc/CC-MAIN-20240524025815-20240524055815-00587.warc.gz
| 900,846,679 | 28,883 |
# How many grams of hydrogen are in 1 gram of water?
Water (H2O) is composed of two hydrogen atoms and one oxygen atom. Since hydrogen has an atomic mass of 1.008 g/mol, and oxygen has an atomic mass of 15.999 g/mol, we can calculate the mass composition of water as follows:
1 mole of H2O contains:
• 2 moles of H (2 atoms x 1.008 g/mol = 2.016 g)
• 1 mole of O (1 atom x 15.999 g/mol = 15.999 g)
So the total molar mass of 1 mole of H2O is:
2.016 g (for H) + 15.999 g (for O) = 18.015 g
Expressed as a percentage:
• Hydrogen: 2.016 g / 18.015 g x 100 = 11.19%
• Oxygen: 15.999 g / 18.015 g x 100 = 88.81%
Therefore, the mass percentage of hydrogen in water is 11.19%.
## Calculating Grams of Hydrogen in 1 Gram of Water
Since we know hydrogen accounts for 11.19% of the mass of water, we can calculate the grams of hydrogen in 1 gram of water as follows:
1 gram H2O x 0.1119 = 0.1119 grams hydrogen
So for every 1 gram of water, there are 0.1119 grams of hydrogen.
## Showing the Calculation
Let’s walk through the calculation step-by-step:
1. 1 gram of water contains 2 hydrogen atoms (H2)
2. Mass of 1 hydrogen atom = 1.008 g/mol
3. So mass of 2 hydrogen atoms = 2 x 1.008 g/mol = 2.016 g/mol
4. Total molar mass of water (H2O) = 2.016 g (for 2 H atoms) + 15.999 g (for 1 O atom) = 18.015 g/mol
5. Mass of hydrogen (2.016 g) divided by total molar mass of water (18.015 g) = 0.1119
6. This means hydrogen accounts for 11.19% of the mass of water
7. For 1 gram of water, the mass of hydrogen is 1 gram x 0.1119 = 0.1119 grams
Therefore, by calculating the mass percentages based on molecular weights, we determine there are 0.1119 grams of hydrogen in 1 gram of water.
## Why This Calculation Works
This calculation relies on the mass relationships between atoms that are defined by their atomic mass (measured in atomic mass units or g/mol). Because we know the atomic mass of hydrogen and oxygen, we can calculate their proportional masses in the water molecule.
Some key points that make this calculation valid:
• The atomic mass of an element is constant and applies to a single atom of that element.
• Water always consists of 2 hydrogen atoms and 1 oxygen atom.
• The mass ratio between hydrogen and oxygen in water is always the same.
• By using molar mass instead of atomic mass units, the values can be scaled up proportionally from a single molecule to grams of substance.
Therefore, by applying the constant atomic mass values to the molecular formula of water, the mass relationships can be derived mathematically and give an accurate result.
## Applications and Examples
Knowing the mass percentage of hydrogen in water is useful for many applications across chemistry, physics, biology and environmental sciences. Here are a few examples of when you may need to use this calculation or data:
• Determining the composition of water sources: The hydrogen level can indicate purity or if water contains dissolved compounds.
• Balancing chemical equations involving water: Use the mass percentages to relate the amounts of hydrogen and oxygen in chemical reactions.
• Calculating the energy content of fuels: The heat released by burning hydrogen fuels can be estimated based on the hydrogen content.
• Designing processes involving electrolysis of water: Efficiency and hydrogen yields depend directly on the hydrogen contained in water.
• Estimating hydrogen production from bioreactors: Many biological processes generate hydrogen through metabolism of organic matter.
In all cases, the fundamental calculation remains the same. By applying the atomic mass ratios, the mass composition of water can be determined.
## Limitations and Considerations
While extremely accurate for pure water under standard conditions, there are some limitations to consider when applying this calculation:
• The calculation assumes 100% pure H2O. Any dissolved compounds or contaminants would alter the mass percentages.
• The calculation is based on the atomic mass of hydrogen on the periodic table. Different hydrogen isotopes would have slightly different masses.
• Water can exist as ice, liquid and vapor. Although the chemical formula remains H2O, the density and mass relationships change between states.
• Extreme temperature and pressure conditions would affect the behavior of the water molecules and atomic masses.
• Quantum effects at the molecular scale may cause slight deviations from the expected mass ratios.
While these limitations exist, they have negligible influence under normal temperature and pressure conditions for typical water sources. The calculation remains an excellent estimation in most practical cases.
## Visualizing the Composition of Water
The mass composition of water can be visualized by depicting the relative proportions of hydrogen and oxygen:
Element Atomic Mass (g/mol) Number of Atoms Mass (g) Percentage
Hydrogen (H) 1.008 2 2.016 11.19%
Oxygen (O) 15.999 1 15.999 88.81%
Total mass 18.015 g/mol
This table summarizes the underlying data used in the calculation. We can see:
• Hydrogen accounts for 2.016 g out of 18.015 g total.
• This equates to 11.19% of the mass.
• Oxygen makes up the remaining 88.81% at 15.999 g.
Visually representing the composition like this provides an at-a-glance view of the mass relationships in water.
## Practical Examples and Calculations
Let’s work through some practical examples using the grams of hydrogen in water calculation:
### Example 1
If a container holds 2.5 liters of water, how many grams of hydrogen does it contain?
Solution:
• 2.5 liters of water = 2500 grams (density is 1 g/mL)
• From previous calculation, 1 gram H2O contains 0.1119 g hydrogen
• So 2500 grams H2O contains 0.1119 x 2500 = 279.75 grams hydrogen
### Example 2
A bioreactor digests organic waste and produces 35 liters/day of hydrogen gas. How many grams of hydrogen is produced if collecting gas for 1 week?
Solution:
• 35 liters/day x 7 days = 245 liters hydrogen gas
• Using density of hydrogen gas (0.089 g/L), 245 L contains 21.805 g
• So over 1 week, bioreactor produces 21.805 grams hydrogen
### Example 3
An industrial electrolyzer passes a 500 amp electrical current through water for 1 hour. If the efficiency is 65%, how many grams of hydrogen are produced?
Solution:
• 500 amps x 1 hour = 500 amp-hours of charge passed
• With 100% efficiency, 1 amp-hour produces 0.1119 g hydrogen (from water composition)
• So 500 amp-hours x 0.1119 g/amp-hr = 55.95 g (with 100% efficiency)
• Accounting for 65% efficiency, amount of hydrogen produced is 0.65 x 55.95 = 36.37 grams
These examples demonstrate practical applications of the hydrogen mass calculation using typical values for water volume, gas density, electrical current and efficiency.
## Conclusion
By examining the molecular makeup of water and the atomic masses of hydrogen and oxygen, we can mathematically determine that there are 0.1119 grams of hydrogen in every 1 gram of water.
Some key points:
• Water is composed of 2 hydrogen atoms (H2) and 1 oxygen atom (O).
• Using the atomic masses of H (1.008 g/mol) and O (15.999 g/mol), the mass percentages can be calculated.
• Hydrogen accounts for 11.19% of the total mass of water.
• Therefore, for 1 gram of water there are 0.1119 grams of hydrogen.
• This composition allows calculations of hydrogen mass from quantities of water.
• The calculation has limitations at extreme conditions, but works well for normal water sources.
Understanding the mass relationships in water has many practical applications across science and engineering fields. Performing calculations using the composition data allows quantification of hydrogen amounts in water for processes like electrolysis, fuel production, and biological digestion.
| 1,859 | 7,761 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875 | 4 |
CC-MAIN-2024-22
|
latest
|
en
| 0.878514 |
https://lazyocho.com/2019/11/01/algebra-2-curriculum-dump-1/?amp=1
| 1,675,243,387,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-06/segments/1674764499919.70/warc/CC-MAIN-20230201081311-20230201111311-00617.warc.gz
| 384,280,436 | 26,913 |
# Algebra 2 curriculum dump #1
I’ve decided to reflect on the interleaved algebra 2 course that I’ve designed and taught over the last couple of years. My hope is that by taking a more outward-facing approach to the curriculum, it will help me make it better. This is the 1st post in the series.
In my last post, I overviewed the course. I’m going to dig into the units now. Yay! Here is the working problem set for the course.
Unit 1 (Problems 1-24)
• Function notation & analysis
• Domain & range graphically
• Function arithmetic
• Trigonometric ratios & circles
• Factoring the greatest common factor
Ah, the opening unit. The point where summer fades away. Students often find graphs to be their mathematical friends, so I lean on them a lot, especially in the first few units. I pair that with some analytical stuff by getting into function notation, function arithmetic, and factoring with GCF. Over 90% of the kids come to me from geometry (the others are taking it concurrently), so I also like jumping right into trig. The ratios are fresh in their minds and they get excited when they see them in the first few problems (e.g. 2, 8, 18). The problems reintroduce sine, cosine, and tangent and we begin thinking about the ratios in the coordinate plane with circles. It’s not explicitly stated in the concepts, but I also go hard with using interval notation when representing domain and range; it sets up so much of our future work on intervals. It’s a small thing, but I like how I frame factoring as “rewriting as multiplication” (problem 7).
Two things that I’m left thinking about: I don’t focus on domain and range in any other representation other than graphically. For this unit, I think this is ok, but as of now, I don’t devote any problems to it later on, either. This is an issue. Also, sadly, the function arithmetic that we do sort of dies at the end of this unit, as I haven’t yet included problems that loop back to it in future units.
In the end, this unit is full of pretty straightforward concepts and the students perform well. This is by design. With such a big contrast in how they learn math as compared to prior years, I find it helpful to build success in for them at the start. It’s a needed confidence booster for the long year ahead.
Unit 2 (problems 25-50)
• Intervals of increasing & decreasing
• Extreme values
• Visual sequences & sequence notation
• Factoring the difference of perfect squares
• Trigonometric ratios & circles
• **Systems of linear equations (2)
By studying where a function is increasing and decreasing and its extreme values, we continue the work on graph analysis that we started in unit 1. Again, interval notation comes in handy here. This problem set also introduces sequences. The intricacies of sequence notation can be tricky for students, but the few problems that I begin with (37, 38) have combated that nicely through the years. We also continue our factoring journey as we study the difference of two perfect squares. Now that I think about it, I wish I had included some difference of squares expressions that had a GCF. And I still don’t know how I feel about obscure expressions like (2x-1)^2 + (3x+8)^2 and asking students to factor them, but hey, that’s why I march to the beat of the Regents exam.
This is an important unit in the development of trigonometry because the kids learn about a unit circle along with defining sine, cosine, and tangent in terms of x and y. We also explore the signs of the three ratios depending on where the terminal side of the angle lies when it’s drawn in standard position. (Annoying: so many kids at this point start using the mnemonic ASTC because of the DeltaMath problems I assign.) Speaking of the terminal side of an angle, lost in all this trig work is the explicit learning of how to sketch an angle in standard position. I assign them problems from DeltaMath, which are really nice, but there’s nothing in the problem set that targets this skill. This results in me creating an opening problem to discuss it with the students at some point during the unit. This is haphazard and sloppy.
**Throughout the course, we come across topics and skills that are included in the problem set, but are not tested on that unit’s exam. I’ve found that there’s no need to test on EVERYTHING that we study during a single unit. Certain concepts, and certain problems, are stepping stones to broader ideas. I simply use them to navigate us to much deeper, more important parts of the curriculum. In this case, wrestling with systems of two linear equations will, over time, eventually lead us to solve systems of three linear equations (along with non-linear systems) in later units.
Unit 3 (problems 50-77)
• End behavior
• Function translations
• Recursive sequences
• Difference & sum of cubes
• **Trigonometric ratios & circles
• **Experimental Design
• **Systems of linear equations (2)
In this unit, we continue our work on graph analysis, sequences, and factoring. Naturally, I am able to spiral in our previous graphing work (domain/range, intervals of increasing/decreasing, extreme values) in almost all of the end behavior problems (e.g. 57, 65). Factoring the difference or sum of two cubes always seems so formulaic and boring to me, but I know of no better way of approaching it. That said, this year I changed problem #59 to include an example of using the formulas, and it really helped. The students also get their first taste of a reciprocal trig function when they learn about cosecant. I define it plainly, which I’m ok with for now. I fold the ratio into several problems in the set to help build familiarity with it, especially in relation to the other ratios. When doing this, we continue to discuss terminal sides of angles and sketching triangles in the coordinate plane. Submerged in this set is rationalization. I touch on it with one problem (61), follow it up with DeltaMath, and bring it in again with problem 74. This basic level stuff is good enough to be able to help us with future problems. We also mix in our first look at transformations of functions when we do problem 53. I use a Desmos activity for this that I like, but I know could be better — especially when it comes to horizontal translations.
The sequence notation that they learned during the last unit is put to good use when I contract a throat virus (which renders me unable to speak) on the day they must investigate recursive sequences in problems 67-69. It’s a fun, memorable lesson. I drop two more linear systems problems in the set for good measure (60, 73). On a related note, I really like problem 72 because it combines function transformations with systems of equations. The equation f(x)=g(x) gets some attention, which is a nice.
I’ve sprinkled in a couple more problems on bias during this unit, which seemed like second nature to the kids. I feel the problems are useful, but they seem to linger with no stronger connection anywhere in sight. This will soon change when we start getting deeper in statistics in upcoming units, but maybe I can adjust for this next year?
bp
{{#pages}} {{/pages}}
%%footer%%
| 1,571 | 7,144 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.84375 | 4 |
CC-MAIN-2023-06
|
latest
|
en
| 0.947671 |
http://www.britannica.com/EBchecked/topic/561675/square-number
| 1,430,028,514,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2015-18/segments/1429246652631.96/warc/CC-MAIN-20150417045732-00057-ip-10-235-10-82.ec2.internal.warc.gz
| 379,659,433 | 12,347 |
# Square number
Mathematics
Thank you for helping us expand this topic!
Once you are finished and click submit, your modifications will be sent to our editors for review.
This topic is discussed in the following articles:
• ## Chinese mathematics
In The Nine Chapters, algorithms for finding integral parts of square roots or cube roots on the counting surface are based on the same idea as the arithmetic ones used today. These algorithms are set up on the surface in the same way as is a division: at the top, the “quotient”; under it, the “dividend”; one row below, the “divisor”;...
• ## definition and properties
Square numbers are the squares of natural numbers, such as 1, 4, 9, 16, 25, etc., and can be represented by square arrays of dots, as shown in Figure 1. Inspection reveals that the sum of any two adjacent triangular numbers is always a square number.
• ## mathematical puzzles
There is a wide variety of puzzles involving coloured square tiles and coloured cubes. In one, the object is to arrange the 24 three-colour patterns, including repetitions, that can be obtained by subdividing square tiles diagonally, using three different colours, into a 4 × 6 rectangle so that each pair of touching edges is the same colour and the entire border of the rectangle is the...
• ## numbers
Other classes of numbers include square numbers—i.e., those that are squares of integers; perfect numbers, those that are equal to the sum of their proper factors; random numbers, those that are representative of random selection procedures; and prime numbers, integers larger than 1 whose only positive divisors are themselves and 1.
• ## relationship to geometrical square
...A square has four axes of symmetry, and its two finite diagonals (as with any rectangle) are equal. Bisection of a square by a diagonal results in two right triangles. If the length of the side of a square is s, then the area of the square is s 2, or “ s squared.” From this relation is derived the algebraic use of the term square, which denotes...
Please select the sections you want to print
MLA style:
"square number". Encyclopædia Britannica. Encyclopædia Britannica Online.
Encyclopædia Britannica Inc., 2015. Web. 26 Apr. 2015
<http://www.britannica.com/EBchecked/topic/561675/square-number>.
APA style:
Harvard style:
square number. 2015. Encyclopædia Britannica Online. Retrieved 26 April, 2015, from http://www.britannica.com/EBchecked/topic/561675/square-number
Chicago Manual of Style:
Encyclopædia Britannica Online, s. v. "square number", accessed April 26, 2015, http://www.britannica.com/EBchecked/topic/561675/square-number.
While every effort has been made to follow citation style rules, there may be some discrepancies.
Please refer to the appropriate style manual or other sources if you have any questions.
Click anywhere inside the article to add text or insert superscripts, subscripts, and special characters.
You can also highlight a section and use the tools in this bar to modify existing content:
Editing Tools:
We welcome suggested improvements to any of our articles.
You can make it easier for us to review and, hopefully, publish your contribution by keeping a few points in mind:
1. Encyclopaedia Britannica articles are written in a neutral, objective tone for a general audience.
2. You may find it helpful to search within the site to see how similar or related subjects are covered.
3. Any text you add should be original, not copied from other sources.
4. At the bottom of the article, feel free to list any sources that support your changes, so that we can fully understand their context. (Internet URLs are best.)
Your contribution may be further edited by our staff, and its publication is subject to our final approval. Unfortunately, our editorial approach may not be able to accommodate all contributions.
MEDIA FOR:
square number
Citation
• MLA
• APA
• Harvard
• Chicago
Email
You have successfully emailed this.
Error when sending the email. Try again later.
| 899 | 3,991 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.03125 | 4 |
CC-MAIN-2015-18
|
latest
|
en
| 0.93876 |
https://digitalvisi.com/study-about-the-family-of-lines-and-general-equation-of-a-line-in-detail/
| 1,716,638,977,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058822.89/warc/CC-MAIN-20240525100447-20240525130447-00417.warc.gz
| 164,012,869 | 15,735 |
Saturday, May [email protected]
# Study About The Family of Lines And General Equation of a Line in Detail!
We all know or have met many types of families throughout our life. In truth, each family has some distinguishing quality that is shared by all members of that family. A family, for example, may all have similar facial features, yet another family may all be sportsmen or foodies. What do you suppose a family of lines in Mathematics would look like?
On that note, let’s learn about what constitutes a family of lines, their similar features and the General Equation of a Line in detail.
## What is a Family of Lines?
is a group of lines that share at least one feature.
A straight line’s equation is given by,
y=mx+c
The two most crucial features of a straight line in this context are:
• the slope of the line
• y−intercept
While the slope defines how much the line rises or falls, the y−intercept indicates the point at which the line and the y−axis intersect.
As a result, there are 2 types of families.
### Types of Family of Lines
Based on their similar features, the family of lines is divided into two categories.
• A group of lines with the same slope
• A group of lines with the same y−intercept.
Let us first learn to write the joint equation of lines before learning to write the general equation of a family of lines.
### Family of Lines with the Same Slope
This family is represented as follows.
Take note of the fact that all of these lines have the same slope. They are, in other words, parallel. Although the lines appear to be identical, they have different y-intercepts. Take note of how the y-intercepts of these lines shift along the yaxis.
The lines have smaller values of b when the y-intercepts are smaller. As the line on the vertical axis rises, so does the value of b. The vertical shift is the name given to this phenomenon.
### Family of Lines with the Same y-intercept
This family can look like the one illustrated below.
Take note that all of these lines converge at (0,2). A family of lines is made up of these plus an endless number of other lines that can be drawn via a specific location. Concurrent lines are formed when three or more lines intersect at a single place. As a result, this is a family of concurrent lines.
### General Equation of a line
The following are the general forms of a line equation:
• Slope-intercept form
• Intercept form
• Normal form
Let us learn all the straight line formulas along with the general equation of a line and different forms to find the equation of a straight line in detail here.
#### General Equation of Different Types of Lines
In Geometry, there are various varieties of lines. Geometry is created from the inspiration of lines. Straight lines are classified into three types: horizontal lines (sleeping lines), vertical lines (sleeping lines), and oblique lines (Slanting lines).
The general equation of a straight line is y = mx + c, where m is the slope and c is the y-intercept. It is the most common sort of line equation in geometry. The equation of a straight line can be expressed in a variety of forms, including point-slope form, slope-intercept form, general form, standard form, and so on.
There are three ways to write general equations: point-slope form, standard form, and slope-intercept form.
The points where a line crosses the x-axis or the y-axis are named the intercepts, given by (a, 0) and (0, b).
• Slope (m) of a non-vertical line going between (x₁, y₁ ) and (x₂, y₂) points
m=(y₂ – y₁)/(x₂ – x₁), x₁ ≠ x₂
• A horizontal line’s equation
y = a or y = -a
• A vertical line’s equation
x = b or x = -b
• Line equation running through locations (x₁, y₁) and (x₂, y₂)
y – y₁= (y₂−y₁)/(x₂−x₁)×(x – x₁)
• The normal type of line equation
x cos α+y sin α = p
Happy
0 %
| 910 | 3,823 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.8125 | 5 |
CC-MAIN-2024-22
|
latest
|
en
| 0.951539 |
https://byjusexamprep.com/area-moment-of-inertia-i
| 1,670,296,421,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-49/segments/1669446711069.79/warc/CC-MAIN-20221206024911-20221206054911-00780.warc.gz
| 183,244,241 | 64,021 |
# Area Moment of Inertia: Definition, Formula, Theorem
By Deepak Yadav|Updated : October 26th, 2022
Before diving into the Area Moment of Inertia details, let's first understand the moment of inertia. The moment of inertia is defined as the quantity expressed by the body resisting angular acceleration, which is the sum of the product of each particle's mass and its square of the distance from the axis of rotation. In simpler terms, it is a quantity determining the amount of torque required for a certain angular acceleration in a rotational axis. The angular mass or rotational inertia are other names for the moment of inertia. kg-m2 is the SI unit for a moment of inertia.
Area Moment of Inertia PDF
The second moment of area, also known as the quadratic moment of area or the area moment of inertia, is a geometrical feature of an area that reflects how its points are distributed concerning an arbitrary axis. The second moment of area is commonly indicated by an I (for an axis in the plane of the area) or a J. (for an axis perpendicular to the plane). It is calculated in both circumstances using a multiple integral over the object in the issue. It has a length of L to the fourth power. Let's take a deep dive into concepts related to area moment of inertia in detail.
## Area Moment of Inertia
The area moment of inertia, also known as the second area moment or the 2nd moment of area, is a feature of a two-dimensional plane form that illustrates how its points are distributed in the cross-sectional plane along an arbitrary axis. This property describes the deflection of a plane shape under a load. The area moment of inertia is essential for the GATE exam as MCQ-based questions can be seen in the exam.
For an axis in a plane, the area moment of inertia is commonly indicated by the symbol I. When the axis is perpendicular to the plane, it is also designated as J. The dimension unit of the second area moment is L4 (length to the power of four). The International System of Units unit of measurement is the meter to the power of four, or m4. It can be inches to the fourth power, in4 if we use the Imperial System of Units.
This concept will appear frequently in the field of engineering. In this context, the area moment of inertia is believed to be a measure of a beam's flexural stiffness. It is a significant feature that is used to calculate the deflection of a beam or to measure its resistance to bending. We'll look at two examples here.
• First, the planar second moment of the area where the force sits perpendicular to the neutral axis can characterize or define a beam's bending resistance.
• Second, when the applied moment is parallel to the beam's cross-section, the polar second moment of the area can be employed to calculate its resistance. It is essentially the beam's resistance to torsion.
## Area Moment of Inertia Formulas
The area moment of inertia formulas can be used to formulate the MSQ-based questions carrying marks scheduled as per the GATE exam pattern. When we consider the area moment of inertia for the x-axis, we get;
Ix = Ixx = ∫y2dxdy
Meanwhile, the area's "product" moment is described by
Ixy = ∫xy dx dy
## Parallel Axis Theorem
The parallel axis theorem is an essential part of the GATE ME syllabus. A body's moment of inertia about any axis is equal to the sum of its moment of inertia about a parallel axis passing through its centre of mass and the product of its mass, and the square of the distance between the two lines.
I = Ig = Md2
Where:
• M = Mass of the body
• D = the perpendicular distance between the two lines
## Perpendicular Axis Theorem
The moment of inertia about a perpendicular axis for a planar body is the sum of the moments of inertia about two perpendicular axes coincident with this perpendicular axis. It sits on the plane of the body. The expression specifies it.
Iz = Ix + Iy
## Area Moments of Inertia for Some Common Shapes
The following is a list of some form's second moments of area. The second moment of area, also known as the area moment of inertia, is a geometrical feature of an area that represents the distribution of its points concerning an arbitrary axis.
Important GATE Notes Work Done By A Force Motion Under Gravity Dynamic Resistance Static Resistance Ideal Diode Bettis Theorem Work Done By A Constant Force Application Layer Protocols Castiglia's Theorem Portal Frames
Get complete information about the GATE exam pattern, cut-off, and all those related things on the Byju Exam Prep official youtube channel.
write a comment
## FAQs about Area Moment of Inertia
• A body's moment of inertia about any axis is equal to the sum of its moment of inertia about a parallel axis passing through its center of mass and the product of its mass and the square of the distance between the two lines.
• The area moment of inertia (also known as the second moment of area) is a geometrical feature of a form that describes how points are distributed around an axis. It is employed as a measure of a body's resistance to bending in classical mechanics.
• A plane area's moment of inertia (MI) about an axis normal to the plane is equal to the sum of the moments of inertia about any two mutually perpendicular axes located in the plane and passing through the specified axis.
• In linear motion, the moment of inertia plays the same role as mass. It is the measurement of a body's resistance to a change in rotational motion. It is constant for a specified rigid frame and axis of rotation.
• The mass moment of inertia is a rotating analog of mass, whereas the area moment of inertia is primarily used in beam equations. The units employed in both moments of inertia are also different.
### ESE & GATE ME
Mechanical Engg.GATEGATE MEHPCLBARC SOESEIES MEBARC ExamISRO ExamOther Exams
| 1,249 | 5,819 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.21875 | 4 |
CC-MAIN-2022-49
|
longest
|
en
| 0.929683 |
https://mathoverflow.net/questions/434967/multivariate-normal-distribution-and-orthonormal-transformation
| 1,721,615,467,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-30/segments/1720763517805.92/warc/CC-MAIN-20240722003438-20240722033438-00812.warc.gz
| 347,377,963 | 21,376 |
# Multivariate normal distribution and orthonormal transformation [closed]
Let $$X$$ be a multivariate normal random variable with mean $$\boldsymbol{\mu}$$ and variance matrix $$\mathrm{\Sigma}$$. Next, define
Suppose that $$Y = AX$$ where $$A$$ is appropriate matrix. Can we say that the distribution of $$Y$$ is same as $$X$$ if and only if $$A$$ is an orthogonal symmetric matrix?
Thank you Iosif for pointing out the mistake. I modified the question.
• If different directions have different variances then rotations can alter the distribution.
– user44143
Commented Nov 20, 2022 at 13:34
## 1 Answer
$$\newcommand\Si\Sigma$$First here, there is no such thing as an orthonormal matrix. So, let us assume that you meant an orthogonal matrix instead.
Then we have this question:
Suppose that $$X\sim N(\mu,\Si)$$ and $$Y = AX$$, where $$A$$ is appropriate matrix. Can we say that the distribution of $$Y$$ is same as $$X$$ if and only if $$A$$ is an orthogonal matrix?
The answer to this is that neither the "if" part of this conjecture nor its "only if" holds in general.
Indeed, the distribution of $$Y=AX$$ is $$N(A\mu,A\Si A^\top)$$. So, the distribution of $$Y$$ is same as $$X$$ if and only if $$A\mu=\mu \tag{1}\label{1}$$ and $$A\Si A^\top=\Si. \tag{2}\label{2}$$
If $$\Si$$ is nonsingular, then condition \eqref{2} can be rewritten as $$A_\Si A_\Si ^\top=I$$, where $$A_\Si:=\Si^{-1/2}A\Si^{1/2}.$$ So, in this case, the distribution of $$Y$$ is same as $$X$$ if and only if \eqref{1} holds and $$A_\Si$$ is an orthogonal matrix.
• The pdf of $Y$ is $$f_{Y}(y) = \frac{1}{2\pi(det(A\Sigma A^T))^{(1/2)}}exp[-(1/2)(AX-A\mu)^T(A\Sigma A^T)]^{-1}(XA-A\mu) = \frac{1}{2\pi(det(\Sigma ))^{(1/2)}}exp[-(1/2)(X-\mu)^T(\Sigma)]^{-1}(X-\mu)$$. Commented Nov 20, 2022 at 14:35
• While there is no such thing as an orthonormal matrix, a good argument can be made that the usual term “orthogonal matrix” was a mistake and it should have been “orthonormal matrix” all along, because a real $n\times n$ matrix is an orthogonal matrix exactly when its columns are an orthonormal basis of $\mathbf R^n$, not just an orthogonal basis. Serge Lang, in the 3rd edition of his Algebra, bemoaned the terminology “orthogonal matrix” because such matrices are not just ones with orthogonal columns: the columns must be orthogonal unit vectors. He suggested the term “real unitary matrix”. Commented Nov 20, 2022 at 16:06
• @KConrad : This is a good point; "orthogonal matrix" is indeed an example of bad generally accepted terminology/notation. Commented Nov 20, 2022 at 16:28
| 770 | 2,576 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 27, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.984375 | 4 |
CC-MAIN-2024-30
|
latest
|
en
| 0.828341 |
https://docsbay.net/vs-is-and-alternating-voltage-this-so-as-to-setup-a-flux-in-the-core
| 1,709,077,022,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-10/segments/1707947474688.78/warc/CC-MAIN-20240227220707-20240228010707-00559.warc.gz
| 208,054,666 | 5,740 |
# Vs Is and Alternating Voltage - This So As to Setup a Flux in the Core
TRANSFORMERS
Vs is and alternating voltage - this so as to setup a flux in the core.
This can be seen from Faraday's law
/ Where e is the voltage set up by the varying flux.
Now we know that the flux varies as a sinusoid.
Therefore = maxsin(t)
Therefore /
From the above analysis we can see that the voltage lags the flux by 90 degrees
A non-ideal transformer has eddy current losses in the core, and resistive losses in its primary and secondary windings. In addition the core of the transformer requires some finite mmf for its magnetization. Also not all the flux links the primary and secondary windings, in other words there exists leakage flux in the windings.
Using all this information on the no ideal transformer we could develop the equivalent circuit shown below for it.
In the above diagram the circuit components used are used to represent the various losses stated above. The relationship between circuit component and loss is detailed below.
Ro => This resistor represents the hysteresis and eddy-current losses. These losses are directly dependent on the input voltage, as can be seen by its position.
Xm => This inductor represents the magnetization reactance - that minimum mmf required for the magnetization of the transformer core.
R1 and R2 - These resistors represent the resistive losses (I2R) present in the windings of the primary and secondary sides of the transformer.
X1 and X2 - These represent the leakage reactance arising from leakage fluxes in the primary and secondary windings.
On No Load conditions we have the familiar relations
To make the equivalent circuit of the transformer a little simpler we can merge the resistive and inductive components of the windings as shown in the diagram below:
In the diagram shown I => current required to magnetize the core (flux per turn)
So for N1 turns on the primary side; total flux = N1
as shown previously (t) = maxsint
Therefore d/dt = maxcost
V1 = N1maxcost
Using the equivalent circuit we can see that the load losses in the transformer can be given by
I12R1 + I22R2
General points to remember: / when going to the HV side, impedances go up
when going to the LV side impedances go down
Any side can be treated as the primary, all the must be done is that the turns ratio should be adjusted to suit.
Paralleling of transformers
Say we had a situation as depicted such. A plant demanded a load of 250kVA. To account for this load the electricity commission installed a 300kVA transformer at the location of the plant, which served well to supply power to the plant. But say in the future the planted decided to expand and logically it demanded a greater load from the lines, say 500kVA. The electricity commission has two choices on how to deal with the plant. It could replace the old transformer with a new one capable of handling the new demand. Or it could add another transformer, which would be able to comfortable handle the extra 200kVA, demanded.
The second option is the more feasible one as a smaller transformer needs to be need be purchased which is much cheaper than the purchase of a brand new big transformer.
In paralleling transformers, however, some basic rules need to be followed.
Obviously the voltage and turns ratio need to be the same otherwise what would occur is that one transformer (the one with the higher secondary voltage) would be driving current into the other transformer. Similar R / X ratio in relation to the transformer size must be observed. And the polarity must be the same for both. The diagram below shows the effect of ignoring polarity.
Base Values
In practice transformers come with various voltage turn ratios and various VA ratings, and these transformers would at times need to be used in the same system. When dealing with situations such as these, analyzing impedance values is problematic. What is done is that a base VA rating, and voltage are used to which all values on all the transformers are referred. The base VA rating is defined as sb, and is expressed in per unit (pu).
So Vb x Ib = sb
From this equation we can get the following relationships.
Zb = Vb / Ib
Ib = sb / Vb
Impedance of T/F pu = Actual Impedance / Base Impedance
For example say we had a 10MVA transformer with turns ratio 66/11 on an 11kV system.
Then Zb = Vb2 / sb = 11x103 / 10x103 = 12.1
Actual Z = 1.21 (say this were given),
Then pu Z = 1.21 / 12.1 = 0.1pu
| 1,067 | 4,505 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.515625 | 4 |
CC-MAIN-2024-10
|
latest
|
en
| 0.918669 |
https://www.statistics-lab.com/category/%E9%9A%8F%E6%9C%BA%E5%9B%BE%E8%AE%BA%E4%BB%A3%E5%86%99/
| 1,716,377,812,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971058542.48/warc/CC-MAIN-20240522101617-20240522131617-00515.warc.gz
| 900,753,676 | 42,155 |
## 数学代写|随机图论代写Random Graph代考|CSL866
statistics-lab™ 为您的留学生涯保驾护航 在代写随机图论Random Graph方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写随机图论Random Graph代写方面经验极为丰富，各种代写随机图论Random Graph相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 数学代写|随机图论代写Random Graph代考|Generalized Random Intersection Graphs
Godehardt and Jaworski [ 389 ] introduced a model which generalizes both the binomial and uniform models of random intersection graphs. Let $P$ be a probability measure on the set ${0,1,2, \ldots, m}$. Let $V={1,2, \ldots, n}$ be the vertex set. Let $M=$ ${1,2, \ldots, m}$ be the set of attributes. Let $S_1, S_7, \ldots, S_n$ be independent random subsets of $M$ such that for any $v \in V$ and $S \subseteq M$ we have $\mathbb{P}\left(S_v=S\right)=P(|S|) /\left(\begin{array}{l}m \ |S|\end{array}\right)$. If we put an edge between any pair of vertices $i$ and $j$ when $S_i \cap S_j \neq \emptyset$, then we denote such a random intersection graph as $G(n, m, P)$, while if the edge is inserted if $\left|S_i \cap S_i\right| \geq s, s \geq 1$, the respective graph is denoted as $G_s(n, m, P)$. Bloznelis [99] extends these definitions to random intersection digraphs.
The study of the degree distribution of a typical vertex of $G(n, m, P)$ is given in [464], [242] and [97], see also [465]. Bloznelis ( see [98] and [100]) shows that the order of the largest component $L_1$ of $G(n, m, P)$ is asymptotically equal to $n \rho$, where $\rho$ denotes the non-extinction probability of a related multi-type Poisson branching process. Kurauskas and Bloznelis [541] study the asymptotic order of the clique number of the sparse random intersection graph $G_s(n, m, P)$.
Finally, a dynamic approach to random intersection graphs is studied by Barbour and Reinert [63], Bloznelis and Karoński [107], Bloznelis and Goetze [104] and Britton, Deijfen, Lageras and Lindholm [166].
One should also notice that some of the results on the connectivity of random intersection graphs can be derived from the corresponding results for random hyperghraphs, see for example [519], [707] and [390].
## 数学代写|随机图论代写Random Graph代考|Random Geometric Graphs
McDiarmid and Müller [588] gives the leading constant for the chromatic number when the average degree is $\Theta(\log n)$. The paper also shows a “surprising” phase change for the relation between $\chi$ and $\omega$. Also the paper extends the setting to arbitrary dimensions. Müller [618] proves a two-point concentration for the clique number and chromatic number when $n r^2=o(\log n)$.
Blackwell, Edmonson-Jones and Jordan [96] studied the spectral properties of the adjacency matrix of a random geometric graph (RGG). Rai [672] studied the spectral measure of the transition matrix of a simple random walk. Preciado and Jadbabaie [665] studied the spectrum of RGG’s in the context of the spreading of viruses.
Sharp thresholds for monotone properties of RGG’s were shown by McColm [582] in the case $d=1$ viz. a graph defined by the intersection of random subintervals. And for all $d \geq 1$ by Goel, Rai and Krishnamachari [391].
First order expressible properties of random points
$\mathscr{X}=\left{X_1, X_2, \ldots, X_n\right}$ on a unit circle were studied by McColm [581]. The graph has vertex set $\mathscr{X}$ and vertices are joined by an edge if and only if their angular distance is less than some parameter $d$. He showed among other things that for each fixed $d$, the set of a.s. FO sentences in this model is a complete noncategorical theory. McColm’s results were anticipated in a more precise paper [388] by Godehardt and Jaworski, where the case $d=1$, i.e., the evolution a random interval graph, was studied.
Diaz, Penrose, Petit and Serna [256] study the approximability of several layout problems on a family of RGG’s. The layout problems that they consider are bandwidth, minimum linear arrangement, minimum cut width, minimum sum cut, vertex separation, and edge bisection. Diaz, Grandoni and Marchetti-Spaccemela [255] derive a constant expected approximation algorithm for the $\beta$-balanced cut problem on random geometric graphs: find an edge cut of minimum size whose two sides contain at least $\beta n$ vertices each.
Bradonjić, Elsässer, Friedrich, Sauerwald and Stauffer [162] studied the broadcast time of RGG’s. They study a regime where there is likely to be a single giant component and show that w.h.p. their broadcast algorithm only requires $O\left(n^{1 / 2} / r+\log n\right)$ rounds to pass information from a single vertex, to every vertex of the giant. They show on the way that the diameter of the giant is $\Theta\left(n^{1 / 2} / r\right)$ w.h.p. Friedrich, Sauerwald and Stauffer [334] extended this to higher dimensions.
# 随机图论代写
## 数学代写|随机图论代写Random Graph代考|Generalized Random Intersection Graphs
Godehardt 和 Jaworski [389] 引入了一个模型，该模型概括了随机相交图的二项式模型和均匀模型。让P是集合上的概率测度0,1,2,…,米. 让在=1,2,…,n是顶点集。让米= 1,2,…,米是属性集。让小号1,小号7,…,小号n是独立的随机子集米这样对于任何在∈在和小号⊆米我们有P(小号在=小号)=P(|小号|)/(米 |小号|). 如果我们在任意一对顶点之间放置一条边一世和j什么时候小号一世∩小号j≠∅，那么我们将这样的随机交叉图表示为G(n,米,P), 而如果边被插入如果|小号一世∩小号一世|≥秒,秒≥1，相应的图表示为G秒(n,米,P). Bloznelis [99] 将这些定义扩展到随机相交有向图。
## 数学代写|随机图论代写Random Graph代考|Random Geometric Graphs
McDiarmid 和 Müller [588] 给出了当平均度数为日(日志n). 该论文还显示了两者之间关系的“令人惊讶”的相变H和哦. 该论文还将设置扩展到任意维度。Müller [618] 证明了团数和色数的两点集中，当nr2=欧(日志n).
Blackwell、Edmonson-Jones 和 Jordan [96] 研究了随机几何图 (RGG) 的邻接矩阵的谱特性。Rai [672] 研究了简单随机游走的转移矩阵的谱测度。Preciado 和 Jadbabaie [665] 在病毒传播的背景下研究了 RGG 的谱。
McColm [582] 在案例中展示了 RGG 的单调特性的尖锐阈值d=1即。由随机子区间的交集定义的图。对于所有人d≥1Goel、Rai 和 Krishnamachari [391]。
\mathscr{X}=\left{X_1, X_2, \ldots, X_n\right}\mathscr{X}=\left{X_1, X_2, \ldots, X_n\right}McColm [581] 在单位圆上进行了研究。该图有顶点集X当且仅当它们的角距离小于某个参数时，顶点才由边连接d. 他表明，除其他外，对于每个固定的d, 该模型中的 as FO 句子集是一个完整的非分类理论。McColm 的结果在 Godehardt 和 Jaworski 的更精确的论文 [388] 中得到了预期，其中案例d=1，即随机区间图的演化，进行了研究。
Diaz、Penrose、Petit 和 Serna [256] 研究了 RGG 族上几个布局问题的近似性。他们考虑的布局问题是带宽、最小线性排列、最小切割宽度、最小和切割、顶点分离和边缘平分。Diaz、Grandoni 和 Marchetti-Spaccemela [255] 为b- 随机几何图上的平衡切割问题：找到最小尺寸的边切割，其两侧至少包含bn每个顶点。
Bradonjić、Elsässer、Friedrich、Sauerwald 和 Stauffer [162] 研究了 RGG 的广播时间。他们研究了一个可能存在单个巨大组件的机制，并表明他们的广播算法只需要欧(n1/2/r+日志n)rounds 将信息从单个顶点传递到巨人的每个顶点。他们在路上表明巨人的直径是日(n1/2/r)whp Friedrich、Sauerwald 和 Stauffer [334] 将其扩展到更高的维度。
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 数学代写|随机图论代写Random Graph代考|Math572
statistics-lab™ 为您的留学生涯保驾护航 在代写随机图论Random Graph方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写随机图论Random Graph代写方面经验极为丰富，各种代写随机图论Random Graph相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 数学代写|随机图论代写Random Graph代考|Binomial Random Intersection Graphs
For $G(n, m, p)$ with $m=n^\alpha, \alpha$ constant, Rybarczyk and Stark [694] provided a condition, called strictly $\alpha$-balanced for the Poisson convergence for the number of induced copies of a fixed subgraph, thus complementing the results of Theorem $12.5$ and generalising Theorem 12.7. (Thresholds for small subgraphs in a related model of random intersection digraph are studied by Kurauskas [540]).
Rybarczyk [696] introduced a coupling method to find thresholds for many properties of the binomial random intersection graph. The method is used to establish sharp threshold functions for $k$-connectivity, the existence of a perfect matching and the existence of a Hamilton cycle.
Stark [725] determined the distribution of the degree of a typical vertex of $G(n, m, p)$, $m=n^\alpha$ and showed that it changes sharply between $\alpha<1, \alpha=1$ and $\alpha>1$.
Behrisch [70] studied the evolution of the order of the largest component in $G(n, m, p)$, $m=n^\alpha$ when $\alpha \neq 1$. He showed that when $\alpha>1$ the random graph $G(n, m, p)$ behaves like $\mathbb{G}_{n, p}$ in that a giant component of size order $n$ appears w.h.p. when the expected vertex degree exceeds one. This is not the case when $\alpha<1$. There is a jump in the order of size of the largest component, but not to one of linear size. Further study of the component structure of $G(n, m, p)$ for $\alpha=1$ is due to Lageras and Lindholm in [542].
Behrisch, Taraz and Ueckerdt [71] study the evolution of the chromatic number of a random intersection graph and showed that, in a certain range of parameters, these random graphs can be colored optimally with high probability using various greedy algorithms.
## 数学代写|随机图论代写Random Graph代考|Uniform Random Intersection Graphs
Uniform random intersection graphs differ from the binomial random intersection graph in the way a subset of the set $M$ is defined for each vertex of $V$. Now for every $k=1,2, \ldots, n$, each $S_k$ has fixed size $r$ and is randomly chosen from the set $M$. We use the notation $G(n, m, r)$ for an $r$-uniform random intersection graph. This version of a random intersection graph was introduced by Eschenauer and Gligor [296] and, independently, by Godehardt and Jaworski [389].
Bloznelis, Jaworski and Rybarczyk [106] determined the emergence of the giant component in $G(n, m, r)$ when $n(\log n)^2=o(m)$. A precise study of the phase transition of $G(n, m, r)$ is due to Rybarczyk [697]. She proved that if $c>0$ is a constant, $r=r(n) \geq 2$ and $r(r-1) n / m \approx c$, then if $c<1$ then w.h.p. the largest component of $G(n, m, r)$ is of size $O(\log n)$, while if $c>1$ w.h.p. there is a single giant component containing a constant fraction of all vertices, while the second largest component is of size $O(\log n)$.
The connectivity of $G(n, m, r)$ was studied by various authors, among them by Eschenauer and Gligor [296] followed by DiPietro, Mancini, Mei, Panconesi and Radhakrishnan [259],
Blackbourn and Gerke [95] and Yagan and Makowski [766]. Finally, Rybarczyk [697] determined the sharp threshold for this property. She proved that if $c>0$ is a constant, $\omega(n) \rightarrow \infty$ as $n \rightarrow \infty$ and $r^2 n / m=\log n+\omega(n)$, then similarly as in $\mathbb{G}_{n, p}$, the uniform random intersection graph $G(n, m, r)$ is disconnected w.h.p. if $\omega(n) \rightarrow \infty$, is connected w.h.p. if $\omega(n) \rightarrow \infty$, while the probability that $G(n, m, r)$ is connected tends to $e^{-e^{-c}}$ if $\omega(n) \rightarrow c$. The Hamiltonicity of $G(n, m, r)$ was studied in [109] and by Nicoletseas, Raptopoulos and Spirakis [636].
If in the uniform model we require $\left|S_i \cap S_j\right|>s$ to connect vertices $i$ and $j$ by an edge, then we denote this random intersection graph by $G_s(n, m, r)$. Bloznelis, Jaworski and Rybarczyk [106] studied phase transition in $G_s(n, m, r)$. Bloznelis and Łuczak [108] proved that w.h.p. for even $n$ the threshold for the property that $G_s(n, m, r)$ contains a perfect matching is the same as that for $G_s(n, m, r)$ being connected. Bloznelis and Rybarczyk [110] show that w.h.p. the edge density threshold for the property that each vertex of $G_s(n, m, r)$ has degree at least $k$ is the same as that for $G_s(n, m, r)$ being $k$-connected (for related results see [771]).
# 随机图论代写
## 数学代写|随机图论代写Random Graph代考|Binomial Random Intersection Graphs
Rybarczyk [696] 引入了一种耦合方法来寻找二项式随机交集图的许多属性的阈值。该方法用于建立尖锐 的阈值函数 $k$-连通性，完美匹配的存在性和哈密顿循环的存在。
Stark [725] 确定了典型顶点的度数分布 $G(n, m, p), m=n^\alpha$ 并表明它在之间急剧变化 $\alpha<1, \alpha=1$ 和 $\alpha>1$
Behrisch [70] 研究了最大分量阶数的演变 $G(n, m, p), m=n^\alpha$ 什么时候 $\alpha \neq 1$. 他表明，当 $\alpha>1$ 随机 图 $G(n, m, p)$ 表现得像 $G_{n, p}$ 那是一个巨大的尺寸订单组成部分 $n$ 当预期的顶点度数超过 1 时出现 whp。 情况并非如此 $\alpha<1$. 最大组件的大小顺序有一个跳跃，但不是线性大小之一。进一步研究的组件结构 $G(n, m, p)$ 为了 $\alpha=1$ 归功于 [542] 中的 Lageras 和 Lindholm。
Behrisch、Taraz 和 Ueckerdt [71] 研究了随机相交图的色数的演变，并表明，在一定的参数范围内，可 以使用各种贪心算法以高概率对这些随机图进行最佳着色。
## 数学代写|随机图论代写Random Graph代考|Uniform Random Intersection Graphs
Bloznelis、Jaworski 和 Rybarczyk [106] 确定了巨大成分在 $G(n, m, r)$ 什么时候 $n(\log n)^2=o(m)$. 相 $r(r-1) n / m \approx c$ ，那么如果 $c<1$ 然后whp最大的组成部分 $G(n, m, r)$ 是大小 $O(\log n)$, 而如果 $c>1$ $w h p$ 有一个巨大的组件包含所有顶点的常数部分，而第二大组件的大小 $O(\log n)$.
Mancini、Mei、Panconesi 和 Radhakrishnan [259]、
Blackbourn 和 Gerke [95] 以及 Yagan 和 Makowski [766]。最后，Rybarczyk [697] 确定了该属性的尖锐 阈值。她证明了如果 $c>0$ 是常数， $\omega(n) \rightarrow \infty$ 作为 $n \rightarrow \infty$ 和 $r^2 n / m=\log n+\omega(n)$, 然后类似于 $\mathbb{G}_{n, p}$, 均匀随机交集图 $G(n, m, r)$ 断开 whp 如果 $\omega(n) \rightarrow \infty$ ， 连接 $w h p$ 如果 $\omega(n) \rightarrow \infty$ ，而概率 $G(n, m, r)$ 连接趋向于 $e^{-e^{-c}}$ 如果 $\omega(n) \rightarrow c$. 的哈密顿性 $G(n, m, r)$ 在 [109] 以及 Nicoletseas、 Raptopoulos 和 Spirakis [636] 中进行了研究。
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
## 数学代写|随机图论代写Random Graph代考|MS-E1603
statistics-lab™ 为您的留学生涯保驾护航 在代写随机图论Random Graph方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写随机图论Random Graph代写方面经验极为丰富，各种代写随机图论Random Graph相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• Advanced Probability Theory 高等概率论
• Advanced Mathematical Statistics 高等数理统计学
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 数学代写|随机图论代写Random Graph代考|RANDOM GEOMETRIC GRAPHS
(c) If $B$ is within distance $100 r$ of two boundary edges of $D$ then $B$ contains no bad cells.
Proof. (a) There are less than $\ell_0^2<n$ such blocks. Furthermore, the probability that a fixed block contains $k_0$ or more bad cells is at most
$$\begin{array}{r} \left(\begin{array}{c} K^2 \ k_0 \end{array}\right)\left(\sum_{i=0}^{i_0}\left(\begin{array}{c} n \ i \end{array}\right)\left(\eta^2 r^2\right)^i\left(1-\eta^2 r^2\right)^{n-i}\right)^{k_0} \ \leq\left(\frac{K^2 e}{k_0}\right)^{k_0}\left(2\left(\frac{n e}{i_0}\right)^{i_0}\left(\eta^2 r^2\right)^{i_0} e^{-\eta^2 r^2\left(n-i_0\right)}\right)^{k_0} \end{array}$$
Here we have used Corollary $23.4$ to obtain the LHS of (12.12).
Now
\begin{aligned} &\left(\frac{n e}{i_0}\right)^{i_0}\left(\eta^2 r^2\right)^{i_0} e^{-\eta^2 r^2\left(n-i_0\right)} \ & \leq n^{O\left(\eta^3 \log (1 / \eta)-\eta^2(1+\varepsilon-o(1)) / \pi\right.} \leq n^{-\eta^2(1+\varepsilon / 2) / \pi}, \end{aligned}
for $\eta$ sufficiently small. So we can bound the RHS of (12.12) by
$$\left(\frac{2 K^2 e n^{-\eta^2(1+\varepsilon / 2) / \pi}}{(1-\varepsilon / 10) \pi / \eta^2}\right)^{(1-\varepsilon / 10) \pi / \eta^2} \leq n^{-1-\varepsilon / 3} .$$
Part (a) follows after inflating the RHS of (12.14) by $n$ to account for the number of choices of block.
(b) Replacing $k_0$ by $k_0 / 2$ replaces the LHS of (12.14) by
$$\left(\frac{4 K^2 e n^{-\eta^2(1+\varepsilon / 2) / \pi}}{(1-\varepsilon / 10) \pi / 2 \eta^2}\right)^{(1-\varepsilon / 10) \pi / 2 \eta^2} \leq n^{-1 / 2-\varepsilon / 6}$$
Observe now that the number of choices of block is $O\left(\ell_0\right)=o\left(n^{1 / 2}\right)$ and then Part (b) follows after inflating the RHS of (12.15) by $o\left(n^{1 / 2}\right)$ to account for the number of choices of block.
(c) Equation (12.13) bounds the probability that a single cell is bad. The number of cells in question in this case is $O(1)$ and (c) follows.
We now do a simple geometric computation in order to place a lower bound on the number of cells within a ball $B(X, r)$.
## 数学代写|随机图论代写Random Graph代考|Chromatic number
We look at the chromatic number of $G_{\mathscr{X}, r}$ in a limited range. Suppose that $n \pi r^2=$ $\frac{\log n}{\omega_r}$ where $\omega_r \rightarrow \infty, \omega_r=O(\log n)$. We are below the threshold for connectivity here. We will show that w.h.p.
$$\chi(G \mathscr{X}, r) \approx \Delta(G \mathscr{X}, r) \approx c l(G \mathscr{X}, r)$$
where will use $c l$ to denote the size of the largest clique. This is a special case of a result of McDiarmid [587].
We first bound the maximum degree.
Lemma 12.13.
$$\Delta\left(G_{\mathscr{X}, r}\right) \approx \frac{\log n}{\log \omega_r} \text { w.h.p. }$$
Proof. Let $Z_k$ denote the number of vertices of degree $k$ and let $Z_{\geq k}$ denote the number of vertices of degree at least $k$. Let $k_0=\frac{\log n}{\omega_d}$ where $\omega_d \rightarrow \infty$ and $\omega_d=$ $o\left(\omega_r\right)$. Then
$$\mathbb{E}\left(Z_{\geq k_0}\right) \leq n\left(\begin{array}{c} n \ k_0 \end{array}\right)\left(\pi r^2\right)^{k_0} \leq n\left(\frac{n e \omega_d \log n}{n \omega_r \log n}\right)^{\frac{\log n}{\omega_d}}=n\left(\frac{e \omega_d}{\omega_r}\right)^{\frac{\log n}{\Phi_d}} .$$
So,
$$\log \left(\mathbb{E}\left(Z_{\geq k_0}\right)\right) \leq \frac{\log n}{\omega_d}\left(\omega_d+1+\log \omega_d-\log \omega_r\right)$$
# 随机图论代写
## 数学代写|随机图论代写Random Graph代考|RANDOM GEOMETRIC GRAPHS
(c) 如果 $B$ 在距离之内 $100 r$ 的两个边界边缘 $D$ 然后 $B$ 不含坏细胞。
$$\left(K^2 k_0\right)\left(\sum_{i=0}^{i_0}(n i)\left(\eta^2 r^2\right)^i\left(1-\eta^2 r^2\right)^{n-i}\right)^{k_0} \leq\left(\frac{K^2 e}{k_0}\right)^{k_0}\left(2\left(\frac{n e}{i_0}\right)^{i_0}\left(\eta^2 r^2\right)^{i_0} e^{-\eta^2 r^2\left(n-i_0\right)}\right)^{k_0}$$
$$\left(\frac{n e}{i_0}\right)^{i_0}\left(\eta^2 r^2\right)^{i_0} e^{-\eta^2 r^2\left(n-i_0\right)} \quad \leq n^{O\left(\eta^3 \log (1 / \eta)-\eta^2(1+\varepsilon-o(1)) / \pi\right.} \leq n^{-\eta^2(1+\varepsilon / 2) / \pi},$$
$$\left(\frac{2 K^2 e n^{-\eta^2(1+\varepsilon / 2) / \pi}}{(1-\varepsilon / 10) \pi / \eta^2}\right)^{(1-\varepsilon / 10) \pi / \eta^2} \leq n^{-1-\varepsilon / 3}$$
(a) 部分是在将 (12.14) 的 RHS 膨胀后的 $n$ 考虑块的选择数量。
(b) 更换 $k_0$ 经过 $k_0 / 2$ 将 (12.14) 的 LHS 替换为
$$\left(\frac{4 K^2 e n^{-\eta^2(1+\varepsilon / 2) / \pi}}{(1-\varepsilon / 10) \pi / 2 \eta^2}\right)^{(1-\varepsilon / 10) \pi / 2 \eta^2} \leq n^{-1 / 2-\varepsilon / 6}$$
(c) 等式 (12.13) 界定了单个电池坏的概率。在这种情况下，有问题的细胞数量是 $O(1)(\mathrm{c})$ 如下。 我们现在做一个简单的几何计算，以便为球内的细胞数量设置一个下限 $B(X, r)$.
## 数学代写|随机图论代写Random Graph代考|Chromatic number
$$\chi(G \mathscr{X}, r) \approx \Delta(G \mathscr{X}, r) \approx \operatorname{cl}(G \mathscr{X}, r)$$
$$\Delta\left(G_{\mathscr{X}, r}\right) \approx \frac{\log n}{\log \omega_r} \text { w.h.p. }$$
$$\mathbb{E}\left(Z_{\geq k_0}\right) \leq n\left(n k_0\right)\left(\pi r^2\right)^{k_0} \leq n\left(\frac{n e \omega_d \log n}{n \omega_r \log n}\right)^{\frac{\log n}{\omega_d}}=n\left(\frac{e \omega_d}{\omega_r}\right)^{\frac{\log n}{\Phi_d}} .$$
$$\log \left(\mathbb{E}\left(Z_{\geq k_0}\right)\right) \leq \frac{\log n}{\omega_d}\left(\omega_d+1+\log \omega_d-\log \omega_r\right)$$
## 有限元方法代写
tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。
| 8,967 | 21,344 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.890625 | 4 |
CC-MAIN-2024-22
|
latest
|
en
| 0.617285 |
https://web2.0calc.com/questions/new-question_12
| 1,596,738,978,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-34/segments/1596439737019.4/warc/CC-MAIN-20200806180859-20200806210859-00196.warc.gz
| 496,882,870 | 7,433 |
+0
# new question
+2
702
3
+533
In rectangle ABCD, points E and F lie on segments AB and CD, respectively, such that AE=AB/3 and CF=CD/2. Segment BD intersects segment EF at P. What fraction of the area of rectangle ABCD lies in triangle EBP? Express your answer as a common fraction.
Jan 13, 2019
#1
+111437
+2
Not a particularly "geometric" solution....but see the following image
Let A = (0,4) B = (6, 4) C = (6,0) D = ( 0, 0) E = (2, 4) F = (3,0)
The segment connecting BD has the slope 2/3
And its equation is y = 2/3x
The segment connecting EF has a slope of [ 4 - 0 ] / [ 2 - 3 ] = -4
So....its equation is y = -4(x - 3) ⇒ y = -4x + 12
The x coordinate for P is
(2/3)x = -4x + 12
(2/3)x + 4x = 12
(14/3)x = 12
x = 36/14 = 18/7
And its y coordinate is y = (2/3)(18 7) = 36/21 = 12/7
So.....the base of EBP = 4
And its height is 4 - 12/7 = 16/7
So...its area = (1/2)(4)(16/7) = 32/7
And the area of the rectangle = 6 * 4 = 24
So....the fractional area of EBP to the area of the rectangle is
(32/7) / 24 = 32 / 168 = 4 / 21
Jan 13, 2019
edited by CPhill Jan 13, 2019
#2
+533
+1
good job cphill!
Jan 13, 2019
#3
+111437
+1
Thanks, asdf...!!!!
Here's a more general solution :
Angle EBP = Angle FDP
And angle BPE = angle DPF
So...by AA congruency.....triangle EBP is similar to triangle FDP
Note that EB is 2/3 of AB = 2/3 of CD
And FD is 1/2 of CD
So.... EB /FD = (2/3)CD / (1/2)CD = 4/3
So....EB is (4/3) FD
So....the height of EBP must be (4/3) the height of FDP
So.....there are 7 equal parts of AD.....and the height of EBP must be 4 of these
(4/21) area of ABCD
Jan 13, 2019
edited by CPhill Jan 13, 2019
edited by CPhill Jan 13, 2019
| 726 | 1,729 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625 | 4 |
CC-MAIN-2020-34
|
latest
|
en
| 0.796742 |
https://gmatclub.com/forum/all-people-prefer-colors-that-they-can-distinguish-easily-to-3799.html?fl=similar
| 1,495,520,392,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-22/segments/1495463607369.90/warc/CC-MAIN-20170523045144-20170523065144-00339.warc.gz
| 743,958,489 | 56,317 |
Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack
It is currently 22 May 2017, 23:19
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# All people prefer colors that they can distinguish easily to
Author Message
TAGS:
### Hide Tags
SVP
Joined: 03 Feb 2003
Posts: 1604
Followers: 9
Kudos [?]: 268 [0], given: 0
All people prefer colors that they can distinguish easily to [#permalink]
### Show Tags
22 Dec 2003, 08:22
00:00
Difficulty:
(N/A)
Question Stats:
80% (01:51) correct 20% (00:44) wrong based on 4 sessions
### HideShow timer Statistics
All people prefer colors that they can distinguish easily to colors that they have difficulty distinguishing. Infants can easily distinguish bright colors but, unlike adults, have difficulty distinguishing subtle shades. A brightly colored toy for infants sells better than the same toy in subtle shades at the same price.
Which one of the following conclusions is most strongly supported by the information in the passage?
(A) Infants prefer bright primary colors to bright secondary colors.
(B) Color is the most important factor in determining which toys an infant will prefer to play with.
(C) Individual infants do now have strong preferences for one particular bright color over other bright colors.
(D) The sales of toys of infants reflect the preferences of infants in at least one respect.
(E) Toy makers study infants to determine what colors the infants can distinguish easily.
If you have any questions
New!
Senior Manager
Joined: 05 May 2003
Posts: 424
Location: Aus
Followers: 2
Kudos [?]: 10 [0], given: 0
### Show Tags
22 Dec 2003, 08:27
A
SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA
Followers: 6
Kudos [?]: 101 [0], given: 0
### Show Tags
20 Jan 2004, 20:59
I vote for D.
Manager
Joined: 29 Aug 2003
Posts: 240
Location: MI
Followers: 1
Kudos [?]: 30 [0], given: 0
### Show Tags
20 Jan 2004, 21:02
Wonder where stolyar is....hasn't been around for a while
SVP
Joined: 16 Oct 2003
Posts: 1805
Followers: 5
Kudos [?]: 154 [0], given: 0
### Show Tags
23 Jan 2004, 21:30
D and B says almost the same thing but D is mild and so is GMATic.
Intern
Joined: 05 Aug 2014
Posts: 10
Followers: 0
Kudos [?]: 0 [0], given: 0
Re: All people prefer colors that they can distinguish easily to [#permalink]
### Show Tags
14 Apr 2016, 05:40
In my opinion D is the most logical one.
Re: All people prefer colors that they can distinguish easily to [#permalink] 14 Apr 2016, 05:40
Similar topics Replies Last post
Similar
Topics:
2 All people prefer colors that they can distinguish easily to colors th 5 24 Mar 2017, 08:36
4 Environmentalist: Many people prefer to live in 4 28 Jan 2017, 02:50
5 Advertisers have learned that people are more easily 2 09 Mar 2015, 02:18
21 All languages known to have exactly six basic color terms 4 05 Sep 2016, 11:57
18 All languages known to have exactly six basic color terms 12 22 Jul 2016, 02:07
Display posts from previous: Sort by
| 971 | 3,536 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.1875 | 4 |
CC-MAIN-2017-22
|
latest
|
en
| 0.900199 |
https://math.stackexchange.com/questions/4558740/xor-equivalent-for-sets
| 1,713,619,815,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-18/segments/1712296817650.14/warc/CC-MAIN-20240420122043-20240420152043-00305.warc.gz
| 346,919,238 | 36,084 |
# XOR Equivalent for Sets
there is a particular set I would like to write but not sure about the correct notation.
Let $$T$$ be the set the satisfies the following:
$$T = \{ (x,y) \, | \, (x \in S \land y \notin S) \lor (x \notin S \land y \in S) \}$$
So for example, if $$S$$ is the set of positive even integers, then for example the pair $$(3,4) \in T$$, and so is $$(4,3) \in T$$, but not $$(10,20) \notin T$$ as well as $$(7,5) \notin T$$, i.e., exactly only one element in the pair is in $$S$$ and the other does not. I can't think of a neat notation to denote this.
In logic, there's the XOR but is there such a thing for sets?
• en.wikipedia.org/wiki/Symmetric_difference
– YSB
Oct 22, 2022 at 4:27
• @YSB it's not exactly what I'm looking for. Since I'm referring to the property of being an element, eg. $x \in S$. Oct 22, 2022 at 4:32
• If an element is not in $S$ then what is it in? There is no universal set. Your examples suggest the integers but you should say that. Oct 22, 2022 at 4:55
• Symmetric difference is the set theory version of XOR. That answers the title of the question. But the question itself is something different: a set of ordered pairs. That is not an "XOR equivalent". Oct 22, 2022 at 6:06
It’s $$(S\times \bar S) \cup (\bar S \times S)$$, where $$\bar S$$ is the complement of $$S$$.
As some have mentioned, your set is related to the symmetric difference, and you could also write it as $$(U \times S) \bigtriangleup (S \times U),$$ where $$\bigtriangleup$$ is the symmetric difference operator and $$U$$ is your "universal set".
Honestly, you’ve almost already written what you want:
$$T = \{ (x,y) \mid x \in S \text{ xor } y \in S\}$$
and you could pick one of the half-dozen symbols for ‘xor’ that are out there, though I couldn’t say which one is most commonly used. $$\oplus$$ is the one that looks the most familiar to me, though more in a computer context than a math context.
I would pick whichever notation best expressed your intentions. It might look more math-y to use the first version, but the set builder notation is also fine, and might make your intentions more immediately clear to your readers.
(If you're going to be doing this for a lot of different sets, and want a brief name, I'll just throw out that this seems like some sort of "anti-diagonal" to me. I don't think that's an existing term, but I kinda like it.)
• Ah I see, but isn't the $\times$ notation for cross product? Oct 22, 2022 at 4:35
• Yes. The cross product is a set of ordered pairs, and it looks you you want to form a set out of certain ordered pairs, no? Oct 22, 2022 at 4:37
| 747 | 2,622 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 16, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.71875 | 4 |
CC-MAIN-2024-18
|
latest
|
en
| 0.918386 |
https://mathlake.com/Constructing-a-Copy-of-Line-Segment
| 1,719,223,098,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198865348.3/warc/CC-MAIN-20240624084108-20240624114108-00693.warc.gz
| 334,916,761 | 4,199 |
# Constructing a Copy of Line Segment
Geometry is an ancient science and an important branch of mathematics. Every day we come across many shapes in daily life, we call them geometrical shapes. The idea of construction came from these shapes. Consider a triangle, a rectangle, a square or any other shape. Every shape begins with a point and then extends to be a line or a line segment or a curve. Number and arrangement of these line segments decide the type of figure. To construct any geometrical figure we need to have an idea to construct a line segment. Let’s learn how to construct a replica of a given line segment.
## Construction in Geometry
In geometry, construction means drawing lines, shapes or angles accurately using pencils, rulers and compasses. This is the pure geometric construction i.e. no measurements, no numbers. So let’s learn pure geometric constructions of a line segments step by step.
A point is a small dot which is the starting point of a line segment. By definition, a line segment is a part of a line which connects two points within a line. Different numbers of line segments give us different figures and such figures may be either open figure or closed figures. A line segment is denoted by its two endpoints. In the figure, m is a line and AB is the line segment on the line m.
### Copy of Line Segments Construction Procedure
We can construct a line segment of a given length either by using a ruler or a combination of ruler and compass. Construction of a copy of a given line segment can also be achieved easily by using either of the above methods. However, the construction of line segments is not that accurate when we use a ruler only. Hence, a ruler-compass combination is always a better approach to construct a copy of line segments. The construction for the copy of the line segment has a place to begin the copy It is necessary to draw the straight line for the given measure in order to start the procedure for copying. Such a kind of line is called the reference line.
Steps for Construction:
Let’s construct a copy of a line segment XY using ruler and compass together.
Step 1: Let be a line segment of unknown length.
Step 2: Fix the compass’ pointer on the point X and pencil pointer on Y gives you the length of XY.
Step 3: Now draw a straight line using a ruler, let’s say line m.
Step 4: Mark a point A and fix the compass pointer (step 2)on A without changing the settings.
Step 5: Fixing compass on A draw an arc on, name the point at which pencil pointer cuts as B.
Now is a copy of i.e. length of AB will same as that of XY (if you didn’t change the compass settings).
### Proof for the Copy of Line Segments
Here, the compass is considered as a measuring tool for the purpose of copying the line segment. Since the original line segment and the copied line segment are of the same measure, the segments are considered as congruent.
For more about geometric constructions and to watch interactive videos on the topic, download BYJU’S – The Learning App. Also, take free tests to practice for exams.
| 655 | 3,079 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.78125 | 5 |
CC-MAIN-2024-26
|
latest
|
en
| 0.931325 |
https://herbie.uwplse.org/demo/72b4740c3e0e9aba8cdff6bc52246925067107d5.85fa6de4adb7ab096c565f35f5d2111e9ff97d70/graph.html
| 1,579,403,561,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2020-05/segments/1579250594101.10/warc/CC-MAIN-20200119010920-20200119034920-00107.warc.gz
| 478,686,999 | 2,315 |
Average Error: 0.0 → 0.0
Time: 13.7s
Precision: 64
$-\log \left(\frac{1}{x} - 1\right)$
$-\left(\log \left(\sqrt{\frac{1}{x} - 1}\right) + \log \left(\sqrt{\frac{1}{x} - 1}\right)\right)$
-\log \left(\frac{1}{x} - 1\right)
-\left(\log \left(\sqrt{\frac{1}{x} - 1}\right) + \log \left(\sqrt{\frac{1}{x} - 1}\right)\right)
double f(double x) {
double r1689458 = 1.0;
double r1689459 = x;
double r1689460 = r1689458 / r1689459;
double r1689461 = r1689460 - r1689458;
double r1689462 = log(r1689461);
double r1689463 = -r1689462;
return r1689463;
}
double f(double x) {
double r1689464 = 1.0;
double r1689465 = x;
double r1689466 = r1689464 / r1689465;
double r1689467 = r1689466 - r1689464;
double r1689468 = sqrt(r1689467);
double r1689469 = log(r1689468);
double r1689470 = r1689469 + r1689469;
double r1689471 = -r1689470;
return r1689471;
}
# Try it out
Results
In Out
Enter valid numbers for all inputs
# Derivation
1. Initial program 0.0
$-\log \left(\frac{1}{x} - 1\right)$
2. Using strategy rm
$\leadsto -\log \color{blue}{\left(\sqrt{\frac{1}{x} - 1} \cdot \sqrt{\frac{1}{x} - 1}\right)}$
4. Applied log-prod0.0
$\leadsto -\color{blue}{\left(\log \left(\sqrt{\frac{1}{x} - 1}\right) + \log \left(\sqrt{\frac{1}{x} - 1}\right)\right)}$
5. Final simplification0.0
$\leadsto -\left(\log \left(\sqrt{\frac{1}{x} - 1}\right) + \log \left(\sqrt{\frac{1}{x} - 1}\right)\right)$
# Reproduce
herbie shell --seed 1
(FPCore (x)
:name "-log(1/x-1)"
(- (log (- (/ 1.0 x) 1.0))))
| 617 | 1,487 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4 | 4 |
CC-MAIN-2020-05
|
latest
|
en
| 0.294665 |
https://homework.cpm.org/category/CCI_CT/textbook/pc3/chapter/1/lesson/1.1.3/problem/1-24
| 1,653,664,134,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-21/segments/1652662658761.95/warc/CC-MAIN-20220527142854-20220527172854-00721.warc.gz
| 367,599,879 | 14,918 |
### Home > PC3 > Chapter 1 > Lesson 1.1.3 > Problem1-24
1-24.
Let $g(x)=x^2-2x$. Evaluate:
Substitute the values into the function.
1. $g\left(-1\right)$
$g(-1)=(-1)^2-2(-1)$
1. $g\left(3\right)$
| 88 | 201 |
{"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.6875 | 4 |
CC-MAIN-2022-21
|
latest
|
en
| 0.253224 |
http://www.show-my-homework.com/2012/10/optics-relativity-gravity.html
| 1,534,733,373,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-34/segments/1534221215523.56/warc/CC-MAIN-20180820023141-20180820043141-00079.warc.gz
| 567,120,424 | 10,016 |
# Optics, Relativity, Gravity
1. The longest wavelength that can be observed in the fifth-order spectrum using a diffraction grating with 4000 slits per centimeter is: (A) 500 nm (B) 550 nm (C) 600 nm (D) 650 nm
2. Suppose two space ships approach each other at the speed of light, and each sends a beam of light to warn the other of its approach. From the frame of reference of either space ship, the light will reach the ships: (A) before they collide (B) at the instant they collide (C) after they collide (D) none of these
3.The acceleration of an asteroid due to the Earth’s gravity at a height above the Earth’s surface equal to the radius of the Earth is: (A) about 2.5 m/s2 (B) about 4.9 m/s2 (C) 9.8 m/s2 (D) the acceleration depends on how the meteor is moving.
1. The longest wavelength in the 5th (K) order spectrum is
$\lambda = distance/n/k =0.01/4000/5 =500 nm$
The correct answer is A) 500 nm
2. The composition of speeds u and v takes place as
$S =((u+v)/1+(u*v/C^2))$
hence there is no speed greater than c
The correct answer is B) at the instant they collide
3. The variation of g with height h is
$g(h)/g0 = 1/(1+h^2/R^2) = 1/(1+1)^2 =0.25$
$g(h) =0.25*g0 = 2.45 m/s^2$
the correct answer is A) about $2.5m/s^2$
| 390 | 1,245 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.640625 | 4 |
CC-MAIN-2018-34
|
latest
|
en
| 0.895113 |
http://familywoodworking.org/forums/showthread.php?29112-radius-question-math-guru-s
| 1,513,006,762,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2017-51/segments/1512948513611.22/warc/CC-MAIN-20171211144705-20171211164705-00115.warc.gz
| 88,782,140 | 14,546 |
1. ## radius question, math guru's
ok how do you folks determine your radius's for a preselected arch,,
example need a arch that is 13.375 long and 1.75 high.. i am sure there is a formula to do it out there but this old noggen has forgotten it and the string trick is getting old.. help also need the radius for a 15.375 arch as well
Why do math when someone else will do it for you? Sigh. I swear the internet will rot your brain (I to used to know how to do this.. at one time..).
Practically speaking I probably wouldn't measure the arc at all, but would just lay it out with a flexible stick pinned at the middle and edges and mark to the curve.
3. Do you know the arc angle? I'm not sure that table will solve the problem unless another variable is known, i.e. the arc angle. In Larry's example I think he's referring to the arc length rather than the chord length, but.......?
Last edited by Al Launier; 12-10-2012 at 11:22 PM.
4. Originally Posted by Al Launier
Do you know the arc angle? I'm not sure that table will solve the problem unless another variable is known, i.e. the arc angle. In Larry's example I think he's referring to the arc length rather than the chord length, but.......?
Yeah good point, a more complete calculator that tries to solve all of the cases is at: http://www.handymath.com/cgi-bin/arc18.cgi?submit=Entry
it will solve either the arc length or the chord length given one or the other and the height.
5. thanks guys after i saw the arc swing marks my dusty brain woke up will make me a example and hang it up for later dates as for laying the spring stick down aqnd tracing it ryan my spring stick doesnt flex equally along its length so that didnt work for me i tried that first.. so that stick can become kindling..
6. Originally Posted by larry merlau
thanks guys after i saw the arc swing marks my dusty brain woke up will make me a example and hang it up for later dates as for laying the spring stick down aqnd tracing it ryan my spring stick doesnt flex equally along its length so that didnt work for me i tried that first.. so that stick can become kindling..
Try the following. I have not tested it out.
H = height of arch
L = length of arch at the base, not arc length
7. Larry,
When I get home, I will check to see if I have a better geometric solution.
8. Or:
L=Arc Length = 13.375
A=Arc Angle = 12.47481
P=Pi=3.1416
L=(R*A*P)/180 or R=(180*L)/A*P
R=(180*13.375)/(61.43030*3.1416)=12.475
9. Originally Posted by Bill Satko
Larry,
When I get home, I will check to see if I have a better geometric solution.
bill, swinging a couple arch's and using the intersection is fine with me.. i got a outfeed table that works well for large arch's just needed to see the picture for me to wake up
10. Ryan, thanks for the handy tables. Didn't know they existed. I saved as a favorite place.
Any others that are as useful? I had to did into my old Machinist Handbook that I bought new in 1960 (16th Edition) to find the formula.
Al
Last edited by Al Launier; 12-10-2012 at 11:55 PM.
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•
| 821 | 3,191 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5 | 4 |
CC-MAIN-2017-51
|
latest
|
en
| 0.942057 |
https://ssconlineexam.com/onlinetest/ssc-cgl-tier-1/quantitative-aptitude/qa-test-86
| 1,643,209,745,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-05/segments/1642320304954.18/warc/CC-MAIN-20220126131707-20220126161707-00214.warc.gz
| 565,822,815 | 8,025 |
# ssc cgl tier 1 :: quantitative aptitude :: qa test 86
## Home ssc cgl tier 1 / quantitative aptitude Questions and Answers
1 .
The average of the terms of the given series (152, 150, 148, 146, 144 .......) is 126. Then, the total number of terms in this series will be
24
23
27
29
2 .
The area of a square field is 4624 $m^ 2$ and there is a path of 3.5 m width around its periphery, the area of the path is
998 $m^2$
976$m^2$
1001$m^2$
1009$m^2$
3 .
The lengths of two adjacent sides of a parallelogram 5 cm and 6 cm respectively. One of its diagonals is 7cm long, the area of the parallelogram is
27.34 $cm^2$
29.39$cm^2$
23.36$cm^2$
34.28$cm^2$
4 .
If the ratio of heights and radii of two cones are respectively 2 : 7 and 3 : 4, the ratio of their volumes is
56 : 9
76 : 9
9 : 56
22 : 56
5 .
A chord of a circle of radius 8 cm makes an angle of 90° at the centre. Area of the minor segment made by the chord is
18.28$cm^2$
19.96$cm^2$
17.24$cm^2$
13.42$cm^2$
6 .
What will be the volume of a prism whose base area is 72 $cm^ 2$ and height is 6 cm?
416$cm^2$
386$cm^2$
432$cm^2$
448$cm^2$
7 .
The average marks of a student in four subjects is 68. If the student obtains 78 marks in the fifth subject, then the new average is
65
70
68
75
8 .
A and B can do a work in 12 days, B and C can do the same work in 16 days while C and A can do it in 24 days. How long will they take to finish it together?
10$2 \over 4$ days
8$2 \over 4$ days
10$2 \over 3$ days
11$7 \over 3$ days
9 .
A is twice as efficient as B. Together they can do a work in 19 days. In how many days will B finish the work?
54 days
57 days
60 days
63 days
10 .Study the pie chart and answer the questions based on it.
Monthly expenditure on education is less than expenditure on rent by
Rs. 250
Rs. 200
Rs. 750
Rs. 500
| 617 | 1,800 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5625 | 4 |
CC-MAIN-2022-05
|
longest
|
en
| 0.877382 |
https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12A_Problems/Problem_23&direction=next&oldid=146154
| 1,620,372,244,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2021-21/segments/1620243988775.25/warc/CC-MAIN-20210507060253-20210507090253-00556.warc.gz
| 128,325,500 | 17,593 |
# 2021 AMC 12A Problems/Problem 23
## Problem
Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?
$\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}$
## Solution 1 (complementary counting)
We will use complementary counting. First, the frog can go left with probability $\frac14$. We observe symmetry, so our final answer will be multiplied by 4 for the 4 directions, and since $4\cdot \frac14=1$, we will ignore the leading probability.
From the left, she either goes left to another edge ($\frac14$) or back to the center ($\frac14$). Time for some casework.
$\textbf{Case 1:}$ She goes back to the center.
Now, she can go in any 4 directions, and then has 2 options from that edge. This gives $\frac12$. --End case 1
$\textbf{Case 2:}$ She goes to another edge (rightmost).
Subcase 1: She goes back to the left edge. She now has 2 places to go, giving $\frac12$
Subcase 2: She goes to the center. Now any move works.
$\frac14 \cdot \frac12 + \frac14 \cdot 1=\frac18 + \frac 14=\frac38$ for this case. --End case 2
She goes back to the center in Case 1 with probability $\frac14$, and to the right edge with probability $\frac14$
So, our answer is $\frac14 \cdot \frac12 + \frac14 \cdot \frac38=\frac14 (\frac12+\frac38)=\frac14 \cdot \frac78 = \frac7{32}$
But, don't forget complementary counting. So, we get $1-\frac7{32}=\frac{25}{32} \implies \boxed{D}$. ~ firebolt360
Video Solution for those who prefer: https://youtu.be/ude2rzO1cmk ~ firebolt360
## Solution 2 (direct counting and probability states)
We can draw a state diagram with three states: center, edge, and corner. Denote center by M, edge by E, and corner by C. There are a few ways Frieda can reach a corner in four or less moves: EC, EEC, EEEC, EMEC. Then, calculating the probabilities of each of these cases happening, we have $1\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot\tfrac{1}{4}\cdot\tfrac{1}{2}+1\cdot\tfrac{1}{4}\cdot1\cdot\tfrac{1}{2}=\tfrac{25}{32}$, so the answer is $\boxed{D}$. ~IceWolf10
## Solution 3 (Similar to Solution 2, but Finds the Numerator and Denominator Separately)
Denominator
There are $4^4=256$ ways to make 4 hops without restrictions.
Numerator (Casework)
Suppose Frieda makes 4 hops without stopping. We perform casework on which hop reaches a corner for the first time.
(1) Hop #2 (Hops #3 and #4 have no restrictions)
The 4 independent hops have 4, 2, 4, 4 options, respectively. So, this case has $4\cdot2\cdot4\cdot4=128$ ways.
(2) Hop #3 (Hope #4 has no restriction)
No matter which direction the first hop takes, the second hop must "wrap around".
The 4 independent hops have 4, 1, 2, 4 options, respectively. So, this case has $4\cdot1\cdot2\cdot4=32$ ways.
(3) Hop #4
Two sub-cases:
(3.1) The second hop "wraps around". It follows that the third hop also "wraps around".
The 4 independent hops have 4, 1, 1, 2 options, respectively. So, this sub-case has $4\cdot1\cdot1\cdot2=8$ ways.
(3.2) The second hop backs to the center.
The 4 independent hops have 4, 1, 4, 2 options, respectively. So, this sub-case has $4\cdot1\cdot4\cdot2=32$ ways.
Together, Case (3) has $8+32=40$ ways.
The numerator is $128+32+40=200.$
Probability
$\frac{200}{256}=\boxed{\textbf{(D)} ~\frac{25}{32}}.$
This problem is quite similar to 1995 AIME Problem 3: https://artofproblemsolving.com/wiki/index.php/1995_AIME_Problems/Problem_3
~MRENTHUSIASM
## Solution 4
Let $C_n$ be the probability that Frieda is on the central square after n moves, $E_n$ be the probability that Frieda is on one of the four squares on the middle of the edges after n moves, and $V_n$ (V for vertex) be the probability that Frieda is on a corner after n moves. The only way to reach the center is by moving in $1$ specific direction out of $4$ total directions from the middle of an edge, so $C_{n+1}=\frac{E_n}{4}$. The ways to reach the middle of an edge are by moving in any direction from the center or by moving in $1$ specific direction from the middle of an edge, so $E_{n+1}=C_n+\frac{E_n}{4}$. The ways to reach a corner are by simply staying there after reaching there in a previous move or by moving in $2$ specific directions from the middle of an edge, so $V_{n+1}=V_n+\frac{E_n}{2}$. Since Frieda always start from the center, $C_0=1$, $E_0=0$, and $V_0=0$. We use the previous formulas to work out $V_4$ and find it to be $\boxed{\textbf{(D)} ~\frac{25}{32}}$.
-SmileKat32
## Solution 5
Imagine an infinite grid of $2$ by $2$ squares such that there is a $2$ by $2$ square centered at $(3x, 3y)$ for all ordered pairs of integers $(x, y).$ It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at $(0, 0)$. (minus the teleportations) Since counting the complement set is easier, we'll count the number of $4$-step paths such that Frieda never reaches a corner point.
In other words, since the reachable corner points are $(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1),$ and $(\pm 2, \pm 2),$ Frieda can only travel along the collection of points included in $S$, where $S$ is all points on $x=0$ and $y=0$ such that $|y|<4$ and $|x|<4$, respectively, plus all points on the big square with side length $6$ centered at $(0, 0).$ We then can proceed with casework:
Case $1$: Frieda never reaches $(0, \pm 3)$ nor $(\pm 3, 0).$
When Frieda only moves horizontally or vertically for her four moves, she can do so in $2^4 - 4 = 12$ ways for each case . Thus, $12 \cdot 2$ total paths for the subcase of staying in one direction. (For instance, all length $4$ combinations of $F$ and $B$ except $FFFF$, $BBBB$, $FFFB$, and $BBBF$ for the horizontal direction.)
There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are $FBUD$, $FBUU$, $UDFB$, and $UDFF.$ Thus, a total of $4 \cdot 4 = 16$ ways for this subcase.
Total for Case $1$: $24 + 16 = 40$
Case $2$: Frieda reaches $(0, \pm 3)$ or $(\pm 3, 0)$.
Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of $4 \cdot 4 = 16$ paths for this case.
Our total number of paths never reaching coroners is thus $16+40=56,$ making for an answer of $$\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.$$
-fidgetboss_4000
## Solution 6 (Casework)
We take cases on the number of hops needed to reach a corner. For simplicity, denote $E$ as a move that takes Freida to an edge, $W$ as wrap-around move and $C$ as a corner move. Also, denote $O$ as a move that takes us to the center.
2 Hops
Then, Freida will have to $(E, C)$ as her set of moves. There are $4$ ways to move to an edge, and $2$ corners to move to, for a total of $4 \cdot 2 = 8$ cases here. Then, there are $4$ choices for each move, for a probability of $\frac{8}{4 \cdot 4} = \frac{1}{2}$.
3 Hops
In this case, Freida must wrap-around. There's only one possible combination, just $(E, W, C)$. There are $4$ ways to move to an edge, $1$ way to wrap-around (you must continue in the same direction) and $2$ corners, for a total of $4 \cdot 1 \cdot 2 = 8$ cases here. Then, there are $4$ choices for each move, for a probability of $\frac{8}{4 \cdot 4 \cdot 4} = \frac{1}{8}$.
4 Hops
Lastly, there are two cases we must consider here. The first case is $(E, O, E, C)$, and the second is $(E, W, W, C)$. For the first case, there are $4$ ways to move to an edge, $1$ way to return to the center, $4$ ways to move to an edge once again, and $2$ ways to move to a corner. Hence, there is a total of $4 \cdot 1 \cdot 4 \cdot 2 = 32$ cases here. Then, for the second case, there are $4$ ways to move to a corner, $1$ way to wrap-around, $1$ way to wrap-around again, and $2$ ways to move to a corner. This implies there are $4 \cdot 1 \cdot 1 \cdot 2 = 8$ cases here. Then, there is a total of $8+32 = 40$ cases, out of a total of $4^4 = 256$ cases, for a probability of $\frac{40}{256} = \frac{5}{32}$.
Then, the total probability that Freida ends up on a corner is $\frac{1}{2} + \frac{1}{8} + \frac{5}{32} = \frac{25}{32}$, corresponding to choice $\boxed{ (D) = \frac{25}{32}}$. ~rocketsri
~ pi_is_3.14
| 2,827 | 9,126 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 117, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.65625 | 5 |
CC-MAIN-2021-21
|
latest
|
en
| 0.862641 |
http://mathhelpforum.com/algebra/78517-need-some-help-please-cheers.html
| 1,529,330,298,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-26/segments/1529267860557.7/warc/CC-MAIN-20180618125242-20180618145242-00075.warc.gz
| 205,496,557 | 8,406 |
1. ## need some help please cheers
got this question:
Pressure P and V are related by c . V cubed = c, where c is a constant.
Determine the approximate % change in c when p is increased by 3% and v is decreased by 1.2%
( P = 1 ) x (V = 1 ) cubed = C (which is 100%)
Therefore (( P + 3%) x (1 - 1.2%)) Cubed = Cn
(1.03 x 0.988) Cubed = 1.053
then you would take Cn from C to get the aproximate % change.
is this right??
2. Hi
$\displaystyle P\:V^3 = c$
$\displaystyle P'\:V'^3 = c'$ where $\displaystyle P' = 1.03\:P$ and $\displaystyle V' = 0.988\:V$
$\displaystyle 1.03\:P\0.988\:V)^3 = c'$
$\displaystyle 1.03\:\:0.988^3\:P\:V^3 = c'$
$\displaystyle 0.9934\:c = c'$
$\displaystyle \frac cc' = 0.9934 = 1-0.66/100$
The approximate % change in c is -0.66%
| 295 | 770 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4 | 4 |
CC-MAIN-2018-26
|
latest
|
en
| 0.881427 |
https://charityschakras.com/2019/11/28/series-in-mathematics-an-in-depth-anaylsis-on-what-works-and-what-doesnt/
| 1,695,311,880,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2023-40/segments/1695233506028.36/warc/CC-MAIN-20230921141907-20230921171907-00546.warc.gz
| 191,010,498 | 11,283 |
## Series in Mathematics at a Glance
I will start with reading a practice question to show you exactly what to do. This post looks at some different kinds of games and the type of mathematical thinking they’re able to develop. In any event it’s very good for you, and your funds satchel! Even in case the obstacles are removed, the issue remains complex. Now now, you might be thinking that I’m making an awfully major deal about a circumstance where we’re speaking about pennies. There did not appear to be a huge issue with the terms.
The typical deviation is utilized to gauge the variability of a data set and it’s an essential value of the confidence calculation. It’s often written as Sn. Figure out the confidence interval.
This probably depends upon the way the subject is taught, though. Learning how to calculate ratios and proportions can allow you to solve many problems in actual life and in math class. It’s the zone of learning where the pupil will have the ability to connect insightfully to new knowledge due to the intrinsic relation between background wisdom and new inputs. In addition, we have that the amount of adults times the cost of an adult admission plus the quantity of children times the price of a youngster’s admission is equivalent to the overall amount collected.
## The Awful Side of Series in Mathematics
Differential calculus is the procedure of finding out the rate of change of a variable in comparison to a different variable. Publishing contracts typically don’t specify only one royalty rate.
## New Step by Step Roadmap for Series in Mathematics
This will unquestionably help make reading easier. It’s possible that you try many methods to turn the pages of eBook to enhance your reading experience. http://en.wikipedia.com/wiki/Envelope An amazing eBook reader needs to be set up. This book is incredible in the feeling it makes you think from the very start.
## Get the Scoop on Series in Mathematics Before You’re Too Late
The very first term of a geometric sequence might not be given. A couple are given below. ‘Now we’re ready to initiate the test. It is equal to 625. It is equal to 81.
Learning is always improved while the relevance of what’s being learned is appreciated. Today, more than every other era in history, there’s an overwhelming pre-occupation with beauty. This practice is called cross-validation in statistics. Some reputed educational brands provide recorded DVDs or CDs together with the study material that makes it straightforward for you to comprehend.
Then you visit the next counting number, 2, and set it in place for the i. They are extremely much like sets but the key distinction is that in a sequence, individual terms can happen repeatedly in a variety of positions. Try to use the mouse if you’re comfy sitting back. The manager may list the forms of work and break them in the comprehensive operations.
You may also use the AutoSum feature to rapidly total a string of values without needing to enter any of them manually in a formula. https://grademiners.com/ Standard is the very best, since it’s the least expensive and it typically arrives in only a day or two. You must only be on the Internet during the download procedure. Memory and clear functions for this normal calculator are given below. Utilize CE to clear out the latest entry. So consider it, then search for the answer inside this week’s Math Dude Video Extra!
Two consecutive Fibonacci numbers don’t have any typical factor, which means they are Co-prime or relatively prime to one another. This new method may be used in instances where the integrand has an essential singularity at and is a significant extension of the preceding method. If that is the exact same and seems to be consistent across the sequence, then it’s indeed a very simple arithmetic progression. When you become familiarized with these basic formulas, you might want to learn more about the way to create complex formulas and try a number of the many functions which are available in Excel.
## What You Don’t Know About Series in Mathematics
This book about math is very good for introducing geometry concepts in an enjoyable and accessible way. You may see the present math calculations in a more compact display that’s below the most important display of the calculator. Brave knights, a strong ruler, and a dilemma that should be solved using math.
## The Honest to Goodness Truth on Series in Mathematics
A critical part of the modeling procedure is the evaluation of whether a given mathematical model describes a system accurately. On the other hand, the method we just utilised to analyze this easy series circuit can be streamlined for much better understanding. It ought to be used ONLY in the event of All Rights Reserved. Throughout history, increasingly more accurate mathematical models are developed.
| 954 | 4,854 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.96875 | 4 |
CC-MAIN-2023-40
|
longest
|
en
| 0.925398 |
http://mathhelpforum.com/pre-calculus/149848-piecewise-defined-function-print.html
| 1,527,460,361,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-22/segments/1526794870470.67/warc/CC-MAIN-20180527205925-20180527225925-00577.warc.gz
| 171,300,182 | 3,043 |
# piecewise defined function
• Jul 1st 2010, 03:29 AM
reiward
piecewise defined function
Help me. Please check if I did the domain range and graph right. Thanks.
1.
http://img541.imageshack.us/img541/2280/math1.jpg
2.
http://img267.imageshack.us/img267/6237/math11.jpg
• Jul 1st 2010, 03:36 AM
Prove It
Quote:
Originally Posted by reiward
Help me. Please check if I did the domain range and graph right. Thanks.
1.
http://img541.imageshack.us/img541/2280/math1.jpg
2.
http://img267.imageshack.us/img267/6237/math11.jpg
1. Everything is correct except your graph.
2. The range is $\displaystyle \mathbf{R}\backslash \{6 \}$.
• Jul 1st 2010, 03:40 AM
reiward
Quote:
Originally Posted by Prove It
1. Everything is correct except your graph.
2. The range is $\displaystyle \mathbf{R}\backslash \{6 \}$.
1. Could you show me how to graph it. Thanks.
2. Oh, does that mean my graph is also wrong?
• Jul 1st 2010, 03:44 AM
Prove It
Quote:
Originally Posted by reiward
1. Could you show me how to graph it. Thanks.
2. Oh, does that mean my graph is also wrong?
1. The line should end at $\displaystyle x = 1$ with a QUADRATIC graph starting at $\displaystyle x = 1$.
2. The graph is almost right, you just need a hole at $\displaystyle (3, 6)$ on the line.
• Jul 1st 2010, 03:58 AM
reiward
Quote:
Originally Posted by Prove It
1. The line should end at $\displaystyle x = 1$ with a QUADRATIC graph starting at $\displaystyle x = 1$.
2. The graph is almost right, you just need a hole at $\displaystyle (3, 6)$ on the line.
1. Sorry to bother but can you draw it? I cant seem to picture it out. Im really not good at this sorry.
2. I get it. Ty! Oh. Is the range = [6] or not equal to 6?
• Jul 1st 2010, 06:12 AM
Prove It
Quote:
Originally Posted by reiward
1. Sorry to bother but can you draw it? I cant seem to picture it out. Im really not good at this sorry.
2. I get it. Ty! Oh. Is the range = [6] or not equal to 6?
$\displaystyle \mathbf{R}\backslash \{6\}$ means all the real numbers except 6.
For the graph, draw the graph of $\displaystyle y = x^2$. Then get rid of everything to the left of $\displaystyle x = 1$.
| 667 | 2,137 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.5625 | 4 |
CC-MAIN-2018-22
|
latest
|
en
| 0.916782 |
http://mathhelpforum.com/advanced-statistics/203053-exponetial-distribution-question.html
| 1,524,300,958,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-17/segments/1524125945082.84/warc/CC-MAIN-20180421071203-20180421091203-00400.warc.gz
| 202,451,886 | 11,058 |
# Thread: Exponetial Distribution Question.
1. ## Exponential Distribution Question.
Have been given a challenge question to try and work out for this week, I have gotten part way through it but Im not really sure how to proceed from where I am...
Questions reads: "Let X be an exponential random variable with rate parameter λ (lambda) > 0. Suppose it is known that X > k where k is a positive constant. Given (conditional on) this, what is the probability that X > k + x? Consequently, what is the distribution of X given X > k?"
So if I denote P() as a probability,
P(X > k+x) = 1 - P(X < k+x)
= 1 - F(k+x) where this is the CDF of X...
Not sure if I'm on the right track so any help would be much appreciated, Thanks!
2. ## Re: Exponential Distribution Question.
Originally Posted by PhoenixFC
Have been given a challenge question to try and work out for this week, I have gotten part way through it but Im not really sure how to proceed from where I am...
So if I denote P() as a probability,
P(X > k+x) = 1 - P(X < k+x)
= 1 - F(k+x) where this is the CDF of X...
Not sure if I'm on the right track so any help would be much appreciated, Thanks!
you are given that X>k, and you are told to find $\displaystyle P(X>k+x | X>k)$
now use the memoryless property of exponential distribution...
3. ## Re: Exponential Distribution Question.
Originally Posted by harish21
you are given that X>k, and you are told to find $\displaystyle P(X>k+x | X>k)$
now use the memoryless property of exponential distribution...
Okay thanks that helped a lot, didn't know about that memoryless property.
So in in the end I worked out
$\displaystyle P(X>k+x | X>k)$ = P(X>k+x)/P(X>K) = e^-(λx) = P(X > x)
So for the last part of the question how would I state the distribution?
"Consequently the distribution of X given X>k is exponential with rate λ and mean = 1/λ?"
Not sure if that's what my lecturer is meaning or not haha.
4. ## Re: Exponential Distribution Question.
Originally Posted by PhoenixFC
Okay thanks that helped a lot, didn't know about that memoryless property.
So in in the end I worked out
$\displaystyle P(X>k+x | X>k)$ = P(X>k+x)/P(X>K) = e^-(λx) = P(X > x)
So for the last part of the question how would I state the distribution?
"Consequently the distribution of X given X>k is exponential with rate λ and mean = 1/λ?" No. this is not exponential
Not sure if that's what my lecturer is meaning or not haha.
$\displaystyle P(X>k+x | X>k) = P(X>x) =1-P(X \leq x) = 1-F_X(x)=e^{-\lambda x}$
oh wait..the last part of your question says "what is the distribution of X given X > k?" which means find P(X=x|X>k)
5. ## Re: Exponential Distribution Question.
Originally Posted by harish21
$\displaystyle P(X>k+x | X>k) = P(X>x) =1-P(X \leq x) = 1-F_X(x)=e^{-\lambda x}$
oh wait..the last part of your question says "what is the distribution of X given X > k?" which means find P(X=x|X>k)
Are you sure...Similar questions in my textbook (without the complication of conditional probability) just state the distribution in words. Why would it not just be exponential.
Also how on earth would you find P(X=x|X>k)?
| 834 | 3,136 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.78125 | 4 |
CC-MAIN-2018-17
|
latest
|
en
| 0.932407 |
https://unitconversion.io/9737-lps-to-galpm
| 1,571,229,212,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-43/segments/1570986668569.22/warc/CC-MAIN-20191016113040-20191016140540-00529.warc.gz
| 732,433,371 | 9,894 |
#### Convert 9737 Liters per Second (l/s) to Gallons per Minute (gal/min)
This is our conversion tool for converting liters per second to gallons per minute.
To use the tool, simply enter a number in any of the inputs and the converted value will automatically appear in the opposite box.
l/s
### =
gal/min
##### How to convert Liters per Second (l/s) to Gallons per Minute (gal/min)
Converting Liters per Second (l/s) to Gallons per Minute (gal/min) is simple. Why is it simple? Because it only requires one basic operation: multiplication. The same is true for many types of unit conversion (there are some expections, such as temperature). To convert Liters per Second (l/s) to Gallons per Minute (gal/min), you just need to know that 1l/s is equal to gal/min. With that knowledge, you can solve any other similar conversion problem by multiplying the number of Liters per Second (l/s) by . For example, 2l/s multiplied by is equal to gal/min.
#### Best conversion unit for 9737 Liters per Second (l/s)
We define the "best" unit to convert a number as the unit that is the lowest without going lower than 1. For 9737 liters per second, the best unit to convert to is .
#### Fast Conversions
1 l/s = gal/min 5 l/s = gal/min 10 l/s = gal/min 15 l/s = gal/min 25 l/s = gal/min 100 l/s = gal/min 1000 l/s = gal/min 1 gal/min = l/s 5 gal/min = l/s 10 gal/min = l/s 15 gal/min = l/s 25 gal/min = l/s 100 gal/min = l/s 1000 gal/min = l/s
| 390 | 1,444 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.609375 | 4 |
CC-MAIN-2019-43
|
latest
|
en
| 0.812477 |
https://www.physicsforums.com/threads/drift-velocity-in-semiconductor.808866/
| 1,716,882,547,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2024-22/segments/1715971059078.15/warc/CC-MAIN-20240528061449-20240528091449-00878.warc.gz
| 807,458,679 | 19,017 |
# Drift velocity in semiconductor
• unscientific
In summary: Is this still true if the hole and electron have the same effective mass?Yes, the effective mass is irrelevant, because the energy of the hole is always greater than the energy of the electron.
unscientific
## Homework Statement
[/B]
(a) Explain the terms intrinsic, extrinsic, mobility and effective mass in semiconductors
(b) What is a hole and explain its mass and charge
(c) Why is mobility of holes often less than mobility of electrons? Find the number density of holes and electrons
(d) Find the mobility of the metal
(e) Find the drift velocities of electrons in the metal and germanium
Only major problem I have is part (e).
## The Attempt at a Solution
Part(a)[/B]
Intrinsic: No impurities, density of holes = density of electrons
Extrinsic: Impurities present, N-doping or P-doping
Mobility: Ease of movement of electrons and holes through semiconductor ##\mu = \frac{v}{E}##.
Effective mass: mass near the bottom of Conduction band or top of valence band
Part(b)
A hole is an absence of an electron. It has opposite charge, and opposite velocity to the electron, since overall charge must be conserved. It has the same effective mass of electrons, due to conservation of momentum.
Part(c)
Holes are surrounded by a sea of bounded electrons in the valence band whereas free electrons are not as exposed in the conduction band due to the absence of states. So it is easier for free electrons to move around inhibited compared to holes.
Using ##J = nev = \sigma E##, we have ##\frac{v}{E} = \frac{\sigma}{ne} = \frac{1}{ne\rho}##.
$$n_e = \frac{1}{e \rho \mu_e} = 6.94 \times 10^{19} m^{-3}$$
$$n_h = \frac{1}{e \rho \mu_h} = 3.47 \times 10^{19} m^{-3}$$
Part(d)
The number density for this metal is ##n = \frac{N}{a^3} = \frac{4}{a^3} = 8.57 \times 10^{28} m^{-3}##.
$$\mu_{metal} = 4.28 \times 10^{-3}$$
It seems that the mobility in this metal is about 100 times less than the mobility in germanium.
Part(e)
Since ##v_d = \mu E##, for a given mobility the drift velocity is dependent on electric field. But since the electric field is ##E = \frac{V}{l}##, if we do not know the length of the metal, how could we figure out the voltage or eletric field across it? Surely an infinitely long metal would dominate the voltage across it than the germanium.
$$v = \mu E = \mu \left(\frac{V}{l}\right)\left( \frac{\rho l}{A} \right) = \frac{\mu V \rho}{A}$$
Can't find the drift velocity without knowing it's voltage across it, which I need to know its length to figure out its resistance to figure out its voltage across it by the potential divider principle.
The mobility of electrons in the metal should be much better than the mobility in germanium. How did you calculate the value?
(e) You can calculate the current flow through the germanium (assuming the resistance of the metal wires is very small in comparison). That allows to find the drift voltage in the metal.
unscientific said:
Holes are surrounded by a sea of bounded electrons in the valence band whereas free electrons are not as exposed in the conduction band due to the absence of states.
Free electrons are surrounded by empty states ("bound holes"), free holes are surrounded by filled states. Where is the difference?
mfb said:
The mobility of electrons in the metal should be much better than the mobility in germanium. How did you calculate the value?
(e) You can calculate the current flow through the germanium (assuming the resistance of the metal wires is very small in comparison). That allows to find the drift voltage in the metal.
Free electrons are surrounded by empty states ("bound holes"), free holes are surrounded by filled states. Where is the difference?
$$\mu = \frac{1}{n e\rho} = \frac{1}{(8.57 \times 10^{28})(1.6 \times 10^{-19})(1.7 \times 10^{-8})} = 4.28 \times 10^{-3}$$So we assume that the potential difference across germanium is ##2V##. Using that, we find the current using ##I = \frac{V}{R}##. Assuming current in metal is same in germanium, using ##I = nevA## we can find the drift velocity in the metal.
Last edited:
mfb said:
Free electrons are surrounded by empty states ("bound holes"), free holes are surrounded by filled states. Where is the difference?
If the electron and hole have the same effective mass, why would electrons have higher mobility?
Ah, forget my comment about electron mobility, sorry.
unscientific said:
So we assume that the potential difference across germanium is ##2V##. Using that, we find the current using ##I = \frac{V}{R}##. Assuming current in metal is same in germanium, using ##I = nevA## we can find the drift velocity in the metal.
Right.
unscientific said:
If the electron and hole have the same effective mass, why would electrons have higher mobility?
Why do you expect the same effective mass?
mfb said:
Ah, forget my comment about electron mobility, sorry.
Right.
Why do you expect the same effective mass?
Consider for an electron near bottom of conduction band:
$$E' = E_0 + \alpha |k-k_{min}|^2 + \cdots$$
$$\alpha = \frac{1}{2} \frac{\partial^2 E}{\partial k^2} = \frac{\hbar^2}{2m^{*}}$$
Similarly for a hole, near the top of a valence band:
$$E' = E_0 - \alpha |k_{max}-k|^2 + \cdots$$
$$\alpha = -\frac{1}{2} \frac{\partial^2 E}{\partial k^2} = \frac{\hbar^2}{2m^{*}}$$
My book says that a hole is at the highest possible energy while the electron is at the lowest possible energy configuration, and that driving a hole away from the maximum is like "pushing a balloon under water". I suppose that is why the mobility of electrons is higher?
## 1. What is drift velocity in semiconductor?
Drift velocity in semiconductor refers to the average velocity of free electrons or holes in a semiconductor material under the influence of an external electric field.
## 2. How is drift velocity calculated?
Drift velocity can be calculated by dividing the applied electric field by the mobility of the charge carriers in the semiconductor.
## 3. What factors affect the drift velocity in semiconductor?
The drift velocity in semiconductor is affected by factors such as the magnitude of the applied electric field, the type and concentration of impurities in the material, and the temperature.
## 4. Why is drift velocity important in semiconductor devices?
Drift velocity is important in semiconductor devices because it determines the speed at which charge carriers move through the material, and therefore affects the overall performance and efficiency of the device.
## 5. How does drift velocity differ from diffusion velocity?
Drift velocity and diffusion velocity are two different mechanisms by which charge carriers move in a semiconductor. While drift velocity is caused by an external electric field, diffusion velocity is due to the concentration gradient of the charge carriers. Additionally, drift velocity is much larger than diffusion velocity in most cases.
Replies
1
Views
893
Replies
1
Views
1K
Replies
4
Views
1K
Replies
7
Views
1K
Replies
1
Views
2K
Replies
7
Views
10K
Replies
1
Views
1K
Replies
3
Views
2K
• High Energy, Nuclear, Particle Physics
Replies
3
Views
909
• Introductory Physics Homework Help
Replies
1
Views
1K
| 1,797 | 7,202 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.78125 | 4 |
CC-MAIN-2024-22
|
latest
|
en
| 0.861522 |
https://people.math.wisc.edu/~roch/mmids/opt-3-convexity.html
| 1,660,821,338,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2022-33/segments/1659882573193.35/warc/CC-MAIN-20220818094131-20220818124131-00634.warc.gz
| 415,786,451 | 50,696 |
# Optimality, convexity, and gradient descent¶
## 3 Convexity¶
Course: Math 535 - Mathematical Methods in Data Science (MMiDS)
Author: Sebastien Roch, Department of Mathematics, University of Wisconsin-Madison
Updated: Oct 10, 2020
Our optimality conditions have only concerned local minimizers. Indeed, in the absence of global structure, local information such as gradients and Hessians can only inform us about the immediate neighborhood of points. Here we consider convexity, under which local minimizers are also global minimizers.
### 3.1 Convex sets¶
Definition (Convex Set): A set $D \subseteq \mathbb{R}^d$ is convex if for all $\mathbf{x}, \mathbf{y} \in D$ and all $\alpha \in [0,1]$
$$(1-\alpha) \mathbf{x} + \alpha \mathbf{y} \in D.$$
$\lhd$
The following is a convex set:
(Source)
The following is not a convex set:
(Source)
Example: Any open ball in $\mathbb{R}^d$ is convex. Indeed, let $\delta > 0$ and $\mathbf{x}_0$. For any $\mathbf{x}, \mathbf{y} \in B_{\delta}(\mathbf{x}_0)$ and any $\alpha \in [0,1]$, we have
\begin{align*} \|[(1-\alpha) \mathbf{x} + \alpha \mathbf{y}] - \mathbf{x}_0\| &= \|(1-\alpha) (\mathbf{x} - \mathbf{x}_0) + \alpha (\mathbf{y} - \mathbf{x}_0)\|\\ &\leq \|(1-\alpha) (\mathbf{x} - \mathbf{x}_0)\| + \|\alpha (\mathbf{y} - \mathbf{x}_0)\|\\ &= (1-\alpha) \|\mathbf{x} - \mathbf{x}_0\| + \alpha \|\mathbf{y} - \mathbf{x}_0\|\\ &\leq (1-\alpha) \delta + \alpha \delta\\ &= \delta \end{align*}
where we used the triangle inequality on the second line. Hence we have established that $(1-\alpha) \mathbf{x} + \alpha \mathbf{y} \in B_{\delta}(\mathbf{x}_0)$. $\lhd$
Exercise: A convex combination of $\mathbf{z}_1, \ldots, \mathbf{z}_m \in \mathbb{R}^d$ is a linear combination of the form
$$\mathbf{w} = \sum_{i=1}^m \alpha_i \mathbf{z}_i$$
where $\alpha_i \geq 0$ for all $i$ and $\sum_{i=1}^m \alpha_i = 1$. Show that a set is convex if and only if it contains all convex combinations of its elements. [Hint: Use induction on $m$.] $\lhd$
Exercise: A set $C \subseteq \mathbb{R}^d$ is a cone if, for all $\mathbf{x} \in C$ and all $\alpha \geq 0$, $\alpha \mathbf{x} \in C$. A conic combination of $\mathbf{z}_1, \ldots, \mathbf{z}_m \in \mathbb{R}^d$ is a linear combination of the form
$$\mathbf{w} = \sum_{i=1}^m \alpha_i \mathbf{z}_i$$
where $\alpha_i \geq 0$ for all $i$. Show that a set $C$ is a convex cone if and only if it contains all conic combinations of its elements. $\lhd$
Exercise: Show that any linear subspace of $\mathbb{R}^d$ is convex. $\lhd$
Exercise: Show that the set
$$\mathbb{R}^d_+ = \{\mathbf{x} \in \mathbb{R}^d\,:\, \mathbf{x} \geq \mathbf{0}\}$$
is convex. $\lhd$
Example: Here is an important generalization of the previous exercise. Think of the space of $n \times n$ symmetric matrices as a linear subspace of $\mathbb{R}^{n^2}$
$$\mathbf{S}^n = \left\{ X \in \mathbb{R}^{n \times n}\,:\, X = X^T \right\}.$$
The dimension of $\mathbf{S}^n$ is $n (n-1)/2$, the number of free parameters under the symmetry assumption. Consider now the set of all positive semidefinite matrices in $\mathbf{S}^n$
$$\mathbf{S}_+^n = \left\{ X \in \mathbf{S}^n \,:\, X \succeq 0 \right\}.$$
Note that $\mathbf{S}_+^n$ is not the same as the set of symmetric matrices with nonnegative elements. Here is a counter-example:
In [1]:
# Julia version: 1.5.1
using LinearAlgebra
In [2]:
A = [0. 1. 0. 1.; 1. 0. 1. 0.; 0. 1. 0. 1.; 1. 0. 1. 0.]
Out[2]:
4×4 Array{Float64,2}:
0.0 1.0 0.0 1.0
1.0 0.0 1.0 0.0
0.0 1.0 0.0 1.0
1.0 0.0 1.0 0.0
In [3]:
F = eigen(A)
F.values
Out[3]:
4-element Array{Float64,1}:
-1.999999999999997
0.0
1.7763568394002505e-15
2.0
We claim that the set $\mathbf{S}_+^n$ is convex. Indeed let $X, Y \in \mathbf{S}_+^n$ and let $\alpha \in [0,1]$. Then by postive semidefiniteness of $X$ and $Y$, for any $\mathbf{v} \in \mathbb{R}^n$
$$\langle \mathbf{v}, [(1-\alpha) X + \alpha Y] \mathbf{v}\rangle = (1-\alpha) \langle \mathbf{v}, X \mathbf{v}\rangle + \alpha \langle \mathbf{v}, Y \mathbf{v}\rangle \geq 0.$$
This shows that $(1-\alpha) X + \alpha Y \succeq$ and hence that $\mathbf{S}_+^n$ is convex. $\lhd$
The following exercise explains the choice of $A$ in the previous example.
Exercise: A graph $G = (V, E)$ is bipartite if there is a bipartition of the vertices $V_1, V_2$ (i.e. $V_1 \cap V_2 = \emptyset$ and $V_1 \cup V_2 = V$) such that all edges are between $V_1$ and $V_2$, that is, for any edge $e = \{u, v\} \in E$ we have that $e \cap V_1 \neq \emptyset$ and $e \cap V_2 \neq \emptyset$. A graph is $\delta$-regular if all its vertices have degree $\delta$. Show that if $G$ is a $\delta$-regular, bipartite graph, then its adjacency matrix has an eigenvalue $-\delta$. [Hint: Try a vector that takes different values on $V_1$ and $V_2$.] $\lhd$
Exercise: Show that $\mathbf{S}_+^n$ is a convex cone. $\lhd$
A number of operations preserve convexity.
Lemma (Operations that Preserve Convexity): Let $S_1$ and $S_2$ be convex sets in $\mathbb{R}^d$. Let $\beta \in \mathbb{R}$ and $\mathbf{b} \in \mathbb{R}^d$. The following sets are also convex:
(a) $\beta S_1 = \{\beta \mathbf{x}\,:\, \mathbf{x} \in S_1\}$
(b) $S_1 + \mathbf{b} = \{\mathbf{x} + \mathbf{b}\,:\, \mathbf{x} \in S_1\}$
(c) $S_1 + S_2 = \{\mathbf{x}_1 + \mathbf{x}_2\,:\, \mathbf{x}_1 \in S_1 \text{ and } \mathbf{x}_2 \in S_2\}$
(d) $S_1 \times S_2 = \{(\mathbf{x}_1, \mathbf{x}_2) \,:\, \mathbf{x}_1 \in S_1 \text{ and } \mathbf{x}_2 \in S_2\}$
(e) $T = \{\mathbf{x}_1\,:\, (\mathbf{x}_1, \mathbf{x}_2) \in S_1 \times \mathbb{R}^d\}$
(f) $S_1 \cap S_2 = \{\mathbf{x}\,:\, \mathbf{x} \in S_1 \text{ and } \mathbf{x} \in S_2\}$
Proof: We only prove (f). The other statements are left as an exercise. Suppose $\mathbf{x}, \mathbf{y} \in S_1 \cap S_2$ and $\alpha \in [0,1]$. Then, by the convexity of $S_1$, $(1-\alpha) \mathbf{x} + \alpha \mathbf{y} \in S_1$ and, by the convexity of $S_2$, $(1-\alpha) \mathbf{x} + \alpha \mathbf{y} \in S_2$. Hence
$$(1-\alpha) \mathbf{x} + \alpha \mathbf{y} \in S_1 \cap S_2.$$
$\square$
Exercise: Prove (a)-(e) from the Operations that Preserve Convexity Lemma. $\lhd$
### 3.2 Convex functions¶
Our main interest is in convex functions.
Definition (Convex Function): A function $f : \mathbb{R}^d \to \mathbb{R}$ is convex if, for all $\mathbf{x}, \mathbf{y} \in \mathbb{R}^d$ and all $\alpha \in [0,1]$
$$f((1-\alpha) \mathbf{x} + \alpha \mathbf{y}) \leq (1-\alpha) f(\mathbf{x}) + \alpha f(\mathbf{y}).$$
More generally, a function $f : D \to \mathbb{R}$ over a convex domains $D \subseteq \mathbb{R}^d$ is convex if the definition above holds over all $\mathbf{x}, \mathbf{y} \in D$. $\lhd$
(Source)
Exercise: Let $f \,:\, \mathbb{R}^d \to \mathbb{R}$ be a function. The eipgraph of $f$ is the set
$$\mathbf{epi} f = \{(\mathbf{x}, y)\,:\, \mathbf{x} \in \mathbb{R}^d, y \geq f(\mathbf{x})\}.$$
Show that $f$ is convex if and only if $\mathbf{epi} f$ is a convex set. $\lhd$
Lemma (Affine Functions are Convex): Let $\mathbf{w} \in \mathbb{R}^d$ and $b \in \mathbb{R}$. The function $f(\mathbf{x}) = \mathbf{w}^T \mathbf{x} + b$ is convex.
Proof: For any $\mathbf{x}, \mathbf{y} \in \mathbb{R}^d$ and $\alpha \in [0,1]$,
$$f((1-\alpha) \mathbf{x} + \alpha \mathbf{y}) = \mathbf{w}^T [(1-\alpha) \mathbf{x} + \alpha \mathbf{y}] + b = (1-\alpha)[\mathbf{w}^T \mathbf{x} + b] + \alpha [\mathbf{w}^T \mathbf{y} + b]$$
which proves the claim. $\square$
Exercise: Let $f_1, \ldots, f_m : \mathbb{R}^d \to \mathbb{R}$ be convex functions and let $\beta_1, \ldots, \beta_m \geq 0$. Show that
$$f(\mathbf{x}) = \sum_{i=1}^m \beta_i f_i(\mathbf{x})$$
is convex. $\lhd$
Exercise: Let $f_1, f_2 : \mathbb{R}^d \to \mathbb{R}$ be convex functions. Show that the pointwise maximum function
$$f(\mathbf{x}) = \max\{f_1(\mathbf{x}), f_2(\mathbf{x})\}$$
is convex. [Hint: First show that $\max\{\alpha + \beta, \eta + \phi\} \leq \max\{\alpha, \eta\} + \max\{\beta, \phi\}$.]$\lhd$
Exercise: Prove the following composition theorem: if $f : \mathbb{R}^d \to \mathbb{R}$ is convex and $g : \mathbb{R} \to \mathbb{R}$ is convex and nondecreasing, then the composition $h = g \circ f$ is convex. $\lhd$
A common way to prove that a function is convex is to look at its Hessian. We start with a first-order condition that will prove useful.
Lemma (First-Order Convexity Condition): Let $f : \mathbb{R}^d \to \mathbb{R}$ be continuously differentiable. Then $f$ is convex if and only if for all $\mathbf{x}, \mathbf{y} \in \mathbb{R}^d$
$$f(\mathbf{y}) \geq f(\mathbf{x}) + \nabla f(\mathbf{x})^T (\mathbf{y}-\mathbf{x}).$$
The condition is illustrated below.
(Source)
Proof (First-Order Convexity Condition): Suppose first that $f(\mathbf{z}_2) \geq f(\mathbf{z}_1) + \nabla f(\mathbf{z}_1)^T (\mathbf{z}_2-\mathbf{z}_1)$ for all $\mathbf{z}_1, \mathbf{z}_2$. For any $\mathbf{x}, \mathbf{y}$ and $\alpha \in [0,1]$, let $\mathbf{w} = (1-\alpha) \mathbf{x} + \alpha \mathbf{y}$. Then taking $\mathbf{z}_1 = \mathbf{w}$ and $\mathbf{z}_2 = \mathbf{x}$ gives
$$f(\mathbf{x}) \geq f(\mathbf{w}) + \nabla f(\mathbf{w})^T (\mathbf{x}-\mathbf{w})$$
and taking $\mathbf{z}_1 = \mathbf{w}$ and $\mathbf{z}_2 = \mathbf{y}$ gives
$$f(\mathbf{y}) \geq f(\mathbf{w}) + \nabla f(\mathbf{w})^T (\mathbf{y}-\mathbf{w}).$$
Multiplying the first inequality by $(1-\alpha)$ and the second one by $\alpha$, and adding them up gives
$$(1-\alpha) f(\mathbf{x}) + \alpha f(\mathbf{y}) \geq f(\mathbf{w}) + \nabla f(\mathbf{w})^T ([(1-\alpha) \mathbf{x} + \alpha \mathbf{y}] - \mathbf{w}) = f(\mathbf{w})$$
proving convexity.
For the other direction, assume that $f$ is convex. For any $\mathbf{x}, \mathbf{y}$ and $\alpha \in (0,1)$, by the Multivariate Mean Value Theorem, for some $\xi \in (0,1)$ it holds that
$$f(\mathbf{w}) = f(\mathbf{x} + \alpha (\mathbf{y} - \mathbf{x})) = f(\mathbf{x}) + \alpha (\mathbf{y} - \mathbf{x})^T \nabla f (\mathbf{x} + \xi \alpha (\mathbf{y} - \mathbf{x}))$$
while convexity implies
$$f(\mathbf{w}) \leq (1-\alpha) f(\mathbf{x}) + \alpha f(\mathbf{y}).$$
Combining, rearranging and dividing by $\alpha$ gives
$$(\mathbf{y} - \mathbf{x})^T \nabla f (\mathbf{x} + \xi \alpha (\mathbf{y} - \mathbf{x})) \leq f(\mathbf{y}) - f(\mathbf{x}).$$
Taking $\alpha \to 0$ gives the claim. $\square$
Lemma (Second-Order Convexity Condition): Let $f : \mathbb{R}^d \to \mathbb{R}$ be twice continuously differentiable. Then $f$ is convex if and only if, for all $\mathbf{x} \in \mathbb{R}^d$, $\mathbf{H}_f(\mathbf{x})$ is positive semidefinite.
Proof: Suppose first that $\mathbf{H}_f(\mathbf{z}_1) \succeq 0$ for all $\mathbf{z}_1$. For any $\mathbf{x}, \mathbf{y}$, by the Multivariate Taylor's Theorem, there is $\xi \in (0,1)$ such that
\begin{align*} f(\mathbf{y}) &= f(\mathbf{x}) + \nabla f(\mathbf{x})^T (\mathbf{y}-\mathbf{x}) + (\mathbf{y}-\mathbf{x})^T \mathbf{H}_f(\mathbf{x} + \xi(\mathbf{y} - \mathbf{x})) \,(\mathbf{y}-\mathbf{x})\\ &\geq f(\mathbf{x}) + \nabla f(\mathbf{x})^T (\mathbf{y}-\mathbf{x}) \end{align*}
where we used the positive semidefiniteness of the Hessian. By the First-Order Convexity Condition, it implies that $f$ is convex.
For the other direction, assume that $f$ is convex. For any $\mathbf{x}, \mathbf{w}$ and $\alpha \in (0,1)$, by the Multivariate Taylor's Theorem again, for some $\xi_\alpha \in (0,1)$ it holds that
\begin{align*} f(\mathbf{x} + \alpha \mathbf{w}) = f(\mathbf{x}) + \alpha \mathbf{w}^T \nabla f (\mathbf{x}) + \alpha^2 \mathbf{w}^T \mathbf{H}_f(\mathbf{x} + \xi_\alpha \alpha \mathbf{w}) \,\mathbf{w} \end{align*}
while the First-Order Convexity Condition implies
$$f(\mathbf{x} + \alpha \mathbf{w}) \geq f(\mathbf{x}) + \alpha \mathbf{w}^T \nabla f (\mathbf{x}).$$
Combining, rearranging and dividing by $\alpha^2$ gives
$$\mathbf{w}^T \mathbf{H}_f(\mathbf{x} + \xi_\alpha \alpha \mathbf{w}) \,\mathbf{w} \geq 0.$$
Taking $\alpha \to 0$ shows that $\mathbf{w}^T \mathbf{H}_f(\mathbf{x}) \,\mathbf{w} \geq 0$. Since $\mathbf{w}$ is arbitrary, this implies that the Hessian is positive semidefinite at $\mathbf{x}$. This holds for any $\mathbf{x}$, which proves the claim. $\square$
Exercise: Show that $f(x) = e^{\beta x}$, where $x, \beta \in \mathbb{R}$, is convex. $\lhd$
$$f(\mathbf{x}) = \frac{1}{2} \mathbf{x}^T P \mathbf{x} + \mathbf{q}^T \mathbf{x} + r.$$
We showed previously that the Hessian is
$$\mathbf{H}_f(\mathbf{x}) = \frac{1}{2}[P + P^T].$$
So $f$ is convex if and only if the matrix $\frac{1}{2}[P + P^T]$ has only nonnegative eigenvalues. $\lhd$
### 3.3 Convexity and unconstrained optimization¶
A key property of convex functions is the following.
Theorem (Global Minimizers of Convex Functions): Let $f : \mathbb{R}^d \to \mathbb{R}$ be a convex function. Then any local minimizer of $f$ is also a global minimizer.
Proof: By contradiction, suppose $\mathbf{x}_0$ is a local minimizer, but not a global minimizer. Then there is $\mathbf{y}$ such that
$$f(\mathbf{y}) < f(\mathbf{x}_0).$$
By convexity, for any $\alpha \in (0,1)$
$$f(\mathbf{x}_0 + \alpha (\mathbf{y} - \mathbf{x}_0)) \leq (1-\alpha) f(\mathbf{x}_0) + \alpha f(\mathbf{y}) < f(\mathbf{x}_0).$$
But that implies that every open ball around $\mathbf{x}_0$ contains a point taking a smaller value than $f(\mathbf{x}_0)$, a contradiction. $\square$
In the continuously differentiable case, we get:
Theorem (First-Order Sufficient Condition for Convex Functions): Let $f : \mathbb{R}^d \to \mathbb{R}$ be a continuously differentiable, convex function. If $\nabla f(\mathbf{x}_0) = \mathbf{0}$ then $\mathbf{x}_0$ is a global minimizer.
Proof: Assume $\nabla f(\mathbf{x}_0) = \mathbf{0}$. By the First-Order Convexity Condition, for any $\mathbf{y}$
$$f(\mathbf{y}) - f(\mathbf{x}_0) \geq \nabla f(\mathbf{x}_0)^T (\mathbf{y} - \mathbf{x}_0) = 0.$$
That proves the claim. $\square$
$$f(\mathbf{x}) = \frac{1}{2} \mathbf{x}^T P \mathbf{x} + \mathbf{q}^T \mathbf{x} + r$$
where $P$ is symmetric and positive definite. The Hessian is then
$$\mathbf{H}_f(\mathbf{x}) = \frac{1}{2}[P + P^T] = P$$
for any $\mathbf{x}$. So $f$ is convex. Further the gradient is
$$\nabla f(\mathbf{x}) = P\mathbf{x} + \mathbf{q}$$
for all $\mathbf{x}$.
Any $\mathbf{x}$ satisfying
$$P\mathbf{x} + \mathbf{q} = \mathbf{0}$$
is a global minimizer. If $P = Q \Lambda Q^T$ is a spectral decomposition of $P$, where all diagonal entries of $\Lambda$ are stricly positive, then $P^{-1} = Q \Lambda^{-1} Q^T$ where the diagonal entries of $\Lambda^{-1}$ are the inverses of those of $\Lambda$. That can be seen by checking that
$$Q \Lambda Q^T Q \Lambda^{-1} Q^T = Q I_{d\times d} Q^T = I_{d \times d}.$$
So the following is a global minimizer
$$\mathbf{x}^* = - Q \Lambda^{-1} Q^T \mathbf{q}.$$
$\lhd$
| 5,679 | 14,738 |
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.0625 | 4 |
CC-MAIN-2022-33
|
latest
|
en
| 0.592865 |
https://oeis.org/A326468
| 1,575,999,467,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2019-51/segments/1575540528457.66/warc/CC-MAIN-20191210152154-20191210180154-00311.warc.gz
| 481,997,004 | 4,244 |
This site is supported by donations to The OEIS Foundation.
Please make a donation to keep the OEIS running. We are now in our 55th year. In the past year we added 12000 new sequences and reached 8000 citations (which often say "discovered thanks to the OEIS"). We need to raise money to hire someone to manage submissions, which would reduce the load on our editors and speed up editing. Other ways to donate
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A326468 Sum of the sixth largest parts of the partitions of n into 9 parts. 9
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 12, 17, 26, 37, 52, 71, 100, 133, 181, 239, 317, 410, 536, 682, 874, 1104, 1392, 1735, 2167, 2670, 3292, 4025, 4911, 5947, 7199, 8645, 10375, 12377, 14736, 17456, 20654, 24307, 28569, 33441, 39071, 45478, 52862 (list; graph; refs; listen; history; text; internal format)
OFFSET 0,12 LINKS FORMULA a(n) = Sum_{q=1..floor(n/9)} Sum_{p=q..floor((n-q)/8)} Sum_{o=p..floor((n-p-q)/7)} Sum_{m=o..floor((n-o-p-q)/6)} Sum_{l=m..floor((n-m-o-p-q)/5)} Sum_{k=l..floor((n-l-m-o-p-q)/4)} Sum_{j=k..floor((n-k-l-m-o-p-q)/3)} Sum_{i=j..floor((n-j-k-l-m-o-p-q)/2)} m. a(n) = A326464(n) - A326465(n) - A326466(n) - A326467(n) - A326469(n) - A326470(n) - A326471(n) - A326472(n) - A326473(n). MATHEMATICA Table[Sum[Sum[Sum[Sum[Sum[Sum[Sum[Sum[m, {i, j, Floor[(n - j - k - l - m - o - p - q)/2]}], {j, k, Floor[(n - k - l - m - o - p - q)/3]}], {k, l, Floor[(n - l - m - o - p - q)/4]}], {l, m, Floor[(n - m - o - p - q)/5]}], {m, o, Floor[(n - o - p - q)/6]}], {o, p, Floor[(n - p - q)/7]}], {p, q, Floor[(n - q)/8]}], {q, Floor[n/9]}], {n, 0, 50}] CROSSREFS Cf. A026815, A326464, A326465, A326466, A326467, A326469, A326470, A326471, A326472, A326473. Sequence in context: A060730 A308928 A308992 * A326593 A123569 A305651 Adjacent sequences: A326465 A326466 A326467 * A326469 A326470 A326471 KEYWORD nonn AUTHOR Wesley Ivan Hurt, Jul 10 2019 STATUS approved
Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.
Last modified December 10 12:30 EST 2019. Contains 329895 sequences. (Running on oeis4.)
| 891 | 2,287 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 3.578125 | 4 |
CC-MAIN-2019-51
|
latest
|
en
| 0.553839 |
https://bestgedclasses.org/mini-math-ged-test-2-10-questions/
| 1,531,867,464,000,000,000 |
text/html
|
crawl-data/CC-MAIN-2018-30/segments/1531676589932.22/warc/CC-MAIN-20180717222930-20180718002930-00277.warc.gz
| 603,404,425 | 11,624 |
# Math Test 10 Questions (set 2)
These questions are written by our Math guru and are similar to the GED Test. Each question has four or five answer choices. Choose the best answer for each question.
Question 1 of 10
1. Solve for x: x + 3 = 12
A.
B.
C.
D.
E.
Question 1 of 10
Question 2 of 10
2. Which is correct?
A.
B.
C.
D.
E.
Question 2 of 10
Question 3 of 10
3. Which is the decimal equivalent of one half?
A.
B.
C.
D.
E.
Question 3 of 10
Question 4 of 10
4. Find the missing factor: 8 x 3 = 6 x ?
A.
B.
C.
D.
E.
Question 4 of 10
Question 5 of 10
5. Solve: 36 / (9 – 3) x 2
A.
B.
C.
D.
E.
Question 5 of 10
Question 6 of 10
6. What is the least common multiple (LCM) among 3, 4, and 6?
A.
B.
C.
D.
E.
Question 6 of 10
Question 7 of 10
7. What is the least common multiple (LCM) among 5, 6, and 12?
A.
B.
C.
D.
E.
Question 7 of 10
Question 8 of 10
8. What is the vehicle's speed when traveling 30 miles in 30 minutes?
A.
B.
C.
D.
E.
Question 8 of 10
Question 9 of 10
9. Car 'A' travels 330 miles on 20 gallons of fuel. Van 'B' uses 12 gallons to go 180 miles.
Which vehicle is more efficient, and by how many more miles per gallon (mpg)?
A.
B.
C.
D.
E.
Question 9 of 10
Question 10 of 10
10. Sam's Bakery makes 3000 donuts in 6 hours, while Annie bakes 4800 donuts per 10-hr day.
Who's more productive, and by how many donuts per hour (dph)?
A.
B.
C.
D.
E.
Question 10 of 10
These practice tests are part of our GED online prep, click here to read how it works. Taking practice tests is very important because they indicate what your knowledge gaps are.
Disclaimer This practice test is not related to the Official GED Practice Test™ produced and distributed by the American Council on Education (ACE) and the GED Testing Service. ACE and GED Testing Service LLC have not approved, authorized, endorsed, been involved in the development of, or licensed the substantive content of this practice test.
| 624 | 1,944 |
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}
| 4.03125 | 4 |
CC-MAIN-2018-30
|
latest
|
en
| 0.901729 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.