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# Thread: Solution to y'' = -W*W*y - Bcos(W*t) ?
1. ## Solution to y'' = -W*W*y - Bcos(W*t) ?
Does anybody know the solution to
y'' = -W*W*y - B cos(W*t)
where W and B are constants and t is time.
y'' is second order derivative of y with respect to t
Thank you.
2. are you sure that it's not -(w^2)t not -(w^2)y
3. Originally Posted by cyal
Does anybody know the solution to
y'' = -W*W*y - B cos(W*t)
where W and B are constants and t is time.
y'' is second order derivative of y with respect to t
Thank you.
What servantes135 said made me got scared there might be a typo...there better not be! This took forever to type.
What we have here is a second order differential equation. I'm much too tired to go through any lengthly explanation, so I won't (unless you need me to). I will use the method of undermined coefficients, hopefully you know it.
$y'' = - W^2 y - B \cos (Wt)$
$\Rightarrow y'' + W^2 y = -B \cos (Wt)$
For the homogeneous solution, assume $y = e^{ \lambda t}$
$\Rightarrow \lambda ^2 + W^2 = 0$
$\Rightarrow \lambda = \frac { \pm \sqrt {-4W^2}}{2} = \pm W ~i$
$\Rightarrow y_h = c_1 \cos (Wt) + c_2 \sin (Wt)$
Now we choose a particular solution of the form:
$y_p = k_1 t \cos (Wt) + k_2 t \sin (Wt)$
$\Rightarrow y_p' = k_1 \cos (Wt) - Wk_1 t \sin (Wt) + k_2 \sin(Wt) + Wk_2 t \cos (Wt)$
$\Rightarrow y_p'' = -Wk_1 \sin (Wt) - Wk_1 \sin (Wt) - W^2 k_1 t \cos (Wt) +$ $W k_2 \cos (Wt) + Wk_2 \cos (Wt) - W^2k_2 t \sin (Wt)$
Plug the respective formulas for the particular solution into the original differential equation, we obtain:
$-Wk_1 \sin (Wt) - Wk_1 \sin (Wt) - W^2 k_1 t \cos (Wt) + W k_2 \cos (Wt) + Wk_2 \cos (Wt)$ $- W^2k_2 t \sin (Wt) + W^2 k_1 t \cos (Wt) + W^2 k_2 t \sin (Wt) = -B \cos (Wt)$
This simplifies to:
$-2Wk_1 \sin (Wt) + 2Wk_2 \cos (Wt) = -B \cos (Wt)$
By equating coefficients we get:
$-2Wk_1 = 0 \implies k_1 = 0$
$2Wk_2 = -B \implies k_2 = - \frac {B}{2W}$
Now our general solution is given by:
$y_g = y_h + y_p$
$\Rightarrow y_g = k_1 \cos (Wt) + k_2 \sin (Wt) - \frac {B}{2W} t \sin (Wt)$
I'm really tired, so you should really double check this
4. Originally Posted by Jhevon
Now our general solution is given by:
$y_g = y_h + y_p$
$\Rightarrow y_g = k_1 t \cos (Wt) + k_2 t \sin (Wt) - \frac {B}{2W} t \sin (Wt)$
I'm really tired, so you should really double check this
Well
$y(t)=- \frac {B}{2W} t \sin (Wt)$
is a particular integral, but you have added in the solution to the homogeneous equation incorrectly, the general solution is:
$y(t) = k_1 \cos (Wt) + k_2 \sin (Wt) - \frac {B}{2W} t \sin (Wt)$
RonL
5. thanks jevon after I posed I started thinking that it might be a homogenous derivative but I was also extreemly tired last night too. That makes so much more sence then what was runnung thought my head.
6. Originally Posted by CaptainBlack
Well
$y(t)=- \frac {B}{2W} t \sin (Wt)$
is a particular integral, but you have added in the solution to the homogeneous equation incorrectly, the general solution is:
$y(t) = k_1 \cos (Wt) + k_2 \sin (Wt) - \frac {B}{2W} t \sin (Wt)$
RonL
ah, yes, sorry. where did my t's come from. i'll correct it.
7. ## Thank you
Thank you everybody for providing the solution
8. ## On a related note..
y'' = - W ^ 2 * y - B cos( K * t )
where K is not equal to W
9. Originally Posted by cyal
y'' = - W ^ 2 * y - B cos( K * t )
where K is not equal to W
this problem is pretty much the same as the last one. in fact, this one is easier than the last one.
find the homogeneous solution as i did, but for the particular solution choose:
$y_p = k_1 \sin (Kt) + k_2 \cos (Kt)$
no product rule is required to differentiate those guys, just a little chain rule, so you'll have a much easier time with the computations
10. ## Thank you very much
Jhevon thank you very much for taking time to answer my question.
11. ## little confused by the solution.
I tried the solution for the 2nd question.
y'' = -w^2y + Bcos(kt) (where k != w)
which was
y = Asin(wt+d) + B/(w^2 - k^2) cos(kt)
I am confused about the 2nd part, seems like as K approaches W, the amplitude will be extreamly high.
If you think of the system as a Simple Oscillator (y'' = -w^2y) with a small driving force (Bcos(kt)), the fact that you get a huge oscialltion when the driving frequency K is close to natural frequency W seems odd. Especially when the amplitude is only growing linearly with time when K = W. Does somebody happen to know any valid explanation for this?
Thank you.
12. Originally Posted by cyal
I tried the solution for the 2nd question.
y'' = -w^2y + Bcos(kt) (where k != w)
which was
y = Asin(wt+d) + B/(w^2 - k^2) cos(kt)
the solution is:
$y = c_1 \cos (Wt) + c_2 \sin (Wt) - \frac {B}{W^2 - K^2} \cos (Kt)$
I am confused about the 2nd part, seems like as K approaches W, the amplitude will be extreamly high.
If you think of the system as a Simple Oscillator (y'' = -w^2y) with a small driving force (Bcos(kt)), the fact that you get a huge oscialltion when the driving frequency K is close to natural frequency W seems odd. Especially when the amplitude is only growing linearly with time when K = W. Does somebody happen to know any valid explanation for this?
Thank you.
No comment
13. Originally Posted by cyal
I tried the solution for the 2nd question.
y'' = -w^2y + Bcos(kt) (where k != w)
which was
y = Asin(wt+d) + B/(w^2 - k^2) cos(kt)
I am confused about the 2nd part, seems like as K approaches W, the amplitude will be extreamly high.
If you think of the system as a Simple Oscillator (y'' = -w^2y) with a small driving force (Bcos(kt)), the fact that you get a huge oscialltion when the driving frequency K is close to natural frequency W seems odd. Especially when the amplitude is only growing linearly with time when K = W. Does somebody happen to know any valid explanation for this?
Thank you.
Resonance.
When the driving function is at the natural frquency of the oscillator it
is pumping energy into the system and so the amplitude goes off to infinity.
Look at the solution to the first ODE where K=W, there is a t infront of
one of the sinusoids, so as t goes to infinity the amplitude goes to
infinity.
RonL
14. Originally Posted by CaptainBlack
Resonance.
When the driving function is at the natural frquency of the oscillator it
is pumping energy into the system and so the amplitude goes off to infinity.
Look at the solution to the first ODE where K=W, there is a t infront of
one of the sinusoids, so as t goes to infinity the amplitude goes to
infinity.
RonL
Right I understand that at Resonance the oscillator will keep gaining energy as the time increases.
But with the solution to the 2nd ODE, time does not need to be infinity, even with in a 1/10th of a second the system can gain humongous energy when K is close to W.
consider K = W + 10^-4
then W^2 - K^2 => -2*W*10^-4
which means the second term in 2nd ODE, becomes
-B/(2*W)*10^4 * cos(kt)
even if you take 't' close to 1 or 2 seconds, that is a huge gain in energy. Even more than what the system would have gained at Resonance within the first two seconds, which is contrary to reality.
Or may be as K approaches W, the solution transforms into the 1st ODE solution for Resonance.
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1. ## Integral
I was doing some practice excersises for some diff eqs and this integral came up...now I got it right, but it was messy, can someone here give a better solution
$\int\sqrt{\frac{1}{x^2}+\frac{1}{x^4}}dx$
Now since the domain is every x greater than zero I rewrote this as
$\int\frac{\sqrt{1+x^2}}{x^2}dx$
Then said, Let $x=\tan(\theta)$
which is not fun, not hard, but not fun.
Is there a cool not obvious (or obvious) trick for this integral?
2. Hi
Originally Posted by Mathstud28
I was doing some practice excersises for some diff eqs and this integral came up...now I got it right, but it was messy, can someone here give a better solution
$\int\sqrt{\frac{1}{x^2}+\frac{1}{x^4}}dx$
Now since the domain is every x greater than zero I rewrote this as
$\int\frac{\sqrt{1+x^2}}{x^2}dx$
Let $x=\sinh t \Longrightarrow \mathrm{d}x=\cosh t \,\mathrm{d}t$
$
\int\frac{\sqrt{1+x^2}}{x^2}\,\mathrm{d}x=\int\fra c{\sqrt{1+\sinh^2t}}{\sinh^2t}\cosh t\,\mathrm{d}t=\int\frac{\cosh^2 t}{\sinh^2 t}\,\mathrm{d}t=\int\frac{1}{\sinh^2 t}+1\,\mathrm{d}t$
hence
$\int\frac{\sqrt{1+x^2}}{x^2}\,\mathrm{d}x=-\frac{1}{\tanh t}+t+C=-\frac{\sqrt{1+x^2}}{x}+\mathrm{asinh} x+C'$
I don't know if it's funnier than your method.
3. Try a reciprocal substitution.
4. Originally Posted by Krizalid
Try a reciprocal substitution.
I am assuming you mean start from here
$\int\frac{\sqrt{1+x^2}}{x^2}dx$
Because that is what most look like when you suggest this method, alright well I will give it a go
Let $x=\frac{1}{u}$
So $dx=\frac{-1}{u^2}$
So we have
$-\int\frac{\frac{1}{u^2}\sqrt{1+\frac{1}{u^2}}}{\fr ac{1}{u^2}}=-\int\sqrt{1+\frac{1}{u^2}}du$
Is this the right start?
5. Yes, now do the remaining algebra and you'll get an easy integral.
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Enter the radius of each balloon, the total number of balloons, and the height of the arch into the calculator to determine the total length of the balloon arch; this calculator can also evaluate any of the variables given the others are known.
## Balloon Arch Formula
The following formula is used to calculate the total length of a balloon arch:
L = (2 * π * r * n) + (2 * h)
Variables:
• L is the total length of the balloon arch
• r is the radius of each balloon
• n is the total number of balloons
• h is the height of the arch
To calculate the total length of the balloon arch, multiply the radius of each balloon by 2π and by the total number of balloons. Add the product to twice the height of the arch.
## What is a Balloon Arch?
A balloon arch is a decorative feature often used at events such as weddings, parties, and corporate functions. It is typically made by inflating numerous balloons, usually of varying colors and sizes, and attaching them to a flexible frame or string in an arch shape. The balloons can be arranged in a uniform pattern or in a more random, organic style, depending on the desired aesthetic. The arch can be used as a focal point, such as a stage backdrop or entranceway, or to frame a specific area, like a photo booth or dessert table. Balloon arches can be further embellished with elements like flowers, foliage, or themed decorations to match the event’s decor.
## How to Calculate Balloon Arch?
The following steps outline how to calculate the Cost Recovery Ratio for a Balloon Arch:
1. First, determine the total cost of materials and labor for the Balloon Arch ($). 2. Next, determine the revenue generated from the Balloon Arch ($).
3. Next, gather the formula from above = CRR = Revenue / Cost.
4. Finally, calculate the Cost Recovery Ratio for the Balloon Arch.
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem :
Use the following variables as an example problem to test your knowledge for a Balloon Arch:
Total cost of materials and labor ($) = 150 Revenue generated from the Balloon Arch ($) = 250
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https://www.coursehero.com/tutors-problems/Managerial-Accounting/43120893-EMV-payoff-probabilities-labels-and-decision-Question-You/
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Question
# EMV, payoff, probabilities, labels, and decision Question: You...
EMV, payoff, probabilities, labels, and decision
Question:
You plan to start your own cafe after graduation. You have to decide to start with a small, medium, or large shop. A consultant's report indicates a .35 probability that demand will be low and a .65 probability that demand will be high.
If you start with a small shop and demand turns out to be low, the payoff will be \$96,000; if demand turns out to be high, you can either subcontract and realize the payoff of \$96,000 or expand for a payoff of \$108,000.
You can start with a medium-size shop as a hedge: If demand turns out to be low, its payoff is estimated at \$54,000; if demand turns out to be high, you could do nothing and realize a payoff of \$98,000, or expand and realize a payoff of \$120,000.
If you start with a large shop and demand is low, the payoff will be -\$45,000, whereas high demand will result in a payoff of \$150,000.
Solved by verified expert
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There are 366 different Starters of The Day, many to choose from. You will find in the left column below some starters on the topic of Enlargements. In the right column below are links to related online activities, videos and teacher resources.
A lesson starter does not have to be on the same topic as the main part of the lesson or the topic of the previous lesson. It is often very useful to revise or explore other concepts by using a starter based on a totally different area of Mathematics.
Main Page
### Enlargements Starters:
Big Bieber: If the dimensions of an object double, its volume increases by a factor of eight.
Middle of Centres: The blue point is exactly in the middle of two red points. What are their coordinates?
Star Wars Day: Estimate the size of an alien given the size of their hand. This could be an introduction to scale factors.
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Transum.org/go/?to=transform
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#### Year 6
Pupils should be taught to solve problems involving similar shapes where the scale factor is known or can be found more...
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### Notes:
When areas and volumes are enlarged the results are far from intuitive. Doubling the dimensions of a rectangle produces a similar shape with four times the area! Doubling the dimensions of a cuboid produces a similar shape with eight times the volume!
The activities provided are intended to give pupils experiences of dealing with enlargements so that they better understand the concept and are able to produce diagrams, make models and answer questions on this subject.
Once positive integer scale factors have been mastered the notion of fractional and negative scale factors await discovery!
### Enlargements Activities:
Transformations: Draw transformations online and have them instantly checked. Includes reflections, translations, rotations and enlargements.
Map Scales: Test your understanding of map scales expressed as ratios with this self marking quiz.
Similar Shapes: Questions about the scale factors of lengths, areas and volumes of similar shapes.
Blow Up: Click on all the points that could be the centre of enlargement of the shape if the image does not go off the grid.
### Enlargements Videos:
Transum's Enlargements Video
Enlargement by a scale factor: A Maths teacher goes through the idea of enlarging a shape from a centre of enlargement.
Enlargements: How to enlarge a shape by a positive scale factor and a centre of enlargement. Includes a scale factor less than 1 and a centre of enlargement inside the shape.
Scale Factors Video: The scale factor, area factor and volume factor of similar shapes are quite different.
Negative Enlargement: A video from MathsWatch about Enlargement by a Negative Scale Factor.
Links to other websites containing resources for Enlargements are provided for those logged into 'Transum Mathematics'. Subscribing also opens up the opportunity for you to add your own links to this panel. You can sign up using one of the buttons below:
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#### Blow Up
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### Teaching Notes:
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SECTION I (50 Marks)
Answer all questions in this section
1.
Use reciprocals, squares and square root tables to evaluate to four significant figures;-
4 marks
2.
Find the area of the quadrilateral.
4 marks
3.
Wekesa earns a basic salary of Ksh.10,000 per month plus commission of 4% on all sales above Ksh.50, 000. Find his income for a month in which he sold goods of value Ksh.90, 000.
3 marks
4.
Find all the integral values that satisfy the inequality 2x + 3 > 5x - 3 > -8.
4 marks
5.
The angles of elevation of the top of a cliff from two boats A and B on the same side of the cliff on the same horizontal level with the foot of the cliff are 30° and 50° respectively. If the distance from the foot of the cliff to boat B is 30m, find the distance between A and 8.
4 marks
6.
A proper fraction is such that its numerator and denominator have a difference of 2. One is added to the denominator and three subtracted from the numerator, the fraction becomes 2/3. Find the fraction.
3 marks
7.
Simplify the expression;-
3 marks
8.
Evaluate;-
3 marks
9.
In the figure below, find the size of angle DCB.
3 marks
10.
Find the centre and radius of the circle whose equation is 4x2-4y2 -104y + 6x = 0.
3 marks
11.
The mean mass of four students was 49.5kg. When the masses of another student and a teacher were included, the mean mass became 53kg. If the mass of the student was 16kg less than that of the teacher, what was the teacher's mass?
3 marks
12.
A translation maps the point P (5,-3) onto P' (2,-5); a) Determine the translation vector T. (1MK)
b) A point R' is the image of (-2,-3) under the same translation. Find the magnitude of P'R. (2mks)
3 marks
13.
A chord in the circle of radius 2.8cm subtends an angle of 35° at the circumference of the circle as shown below. Calculate the area of the shaded region.
3 marks
14.
Two similar right angled triangles have areas of 36cm2 and 9cm2 respectively. If the base of the smaller triangle is 2cm, find the height of the bigger triangle
3 marks
15.
The diagram below shows a histogram representing marks obtained in mathematics test;-
From the histogram, develop a frequency distribution table.
2 marks
16.
The interior angles of a hexagon are x+5, 4y-5, 2x+5, 3x, 3x - 20 and 2x. Find the value of x.
2 marks
SECTION II (50 Marks)
Answer any FIVE questions in this section
17.
Nekesa planned to use sh. 16,800 to buy some text books. When she went to the bookshop, she discovered that the price of the books had increased by sh.200 per book. She could now afford to buy two books less than she had desired to buy with the same amount of money,
a) Determine the number of books she had planned to buy. (6mks)
b) She later sold the books at sh.750 each. Find the percentage profit she made. (4mks)
10 marks
18.
School A is 250km on a bearing of 040° from school B. School C is 200km from A and due East of school B. A fourth school D is on a bearing of 220° from C and due South of school B.
a) Using a scale of 1cm represents 50km; draw an accurate scale drawing showing the positions of schools A, B, C and D. (5mks)
b) By measuring from the scale drawing in (a) above, determine;-
i) The distance BC (1mk)
ii) The distance BD (1mk)
iii) The distance CD (1mk)
iv) The bearing of A from C ( 2mks)
10 marks
19.
The table below shows marks scored by 50 Form four students in mathematics CAT.
a) Find the value of n in the table. (1mk)
b) State the limits of the modal class. (1mk)
c) Calculate the median. (3mks)
d) Calculate the mean of the data (5mks)
10 marks
20.
Easy Coach bus left Bungoma at 10.45am towards Nairobi moving at an average speed of 60km/hr. A shuttle left Bungoma at 1.15pm on the same road to Nairobi at an average speed of 100km/hr, the distance between Bungoma and Nairobi is 400km.
a) Determine the time of the day when the shuttle overtook the bus. (6mks)
b) Both vehicles moved non-stop to Nairobi. Find how long the shuttle waited for the bus to arrive. (4mks)
10 marks
21.
On the grid provided;-
a) Draw triangle XYZ whose coordinates are X (8, 6), Y (6, 10) and Z (10, 12) and its image is X'Y'Z' after undergoing a reflection in the line y=x. (4mks)
b) Triangle X'Y'Z' undergoes an enlargement centre (0, 0) scale factor 1/2 to form triangle X"Y"Z". Draw X"Y"Z". (3mks)
c) Triangle X"Y"Z" is rotated about (2, -1) through angle of negative ninty to X'"Y'"Z'". state the coordinates of X'"Y'"Z'". (3mks)
10 marks
22.
In the figure below, OABC is a trapezium. AB is parallel to OC and OC = 5AB D is a point on OC such that OD: DC=3:2.
a) Given that OA=p and AB=q, express in terms of p and q;-
i) OB (1mk)
ii) AB (2mks)
iii) CD (2mks)
b) Lines OB and AD intersect at point x such that AX=kAD and OX=rOB, where k and r are scalers. Determine the values of k and r.
10 marks
23.
The figure below represents a rectangle PQRS inscribed in a circle with centre O.
Given that the radius OP=17cm and PQ=16cm, calculate;-
a) The length QR of the rectangle. (3mks)
b) The angle ROS. (3mks)
c) The area of the shaded region. (4mks)
10 marks
24.
Wafula bought 3 brands of rice A, B and C. The cost prices were sh.25, sh.30 and sh.45 per kilogram respectively. He mixed the three in the ratio 5:2:1 respectively. After selling the mixture, he made a profit of 20%.
a) How much profit did he make per kilogram of the mixture? (4mks)
b) After one year, the cost price of each brand was increased by 12%.
i) For how much did he sell one kilogram of the mixture to make a profit of 20%. (3mks)
ii) What would have been his percentage profit if he sold one kilogram of the mixture at sh.40.25? (3mks)
10 marks
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Chap2 Section5
# Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)
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Section 2.5 Vertical Geometric Transformations 183 Version: Fall 2007 2.5 Exercises Pictured below is the graph of a function f . x 10 y 10 f The table that follows evaluates the function f in the plot at key values of x . Notice the horizontal format, where the first point in the table is the ordered pair ( 4 , 0) . x 4 3 0 2 5 6 f ( x ) 0 4 4 4 4 0 Use the graph and the table to complete each of following tasks for Exercises 1 - 10 . i. Set up a coordinate system on graph paper. Label and scale each axis, then copy and label the original graph of f onto your coordinate system. Re- member to draw all lines with a ruler. ii. Use the original table to help com- plete the table for the given function in the exercise. iii. Using a different colored pencil, plot the data from your completed table on the same coordinate system as the original graph of f . Use these points Copyrighted material. See: 1 to help complete the graph of the given function in the exercise, then label this graph with its equation given in the exercise. 1. y = 2 f ( x ) . x 4 3 0 2 5 6 y 2. y = (1 / 2) f ( x ) . x 4 3 0 2 5 6 y 3. y = f ( x ) . x 4 3 0 2 5 6 y 4. y = f ( x ) 2 . x 4 3 0 2 5 6 y 5. y = f ( x ) + 4 . x 4 3 0 2 5 6 y 6. y = 2 f ( x ) . x 4 3 0 2 5 6 y
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184 Chapter 2 Functions Version: Fall 2007 7. y = ( 1 / 2) f ( x ) . x 4 3 0 2 5 6 y 8. y = f ( x ) + 3 . x 4 3 0 2 5 6 y 9. y = f ( x ) 2 . x 4 3 0 2 5 6 y 10. y = ( 1 / 2) f ( x ) + 3 . x 4 3 0 2 5 6 y 11. Use your graphing calculator to draw the graph of y = x . Then, draw the graph of y = x . In your own words, explain what you learned from this exer- cise.
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222
## Balance Sheet \$type=three\$count=6\$author=hide\$comment=hide\$label=hide\$date=hide\$show=home\$s=0
In current ratio analysis, we will explain, how will current ratio affect to our business decisions. But before this, we will explain the simple meaning of current ratio.
Current ratio is a relationship between current assets and current liabilities. Just dividing current assets with current liabilities, we can calculate current ratio. Suppose, current assets are Rs. 250000 and current liabilities are Rs. 10000. Result will be 2.: 1. It means we have 2.5 times more current assets than current liabilities. In current assets, we can include cash, bank, prepaid expenses, marketable investment, sundry debtors, bill receivables and inventories.
In current liabilities, we can include outstanding expenses, sundry creditors, bank overdraft, short term taken loans and advances. Now, we analyze current ratio.
Main Point of Analysis
1. Rule of Thumb of Current Ratio 2:1
As a rule of thumb, current ratio should be 2:1. Proportion of current assets in current ratio should not be very high or very low than what has been given in the rule of thumb.
2. If current ratio is less than 2:1
If current ratio is less than 2:1, then it means we have no sufficient current assets to pay our current liabilities. This will not show a good short term position. So, our current ratio must be 2:1.
3. If current ratio is more than 2:1
If our current ratio is more than 2:1, it means, we are misusing our current assets. Suppose, our current ratio is 5:1. This is because of more Rs. 100000 cash in hand. But there is no investment use of this cash. Suppose, we have more 400000 what is necessary for minimum reserve but this bank balance will giving just saving interest earning if we have saving account in bank. But, if we invest this money in business, this money can give us high return. It may be 20% or more. Like this, if we have very high level of debtors. More debtors means more risk of bad debt. If we have high level of inventory, it will increase our cost of storing that more inventory. More inventory in our stock will also increase the risk of obsolescence loss.
4. Analysis of Exceptions of Rule of Thumb
Sometime, we may keep our current ratio more or less than rule of thumb that is 2:1. But before taking this decision, we have to analysis important factors:
a) Good Reputation
If company's reputation in market is very good. At that time, company can operate with lower current ratio because company can sell his product on credit at high level and creditor may give loan for long period.
b) Seasonal Influence
Some current assets may be seasonal. At that time, current ratio may be high at that season. For example, at Diwali, sweet shops buy high quantity of raw material because at that time, it is expected that sale will be very high.
5. Analysis of Limitation of Current Ratio
If you are creditor or any other interested party, you should not only analyze above points of current ratio, but you should also analyze the limitation of current ratio. One of limitation of window dressing. For changing the current ratio, company may change the value of current assets or current liabilities.
Following things may happen.
Related : Introduction of Ratio Analysis
Name
ltr
item
Accounting Education: Current Ratio Analysis
Current Ratio Analysis
http://3.bp.blogspot.com/-sKrON6STxfs/TpvCLd4wBAI/AAAAAAAAG6M/6_5kP1Nj82w/s640/limitation+of+ratio.PNG
http://3.bp.blogspot.com/-sKrON6STxfs/TpvCLd4wBAI/AAAAAAAAG6M/6_5kP1Nj82w/s72-c/limitation+of+ratio.PNG
Accounting Education
https://www.svtuition.org/2011/10/current-ratio-analysis.html
https://www.svtuition.org/
https://www.svtuition.org/
https://www.svtuition.org/2011/10/current-ratio-analysis.html
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## Calculus (3rd Edition)
$y'=\frac{y(y^{2}-x^{2})}{x(y^{2}-x^{2}-2xy^{2})}$
Differentiating $\frac{y}{x}+\frac{x}{y}=2y$ with respect to $x$, we have $\frac{y'\times x-y\times1}{x^{2}}+\frac{1\times y-x\times y'}{y^{2}}=2y'$ $\implies \frac{y^{2}(y'x-y)+(y-xy')x^{2}}{x^{2}y^{2}}=2y'$ $\implies y^{2}y'x-y^{3}+yx^{2}-x^{3}y'=2y'x^{2}y^{2}$ $\implies y^{2}y'x-x^{3}y'-2y'x^{2}y^{2}=y^{3}-yx^{2}$ $\implies y'(y^{2}x-x^{3}-2x^{2}y^{2})=y^{3}-yx^{2}$ $\implies y'= \frac{y^{3}-yx^{2}}{y^{2}x-x^{3}-2x^{2}y^{2}}=\frac{y(y^{2}-x^{2})}{x(y^{2}-x^{2}-2xy^{2})}$
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# 2 people in a canoe, switch places, canoe moves, find mass of one person
1. Oct 24, 2008
### musicfairy
Ricardo, mass 80 kg, and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 20 kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0 m apart and symmetrically located with respect to the canoe's center. Ricardo notices that the canoe moved 45 cm relative to a submerged log during the exchange and calculates Carmelita's mass, which she has not told him. What is it?
I don't really get those sort of problems, so I tried cutting and pasting into a problem similar to this one. This is what I tried to do:
xcom = xcom
80(0) + 20(1.5) + 3m = 80(3-x) + 20(1.5-x) + m(3-x)
After plugging 0.45 for x, I ended up with m = 500 kg.
What's the right way to do this problem?
Last edited: Oct 24, 2008
2. Oct 24, 2008
### Staff: Mentor
You seem to understand that the center of mass of the system does not change. What I'd do is find the change in the center of mass (mΔx) due to each object's movement--the three contributions must add to zero. Hint: Measure the movements with respect to the submerged log.
3. Oct 24, 2008
### musicfairy
So I tried to find the change in position of com. It didn't work to well. This is what I tried:
xcom = (80(0) + 20(1.5) + 3m) / (100 + 3m)
xcom = (80(0.45) + 20(1.5 + 0.45) + 3.45m) / (100 + 3m)
The I subtracted the first from the second and set it equal to 0.45.
I guess that was a bad attempt. What hints can you give me to get on the right track?
4. Oct 24, 2008
### Staff: Mentor
Just what I told you: Find mΔx for each object, measuring Δx with respect to the submerged log. Also, since Ricardo is heavier, if he moves to the right, which way does the canoe move?
5. Oct 24, 2008
### musicfairy
I had the same setup equation as before, except this time I changed the positions.
I assume that at Ricando is originally at x = 0, and the log is also there. When Ricardo and Carmelita switched places, the canoe moves left 0.45 m. So this time I made the equation:
(80(0.45) + 20(1.5-0.45) + 0.45m) / (100 + m) = (225 - 0.45m) / (100 + m)
The original equation is: xcom = (80(0) + 20(1.5) + 3m) / (100 + m)
I subtracted the original from the switched
(225 - 0.45m - 30 - 3m) / 100 = 0.45
195 - 3.45m = 45 + 0.45m
m = 38.46 kg
Is this what you mean by mΔx?
6. Oct 24, 2008
### Staff: Mentor
Here's what I mean by mΔx:
The canoe (m = 20) moves 45cm left with respect to the log (Δx = -0.45), so mΔx = - 20*0.45.
Ricardo moves 3 m to the right with respect to the canoe, so how far does he move with respect to the log? Use that logic to figure out mΔx for him.
Then do similar thinking for Carmelita.
Note: The change in center of mass of the system (canoe and two people) will equal ∑(mΔx)/(Total mass). But since we know that the change is zero, we can just say ∑(mΔx) = 0. (I hope that makes sense.)
7. Oct 24, 2008
### musicfairy
I assumed that Δx would be the original - 0.45. I can't think of it in any other way. Then I guess my first set up is correct since I figured what you mean by com doesn't change.
So would Ricardo's Δx be 3.0 - 0.45?
So is the equation
80(0) + 20(1.5) + 3m = 80(2.55) + 21 - 0.45m
m = 56.52 kg
?
8. Oct 24, 2008
### Staff: Mentor
Looks good.
Here's how I would have done it:
canoe: mΔx = -(20)*(0.45)
man: mΔx = (80)*(3-0.45)
woman: mΔx = (m)*(-3.45)
Add it up and set equal to zero. (Equivalent to what you did.)
9. Oct 24, 2008
### musicfairy
Thank you so much! I understand this type of problem a lot better now.
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# Differentiation using the quotient rule formula. How to differentiate an algebraic fraction.
The quotient rule can be used to differentiate an equation when you have one function being divided by another function. All you need to do is look for an algebraic fraction (a function of a x on the numerator and a function of x on the denominator).
So if your equation y takes the form y = f(x)/g(x),
then dy/dx = [f’(x)g(x) – f(x)g’(x)]/g2(x)
Let’s go over some examples that involve the quotient rule.
Example 1
Use the quotient rule to differentiate y = 7x3/cos3x
As you can see you have an algebraic fraction is it is clearly a quotient rule question.
First of all write down the functions f(x) and g(x):
f(x) = 7x3 and g(x) = cos3x
Next work out the derivatives of these two functions:
f’(x) = 21x2 and g’(x) = -3sin3x
All you need to do next is substitute these into the quotient rule formula:
dy/dx = [f’(x)g(x) – f(x)g’(x)]/g2(x)
dy/dx = [(21x2)(cos3x) – (7x3)(-3sin3x)]/(cos3x)2
All you need to do next is tidy up this derivative:
dy/dx = [21x2cos3x + 21x3sin3x]/cos23x]
Example 2
Use the quotient rule to differentiate y = e4x+1/sinx
Again you have an algebraic fraction so use the quotient rule:
First of all write down the functions f(x) and g(x):
f(x) = e4x+1 and g(x) = sinx
Next work out the derivatives of these two functions:
f’(x) = 4e4x+1 and g’(x) = cosx
All you need to do next is substitute these into the quotient rule formula:
dy/dx = [f’(x)g(x) – f(x)g’(x)]/g2(x)
dy/dx = [(4e4x+1)( sinx) – (e4x+1)(cosx)]/(sinx)2
All you need to do next is tidy up this derivative:
dy/dx = [4e4x+1.sinx - e4x+1.cosx]/sin2x]
You can also factorise the top and divide the numerator and dive the numerator and denominator by sinx:
dy/dx = [e4x+1(4-cotx)]/sinx
Let’s take a look at one last example that involves using the quotient rule.
Example 3
Use the quotient rule to differentiate y = ln(7x+4)/(5x2+3)
Again you have one function of x dividing another function x so the quotient rule can be applied to this question:
First of all write down the functions f(x) and g(x):
f(x) = ln(7x+4) and g(x) = 5x2+3
Next work out the derivatives of these two functions:
f’(x) = 7/(7x+4) and g’(x) = 10x
All you need to do next is substitute these into the quotient rule formula:
dy/dx = [f’(x)g(x) – f(x)g’(x)]/g2(x)
dy/dx = [(7/(7x+4)).(5x2+3) – (ln(7x+4).10x]/(5x+3)2
All you need to do next is tidy up this derivative:
dy/dx = [(35x2+21)/(7x+4)-10x.ln(7x+4)]/(5x+3)2
Summary
So as you can see from the last 3 examples the quotient rule can be applied when you have one function being divided by another function. If this is the case, then write down formulas for f(x), g(x), f’(x) and g’(x) and then substitute these into the formula dy/dx = [f’(x)g(x) – f(x)g’(x)]/g2(x). Be careful when simplifying your answer and use plenty of brackets to help you distinguish between the different functions.
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My Math Forum coordinate geometry problem solving
Geometry Geometry Math Forum
March 9th, 2013, 11:24 PM #1 Member Joined: Aug 2012 Posts: 71 Thanks: 1 coordinate geometry problem solving I have absolutely no idea how to solve this. Your help would be very appreciated. A(5,-3) and B(1,k) are two points. The perpendicular bisector of AB cuts the x-axis at -2. Find the two possible values for k. Many thanks.
March 10th, 2013, 05:04 AM #2 Newbie Joined: Feb 2013 From: Australia Posts: 2 Thanks: 0 Re: coordinate geometry problem solving Gradient of AB = (k - -3)/(1 - 5) = (k + 3)/-4 Gradient of any perpendicular to AB = negative reciprocal of this = 4/ (k + 3). Call this "m" Midpoint of AB = ((5 + 1)/2 , (-3 + k)/2) = (3, (k-3)/2) Call this (x1, y1) Equation of this line using y - y1 = m * (x - x1) is y - (k - 3)/2 = 4/(k - 3) * (x - 3) And we're told the point this crosses the x-axis is (-2, 0) so we substitute x = -2 and y = 0 0 - (k - 3)/2 = 4/(k - 3) * (-5) i.e. (3 - k)/2 = -20/(k - 3) i.e. (3 - k)/2 = 20/(3 - k) Cross-multiplying (3 - k)^2 = 40 or same as (k - 2)^2 = 40 taking square root k - 2 = Sqrt40 or - sqrt 40 k - 2 = 2 sqrt 10 or - 2 sqrt 10 adding 2, k = 2 + 2sqrt 10 or 2 - 2sqrt 10 k = 8.32 or - 4.32
March 10th, 2013, 07:36 AM #3 Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 Using a long-winded method makes careless errors easy to miss. For example, (3 - k)² was changed to (k - 2)² instead of (k - 3)². However, there was an earlier mistake. The equation (3 - k)/2 = -20/(k - 3) should have been (3 - k)/2 = -20/(k + 3); cross-multiplying then gives 3² - k² = -40, so k² = 49, i.e. k = ±7. Instead: as B and A are equidistant from (-2, 0), 3² + k² = 7² + 3², and so k = ±7.
March 10th, 2013, 09:33 AM #4 Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,612 Thanks: 845 Re: coordinate geometry problem solving Code: B(1,k) C(-2,0) A(5,-3) That solution had me confused too! Same idea as Skip's: BC = SQRT[k^2 + (-2 - 1)^2] = SQRT(k^2 + 9) AC = SQRT[(5 - (-2))^2 + (-3 - 0)^2] = SQRT(54)**** Since BC = AC: k^2 + 9 = 54**** k^2 = 49 k = +- 7 EDIT: **** should be 58
March 10th, 2013, 11:31 AM #5 Member Joined: Aug 2012 Posts: 71 Thanks: 1 Re: coordinate geometry problem solving Thanks everyone for the prompt replies.
March 10th, 2013, 01:47 PM #6 Math Team Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: coordinate geometry problem solving [color=#000000]Let C be the point (-2,0), then $M=\frac{A+B}{2}=\left(3,\frac{k-3}{2}\right)$. $\vec{BM}=\left<2,\frac{k-3}{2}-k\right=>=\left<2,-\frac{k+3}{2}\right=>=$ $\vec{CM}=\left<5,\frac{k-3}{2}\right=>=$ and $\vec{BM}\cdot \vec{CM}=0\Leftrightarrow 10-\frac{(k-3)(k+3)}{4}=0\Leftrightarrow k^2=49\Leftrightarrow k=\pm 7$.[/color]
March 10th, 2013, 02:22 PM #7 Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 In lengthening my explanation, Denis introduced a careless error in the arithmetic. I rest my face.
March 10th, 2013, 04:41 PM #8
Math Team
Joined: Oct 2011
Posts: 12,612
Thanks: 845
Re:
Quote:
Originally Posted by skipjack In lengthening my explanation, Denis introduced a careless error in the arithmetic. I rest my face.
Yikes...49 + ..1 finger,2 fingers... 9 fingers = 58.
Edited; then recited a full 50beads rosary as penance...
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Explore BrainMass
# Collision Involving a Photon
This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!
An object moving to the right at 0.8c is struck head-on by a photon of wavelength lambda, moving to the left. The object absorbs the photon (i.e. the photon disappears) and is afterward moving to the right at 0.6c.
(a) Determine the ratio of the object's mass after the collision to its mass before the collision.
(b) Does kinetic energy increase or decrease?
https://brainmass.com/physics/gamma/collision-involving-photon-173965
#### Solution Preview
It is convenient to use c = 1 units. We are going to evaluate the ratio of the masses and the ratio of the kinetic energies which are dimensionless, so they take the same values in any system of units. Velocities are dimensionless in c = 1 units. The numerical value of a velocity in c = 1 units is equal to the ratio of the velocity and the speed of light in any other system of units.
Before we start, let's review some relativity theory, in particular four-momentum vectors. The four-momentum of a particle is given (in c = 1 units) by (E, Px, Py,Pz), where E is the energy and Px, Py, Pz are the three components of the momentum. The inner product between two four-vectors is defined as:
(V0, V1, V2, V3) dot (W0, W1, W2, W3) = V0 W0 - (V1 W1 + V2 W2 + V3 W3)
This definition makes the inner product between four-vectors invariant w.r.t. Lorentz transformations. We can derive the mass energy relation using this inner product as follows. Suppose a particle has an energy E and momentum P in one frame. In that frame the four-momentum vector is (E,P). The inner product of this vector with itself (in the following this will be called "square of the four-vector") is E^2 - P^2.
In some other ...
#### Solution Summary
A detailed self-contained solution is given. Four-momentum algebra is first explained and then applied to the problem.
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# 3 Dimensional Geometry
Greedy Geoff sawed off a corner of a brick shaped block of Christmas cake, exposing a triangular fresh face of moist rich delicious gateau. He placed the tetrahedral fragment on the table, with its fresh face downwards. He mused through a port laden haze that it looked very stable, just like a mountain in fact, with its summit above a point inside its (not necessarily equilateral) triangular footprint $ABC$. He decided to decorate it, and took a UKMT pennant flying from a toothpick, and stuck it at the summit, with the flagpole perfectly vertical. Of course, the port was still at work and he is a bit clumsy, so he jammed the toothpick right through the cake, stabbing it into the tablecloth at a point $X$. Show that the circles $ABX$, $BCX$ and $CAX$ all have the same radius.
-
I thought gato meant cat. – Will Jagy Nov 22 '12 at 20:26
Damn it, contest problem: mathcomp.leeds.ac.uk – Will Jagy Nov 22 '12 at 21:50
Could you please make the title a little more descriptive? – Rahul Nov 23 '12 at 3:19
But of course the OP should say (if he knows) "contest problem from 2005". Then Will (and others, like me) won't get all excited about it's appropriateness for this forum. – GEdgar Nov 23 '12 at 16:56
Call summit vertex $S$.
Now, put the piece of cake back where it came from --use the toothpick to hold it in place-- and let's assume that vertex $A$ lies along the vertical edge of the cake. Take the knife and make a vertical cut along $SA$ that slices perpendicularly through $BC$ at, say, $F$. The plane of the cut, being perpendicular to a line in face $ABC$, is necessarily perpendicular to face $ABC$ itself; the cut must have split the toothpick. Therefore, the projection into face $ABC$ of segment $SA$ --together with the projection of $SF$-- forms an altitude of $\triangle ABC$. The same is true of similar cuts through $SB$ and $SC$: the foot of the toothpick lies on all three altitudes, so that $X$ must be the orthocenter of $\triangle ABC$.
From here, proof is fairly straightforward using the plane geometry of $\triangle ABC$: With $X$ the common point on altitudes dropped from $A$, $B$, and $C$, one can show (for instance) that $\angle BXC = 180^\circ-\angle BAC$. Consequently, by the Law of Sines in $\triangle ABC$ and $\triangle XBC$,
$$\text{circumdiameter of } \triangle ABC = \frac{|BC|}{\sin\angle BAC} = \frac{|BC|}{\sin\angle BXC} = \text{circumdiameter of } \triangle XBC$$
Thus, not only are the circumcircles of $\triangle XBC$, $\triangle XCA$, and $\triangle XAB$ congruent to each other, they're congruent to the circumcircle of $\triangle ABC$ itself.
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# Do all interior angles add to 360?
## Do all interior angles add to 360?
180° × 2 = 360° and so the interior angles of any quadrilateral must always equal 360°.
Do all triangle angles add up to 180?
The three interior angles of a triangle will always have a sum of 180°. A triangle cannot have an individual angle measure of 180°, because then the other two angles would not exist (180°+0°+0°).
What is a sum of interior angles of a triangle?
180°Triangle / Sum of interior angles
### What angles add up to 360?
The General Rule
Shape Sides Sum of Interior Angles
Pentagon 5 540°
Hexagon 6 720°
Heptagon (or Septagon) 7 900°
Do triangles add up to 180 or 360?
Angles in a triangle sum to 180° proof.
Do triangles add up to 90?
Right triangles are triangles in which one of the interior angles is 90 degrees, a right angle. Since the three interior angles of a triangle add up to 180 degrees, in a right triangle, since one angle is always 90 degrees, the other two must always add up to 90 degrees (they are complementary).
## Does supplementary mean 180?
Two angles are called supplementary when their measures add up to 180 degrees.
Do 4 angles add up to 360?
A quadrilateral is a polygon which has 4 vertices and 4 sides enclosing 4 angles and the sum of all the angles is 360°. When we draw a draw the diagonals to the quadrilateral, it forms two triangles. Both these triangles have an angle sum of 180°. Therefore, the total angle sum of the quadrilateral is 360°.
Which angles add up to 180 degrees?
### What is the sum of interior and exterior angles?
Exterior Angles The sum of an adjacent interior angle and exterior angle for any polygon is equal to 180 degrees since they form a linear pair. Also, the sum of exterior angles of a polygon is always equal to 360 degrees.
How the sum of triangle is 180?
The angle sum property of a triangle states that the sum of interior angles of a triangle is 180°. Because one outside angle equals the total of the other two angles in the triangle, the angles of a triangle add up to 180 degrees.
Are all triangles equal to 180?
It is no longer true that the sum of the angles of a triangle is always 180 degrees. Very small triangles will have angles summing to only a little more than 180 degrees (because, from the perspective of a very small triangle, the surface of a sphere is nearly flat).
## Are all triangles 180?
Keep on reading to find out! Why Does a Triangle Have 180 Degrees? A triangle’s angles add up to 180 degrees because one exterior angle is equal to the sum of the other two angles in the triangle.
Do all right triangles equal 180?
Which pair of angles add up to 180 degrees?
Supplementary Angles
Supplementary angles are two angles whose measures add up to 180° . The two angles of a linear pair , like ∠1 and ∠2 in the figure below, are always supplementary.
### What is the supplement of a 170 degree angle?
The supplement of 170° is the angle that when added to 170° forms a straight angle (180° ).
What are angles called that add up to 180 degrees?
Supplementary angles are two angles with a sum of 18 0 ∘ 180 ^\circ 180∘ . A common case is when they lie on the same side of a straight line.
What angles add up to 90?
In geometry, complementary angles are defined as two angles whose sum is 90 degrees. In other words, two angles that add up to 90 degrees are known as complementary angles. For example, 60° and 30°.
## Is the sum of exterior angles always 360?
The sum of exterior angles in a polygon is always equal to 360 degrees. Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon.
What is the sum of all exterior angles of a triangle?
360 degrees
The exterior angle of a triangle is defined as the angle formed between one of its sides and its adjacent extended side. The sum of exterior angles of a triangle is equal to 360 degrees.
What is the sum of the interior angles of a triangle?
The sum of interior angles of a triangle is always equal to 180°. This is because, if we join the three interior angles of the triangle, we will form a straight line. This property can be used to find the different measures of the three interior angles of the triangle.
### How to find the exterior angles of a triangle?
Every triangle has six exterior angles (two at each vertex are equal in measure).
• The exterior angles,taken one at each vertex,always sum up to 360°.
• An exterior angle is supplementary to its adjacent triangle interior angle.
• How do you calculate the exterior angles of a triangle?
sin (x) = opposite/hypothenuse
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Overview
# chain rule
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## Quick Reference
The following rule that gives the derivative of the composition of two functions: If h(x)=(fg)(x)=f(g(x) for all x, then h′(x)=f′(g(x)g′(x). For example, if h(x)=(x2+1)3, then h=fg, where f(x)=x3 and g(x)=x2+1. Then f′(x)=3x2 and g′(x)=2x. So h′(x)=3(x2+1)22x=6x(x2+1)2. Another notation can be used: if y=f(g(x), write y=f(u), where u=g(x). Then the chain rule says that dy/dx=(dy/du)(du/dx). As an example of the use of this notation, suppose that y=(sin x)2. Then y=u2, where u=sin x. So dy/du=2u and du/dx=cos x, and hence dy/dx=2 sin x cos x.
Subjects: Mathematics.
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Which steps can be taken to translate the phrase “the height of a tree is increased by seven inches” into an algebraic expression? Check all that apply.
Replace “the height of a tree” with a variable, x.
Replace “seven inches” with a variable, x.
Replace “increased by” with an addition symbol.
Replace “seven inches” with 7.
Replace “increased by” with a multiplication symbol.
Replace “the height of a tree” with 7.
Write the expression .
Write the expression .
2. Substitute and to determine if the two expressions are equivalent.
Which statements are true? Check all that apply.
The value of both expressions when is 33.
The value of both expressions when is 23.
The value of both expressions when is 41.
The value of both expressions when is 51.
The two expressions are equivalent.
The two expressions are not equivalent.
Juan used algebraic properties to make an equivalent expression. He listed the properties in the order that he used them.
Step Property Used 1 Associative 2 Distributive 3 Addition
Which steps match the properties used?
Fiona doubled the original amount in her savings account, s. Which expression represents her new balance and what is that new balance if ?
4.
5. Dan’s new appliance arrived in a cardboard box in the shape of cube. What is the surface area of the cardboard box?
Use the formula , where SA is the surface area of the cube and s is the length of each side.
6. Which expression is equivalent to ?
7. Which expression correctly represents “ten times the difference of a number and six”?
8. Misha’s family bought a large cattle farm. The number of acres of the cattle farm is the cube of the acres they used to own plus 10 acres. If a represents the number of acres the family used to own, which expression represents the number of acres they own now?
9. Jillian was asked to translate the expression into words. She wrote “a number increased by nine.” Which best describes the accuracy of Jillian’s answer?
10. Which scenario could be represented by the algebraic expression ?
1. 1 3 4 7
2. 1 4 5
3. 3
4. 3
5. 3
6. 2
7. 1
8. 3
9. 3
10. 1
11.
by Top Rated User (639k points)
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# Probability (Lec 1) (Probability and Distributions) Engineering Mathematics Notes | EduRev
## Engineering Mathematics : Probability (Lec 1) (Probability and Distributions) Engineering Mathematics Notes | EduRev
``` Page 1
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 1
MODULE 1
PROBABILITY
LECTURE 5
Topics
1.3.2 Bayes’ Theorem
1.3.2 Bayes’ Theorem
The following theorem provides a method for finding the probability of occurrence of an
event in a past trial based on information on occurrences in future trials.
1.3.2 Theorem 3.4 (Bayes’ Theorem)
Let ?? ,F,?? be a probability space and let ?? ?? : ?? ??? be a countable collection of
mutually exclusive and exhaustive events with ?? ?? ?? > 0,?? ??? . Then, for any event
?? ?F with ?? ?? > 0, we have
?? ?? ?? |?? =
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? , ?? ??? ·
Proof. We have, for ?? ??? ,
?? ?? ?? |?? =
?? ?? ?? n??
?? (?? )
=
?? ?? |?? ?? ?? ?? ??
?? (?? )
=
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? using Theorem of Total Probability . _
Remark 3.2
(i) Suppose that the occurrence of any one of the mutually exclusive and
exhaustive events ?? ?? ,?? ??? , causes the occurrence of an event ?? . Given that the
event ?? has occurred, Bayes’ theorem provides the conditional probability that
the event ?? is caused by occurrence of event ?? ?? ,?? ??? .
Page 2
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 1
MODULE 1
PROBABILITY
LECTURE 5
Topics
1.3.2 Bayes’ Theorem
1.3.2 Bayes’ Theorem
The following theorem provides a method for finding the probability of occurrence of an
event in a past trial based on information on occurrences in future trials.
1.3.2 Theorem 3.4 (Bayes’ Theorem)
Let ?? ,F,?? be a probability space and let ?? ?? : ?? ??? be a countable collection of
mutually exclusive and exhaustive events with ?? ?? ?? > 0,?? ??? . Then, for any event
?? ?F with ?? ?? > 0, we have
?? ?? ?? |?? =
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? , ?? ??? ·
Proof. We have, for ?? ??? ,
?? ?? ?? |?? =
?? ?? ?? n??
?? (?? )
=
?? ?? |?? ?? ?? ?? ??
?? (?? )
=
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? using Theorem of Total Probability . _
Remark 3.2
(i) Suppose that the occurrence of any one of the mutually exclusive and
exhaustive events ?? ?? ,?? ??? , causes the occurrence of an event ?? . Given that the
event ?? has occurred, Bayes’ theorem provides the conditional probability that
the event ?? is caused by occurrence of event ?? ?? ,?? ??? .
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 2
(ii) In Bayes’ theorem the probabilities ?? ?? ?? ,?? ??? , are referred to as prior
probabilities and the probabilities ?? ?? ?? |?? ,?? ??? , are referred to as posterior
probabilities. _
To see an application of Bayes’ theorem let us revisit Example 3.4.
Example 3.5
Urn ?? 1
contains 4 white and 6 black balls and urn ?? 2
contains 6 white and 4 black balls.
A fair die is cast and urn ?? 1
is selected if the upper face of die shows five or six dots.
Otherwise urn ?? 2
is selected. A ball is drawn at random from the selected urn.
(i) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 1
;
(ii) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 2
.
Solution. Define the events:
?? : drawn ball is white;
?? 1
: urn ?? 1
is selected
?? 2
: urn ?? 2
is selected
mutually exclusive & exhaustive events
(i) We have
?? ?? 1
|?? =
?? ?? |?? 1
?? ?? 1
?? ?? |?? 1
?? ?? 1
+?? ?? |?? 2
?? ?? 2
=
4
10
×
2
6
4
10
×
2
6
+
6
10
×
4
6
=
1
4
·
(ii) Since ?? 1
and ?? 2
are mutually exclusive and ?? ?? 1
??? 2
|?? =?? ?? |?? = 1,we
have
?? ?? 2
|?? = 1-?? ?? 1
|??
=
3
4
· _
In the above example
?? ?? 1
|?? =
1
4
<
1
3
=?? ?? 1
,
Page 3
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 1
MODULE 1
PROBABILITY
LECTURE 5
Topics
1.3.2 Bayes’ Theorem
1.3.2 Bayes’ Theorem
The following theorem provides a method for finding the probability of occurrence of an
event in a past trial based on information on occurrences in future trials.
1.3.2 Theorem 3.4 (Bayes’ Theorem)
Let ?? ,F,?? be a probability space and let ?? ?? : ?? ??? be a countable collection of
mutually exclusive and exhaustive events with ?? ?? ?? > 0,?? ??? . Then, for any event
?? ?F with ?? ?? > 0, we have
?? ?? ?? |?? =
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? , ?? ??? ·
Proof. We have, for ?? ??? ,
?? ?? ?? |?? =
?? ?? ?? n??
?? (?? )
=
?? ?? |?? ?? ?? ?? ??
?? (?? )
=
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? using Theorem of Total Probability . _
Remark 3.2
(i) Suppose that the occurrence of any one of the mutually exclusive and
exhaustive events ?? ?? ,?? ??? , causes the occurrence of an event ?? . Given that the
event ?? has occurred, Bayes’ theorem provides the conditional probability that
the event ?? is caused by occurrence of event ?? ?? ,?? ??? .
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 2
(ii) In Bayes’ theorem the probabilities ?? ?? ?? ,?? ??? , are referred to as prior
probabilities and the probabilities ?? ?? ?? |?? ,?? ??? , are referred to as posterior
probabilities. _
To see an application of Bayes’ theorem let us revisit Example 3.4.
Example 3.5
Urn ?? 1
contains 4 white and 6 black balls and urn ?? 2
contains 6 white and 4 black balls.
A fair die is cast and urn ?? 1
is selected if the upper face of die shows five or six dots.
Otherwise urn ?? 2
is selected. A ball is drawn at random from the selected urn.
(i) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 1
;
(ii) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 2
.
Solution. Define the events:
?? : drawn ball is white;
?? 1
: urn ?? 1
is selected
?? 2
: urn ?? 2
is selected
mutually exclusive & exhaustive events
(i) We have
?? ?? 1
|?? =
?? ?? |?? 1
?? ?? 1
?? ?? |?? 1
?? ?? 1
+?? ?? |?? 2
?? ?? 2
=
4
10
×
2
6
4
10
×
2
6
+
6
10
×
4
6
=
1
4
·
(ii) Since ?? 1
and ?? 2
are mutually exclusive and ?? ?? 1
??? 2
|?? =?? ?? |?? = 1,we
have
?? ?? 2
|?? = 1-?? ?? 1
|??
=
3
4
· _
In the above example
?? ?? 1
|?? =
1
4
<
1
3
=?? ?? 1
,
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 3
and ?? ?? 2
|?? =
3
4
>
2
3
=?? ?? 2
,
i.e.,
(i) the probability of occurrence of event ?? 1
decreases in the presence of the
information that the outcome will be an element of ?? ;
(ii) the probability of occurrence of event ?? 2
increases in the presence of information
that the outcome will be an element of ?? .
These phenomena are related to the concept of association defined in the sequel.
Note that
?? ?? 1
|?? <?? ?? 1
??? ?? 1
n?? <?? ?? 1
?? (?? ),
and
?? ?? 2
|?? >?? ?? 2
??? ?? 2
n?? >?? ?? 2
?? (?? ).
Definition 3.2
Let ?? ,F,?? be a probability space and let ?? and ?? be two events. Events ?? and ?? are
said to be
(i) negatively associated if ?? ?? n?? <?? ?? ?? ?? ;
(ii) positively associated if ?? ?? n?? >?? ?? ?? ?? ;
(iii) independent if ?? ?? n?? =?? ?? ?? ?? . _
Remark 3.3
(i) If ?? ?? = 0 then ?? ?? n?? = 0 =?? ?? ?? ?? ,??? ?F, i.e., if ?? ?? = 0 then
any event ?? ?F and ?? are independent;
(ii) If ?? ?? > 0 then ?? and ?? are independent if, and only if, ?? ?? |?? =?? (?? ),
i.e., if ?? ?? > 0, then events ?? and ?? are independent if, and only if, the
availability of the information that event ?? has occurred does not alter the
probability of occurrence of event ?? . _
Now we define the concept of independence for arbitrary collection of events.
Page 4
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 1
MODULE 1
PROBABILITY
LECTURE 5
Topics
1.3.2 Bayes’ Theorem
1.3.2 Bayes’ Theorem
The following theorem provides a method for finding the probability of occurrence of an
event in a past trial based on information on occurrences in future trials.
1.3.2 Theorem 3.4 (Bayes’ Theorem)
Let ?? ,F,?? be a probability space and let ?? ?? : ?? ??? be a countable collection of
mutually exclusive and exhaustive events with ?? ?? ?? > 0,?? ??? . Then, for any event
?? ?F with ?? ?? > 0, we have
?? ?? ?? |?? =
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? , ?? ??? ·
Proof. We have, for ?? ??? ,
?? ?? ?? |?? =
?? ?? ?? n??
?? (?? )
=
?? ?? |?? ?? ?? ?? ??
?? (?? )
=
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? using Theorem of Total Probability . _
Remark 3.2
(i) Suppose that the occurrence of any one of the mutually exclusive and
exhaustive events ?? ?? ,?? ??? , causes the occurrence of an event ?? . Given that the
event ?? has occurred, Bayes’ theorem provides the conditional probability that
the event ?? is caused by occurrence of event ?? ?? ,?? ??? .
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 2
(ii) In Bayes’ theorem the probabilities ?? ?? ?? ,?? ??? , are referred to as prior
probabilities and the probabilities ?? ?? ?? |?? ,?? ??? , are referred to as posterior
probabilities. _
To see an application of Bayes’ theorem let us revisit Example 3.4.
Example 3.5
Urn ?? 1
contains 4 white and 6 black balls and urn ?? 2
contains 6 white and 4 black balls.
A fair die is cast and urn ?? 1
is selected if the upper face of die shows five or six dots.
Otherwise urn ?? 2
is selected. A ball is drawn at random from the selected urn.
(i) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 1
;
(ii) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 2
.
Solution. Define the events:
?? : drawn ball is white;
?? 1
: urn ?? 1
is selected
?? 2
: urn ?? 2
is selected
mutually exclusive & exhaustive events
(i) We have
?? ?? 1
|?? =
?? ?? |?? 1
?? ?? 1
?? ?? |?? 1
?? ?? 1
+?? ?? |?? 2
?? ?? 2
=
4
10
×
2
6
4
10
×
2
6
+
6
10
×
4
6
=
1
4
·
(ii) Since ?? 1
and ?? 2
are mutually exclusive and ?? ?? 1
??? 2
|?? =?? ?? |?? = 1,we
have
?? ?? 2
|?? = 1-?? ?? 1
|??
=
3
4
· _
In the above example
?? ?? 1
|?? =
1
4
<
1
3
=?? ?? 1
,
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 3
and ?? ?? 2
|?? =
3
4
>
2
3
=?? ?? 2
,
i.e.,
(i) the probability of occurrence of event ?? 1
decreases in the presence of the
information that the outcome will be an element of ?? ;
(ii) the probability of occurrence of event ?? 2
increases in the presence of information
that the outcome will be an element of ?? .
These phenomena are related to the concept of association defined in the sequel.
Note that
?? ?? 1
|?? <?? ?? 1
??? ?? 1
n?? <?? ?? 1
?? (?? ),
and
?? ?? 2
|?? >?? ?? 2
??? ?? 2
n?? >?? ?? 2
?? (?? ).
Definition 3.2
Let ?? ,F,?? be a probability space and let ?? and ?? be two events. Events ?? and ?? are
said to be
(i) negatively associated if ?? ?? n?? <?? ?? ?? ?? ;
(ii) positively associated if ?? ?? n?? >?? ?? ?? ?? ;
(iii) independent if ?? ?? n?? =?? ?? ?? ?? . _
Remark 3.3
(i) If ?? ?? = 0 then ?? ?? n?? = 0 =?? ?? ?? ?? ,??? ?F, i.e., if ?? ?? = 0 then
any event ?? ?F and ?? are independent;
(ii) If ?? ?? > 0 then ?? and ?? are independent if, and only if, ?? ?? |?? =?? (?? ),
i.e., if ?? ?? > 0, then events ?? and ?? are independent if, and only if, the
availability of the information that event ?? has occurred does not alter the
probability of occurrence of event ?? . _
Now we define the concept of independence for arbitrary collection of events.
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 4
Definition 3.3
Let ?? ,F,?? be a probability space. Let ?? ?R be an index set and let ?? ?? :?? ??? be a
collection of events in F.
(i) Events ?? ?? :?? ??? are said to be pairwise independent if any pair of events ?? ??
and ?? ?? ,?? ??? in the collection ?? ?? :?? ??? are independent. i.e., if ?? ?? ?? n
?? ?? =?? ?? ?? ?? ?? ?? , whenever ?? ,?? ??? and ?? ??? ;
(ii) Let ?? = 1, 2,… , n , for some ?? ?N, so that ?? ?? :?? ??? = ?? 1
,… ,?? ?? is a
finite collection of events in F. Events ?? 1
,… ,?? ?? are said to be independent if,
for any sub collection ?? ?? 1
,… ,?? ?? ?? of ?? 1
,… ,?? ?? ?? = 2,3,… ,??
?? ?? ?? ?? ?? ?? =1
= ?? ?? ?? =1
?? ?? ?? . (3.6)
(iii) Let ?? ?R be an arbitrary index set. Events ?? ?? :?? ??? are said to be
independent if any finite sub collection of events in ?? ?? :?? ??? forms a
collection of independent events. _
Remark 3.4
(i) To verify that?? events ?? 1
,… ,?? ?? ?F are independent one must verify
2
?? -?? - 1 =
?? ??
?? ?? =2
conditions in (3.6). For example, to conclude that
three events ?? 1
,?? 2
and ?? 3
are independent, the following 4 = 2
3
- 3- 1
conditions must be verified:
?? ?? 1
n?? 2
=?? ?? 1
?? ?? 2
;
?? ?? 1
n?? 3
=?? ?? 1
?? ?? 3
;
?? ?? 2
n?? 3
=?? ?? 2
?? ?? 3
;
?? ?? 1
n?? 2
n?? 3
=?? ?? 1
?? ?? 2
?? ?? 3
.
(ii) If events ?? 1
,… ,?? ?? are independent then, for any permutation ?? 1
,… ,?? ?? of
1,… ,?? , the events ?? ?? 1
,… ,?? ?? ?? are also independent. Thus the notion of
independence is symmetric in the events involved.
(iv) Events in any subcollection of independent events are independent. In
particular independence of a collection of events implies their pairwise
independence. _
Page 5
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 1
MODULE 1
PROBABILITY
LECTURE 5
Topics
1.3.2 Bayes’ Theorem
1.3.2 Bayes’ Theorem
The following theorem provides a method for finding the probability of occurrence of an
event in a past trial based on information on occurrences in future trials.
1.3.2 Theorem 3.4 (Bayes’ Theorem)
Let ?? ,F,?? be a probability space and let ?? ?? : ?? ??? be a countable collection of
mutually exclusive and exhaustive events with ?? ?? ?? > 0,?? ??? . Then, for any event
?? ?F with ?? ?? > 0, we have
?? ?? ?? |?? =
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? , ?? ??? ·
Proof. We have, for ?? ??? ,
?? ?? ?? |?? =
?? ?? ?? n??
?? (?? )
=
?? ?? |?? ?? ?? ?? ??
?? (?? )
=
?? ?? |?? ?? ?? ?? ??
?? ?? |?? ?? ?? ?? ??
?? ??? using Theorem of Total Probability . _
Remark 3.2
(i) Suppose that the occurrence of any one of the mutually exclusive and
exhaustive events ?? ?? ,?? ??? , causes the occurrence of an event ?? . Given that the
event ?? has occurred, Bayes’ theorem provides the conditional probability that
the event ?? is caused by occurrence of event ?? ?? ,?? ??? .
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 2
(ii) In Bayes’ theorem the probabilities ?? ?? ?? ,?? ??? , are referred to as prior
probabilities and the probabilities ?? ?? ?? |?? ,?? ??? , are referred to as posterior
probabilities. _
To see an application of Bayes’ theorem let us revisit Example 3.4.
Example 3.5
Urn ?? 1
contains 4 white and 6 black balls and urn ?? 2
contains 6 white and 4 black balls.
A fair die is cast and urn ?? 1
is selected if the upper face of die shows five or six dots.
Otherwise urn ?? 2
is selected. A ball is drawn at random from the selected urn.
(i) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 1
;
(ii) Given that the drawn ball is white, find the conditional probability that it came
from urn ?? 2
.
Solution. Define the events:
?? : drawn ball is white;
?? 1
: urn ?? 1
is selected
?? 2
: urn ?? 2
is selected
mutually exclusive & exhaustive events
(i) We have
?? ?? 1
|?? =
?? ?? |?? 1
?? ?? 1
?? ?? |?? 1
?? ?? 1
+?? ?? |?? 2
?? ?? 2
=
4
10
×
2
6
4
10
×
2
6
+
6
10
×
4
6
=
1
4
·
(ii) Since ?? 1
and ?? 2
are mutually exclusive and ?? ?? 1
??? 2
|?? =?? ?? |?? = 1,we
have
?? ?? 2
|?? = 1-?? ?? 1
|??
=
3
4
· _
In the above example
?? ?? 1
|?? =
1
4
<
1
3
=?? ?? 1
,
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 3
and ?? ?? 2
|?? =
3
4
>
2
3
=?? ?? 2
,
i.e.,
(i) the probability of occurrence of event ?? 1
decreases in the presence of the
information that the outcome will be an element of ?? ;
(ii) the probability of occurrence of event ?? 2
increases in the presence of information
that the outcome will be an element of ?? .
These phenomena are related to the concept of association defined in the sequel.
Note that
?? ?? 1
|?? <?? ?? 1
??? ?? 1
n?? <?? ?? 1
?? (?? ),
and
?? ?? 2
|?? >?? ?? 2
??? ?? 2
n?? >?? ?? 2
?? (?? ).
Definition 3.2
Let ?? ,F,?? be a probability space and let ?? and ?? be two events. Events ?? and ?? are
said to be
(i) negatively associated if ?? ?? n?? <?? ?? ?? ?? ;
(ii) positively associated if ?? ?? n?? >?? ?? ?? ?? ;
(iii) independent if ?? ?? n?? =?? ?? ?? ?? . _
Remark 3.3
(i) If ?? ?? = 0 then ?? ?? n?? = 0 =?? ?? ?? ?? ,??? ?F, i.e., if ?? ?? = 0 then
any event ?? ?F and ?? are independent;
(ii) If ?? ?? > 0 then ?? and ?? are independent if, and only if, ?? ?? |?? =?? (?? ),
i.e., if ?? ?? > 0, then events ?? and ?? are independent if, and only if, the
availability of the information that event ?? has occurred does not alter the
probability of occurrence of event ?? . _
Now we define the concept of independence for arbitrary collection of events.
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 4
Definition 3.3
Let ?? ,F,?? be a probability space. Let ?? ?R be an index set and let ?? ?? :?? ??? be a
collection of events in F.
(i) Events ?? ?? :?? ??? are said to be pairwise independent if any pair of events ?? ??
and ?? ?? ,?? ??? in the collection ?? ?? :?? ??? are independent. i.e., if ?? ?? ?? n
?? ?? =?? ?? ?? ?? ?? ?? , whenever ?? ,?? ??? and ?? ??? ;
(ii) Let ?? = 1, 2,… , n , for some ?? ?N, so that ?? ?? :?? ??? = ?? 1
,… ,?? ?? is a
finite collection of events in F. Events ?? 1
,… ,?? ?? are said to be independent if,
for any sub collection ?? ?? 1
,… ,?? ?? ?? of ?? 1
,… ,?? ?? ?? = 2,3,… ,??
?? ?? ?? ?? ?? ?? =1
= ?? ?? ?? =1
?? ?? ?? . (3.6)
(iii) Let ?? ?R be an arbitrary index set. Events ?? ?? :?? ??? are said to be
independent if any finite sub collection of events in ?? ?? :?? ??? forms a
collection of independent events. _
Remark 3.4
(i) To verify that?? events ?? 1
,… ,?? ?? ?F are independent one must verify
2
?? -?? - 1 =
?? ??
?? ?? =2
conditions in (3.6). For example, to conclude that
three events ?? 1
,?? 2
and ?? 3
are independent, the following 4 = 2
3
- 3- 1
conditions must be verified:
?? ?? 1
n?? 2
=?? ?? 1
?? ?? 2
;
?? ?? 1
n?? 3
=?? ?? 1
?? ?? 3
;
?? ?? 2
n?? 3
=?? ?? 2
?? ?? 3
;
?? ?? 1
n?? 2
n?? 3
=?? ?? 1
?? ?? 2
?? ?? 3
.
(ii) If events ?? 1
,… ,?? ?? are independent then, for any permutation ?? 1
,… ,?? ?? of
1,… ,?? , the events ?? ?? 1
,… ,?? ?? ?? are also independent. Thus the notion of
independence is symmetric in the events involved.
(iv) Events in any subcollection of independent events are independent. In
particular independence of a collection of events implies their pairwise
independence. _
NPTEL- Probability and Distributions
Dept. of Mathematics and Statistics Indian Institute of Technology, Kanpur 5
The following example illustrates that, in general, pairwise independence of a collection
of events may not imply their independence.
Example 3.6
Let ?? = 1, 2, 3, 4 and let F =?? ?? , the power set of ?? . Consider the probability space
?? ,F, P , where ?? ?? =
1
4
,?? = 1, 2, 3, 4 . Let ?? = 1, 4 , ?? = 2, 4 and ?? = 3, 4 .
Then,
?? ?? =?? ?? =?? ?? =
1
2
,
?? ?? n?? =?? ?? n?? =?? ?? n?? =?? {4} =
1
4
,
and ?? ?? n?? n?? =?? 4 =
1
4
·
Clearly,
?? ?? n?? =?? ?? ?? ?? ;?? ?? n?? =?? ?? ?? (?? ), and ?? ?? n?? =?? ?? ?? (?? ),
i.e., ?? ,?? and ?? are pairwise independent.
However,
?? ?? n?? n?? =
1
4
??? ?? ?? ?? ?? (?? ).
Thus ?? ,?? and ?? are not independent. _
```
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# Mathematical Solution: Dividend Yield Ratio
Dividend yield ratio explains what percentage of the market price of a share a firm yearly pays to be able to its stockholders as dividends. It can be calculated by dividing the annual dividend each share by industry value per discuss. The ratio is usually expressed in percentage form which is sometimes called dividend generate percentage.
Since dividend yield ratio is utilized to measure their bond between the annual quantity of dividend per share and also the current market price of a share, it is certainly caused by used by investors seeking dividend income upon continuous basis.
Formula:
The following formula is used to calculated dividend yield ratio:
Example 1 – simple computation
Suppose a company declares dividend at \$1.70 per share. The par value of a share of the company is \$15 and the market price per share is \$20. The dividend yield ratio would be computed as follows:
The dividend yield ratio is 8%. It describe an investor earns 8% on his investment (in the term of dividends) if he purchases the common stock of the business at present market price.
Generally, the famous and good established companies are in a higher position to pay a good percentage to the stockholders on their investment in the form of annual dividends as compared to another ones. Consider the following example:
Example 2 – comparison of two companies:
The following information is related to ASD company and XXC company for the year 2013:
An investor should desire the ASD company because its dividend yield ratio is significantly better than that of ZXC company. PQR is an old and well established company with a stable dividend distribution history. Also there are well chances of the appreciation in the market value of the stock of ASD. Because of these matter, ASD is a more reliable and less risky company for investment portfolio as compared to ZXC.
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# pre algebra 2 1
Table of Contents
## Question 1 (5 points)
Graph each linear equation.
Question 1 options:
## Question 2 (5 points)
Use the vertical line test to determine which of the following relations is a function.
Question 2 options:
## Question 3 (5 points)
Find the slope of the line.
Question 3 options:
## Question 4 (5 points)
Graph each linear equation.
Question 4 options:
## Question 5 (5 points)
Determine if there is a proportional linear relationship. Explain your reasoning.
Question 5 options:
decrease of 16 oz/lb increase of 16 oz/lb increase of 3 lb/oz increase of 32 oz/lb
## Question 6 (5 points)
Determine if there is a proportional linear relationship. Explain your reasoning.
Time (h) Temperature (°F) x y 0 52 1 54 2 56 3 58
Question 6 options:
decrease of 2 °F/h increase of 2 h/°F increase of 3 °F/h increase of 2 °F/h
## Question 7 (5 points)
There are a total of 21 students in the choir and the drama club. There are 7 more students in the choir than the drama club. Write and solve a system of equations to represent this solution. Let x represent the number of students in the choir, and let y represent the number of students in the drama club. Interpret the solution.
Question 7 options:
(7, 14); There are 14 students in the choir and 7 students in the drama club. (7, 14); There are 7 students in the choir and 14 students in the drama club. (14, 7); There are 14 students in the choir and 7 students in the drama club. (14, 7); There are 7 students in the choir and 14 students in the drama club.
## Question 8 (5 points)
Graph each linear equation.
Question 8 options:
## Question 9 (5 points)
Determine whether the system of equations has one solution, no solution, or infinitely many solutions by graphing. If the system has one solution, name it.
Question 9 options:
one solution (2, 2) infinitely many solutions no solutions no solutions
## Question 10 (5 points)
The length of a rectangular picture frame is twice the width. The frame has a perimeter of 96 inches. Write and solve a system of equations to represent this solution. Let x represent the width of the frame, and let y represent the length of the frame. Interpret the solution.
Question 10 options:
(32, 16); The width of the frame is 32 inches, and the length of the frame is 16 inches. (32, 16); The width of the frame is 16 inches, and the length of the frame is 32 inches. (16, 32); The width of the frame is 32 inches, and the length of the frame is 16 inches. (16, 32); The width of the frame is 16 inches, and the length of the frame is 32 inches.
## Question 11 (5 points)
Graph each equation using the slope and the y-intercept.
y = –4
x – 2.5
Question 11 options:
## Question 12 (5 points)
The amount of sales tax varies directly with the cost of the purchase. If the sales tax is \$3.80 on a purchase of \$76, what would be the sales tax on a purchase of \$46?
Question 12 options:
\$1.77 \$2.30 \$2.94 \$3.47
## Question 13 (5 points)
The amount a spring will stretch varies directly with the amount of weight attached to the spring. If a spring stretches 2.1 inches when 80 pounds is attached, how far will it stretch when 40 pounds is attached?
Question 13 options:
1.05 in. 2.1 in. 4.2 in. 0.95 in.
## Question 14 (5 points)
The cost of tickets varies directly with the number of tickets. Suppose 7 tickets cost \$126. How much would 12 tickets cost?
Question 14 options:
\$180 \$216 \$252 \$18
## Question 15 (5 points)
Graph each equation using the slope and the y-intercept.
y = –2
x + 0.5
Question 15 options:
## Question 16 (5 points)
Find the slope of the line that passes through the pair of points.
(1, 4), (–5, 4)
Question 16 options:
undefined –2 0
## Question 17 (5 points)
Solve the system of equations.
y = 4
x – 4
y = 5
x
Question 17 options:
(–4, –20) (–4, 20) (–20, –4) (4, –20)
## Question 18 (5 points)
Find four solutions of the given function. Write the solutions as ordered pairs.
4
x –
y = 4
Question 18 options:
{(8, 28), (–6, –28), (1, 0), (–2, –12)} {(8, 28), (–2, –28), (1, 0), (–6, –12)} {(8, 32), (–6, –28), (1, 4), (–2, –12)} {(7, 28), (–6, –28), (1, 32), (8, –12)}
## Question 19 (5 points)
Solve the system of equations by graphing.
Question 19 options:
## Question 20 (5 points)
Find the function value.
f(
x)
3
x – 6,
f(–4)
Question 20 options:
–18 –6 –7 5
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Emmy Combs
2022-01-29
Matrix transformation
$f:{\mathbb{R}}^{3}\to {\mathbb{R}}^{3}$
$\left(\begin{array}{ccc}4& 1& 3\\ 2& -1& 3\\ 2& 2& 0\end{array}\right)$
Establish x,y and z such that,
$f\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$
Do I just need to multiply the values of f for the $3×3$ matrix? What does this mean overall?
dodato0n
Step 1 You need to solve the non-homogeneous system $\left(\begin{array}{ccc}4& 1& 3\\ 2& -1& 3\\ 2& 2& 0\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}4\\ 5\\ -1\end{array}\right)$ Form then the augmented coefficients matrix and reduce it by rows: $\left(\begin{array}{ccccc}4& 1& 3& :& 4\\ 2& -1& 3& :& 5\\ 2& 2& 0& :& -1\end{array}\right)\stackrel{{R}_{1}↔{R}_{3}\cdot \left(\frac{1}{2}\right)}{⟶}\left(\begin{array}{ccccc}1& 1& 0& :& -1/2\\ 2& -1& 3& :& 5\\ 4& 1& 3& :& 4\end{array}\right)\stackrel{{R}_{2}-2{R}_{1}\phantom{\rule{0ex}{0ex}}{R}_{3}-4{R}_{1}}{⟶}$ $\left(\begin{array}{ccccc}1& 1& 0& :& -1/2\\ 0& -3& 3& :& 6\\ 0& -3& 3& :& 6\end{array}\right)\stackrel{{R}_{3}-{R}_{2}}{⟶}\left(\begin{array}{ccccc}1& 1& 0& :& -1/2\\ 0& -3& 3& :& 6\\ 0& 0& 0& :& 0\end{array}\right)$ Well, there are infinite solutions to the above system: begin with row 3 , and remember the columns represent x,y,z from left to right, so: ${R}_{2}:\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}-3y+3z=6\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}y=z-2\phantom{\rule{0ex}{0ex}}{R}_{1}:\phantom{\rule{0.278em}{0ex}}\phantom{\rule{0.278em}{0ex}}x+y=-\frac{1}{2}\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}x=-\frac{1}{2}-y=-\frac{1}{2}-z+2=\frac{3}{2}-z$ Now just choose a nice z , say $z=0\phantom{\rule{0.278em}{0ex}}⟹\phantom{\rule{0.278em}{0ex}}y=-2\phantom{\rule{0.167em}{0ex}},\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}x=\frac{3}{2}$, and one solution to your problem is $\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}3/2\\ -2\\ 0\end{array}\right)$ as you can easily check.
kumewekwah0
Step 1
You are given that
$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{ccc}4& 1& 3\\ 2& -1& 3\\ 2& 2& 0\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)$
Of course,
$\left(\begin{array}{ccc}4& 1& 3\\ 2& -1& 3\\ 2& 2& 0\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}4x+y+3z\\ 2x-y+3z\\ 2x+2y\end{array}\right)$
So we have
$\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}4x+y+3z\\ 2x-y+3z\\ 2x+2y\end{array}\right)$
That gives us three equations:
$x=4x+y+3z\phantom{\rule{1em}{0ex}}\text{which reduces to}\phantom{\rule{1em}{0ex}}3x+y+3z=0.$
$y=2x-y+3z\phantom{\rule{1em}{0ex}}\text{which reduces to}\phantom{\rule{1em}{0ex}}2x-2y+3z=0.$
$z=2x+2y\phantom{\rule{1em}{0ex}}\text{which reduces to}\phantom{\rule{1em}{0ex}}2x+2y-z=0.$
An obvious solution to that is $x=y=z=0$. Notice that the determinant is $\begin{array}{rl}|\begin{array}{ccc}3& 1& 3\\ 2& -2& 3\\ 2& 2& -1\end{array}|& =3|\begin{array}{cc}-2& 3\\ 2& -1\end{array}|-|\begin{array}{cc}2& 3\\ 2& -1\end{array}|+3|\begin{array}{cc}2& -2\\ 2& 2\end{array}|\\ & =3\left(2-6\right)-\left(-2-6\right)+3\left(4+4\right)\\ & =-12+8+24\\ & =20\end{array}$
Since that is not $0,x=y=z=0$ is the only solution.
Do you have a similar question?
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# QUESTION BASED ON COULOMB’S LAW
ELECTROSTATICS
exercise # 1
Question based on coulomb’s law
1 An electron at rest has a charge of 1.6 × 10–19 C. It starts moving with a velocity v = c/2, where c is the speed of light, then the new charge on it is –
(1) 1.6 × 10–19 Coulomb
(2) 1.6 × 10–19 Coulomb
(3) 1.6 × 10–19 Coulomb (4) Coulomb
2 Two similar charge of +Q , as shown in figure are placed at A and B. –q charge is placed at point C midway between A and B. –q charge will oscillate if
A c B
+Q———- -q ————- +Q
(1) It is moved towards A. (2) It is moved towards B.
(3) It is moved upwards AB.
(4) Distance between A and B is reduced.
3 When the distance between two charged particle is halved, the force between them becomes –
(1) One fourth (2) One half (3) Double (4) Four times
4 Two charges are at distance (d) apart in air. Coulomb force between them is F. If a dielectric material of dielectric constant (K) is placed between them, the coulomb force now becomes.
(1) F/K (2) FK (3) F/K2 (4) K2F
5 A certain charge Q is divided at first into two parts, (q) and (Q-q). Later on the charges are placed at a certain distance. If the force of interaction between the two charges is maximum then-
(1) (Q/q) = (4/1) (2) (Q/q) = (2/1)
(3)(Q/q) = (3/1) (4) (Q/q) = (5/1)
6 The three charges each of 5 × 10–6 coloumb are placed at vertex of an equilateral triangle of side 10cm. The force exerted on the charge of 1 m C placed at centre of triangle in newton will be
(1) 13.5 (2) zero
(3) 4.5 (4) 6.75
7 ABC is a right angle triangle AB=3cm, BC=4cm charges + 15, +12, –12 esu are placed at A, B and C respectively. The magnitude of the force experienced by the charge at B in dyne is-
(1) 125 (2) 35
(3) 22 (4) 0
8 Five point charges, each of value +q coulomb, are placed on five vertices of a regular hexagon of side L metre. The magnitude of the force on a point charge of value -q coul. placed at the centre of the hexagon is –
(1) (2)
(3) (4) Zero
9 Two charged spheres A and B are charged with the charges of +10 and +20 coul. respectively and separated by a distance of 80cm. The electric field at a point on the line joining the centres of the two sphers will be zero at a distance from sphere A.
(1) 20 cm (2) 33 cm (3) 55 cm (4) 60 cm.
10 Four charges +q, +q, –q and –q are placed respectively at the corners A, B, C and D of a square of side (a), arranged in the given order. Calculate the intensity at (O) the centre of the square .
(1) (2) (3) (4)
11 The electric potential V at any point (x, y, z) in space is given by V = 4×2 volt. The electric field E in V/m at the point (1, 0, 2) is –
(1) +8 in x direction (2) 8 in –x direction (3) 16 in + x direction (4) 16 in –x direction
12 Charges of + × 10–9 are placed at each of the four corners of a square of side 8cm. The potential at the intersection of the diagonals is
(1) 150 Volt (2) 1500 Volt
(3) 900 Volt (4) 900 Volt
13 The electron potential (V) as a function of distance (x) [in meters] is given by
V = (5×2 + 10 x – 9)Volt.
The value of electric field at x =1m would be-
(1) 20 Volt/m (2) 6 Volt/m (3) 11 Volt/m (4) –23 Volt/m
14 A – particle moves towards a rest nucleus, if kinetic energy of -particle is 10 MeV and atomic number of nucleus is 50. The closest approach will be –
(1) 1.44 × 10–14 m (2) 2.88 × 10–14 m (3) 1.44 × 10–10 m (4) 2.88 × 10–10 m
15 A charge of Q coloumb is located at the centre of a cube. If the corner of the cube is taken as the origin, then the flux coming out from the faces of the cube in the direction of X- axis will be-
(1) 4 (2) Q/6 (3) Q/3 (4) Q/4
16 A rectangular surface of 2 metre width and 4 metre length, is placed in an electric field of intensity 20 newton/C, there is an angle of 60º between the perpendicular to surface and electrical field intensity. Then total flux emitted from the surface will be- (In Volt- metre)
(1) 80 (2) 40
(3) 20 (4) 160
17 A square of side 20cm. is enclosed by a surface of sphere of 80 cm. radius . square and sphere have the same centre. four charges +2 × 10–6 c, –5 × 10–6 c, –3 × 10–6 c, +6 × 10–6c are located at the four corners of a square, Then out going total flux from spherical surface in N-m2/c will be
(1) zero (2) (16p) × 10–6
(3) (8p) × 10–6 (4) (36 p) × 10–6
18 A charge Q is distributed over two concentric hollow spheres of radii (r) and (R) > (r) such the surface densities are equal. Find the potential at the common centre.
(1) (2)
(3) (4) none of these
19 A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of 3Q the new potential difference between the same two surfaces is
(1) V (2) 2V
(3) 4V (4) –2V
20 An electric dipole consists of two opposite charges each of magnitude 1 × 10–6 C separated by a distance 2cm. The dipole is placed in an external field of 10 × 105N/C. The maximum torque on the dipole is –
(1) 0.2 × 10–3 N-m
(2) 1.0 × 10–3 N-m
(3) 20 × 10-3 N-m
(4) 4 × 10–3 N-m
21 If an electric field is given by , calculate the electric flux through a surface of area 10 units lying in yz plane-
(1) 100 units (2) 10 units
(3) 30 units (4) 40 units
22 Two long thin charged rods with charge density l each are placed parallel to each other at a distance d apart. The force per unit length exerted on one rod by the other will be-
(1) (2)
(3) (4)
23 The electric field intensity due to a thin infinite long straight wire of uniform linear charge density l at O is –
(1) (2)
(3) (4) Zero
24 Figure shows a set of euipotential surfaces. The magnitude and direction of electric field that exists in the region is-
(1) V/m at 45º with x-axis
(2) V/m at –45º with x-axis
(3) V/m at 45º with x-axis
(4) V/m at –45º with x-axis
25 Determine the electric field strength vector if the potential of this field depends on x, y coordinates as V = 10 axy –
(1) (2)
(3) (4)
26 An electric dipole of length 2 cm is placed with its axis making an angle of 30º to a uniform electric field 105 N/C. If it experiences a torque of Nm, then potential energy of the dipole-
(1) –10 J (2) –20 J
(3) – 30 J (4) –40 J
7 Two isolated metallic solid spheres of radii R and 2R are charged, such that both of these have same charge density s. The spheres are located far away from each other and connected by a thin conducting wire. The new charge density on the bigger sphere is-
(1) (2)
(3) (4) .
28 Electric potential in an electric field is given as V= K/r, (K being constant), if position vector then electric field will be
(1) (2) (3) (4)
29 At any point ( x,0,0) the electric potential V is volt, then electric field at x = 1 m –
(1) (2)
(3) (4)
30 8 small droplets of water of same size and same charge form a large spherical drop. The potential of the large drop, in comparision to potential of a small drop will be –
(1) 2 times (2) 4 times (3) 8times (4) same
31 As per this diagram a point charge +q is placed at the origin O. Work done in taking another pont charge –Q from the point A [co-ordinates (0, a)] to another point B [co-ordinates (a,0)] along the straight path AB is
(1) Zero (2)
(3) (4)
32 Determine dimensions of e0 (permitivity of free space) –
(1) [M–1L–3T4A2] (2) [M–1L–3T2A4] (3) [ML3T–4A–2] (4) [M–1L–3T2A2]
33 If in Millikan’s oil drop experiment charges on drops are found to be 8µC, 12µC, 20µC, then quanta of charge is-
(1) 8µC (2) 4µC
(3) 20µC (4) 12µC
34 Force between two identical spheres charged with same charge is F. If 50% charge of one sphere is transffered to second sphere then new force will be-
(1) (2)
(3) (4) none of these
35 In the electric field of charge Q, another charge is carried from A to B, A to C, A to D and A to E, then work done will be-
(1) minimum along path AB (2) minimum along path AD
(3) minimum along path AE (4) zero along all the paths
6 The total flux associated with given cube will be where ‘a’ is side of cube –
(=4p × 9 ×109)
(1) 162p × 10–3 Nm2/C (2) 162p × 103 Nm2/C (3) 162p × 10–6 Nm2/C (4) 162p × 106 Nm2/C
37 A sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere is charged to a potential 3 e.s.u. When the outer sphere is earthed. The charge on the inner sphere is –
(1) 54 e.s.u. (2) e.s.u.
(3) 30 e.s.u. (4) 36 e.s.u.
38 Two identical small spheres carry charge of Q1 and Q2 with Q1 >> Q2. The charges are d distance apart. The force they exert on one another is F1. The spheres are made to touch one another and then separated to distance d apart. The force they exert on one another now is F2. Then F1/F2 is-
(1) (2) (3) (4)
39 A point particle of mass M is attached to one end of a massless rigid non-conducting rod of length L. Another point particle of same mass is attached to the other end of the rod. The two particles carry charges +q and –q respectively. This arrangement is held in a region of uniform electric field E such that the rod makes a small angle q(<5º) with the field direction. The minimum time needed for the rod to become parallel to the field after it is set free.( rod rotates about centre of mass)
(1) (2)
(3) (4)
in the figure. A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is , where k is-
(1) 8q2 (2) 6q2 (3) 8q1 (4) 6q1
41 The electric potential at a point (x, y, z) is given by
V = –x2y – xz3 + 4
The electric field at that point is-
(1) (2)
(3)
(4)
42 Three concentric spherical shells have radii a, b and c(a < b < c) and have surface charge densities s, –s and s respectively. If VA, VB and VC denote the potentials of the three shells, then, for
c = a + b, we have-
(1) VC = VB = VA (2) VC = VA ¹ VB
(3) VC = VB ¹ VA (4) VC ¹ VB ¹ VA
43 The figure shows some of the electric field lines corresponding to an electric field.
The figure suggests –
(1) EA > EB > EC
(2) EA = EB = EC
(3) EA = EC > EB
(4) EA = EC < EB
4 Two identical thin rings, each of radius R meters, are coaxially placed at a distance R meters apart. If Q1 coulomb and Q2 coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of other is-
(1) zero
(2)
(3)
(4)
45 Three charges –q1, +q2 and –q3 are placed as shown in the figure. The x-component of the force on –q1 is proportional to-
(1) (2)
(3) (4)
Q.N. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Ans. 1 3 4 1 2 2 3 1 2 2 2 2 1 1 3 1 1 3 1 3 1 2 1 1
Q.N 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
ANS. 2 3 2 2 2 2 1 1 2 1 4 2 4 3 3 1 2 2 3 2 3
ELECTROSTATICS
exercise # 1
Question based on coulomb’s law
1 An electron at rest has a charge of 1.6 × 10–19 C. It starts moving with a velocity v = c/2, where c is the speed of light, then the new charge on it is –
(1) 1.6 × 10–19 Coulomb
(2) 1.6 × 10–19 Coulomb
(3) 1.6 × 10–19 Coulomb (4) Coulomb
2 Two similar charge of +Q , as shown in figure are placed at A and B. –q charge is placed at point C midway between A and B. –q charge will oscillate if
A c B
+Q———- -q ————- +Q
(1) It is moved towards A. (2) It is moved towards B.
(3) It is moved upwards AB.
(4) Distance between A and B is reduced.
3 When the distance between two charged particle is halved, the force between them becomes –
(1) One fourth (2) One half (3) Double (4) Four times
4 Two charges are at distance (d) apart in air. Coulomb force between them is F. If a dielectric material of dielectric constant (K) is placed between them, the coulomb force now becomes.
(1) F/K (2) FK (3) F/K2 (4) K2F
5 A certain charge Q is divided at first into two parts, (q) and (Q-q). Later on the charges are placed at a certain distance. If the force of interaction between the two charges is maximum then-
(1) (Q/q) = (4/1) (2) (Q/q) = (2/1)
(3)(Q/q) = (3/1) (4) (Q/q) = (5/1)
6 The three charges each of 5 × 10–6 coloumb are placed at vertex of an equilateral triangle of side 10cm. The force exerted on the charge of 1 m C placed at centre of triangle in newton will be
(1) 13.5 (2) zero
(3) 4.5 (4) 6.75
7 ABC is a right angle triangle AB=3cm, BC=4cm charges + 15, +12, –12 esu are placed at A, B and C respectively. The magnitude of the force experienced by the charge at B in dyne is-
(1) 125 (2) 35
(3) 22 (4) 0
8 Five point charges, each of value +q coulomb, are placed on five vertices of a regular hexagon of side L metre. The magnitude of the force on a point charge of value -q coul. placed at the centre of the hexagon is –
(1) (2)
(3) (4) Zero
9 Two charged spheres A and B are charged with the charges of +10 and +20 coul. respectively and separated by a distance of 80cm. The electric field at a point on the line joining the centres of the two sphers will be zero at a distance from sphere A.
(1) 20 cm (2) 33 cm (3) 55 cm (4) 60 cm.
10 Four charges +q, +q, –q and –q are placed respectively at the corners A, B, C and D of a square of side (a), arranged in the given order. Calculate the intensity at (O) the centre of the square .
(1) (2) (3) (4)
11 The electric potential V at any point (x, y, z) in space is given by V = 4×2 volt. The electric field E in V/m at the point (1, 0, 2) is –
(1) +8 in x direction (2) 8 in –x direction (3) 16 in + x direction (4) 16 in –x direction
12 Charges of + × 10–9 are placed at each of the four corners of a square of side 8cm. The potential at the intersection of the diagonals is
(1) 150 Volt (2) 1500 Volt
(3) 900 Volt (4) 900 Volt
13 The electron potential (V) as a function of distance (x) [in meters] is given by
V = (5×2 + 10 x – 9)Volt.
The value of electric field at x =1m would be-
(1) 20 Volt/m (2) 6 Volt/m (3) 11 Volt/m (4) –23 Volt/m
14 A – particle moves towards a rest nucleus, if kinetic energy of -particle is 10 MeV and atomic number of nucleus is 50. The closest approach will be –
(1) 1.44 × 10–14 m (2) 2.88 × 10–14 m (3) 1.44 × 10–10 m (4) 2.88 × 10–10 m
15 A charge of Q coloumb is located at the centre of a cube. If the corner of the cube is taken as the origin, then the flux coming out from the faces of the cube in the direction of X- axis will be-
(1) 4 (2) Q/6 (3) Q/3 (4) Q/4
16 A rectangular surface of 2 metre width and 4 metre length, is placed in an electric field of intensity 20 newton/C, there is an angle of 60º between the perpendicular to surface and electrical field intensity. Then total flux emitted from the surface will be- (In Volt- metre)
(1) 80 (2) 40
(3) 20 (4) 160
17 A square of side 20cm. is enclosed by a surface of sphere of 80 cm. radius . square and sphere have the same centre. four charges +2 × 10–6 c, –5 × 10–6 c, –3 × 10–6 c, +6 × 10–6c are located at the four corners of a square, Then out going total flux from spherical surface in N-m2/c will be
(1) zero (2) (16p) × 10–6
(3) (8p) × 10–6 (4) (36 p) × 10–6
18 A charge Q is distributed over two concentric hollow spheres of radii (r) and (R) > (r) such the surface densities are equal. Find the potential at the common centre.
(1) (2)
(3) (4) none of these
19 A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of 3Q the new potential difference between the same two surfaces is
(1) V (2) 2V
(3) 4V (4) –2V
20 An electric dipole consists of two opposite charges each of magnitude 1 × 10–6 C separated by a distance 2cm. The dipole is placed in an external field of 10 × 105N/C. The maximum torque on the dipole is –
(1) 0.2 × 10–3 N-m
(2) 1.0 × 10–3 N-m
(3) 20 × 10-3 N-m
(4) 4 × 10–3 N-m
21 If an electric field is given by , calculate the electric flux through a surface of area 10 units lying in yz plane-
(1) 100 units (2) 10 units
(3) 30 units (4) 40 units
22 Two long thin charged rods with charge density l each are placed parallel to each other at a distance d apart. The force per unit length exerted on one rod by the other will be-
(1) (2)
(3) (4)
23 The electric field intensity due to a thin infinite long straight wire of uniform linear charge density l at O is –
(1) (2)
(3) (4) Zero
24 Figure shows a set of euipotential surfaces. The magnitude and direction of electric field that exists in the region is-
(1) V/m at 45º with x-axis
(2) V/m at –45º with x-axis
(3) V/m at 45º with x-axis
(4) V/m at –45º with x-axis
25 Determine the electric field strength vector if the potential of this field depends on x, y coordinates as V = 10 axy –
(1) (2)
(3) (4)
26 An electric dipole of length 2 cm is placed with its axis making an angle of 30º to a uniform electric field 105 N/C. If it experiences a torque of Nm, then potential energy of the dipole-
(1) –10 J (2) –20 J
(3) – 30 J (4) –40 J
7 Two isolated metallic solid spheres of radii R and 2R are charged, such that both of these have same charge density s. The spheres are located far away from each other and connected by a thin conducting wire. The new charge density on the bigger sphere is-
(1) (2)
(3) (4) .
28 Electric potential in an electric field is given as V= K/r, (K being constant), if position vector then electric field will be
(1) (2) (3) (4)
29 At any point ( x,0,0) the electric potential V is volt, then electric field at x = 1 m –
(1) (2)
(3) (4)
30 8 small droplets of water of same size and same charge form a large spherical drop. The potential of the large drop, in comparision to potential of a small drop will be –
(1) 2 times (2) 4 times (3) 8times (4) same
31 As per this diagram a point charge +q is placed at the origin O. Work done in taking another pont charge –Q from the point A [co-ordinates (0, a)] to another point B [co-ordinates (a,0)] along the straight path AB is
(1) Zero (2)
(3) (4)
32 Determine dimensions of e0 (permitivity of free space) –
(1) [M–1L–3T4A2] (2) [M–1L–3T2A4] (3) [ML3T–4A–2] (4) [M–1L–3T2A2]
33 If in Millikan’s oil drop experiment charges on drops are found to be 8µC, 12µC, 20µC, then quanta of charge is-
(1) 8µC (2) 4µC
(3) 20µC (4) 12µC
34 Force between two identical spheres charged with same charge is F. If 50% charge of one sphere is transffered to second sphere then new force will be-
(1) (2)
(3) (4) none of these
35 In the electric field of charge Q, another charge is carried from A to B, A to C, A to D and A to E, then work done will be-
(1) minimum along path AB (2) minimum along path AD
(3) minimum along path AE (4) zero along all the paths
6 The total flux associated with given cube will be where ‘a’ is side of cube –
(=4p × 9 ×109)
(1) 162p × 10–3 Nm2/C (2) 162p × 103 Nm2/C (3) 162p × 10–6 Nm2/C (4) 162p × 106 Nm2/C
37 A sphere of 4 cm radius is suspended within a hollow sphere of 6 cm radius. The inner sphere is charged to a potential 3 e.s.u. When the outer sphere is earthed. The charge on the inner sphere is –
(1) 54 e.s.u. (2) e.s.u.
(3) 30 e.s.u. (4) 36 e.s.u.
38 Two identical small spheres carry charge of Q1 and Q2 with Q1 >> Q2. The charges are d distance apart. The force they exert on one another is F1. The spheres are made to touch one another and then separated to distance d apart. The force they exert on one another now is F2. Then F1/F2 is-
(1) (2) (3) (4)
39 A point particle of mass M is attached to one end of a massless rigid non-conducting rod of length L. Another point particle of same mass is attached to the other end of the rod. The two particles carry charges +q and –q respectively. This arrangement is held in a region of uniform electric field E such that the rod makes a small angle q(<5º) with the field direction. The minimum time needed for the rod to become parallel to the field after it is set free.( rod rotates about centre of mass)
(1) (2)
(3) (4)
in the figure. A third charge q3 is moved along the arc of a circle of radius 40 cm from C to D. The change in the potential energy of the system is , where k is-
(1) 8q2 (2) 6q2 (3) 8q1 (4) 6q1
41 The electric potential at a point (x, y, z) is given by
V = –x2y – xz3 + 4
The electric field at that point is-
(1) (2)
(3)
(4)
42 Three concentric spherical shells have radii a, b and c(a < b < c) and have surface charge densities s, –s and s respectively. If VA, VB and VC denote the potentials of the three shells, then, for
c = a + b, we have-
(1) VC = VB = VA (2) VC = VA ¹ VB
(3) VC = VB ¹ VA (4) VC ¹ VB ¹ VA
43 The figure shows some of the electric field lines corresponding to an electric field.
The figure suggests –
(1) EA > EB > EC
(2) EA = EB = EC
(3) EA = EC > EB
(4) EA = EC < EB
4 Two identical thin rings, each of radius R meters, are coaxially placed at a distance R meters apart. If Q1 coulomb and Q2 coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of other is-
(1) zero
(2)
(3)
(4)
45 Three charges –q1, +q2 and –q3 are placed as shown in the figure. The x-component of the force on –q1 is proportional to-
(1) (2)
(3) (4)
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# Algorithms Quiz | SP Contest 1 | Question 10
• Difficulty Level : Easy
• Last Updated : 06 Sep, 2021
Here are the two concurrent process A, B with respective codes:
Code A:
```while (true) // infinite condition
{
M :____;
printf("%c", b);
printf("%c", b);
N:____;
}
```
Code B:
```while (true) // infinite condition
{
O:____;
printf("%c", a);
printf("%c", a);
P:____;
}
```
What should be the binary semaphore operation on M, N, O, P respectively and what must be the initial values of semaphore X, Y in order to get the output bbaabbaabbaa . . . ?
Where P is down and V is up operation respectively.
(A) M = P(Y), N = V(X), O = P(X), P = V(Y); X = 0, Y = 1;
(B) M = P(Y), N = V(X), O = P(X), P = P(Y); X = Y = 1;
(C) M = P(Y), N = V(Y), O = P(X), P = V(X); X = 1, Y = 0;
(D) M = P(Y), N = V(Y), O = P(X), P = V(X); X = Y = 1;
Explanation: In semaphore up is always a successful operation but down is not always successful.
In following concurrent process Operations are:
A: code
```while (true) // infinite condition
{
M :P(Y); // Y become 0 successful down operation.
printf("%c", b);
printf("%c", b);
N:V(X); // X become 1 successful up operation.
}
```
B code:
```while (true) // infinite condition
{
O:P(X); // X become 0 successful down operation.
printf("%c", a);
printf("%c", a);
P:V(Y); // Ybecome 1 successful up operation.
}```
Here all operation are successful with initial values of X and Y are 0 and 1 respectively.
So, option (A) is correct.
Quiz of this Question
My Personal Notes arrow_drop_up
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## Group mod a subgroup
Suppose G is a group and H is a subgroup of G. Then is G\H a subgroup itself? My feeling is it shouldn't be since 1$$\in$$H, therefore 1$$\notin$$G\H?
I'm getting a bit confused about this because I'm doing a homework sheet and the question deals with a homomorphism from G $$\rightarrow$$ G\H and I'm wondering what happens to the identity element...
Thanks!
PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
if H is a normal subgroup, then G/H is a group. It's not a subgroup of G though since its elements are the cosets $eH=H, g_1H, g_2H, ...$, which are the images of the elements of G under that mapping. maybe I misunderstood completely though, did you get / mixed up with \ ? because G\H usually means the set {g in G | g not in H}, while G/H means quotient group
Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus I think your confusing some notation. G/H is different from G\H. G\H is the set-difference and it is not a group. And certainly not a subgroup. G/H is the quotient group, and it is a group (when H is normal). But it's not a subgroup (in general).
Recognitions:
## Group mod a subgroup
when H is a normal subgroup of G, G/H can be made into a group by setting coset multiplication as: (Hx)(Hy) = H(xy).
the condition of normality is necessary, because Hx is not uniquely determined by x (it has other members of G in it, as well), and if Hx ≠ xH (equivalently if xHx^-1 ≠ H), then the multiplication will not be well-defined.
there is a "canonical" homomorphism G-->G/H for any quotient group G/H of G, which sends the element g of G to the right (= left) coset Hg. the identity of G/H is He = H, which is also the kernel of the canonical homomorphism.
one way to look at such quotient (or factor) groups is to think of all the members of H being set arbitrarily to the identity, e. since the identity commutes with everything, the normality condition is akin to saying H has to "commute with every g" (this is NOT to say that hg = gh for every h in H. remember, we're "shrinking H to a point" so we can't really tell the difference between h and h' in H. so hg = gh' is the best we can say, they both get packed down into the coset Hg).
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# Compound interest with an equal monthly investment
The calculation of the accumulated amount with the monthly investiment
Сomply with request of user frouzen, who asked to make /571/ - calculating accreted amount when using compound interest and additional monthly investment. The calculation of interest is also expected to be monthly (most favorable case).
In order to do not distract the user from the calculator itself it's located below. Also there is a little bit of theory and formulas for those who need it.
Calculator
### Compound interest with an equal monthly investment
Digits after the decimal point: 2
Accreted amount
Save the calculation to reuse next time or share with friends.
The formula of compound interest, accruing a several times during the year is $S=P(1 + \frac{j}{m})^{mn}$, where m in our case is 12 and n - is a deposit period in years.
That's the simplest case when you make a contribution immediately and without further investment to it.
Now for the more complicated case - deposit replenishment with equal monthly installments..
Note that the factor degree mn nothing more than the number of periods of interest accrual.
Thus, for the very first deposit the accreted amount for the first few years will be the same.
$S_1=P(1 + \frac{j}{m})^{mn}$
For the deposit that has been made at the end of the first month, the amount of periods of interest accrual is one fewer and the formula would look like this
$S_2=P(1 + \frac{j}{m})^{mn-1}$,
for the third deposit - like this
$S_3=P(1 + \frac{j}{m})^{mn-2}$,
...
and for the last deposit, i.e. made in the las month before the end of the term - like this
$S_{mn}=P(1 + \frac{j}{m})$,
General result is a sum of all these expressions. And there are similarity in these expressions - they are terms of geometric progression in which the first term is equal to
$P(1 + \frac{j}{m})$ and the common ratio of a geometric progression is $1 + \frac{j}{m}$.
About the geometric progression see Geometric progression
Thus,the required amount of the sum of the geometric progression formula is
$S=\frac{a_nq-a_1}{q-1}=\frac{P(1 + \frac{j}{m})^{mn}(1 + \frac{j}{m})-P(1 + \frac{j}{m})}{\frac{j}{m}}$
That's all for today
Update
Added the ability to specify individual size of the first installment(according to the user's request).
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You are Here: Home >< Maths
# M2 Statics of rigid bodies
1. ABCD is a uniform square lamina of side 4m and mass 8Kg. It is hinged at A so that it is free to move in a vertical plane. It is maintained in equilibrium with B vertically below A by a horizontal force acting at C and a vertical force acting at D, each of magnitude F newtons. Find:
(a) the value of F
(b) the magnitude and direction of the force exerted by the hinge on the lamina.
I could do b if it weren't for a.
Any help?? - I have tried resolving around A to remove the hinges forces and I get
I may post some more up later.
Thanks
-Sohan-
2. You've done it correctly, but you've forgotten to take into account the direction in which the forces are acting. The first and last terms in your moment-taking above should be resolved, if you know what I mean. If you don't understand, I'll try and post some working.
3. (Original post by tommmmmmmmmm)
You've done it correctly, but you've forgotten to take into account the direction in which the forces are acting. The first and last terms in your moment-taking above should be resolved, if you know what I mean. If you don't understand, I'll try and post some working.
I'm confused. I missed the week we did this in class and I totally have no idea what to do. Some working would be great. Thanks.
4. Let me draw a diagram on paint, I'll be a few minutes
5. Did you mean statics?
6. (Original post by Mush)
Did you mean statics?
yes I did
7. There you go.
Do you see why the sin(theta)s are needed?
Attached Images
8. Am i correct in saying, everything has to be at right angles to the line connecting it with the moment point??
And why can it not be parrallel?
9. You need to use the component of the force resolved perpendicular to the line connecting it with the moment point.
If it was parallel, there would be no moment. Imagine if you had a sheet of square paper which was attached to the table at one corner with a drawing pin, if you pulled on the opposite corner in the direction of the diagonal from the pin to your hand, then there would be no moment clockwise or anti-clockwise. If you changed the direction you were pulling in, how would this affect the moment of your hand's force on the paper?
10. (Original post by tommmmmmmmmm)
You need to use the component of the force resolved perpendicular to the line connecting it with the moment point.
If it was parallel, there would be no moment. Imagine if you had a sheet of square paper which was attached to the table at one corner with a drawing pin, if you pulled on the opposite corner in the direction of the diagonal from the pin to your hand, then there would be no moment clockwise or anti-clockwise. If you changed the direction you were pulling in, how would this affect the moment of your hand's force on the paper?
I'd rep but I already did 2 days ago. Maybe I'll do what I have with Bruceleej and make you wait for your next one
Thanks so much.
-Sohan-
11. No problem Sohan
12. Another one: I get so many variables here though.
A uniform ladder rests with one end on rough horizontal ground and the other end against a rough vertical wall. The coefficient of friction between the ground and the ladder is and the coefficient of friction between the wall and the ladder is . The ladder is on the point of slipping when it makes an angle with the horizontal. Find .
13. At 2 in the morning?
Anyway, you should get five forces on your diagram:
- the normal reaction from the floor
- the normal reaction from the wall
- the frictional force exerted by the floor (make sure it's in the right direction)
- frictional force exerted by wall (ditto)
Let the length of the ladder = 2a or something similar.
Take moments (remember to use the sine of the acute angle that the force forms with the line of action of the moment), and resolve vertically and horizontally. You should get a nice set of simultaneous equations to solve
14. is this correct for a diagram?
Attached Images
15. (Original post by tommmmmmmmmm)
At 2 in the morning?
Oh and what the hell are you doing at 2 in the morining here?
I you really. If you weren't here then I'd have to wait a long long time lol.
Thanks sooooo much. Bruceleej's Rep timetable applies to you too now. I shall rep you every 28 days for a long long time.
And you shall go on my Sig.
-Sohan-
16. Nearly, you've got the direction of the frictional force at the top of the ladder wrong, though. Think about which way the ladder would move if it slipped; the frictional force must be opposed to it. It helps to picture the situation in your head.
17. (Original post by tommmmmmmmmm)
Nearly, you've got the direction of the frictional force at the top of the ladder wrong, though. Think about which way the ladder would move if it slipped; the frictional force must be opposed to it. It helps to picture the situation in your head.
I still cannot solve
with
18. Erm, no idea why I'm here at this time
I don't mind doing any of this, keep the questions coming because I've got to prepare for M2 as well (5 weeks and 3 days, scary thought, eh?)
Anyway, it's bedtime for me, much man-love
19. I'll leave you with a hint: divide through by cos(alpha)
20. love you in a very non homosexual manner
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# The Definition of Improper Fractions
Are you one of those students who hate studying algebra and improper fractions ? If your answer is positive, you should know that you’re not alone and many people agree that it seems completely pointless to translate word problems into math and use different alphabets to solve them. This discipline involves a number of subjects, such as improper fractions, but studying them is not as hard as you think. That’s because there are simple and effective tips that will help you do your algebra homework successfully, so stop wondering where you use it in your real life. Once you master basic skills, you will understand that this subject is quite easy, but learning improper fractions still requires enough practice. It’s only up to you if they will become your friends or nightmares, so be sure to learn how to identify their different types. Don’t forget that you can always get professional help because it’s provided by talented and trusted freelancers online. They will help you with anything, including narrative essays, at reasonable rates and fast speed.
Let’s imagine a situation where you and your friends are trying to decide what you will do the next weekend. For example, some of your friends like ice skating, while others prefer fishing or other recreational activities. This means that you’re at an impasse: a few votes for ice skating, and a few votes for fishing. A half of your group likes one thing, while others prefer something else. You will think about either proper or improper fractions once you start dividing this group into halves, regardless of whether you understand that or not.
Basically, fractions are used to describe how one part of this group related to the entire group. To illustrate this example clearly, it’s advisable to think about one relevant word «fracture». When you drop a plate, it will fracture into multiple pieces so that you may be concerned with picking up each one to recreate this place, thus ensuring that there are no pieces left on the ground. It’s obvious that this plate fractured into pieces, but you are still able to imagine it as the whole unit. Likewise, when writing assignments on this subject, keep in mind that fractions are used to represent the complete groups that have been broken apart or fractured in any way. They can help you understand how these pieces fit into an original group.
## Different Types of Fractions
What are improper fractions all about? To answer this question, you need to start with defining what a fraction is. As you already know, it can tell you how many parts of a whole you have, and it’s easy to recognize fractions by the slash written between 2 numbers. Pay attention to a numerator (a top number) and a denominator (a bottom number). As an example, ½ is a fraction, so feel free to write it with the slanted slash, and this is where one is a numerator and two is a denominator. You should use this knowledge if you don’t know how to start a thesis.
What do fractions mean? Imagine a pie and a bottom number tells you how many slices to cut it, while a top number tells you how many of them you can get. This means that ½ can tell you that you sliced this pie into 2 slices and you took only one of them, and it’s a half of a given pie. Besides, you need to understand that there are different types, including improper fractions.
Let’s start with proper and improper fractions. When dealing with the proper ones, you should realize that their numerators must be less than denominators. What about improper fractions ? Their numerators are greater than denominators (for instance, 8/7 is an improper type, while 7/8 is the proper one). You should go back to a pie and imagine yourself taking slices from it. When it comes to proper fractions, it’s possible to take all of them, but with improper fractions, you will need more than 1 pie to end up with the necessary number of slices. 7/8 tells you that you can take seven slices out of 1 pie that has eight slices. However, 8/7 tells you that you require eight slices, but a pie has only seven slices so that you can take only this number from one pie. To get the 8th slice, you require the second pie with the equal number of slices. Many math students think that improper fractions are quite greedy because they need more than 1 pie to get the necessary number, while the proper ones are easy to get from 1 pie.
Next, you should learn more about like and unlike fractions if your turabian paper is about them. Interested in like fractions? Then you should know that they are the same, while unlike fractions are the ones that are different. For instance, ½ and 2/4 are like fractions because they are the same. How is it possible? Imagine yourself cutting one pie into two slices and taking one slice, and then cut another pie into four slices and take two slices. How much of both pies did you take? It’s obvious that you took a half of the pie in both cases so that they are the same. You can also call them equivalent fractions. On the other hand, unlike fractions are the ones that are different (for example, 2/4 and 6/9). That’s because you end up with different fractions even after simplifying them.
## The Detailed Explanation of Improper Fractions
Do you know anyone who has broad shoulders and skinny legs? Then you should understand that this person looks quite similar to improper fractions. They are those fractions that have larger numbers on their top, not a bottom, and this number is called a numerator, while the number of their bottom is called a denominator. So, all improper fractions have greater numerators than denominators, and they are quite different from the proper ones that have greater denominators and smaller numerators. Are there any examples of improper fractions ? Once you find them, you will realize that fractions are considered improper if their numerators are bigger than denominators, no matter how large or small they are.
## Basic Questions about Improper Fractions
What do they mean? If your algebra homework includes improper fractions, keep in mind that their numerators are easy to translate as the amount of pieces or parts that you have, while denominators are the numbers of parts that something whole is divided into. For instance, if your fraction is 13/4, this means that have thirteen parts out of four pieces of the whole. Another method that you can use when looking at improper fractions is via the example of key lime pieces (13/4 means that you have thirteen pieces of them).
What about mixed and improper fractions ? The first one is a fraction and a whole number that are put together. Sometimes, you may need to convert improper fractions into the mixed ones, and that’s why you should learn how to complete this academic task successfully and easily. If it’s necessary to convert 13/4 to a mixed one, make sure that you use division. Divide a denominator into a numerator to do that, but you will end up with a remainder one. Whenever you see it, you need to put it over the denominator of improper fractions, thus making them mixed. If you have no remainder, you will end up with a whole number.
## Helpful Tips on How to Simplify Improper Fractions
To get this knowledge, you should start with understanding the key fundamentals, and there are certain steps involved. First, you need to recognize all parts of improper fractions, but remember that they all have 3 parts, including a denominator, a fraction bar, and a numerator. As you already know, the latter one is a top number and it represents the number of parts of a whole unit. A denominator is the bottom number of improper fractions, and its basic function is to represent the number of equal parts that this unit has. A fraction bar is either a diagonal or horizontal line that separates a denominator and a numerator. You can call it a division symbol so that all improper fractions can be viewed as specific division problems in algebra.
You also need to get a better understanding of proper fractions that represent numbers less than one and their numerators are always smaller than denominators. Be sure to determine improper fractions, and this task is easy if you know that they represent numbers greater than one, while their numerators are larger than denominators. It’s also advisable to learn how mixed numbers work, so don’t forget that they are those improper fractions that are rewritten like natural numbers and proper fractions.
Once you learn the above-mentioned things, you’re ready to find out more about simplifying improper fractions. Start with determining whether your given ones are really improper, and it’s easy to do that by checking their numerators and denominators. Ensure that you simplify them according to the following steps.
You need to divide numerators by denominators, but remember that fraction bars should be considered as specific division symbols. This knowledge will come in handy when dealing with improper fractions because their basic function is to represent those numbers that are larger than 1. Once you start dividing, it becomes possible to express these fractions as the right combination of its parts and wholes.
The next step that should be taken is calculating a remainder. If numerators were not divided evenly by given denominators, then you will end up with a remainder. If you want to calculate it properly, it’s necessary to multiply the number of whole pieces by denominators and subtract the result you get from the numerators of improper fractions. Don’t forget to write them as mixed numbers. Once you succeed to calculate remainders, you will be able to rewrite given improper fractions as mixed numbers. To achieve this goal, you only need to write wholes first and follow them with parts.
## Basic Rules for Subtracting and Adding Improper Fractions
Before you will become concerned with memorizing a completely new set of rules for all improper fractions, you should keep in mind that they involve the same set of rules as the ones applied for standard and proper fractions. When you need to subtract or add improper fractions, it’s necessary to start with finding common denominators between them. Look for the smallest number that all denominators will go into evenly. As a rule of thumb, it’s possible to multiply them together and use the result you get as a new denominator. If you think that this task is a bit confusing, entrust it to reliable and talented freelancers who offer their quality and affordable dissertation writing services.
## How to Complete Your Algebra Assignments
You should realize that this discipline is hard to memorize because all sets of problems, including improper fractions, are quite different so that they should be approached in different ways. The most effective thing that you can do is to understand how to approach a specific category. You need to use different resources when studying specific subjects, such as improper fractions. One of the best ways to learn important algebraic concepts is to use relevant interactive videos.
Make sure that you’re focused on these concepts because everything will come later. This subject is largely based on your personal understanding of concepts, so don’t try to solve given problems until you learn them. Remember that examples are your helpful tools, especially when studying any new topic, so look for them to learn how to solve a particular algebraic problem. Most professors advise their students to take into consideration at least a few approaches to end up with better results. Finally, you need to learn in terms of patterns.
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A075546 Perfect powers n such that (n-9)/2 is prime. 1
343, 1331, 12167, 29791, 42875, 79507, 103823, 300763, 357911, 857375, 1520875, 2248091, 2924207, 6436343, 9393931, 11089567, 11697083, 15069223, 15813251, 19487171, 19902511, 20796875, 22665187, 30080231, 51064811, 65450827, 77854483, 80062991, 99252847 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS n must be an odd power of a number congruent to 3 modulo 4. - Charlie Neder, Feb 11 2019 LINKS Charlie Neder, Table of n, a(n) for n = 1..841 EXAMPLE a(1)=7^3, a(2)=11^3, a(3)=23^3, a(4)=31^3, a(5)=35^3, a(6)=43^3, a(7)=47^3, a(8)=67^3, a(9)=71^3, a(10)=95^3, a(11)=115^3, a(12)=131^3, a(13)=143^3, a(14)=23^5 and a(15)=211^3; most are cubes. MATHEMATICA pp = Join[{1}, Select[ Range[10^7], Apply[GCD, Last[ Transpose[ FactorInteger[ # ]]]] > 1 & ]]; Select[pp, PrimeQ[(# - 9)/2] & ] CROSSREFS Sequence in context: A179147 A047739 A052084 * A167733 A167730 A250138 Adjacent sequences: A075543 A075544 A075545 * A075547 A075548 A075549 KEYWORD easy,nonn AUTHOR Zak Seidov, Oct 11 2002 EXTENSIONS Extended by Robert G. Wilson v, Oct 14 2002 a(16)=a(29) from Charlie Neder, Feb 11 2019 STATUS approved
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Last modified April 14 12:11 EDT 2021. Contains 342949 sequences. (Running on oeis4.)
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# Graphing tangent functions pdf
2019-09-15 13:12
Graphing Trig Functions Date Period Using degrees, find the amplitude and period of each function. Then graph. 1) y sin 3 60 120 180 240 300 360 Using radians, find the amplitude and period of each function. Then graph. 7)56 Graphing Basic Trigonometric Functions 393 more insight into the nature of these functions by observing how y sin x b varies as the circular point P(a, b) moves around the unit circle. graphing tangent functions pdf
The Graph of the Sine Function and the Unit Circle Recall from Chapter 9 that if ROP is an angle in standard position with mea sure u and P ( p, q ) is a point on the unit circle, then (
The tangent function has a parent graph just like any other function. Using the graph of this function, you can make the same type of transformation that applies to the parent graph of any function. The easiest way to remember how to graph the tangent function is to remember that some interesting Graphs of the Sine andCosine Functions To help us graph the sine and cosine functions, we rst observe that these functions repeat their The graph of the sine function is symmetric with respect to the origin. This is as expected, since sine is an odd function. Since the cosine function is an even function, its graph is symmetric graphing tangent functions pdf 4 Graph of the Tangent Function The tangent function is odd, tan(x) tan x. The graph of y tan x is symmetric with respect to the origin. tan x sin x cos x tangent is undefined for values at which cos x 0. Two such values are x 2 1. 5708.
Graphs of the Six Trigonometric Functions y sinx y cosx Domain: Domain: All Reals All Reals Range: Range: [1; 1 [1; 1 Period: 2 Period: 2 graphing tangent functions pdf in Section1. 4. For the function f(x) 1 5cos(2x ), the argument of fis x. We shall have occasion, however, to refer to the argument of the cosine, which in this case is 2x. Loosely stated, the argument of a trigonometric function is the expression inside the function. Example. Graph one cycle of the following functions. 832 Chapter 14 Trigonometric Graphs, Identities, and Equations For a0 and b0, the graphs of yasinbxand yacosbx each have five keyxvalues on the interval 0x2 b: the xvalues at which the maximumand minimumvalues occur and the xintercepts. Graphing Sine and Cosine Functions Graph the function. a. y 2 sin x b. y cos 2x SOLUTION a.
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# math i need help 3 questions
posted by .
1. The numerator and denominator of a fraction are in the ratio of 3 to 4. If 4 is added to the numerator and subtracted from the denominator, the value of the resulting fraction is 4/3. The numerator of the original fraction is
A. 12
B.
C. 3
D. 4
2. If it takes b hours to walk a certain distance at the rate of 3 miles an hour, the number of hours it takes to return the same distance at 4 miles an hour is
A. 4/3
B. 4b/3
C. 3b/4
D. 3/4
3. The perimeter of a rectangle is 20 inches and its length is 6 inches. The area of the rectangle in square is
A. 20
B. 24
C. 60
D. 48
• math i need help 3 questions -
number 1 choice b is 9
• math i need help 3 questions -
I'll do #1. How about you show some effort on the others? Just write the words in symbols.
If the fraction is n/d, then we are told
n/d = 3/4
(n+4)/(d-4) = 4/3
Work through all that, and you have n=12,d=16
check:
12+4 / 16-4 = 16/12 = 4/3
OK
## Similar Questions
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The numerator of a fraction is 7 less than the denominator if three is added to the numerator and 9 is subtracted from the denominator the new fraction is equal to 3/2. Find the original fraction.
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10. ### math
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#### Harmonograph Video Demo
07/09/2011
A video demonstration of the virtual harmonograph running through a series of sixteen preset patterns. [sociallocker][/sociallocker]
#### A 3-pendulum Animated Harmonograph Model – video preview
07/08/2011
A video review of the 3-pendulum animated harmonograph model together with the tutorials (part #1, part #2). [sociallocker][/sociallocker]
#### Instant 3D-2D Perspective Mapping – the Navigator Functions – a convenient pair of user defined functions
02/02/2011
This tutorial explains a pair of important user defined functions, the “Navigator_u()” and the “Navigator_v()”. These functions save the user nine columns of formulas by calculating the effects of: 3-dimenssional shift, rotations around the three axes of coordinates and 3D-2D perspective mapping. These user defined functions are also easy to use compared to writing all the perspective mapping equations from scratch. In the previous 3D perspective tutorials we took…
#### 3D-2D Perspective Mapping in Excel – part #2
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The first half of this presentation dealt with mapping a 3D scene onto a 2D surface which can be a computer screen, a projection screen or the retina of the eye. By doing so we preserve much of the feel of depth of the scene. Of course when we look at an picture we get only part of the feel of depth because our both eyes see the same image….
#### 3D-2D Perspective Mapping in Excel – part #1
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In computer graphics we often need to be able to display a three-dimensional image in two dimensions and preserve the perspective appearance. If we walk on a straight road, it appears that the road narrows with the distance. This is the perspective effect and it is a result of mapping a three-dimensional image on a two-dimensional surface (i.e. a computer monitor, a screen, or the retina of the eye). This tutorial…
#### Modeling a Three-Pendulum Harmonograph – tutorial: part #2
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Hi guys, here is the second part of a tutorial describing the matematical equations used in modeling a three-pendulum harmonograph (automatic drawing machine). It pertains to the second version of the model. This section describes the kinematic equations involved in the articulated linkage mechanism on the top of the table, the custom spreadsheet functions used in the model and some overall considerations about the partition of the spreadsheet. This is an…
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# Question Video: Calculating the Equilibrium Constant of the Weak Acid HPO₄²⁻
Consider the equilibrium of the HPO₄²⁻ ion acting as a weak base. HPO₄²⁻(aq) + H₂O(l) ⇌ H₂PO₄⁻(aq) + OH⁻(aq), What is the equilibrium constant for this reaction under these conditions? [OH⁻] = 1.3 × 10⁻⁶ M, [H₂PO₄⁻] = 0.042 M, [HPO₄²⁻] = 0.341 M.
03:20
### Video Transcript
Consider the equilibrium of the HPO₄²⁻ ion acting as a weak base. HPO₄²⁻ aqueous plus H₂O liquid in equilibrium with H₂PO₄⁻ aqueous plus OH− aqueous. What is the equilibrium constant for this reaction under these conditions? Concentration of our each minus equal to 1.3 times 10 to the minus six molars, concentration of H₂PO₄⁻ equal to 0.042 molars, and concentration of HPO₄²⁻ equal to 0.341 molars.
Before we go any further, it might be helpful to understand the names for a couple of these ions. Hydrogen phosphate and dihydrogen phosphate all have at their core the phosphate anion, PO₄³⁻. You can get the full name by just adding the number of hydrogens to the beginning: hydrogen for one or dihydrogen for two. In this equilibrium, hydrogen phosphate is acting as a weak base, taking a proton from a water molecule to form dihydrogen phosphate and hydroxide. our job is to figure out the equilibrium constant 𝑘 𝑐 for this reaction, where the concentrations of the components are specified.
Since here hydrogen phosphate is acting as a base, we can also describe this equilibrium as 𝑘 𝑏 of the hydrogen phosphate anion. For the equilibrium where chemicals 𝐴 and 𝐵 are in equilibrium with 𝐶 and 𝐷, this is the equilibrium constant expression. The presence of any pure liquids or pure solids is ignored. Only gases or solutes will appear in the equilibrium constant expression.
We start constructing our equilibrium constant expression by taking the concentrations of our products and multiplying them together. All the components of this equilibrium are in a one-to-one ratio with one another. So we don’t have any funny powers. We finished the expression by adding in the concentration of the hydrogen phosphate anion at the bottom, ignoring the concentration of liquid water. Now all that remains is to substitute in our concentration values, giving us 0.042 molars times 1.3 times 10 to the minus six molars all over 0.341 molars. This evaluates to 1.60117 times 10 to the minus seven molars.
The least significant value in our calculation was given to two significant figures. So we should give the answer to the same precision, meaning our final answer is 1.6 times 10 to the minus seven molars. However, equilibrium constants are conventionally given without units. So our final answer is simply 1.6 times 10 to the minus seven.
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# Thread: Proving that a trig equation follows from another given trig equation
1. ## Proving that a trig equation follows from another given trig equation
$\displaystyle \tan({\frac{\theta}{2}})=\sqrt{\frac{1-e}{1+e}}\tan({\frac{\phi}{2}})$
Then prove that $\displaystyle \cos\phi=\frac{\cos\theta-e}{1-e\cos\theta}$
2. Hello,
Let $\displaystyle t=\tan\tfrac \theta2$ and $\displaystyle p=\tan\tfrac\phi2$
We have $\displaystyle t=\sqrt{\frac{1-e}{1+e}} \cdot p$
Square it : $\displaystyle t^2=\frac{1-e}{1+e} \cdot p^2$
We know that $\displaystyle \cos(x)=\frac{1-\tan^2\tfrac x2}{1+\tan^2\tfrac x2}$ (that's used in Weierstrass substitution - go google)
So :
\displaystyle \begin{aligned}\cos \phi &=\frac{1-p^2}{1+p^2} \\ &=\frac{1-\frac{1+e}{1-e}\cdot t^2}{1+\frac{1+e}{1-e}\cdot t^2} \\ &=\frac{1-e-(1+e) t^2}{1-e+(1+e) t^2} \\ &=\frac{1-t^2-e(1+t^2)}{1+t^2-e(1-t^2)}\end{aligned}
$\displaystyle ...=\frac{\frac{1-t^2}{1+t^2}-e}{1-e \cdot\frac{1-t^2}{1+t^2}}$
$\displaystyle ...=\frac{\cos\theta-e}{1-e\cos\theta}$
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# Really Basic A level Statistics Test. (Multiple Choice)
#### Alice15492
##### New Member
1) ‘Cluster sampling’ is a term given to one method of
(a) Selecting random numbers using excel
(b) Probabilistic sample selection
(c) Snowballing
(e) Selecting a population
2) ‘Stratified sampling’ may provide greater precision of population estimates, compared to ‘simple random sampling’ because it
(a) only applies to collecting secondary data
(b) requires the whole population to be sampled
(c) makes use of supplementary information
(d) always divides the population into geographic areas
(e) all of the above
Data for questions 5, 6 and 7
The data below describes the test marks from a class of fifteen students. They represent the marks from the full group, ie the total population. Use this data for questions 5 and 6 and 7.
Marks: 38, 60, 63, 45, 47, 45, 72, 75, 31, 51, 72, 66, 62, 54, 59.
5) Are the population mean and the median (to 2dp):
(a) 56 and 56.5
(b) 56 and 59
(c) 58 and 59.5
(d) 58 and 60
(e) None of the above
Continued…
6) Are the range and the average deviation (to 2dp):
(a) 44 and 0
(b) 44 and 10.8
(c) 37 and 0
(d) 75 and 10.8
(e) None of the above
7) Are the standard deviation and coefficient of variation from this population (to 2 dp):
(a) 12.61 and 0.23
(b) 12.61 and 4.44
(c) 13.05 and 0.23
(d) 13.05 and 4.29
(e) None of the above
8) A population was found to have a standard deviation of 3, a mean of 14, a mode of 8, and a median of 10. Is the frequency distribution:
(a) Positively skewed
(b) Negatively skewed
(c) A completely symmetrical distribution
(d) Bimodal
(e) Cannot be described without further information
Use the data below for questions 9, 10, 11 and 12. The data describes the ages of the whole population on a mailing list. It is being evaluated by a company that is considering buying the list.
The ages (in years) of the 50 individuals on the mailing list, in ascending order are:
10,11,13,14,14,15,15,17,18,20,20,21,22,24,26,26,26,27,29,30,31,31,33,34,36,36,36,37,38,
39,40,41,41,43,44,45,45,46,47,48,49,51,53,55,57,58,59,59,60,60.
These 50 ages average 35.
9) Is the variance of this raw data (to two decimal places):
(a) 14.86
(b) 216.4
(c) 220.522
(d) 10820.0
(e) none of the above
Put the raw data into class intervals of:
Class Mid Point (xi)
0 -< 16 12
16 -< 21 18.5
21 -< 30 25.5
30 -< 40 35
40 -< 60 50
> 60 65
10) Is the mean of the grouped data (to 2 dp):
(a) 8.63
(b) 35.0
(c) 35.54
(d) 206.0
(e) None of the above
11) You wish to consider the accuracy of your estimate in the previous
question. The percentage error for the estimate of the mean of the
grouped data is (to 2 dp):
(a) – 0.54%
(b) – 0.2
(c) 0%
(d) + 0.02%
(e) None of the above
12) Is the median of the grouped data (to2 dp):
(a) 32.17
(b) 35.32
(c) 35.91
(d) 45.91
(e) None of the above
13) When calculating the median from grouped data which of the
following is assumed:
(a) All observations are integers
(b) All observations are spread evenly throughout their class
(c) All classes are of the same width
(d) All classes are of different widths
(e) All observations are located at the class mid point
14) For a standard normal (Z) variable ie. Z~N(0,12), p(1.6<Z<1.8) is (to accuracy of 4dp):
(a) 0.0189
(b) 0.4811
(c) 0.5189
(d) 0.9093
(e) None of the above
15) For a standard normal (Z) variable ie. Z~N(0,12), p(Z > -2.22) is (to accuracy of 4dp):
(a) 0.0132
(b) 0.4861
(c) 0.4868
(d) 0.9861
(e) 0.9868
#### Dason
Hi! :welcome: We are glad that you posted here! This looks like a homework question though. Our homework help policy can be found here. We mainly just want to see what you have tried so far and that you have put some effort into the problem. I would also suggest checking out this thread for some guidelines on smart posting behavior that can help you get answers that are better much more quickly.
#### Alice15492
##### New Member
I tried some. i want to double check my answers.
#### Dason
Then tell us what you got for the answers and your reasoning.
#### Alice15492
##### New Member
1B 2C 5B 6B 7C 8A 9C 10C 11A 12E 13C 14 A 15 A
This is due tomorrow. i would really appreciate if you could check my answers.
this counts as 10% of my final mark!
#### Alice15492
##### New Member
its a multiple choice question. they don't need my reasoning. but u want me to spent my time typing it here?
i thought this was a forum that would help me. not ppl that judge me and i have to prove my self first to get advice.
ill say it again that this is for tomorrow morning. i dnt have the time to satisfy you.
im really disappointed and sad with your replies and the way u talk to me.
#### CB
##### Super Moderator
If you aren't willing to take the time to read and follow our homework policy, do you think it's really fair to demand us to take time out of our days to help you?
#### trinker
##### ggplot2orBust
This is what I got:
1H 2C 5W 6B 7R 8A 9C 10B 11A 12E 13L 14 A 15 T
But I didn't use any reasoning because it's multiple choice.
Alice15492 we really want to help people understand stats better. We love stats and want other people to be passionate too. When you post an entire assignment and ask us to check it it really cheapens what we do. Many of us are actually professionals some even profs and providing you with a straight up check of your HW would be at best unethical. Certain questions or concepts? Sure we're here for that but checking answers to an assignment really isn't what we do.
Are there particular problem's you're struggling with? Concepts?
#### Alice15492
##### New Member
i had 24 hours. anw doesnt matter.
thank you
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###### Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
##### Thank you for watching the video.
To unlock all 5,300 videos, start your free trial.
# Factoring Trinomials, a = 1 - Problem 2
Alissa Fong
###### Alissa Fong
MA, Stanford University
Teaching in the San Francisco Bay Area
Alissa is currently a teacher in the San Francisco Bay Area and Brightstorm users love her clear, concise explanations of tough concepts
Share
Here I’m given a trinomial with the leading coefficient as one. But I have something that’s’ a little tricky here I have a minus sign so when I’m asked to factor this, this tells me the product of two numbers gives me the answer -10. If you have 2 numbers and the answer is negative that means one of your numbers has to be a negative value. I’m going to keep that in mind while I’m looking for things that multiply to -10 and add up to +3. So numbers that multiply to 10 might be a pair of 1 and 10 or 2 and 5 and that’s it, those are the only pairs. So it’s going to be some combination of these with some pluses and minuses. One has to be negative; one has to be positive so that when I add them up the answer is positive three. A lot of you guys can do that pretty quickly.
Notice that if I were to use +5 and -2 the product would be -10 and the sum would be +3. So there’s my factors, p minus 2 and p plus 5. That’s the factored form of this trinomial. Check your work by foiling just to make sure firsts, outers, inners, lasts combine this and you’ll see that we did do it correctly. We have 3p as the middle term there and we know we did it right.
So what I want to leave you with is if you see a negative signs before your c term that means one of your numbers that go into the factors is going to be negative the other one’s going to be positive. So you want to keep that in mind when you’re looking for your factor pairs for the c term.
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# What is the piecewise function formula?
## What is the piecewise function formula?
A piecewise function is a function built from pieces of different functions over different intervals. For example, we can make a piecewise function f(x) where f(x) = -9 when -9 < x ≤ -5, f(x) = 6 when -5 < x ≤ -1, and f(x) = -7 when -1
## What are the different types of piecewise functions?
Piecewise Functions
• Absolute value functions.
• Floor function.
• Ceiling function.
• Sign function.
What are the 8 types of functions?
The eight types are linear, power, quadratic, polynomial, rational, exponential, logarithmic, and sinusoidal.
### Which is an example of a piecewise function?
A piecewise function is a function built from pieces of different functions over different intervals. For example, we can make a piecewise function f (x) where f (x) = -9 when -9 < x ≤ -5, f (x) = 6 when -5 < x ≤ -1, and f (x) = -7 when -1
### Which is the domain of a piecewise function?
The Domain (all the values that can go into the function) is all Real Numbers up to and including 6, which we can write like this: What is h (−1)? What is h (1)? What is h (4)? Piecewise functions let us make functions that do anything we want! Example: A Doctor’s fee is based on the length of time.
Is the absolute value function a piecewise function?
The Absolute Value Function. The Absolute Value Function is a famous Piecewise Function. It has two pieces: below zero: -x; from 0 onwards: x; f(x) = |x| The Floor Function. The Floor Function is a very special piecewise function. It has an infinite number of pieces: The Floor Function
#### Which is the first piecewise function on Ka?
Hey! Algebra II is the first time piecewise functions are explained on KA. The playlist ‘Domain and Range’ (Which includes the exercise ‘Domain of a Function’) is on both Algebra I & II. Comment on Clare’s post “Hey!
26/01/2021
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0
# What is 2c plus 4-3c equals -9 plus c plus 5?
Updated: 4/28/2022
Wiki User
ā 12y ago
2c + 4 - 3c = -9 + c + 5
-c + 4 = c - 4
-2c = -8
2c = 8
c = 4
Wiki User
ā 12y ago
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# The velocity-time graph of a particle in one-dimensional motion is shown in Fig. Which of the following formulae are correct for describing the motion of the particle over the time-interval t1 to t2 : (a) x(t2)=x(t1)+υ(t1)(t2−t1)+(1/2)a(t2−t1)2 (b) υ(t2)=υ(t1)+a(t2−t1) (c ) υaverage=(x(t2)−x(t1))/(t2−t1) (d) aaverage=(υ(t2)−υ(t1))/(t2−t1) (e ) x(t2)=x(t1)+υaverage(t2−t1)+(1/2)aaverage(t2−t1)2 (f) x(t2)−x(t1)= area under the υ−t curve bounded by the t-axis and the dotted line shown.
Text Solution
Verified by Experts
## (c ), (d), (f)
|
Updated on:21/07/2023
### Knowledge Check
• Question 1 - Select One
## In any progression, if t2t3t1t4=t2+t3t1+t4=3(t2−t3t1−t4), then t1,t2,t3,t4 are in
AA.P.
BG.P.
CH.P.
Dnone
• Question 2 - Select One
## For any arbitrary motion in space, which of the following relations are true? a) vaverage=(1/2)(v(t1+v(t2) b) vaverage=[r(t2)−rt1t2−t1 v(t)=v(0)+at d) aaverage=[v(t2)−vt1t2−t1 The average stands for average of the quantity over time interval t1 to t_(2)
Avaverage=12[v(t1)+v(t2)]
Bvaverage=r(t2)r(t1)t2t1
Cv(t)=v(0)+at
Dr(t)=r(0)+v(0)t+12at2
• Question 3 - Select One
## If (12−t1)+(22−t2)++(n2−tn)+=n(n2−1)3 , then tn is equal to n2 b. 2n c. n2−2n d. none of these
An/2
Bn - 1
Cn + 1
Dn
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Wierzycaz
2020-10-27
If possible, find scalars so that
${c}_{1}\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]+{c}_{2}\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]+{c}_{3}\left[\begin{array}{c}-1\\ 4\\ -1\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]$
Delorenzoz
Step 1
Consider the equation:
${c}_{1}\left[\begin{array}{c}1\\ 2\\ -3\end{array}\right]+{c}_{2}\left[\begin{array}{c}-1\\ 1\\ 1\end{array}\right]+{c}_{3}\left[\begin{array}{c}-1\\ 4\\ -1\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]\dots \left(1\right)$
Step 2
Now, when a matrix is multiplied by a scalar quantity, then all elements of matrix is multiplied by that scalar quantity.
Using this principle, rewrite equation (1) as:
$\left[\begin{array}{c}{c}_{1}\\ 2{c}_{1}\\ -3{c}_{1}\end{array}\right]+\left[\begin{array}{c}-{c}_{2}\\ {c}_{2}\\ {c}_{2}\end{array}\right]+\left[\begin{array}{c}-{c}_{3}\\ 4{c}_{3}\\ -{c}_{3}\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]\dots \left(2\right)$
step 3
Now again,
Addition of matrices can be done by adding same position of elements together, thus creating a new single matrix.
From (2):
$\left[\begin{array}{c}{c}_{1}-{c}_{2}-{c}_{3}\\ 2{c}_{1}+{c}_{2}+4{c}_{3}\\ -3{c}_{1}+{c}_{2}-{c}_{3}\end{array}\right]=\left[\begin{array}{c}2\\ -2\\ 3\end{array}\right]$
Step 4
Equating rows from both sides of matrices:
${c}_{1}-{c}_{2}-{c}_{3}=2$
${c}_{1}=2+{c}_{2}+{c}_{3}\dots \left(3\right)$
and
$2{c}_{1}+{c}_{2}+4{c}_{3}=-2$
$3{c}_{2}+6{c}_{3}=-6$
${c}_{2}+2{c}_{3}=-3\dots \left(4\right)$
Step 5
and
$-3{c}_{1}+{c}_{2}-{c}_{3}=3$
$-2{c}_{2}-4{c}_{3}=9$
${c}_{2}+2{c}_{3}=-4.5\dots \left(5\right)$
From (4) and (5): Since L.H.S. of both equations are same but R.H.S. are not same, It is impossible to find the scalar quantity .
Jeffrey Jordon
Answer is given below (on video)
Jeffrey Jordon
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Next: V-filtration and spectral numbers Up: Effective Construction of Algebraic Previous: Computing a basis for
## Example
The above algorithms are implemented in the library brnoeth.lib of SINGULAR, together with procedures for coding and decoding and are distributed with version 2.0. To compute an example, we first have to load the library.
LIB "brnoeth.lib";
Let be the absolutely irreducible plane projective curve given by the affine equation . We compute all places up to degree , by
ring r=2,(x,y),dp;
poly f=y2+y+x5;
==> The genus of the curve is 2.
CURVE=NSplaces(3,CURVE); // places up to degree 4=1+3
We can consider the curve as being defined over in order to get many rational places.
CURVE=extcurve(4,CURVE);
==> Total number of rational places : NrRatPl = 33
The degree of the computed (conjugacy classes of) places is displayed by
list L=CURVE[3]; L;
==> [1]: 1,1 [2]: 1,2 [3]: 1,3
==> [4]: 2,1
==> [5]: 3,1 [6]: 3,2
==> [7]: 4,1 [8]: 4,2 [9]: 4,3 [10]: 4,4
==> [11]: 4,5 [12]: 4,6 [13]: 4,7
In particular, besides the rational places over there are closed places of degree . The adjunction divisor is given by , where is the unique (rational) point on mapping to the singular point . This can be read off as follows:
CURVE[4]; // the mult's d_Q at L[1],L[2],...
// (zeroes omitted)
==> 8
def r1=CURVE[5][1][1];
setring r1;
POINTS[1]; // coordinates of the base point of L[1]
==> [1]: 0 [2]: 1 [3]: 0
PARAMETRIZATIONS[1]; // parametrization of L[1]
==> [1]: _[1]=t3+t8
_[2]=t5+t15
// exact up to order:
[2]: 8,10
We construct the evaluating AG-code where all rational points of appear in the support of and .
intvec G=0,0,0,0,1,1,0,0,0,0,0,0,0;
intvec D=1..33;
def R=CURVE[1][4];
setring R;
matrix CODE=AGcode_L(G,D,CURVE);
The echelon form of the resulting -matrix is
Note that the constructed code has block length 33, dimension 5 and designed distance . On the other hand, the first row corresponds to a word of weight 27, whence the designed distance coincides with the minimal distance. As a result, we get that is a -code. Note that the parameters, that is, the information rate and the relative minimum distance , lie above the Gilbert-Varshamov bound.
Next: V-filtration and spectral numbers Up: Effective Construction of Algebraic Previous: Computing a basis for
Christoph Lossen
2001-03-21
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## Integrating polynomials over polygonal domains in 2D
Posted by Diego Assencio on 2014.03.27 under Computer science (Numerical methods)
In a previous post, I have described how one can integrate polynomials over polygonal curves in two dimensions. This post extends the previous one by describing how one can integrate polynomials over polygonal domains (whose boundaries do not self-intersect).
Let $p_d(x,y)$ be a polynomial of degree $d$: $$p_d(x,y) = \sum_{m,n = 0}^{d} c_{mn}x^m y ^n$$ For instance, if $d=2$: $$p_2(x,y) = c_{00} + c_{10}x + c_{01}y + c_{20}x^2 + c_{11}xy + c_{02}y^2 \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_polyn_deg_2}$$ Let $\Omega$ be a two-dimensional polygonal domain and let $S$ be the positively oriented boundary of $\Omega$. We wish to compute the area integral of $p_d(x,y)$ over $\Omega$: $$\int_{\Omega} p_d(x,y) dxdy \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_int_p_omega}$$ The curve $S$ is a chain of $n$ segments $S_i$, $i = 1,2,\ldots,n$. Let $L_i$ be the length of $S_i$ and let $(x_i,y_i)$ and $(x_i + \Delta{x_i},y_i + \Delta{y_i})$ be the coordinates of its end nodes (see figure 1).
Fig. 1: A polygonal domain $\Omega$ and its positively oriented boundary $S$; $(x_{i+1},y_{i+1}) = (x_i + \Delta{x_i}, y_i + \Delta{y_i})$ for $i = 1,2,\ldots,6$ (assuming that $(x_7,y_7) = (x_1,y_1)$).
To compute $\int_{\Omega} p_d(x,y) dxdy$, we can use the divergence theorem. First notice that: $$p_d(x,y) = \frac{d}{dx}\left(\sum_{m,n = 0}^{d} c_{mn} \frac{x^{m+1} }{m+1}y^n\right) = \nabla\cdot\binom{q_{d+1}(x,y)}{0} \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_pd_div}$$ where $$q_{d+1}(x,y) = \sum_{m,n = 0}^{d} c_{mn} \frac{x^{m+1} }{m+1}y^n \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_def_qdp1}$$ is a polynomial of degree $(d+1)$. Integrating both sides of equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_pd_div} yields: $$\begin{eqnarray} \int_{\Omega} p_d(x,y) dxdy &=& \int_{\Omega} \nabla\cdot\binom{q_{d+1}(x,y)}{0} dxdy \nonumber\\[5pt] &=& \int_{S} \binom{q_{d+1}(x,y)}{0} \cdot {\bf n}(x,y) dl \nonumber\\[5pt] &=& \int_{S} q_{d+1}(x,y)n^1(x,y) dl \nonumber\\[5pt] &=& \sum_{i=1}^{n} \int_{S_i} q_{d+1}(x,y)n^1(x,y) dl \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_int_qdp1n1} \end{eqnarray}$$ where ${\bf n}(x,y)$ is the outward unit normal to the surface $S$ (remember $S$ is positively oriented) and $n^1(x,y)$ is its $x$ component. Over the extension of each segment, the normal ${\bf n}(x,y)$ is constant. Denoting the normal along the segment $S_i$ by ${\bf n}_i$, we can then write equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_int_qdp1n1} as: $$\int_{\Omega} p_d(x,y) dxdy = \sum_{i=1}^{n} n^1_i \int_{S_i} q_{d+1}(x,y) dl \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_int_qdp1}$$ An expression for ${\bf n}_i$ can be obtained by rotating the vector $(\Delta{x_i},\Delta{y_i})$ by $\pi/2$ in the clockwise direction (remember that $(\Delta{x_i},\Delta{y_i}) = (x_{i+1},y_{i+1}) - (x_i,y_i)$ is the vector which connects the two end points of $S_i$) and normalizing the resulting vector: $$\displaystyle{\bf n}_i = \frac{(\Delta{y_i},-\Delta{x_i})}{\sqrt{\Delta{y_i}^2 + \Delta{x_i}^2}} \Longrightarrow n^1_i = \frac{\Delta{y_i}}{\sqrt{\Delta{x_i}^2 + \Delta{y_i}^2}} = \frac{\Delta{y_i}}{L_i} \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_n_i}$$ Although the right-hand side of equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_int_qdp1} can be computed using the method described in the previous post mentioned above, we will use another approach which yields a simpler expression for the area integral we wish to compute.
To begin, notice that segments having $\Delta{y_i} = 0$ (horizontal segments) do not contribute to the sum in equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_int_qdp1} since for these segments $n^1_i = 0$. In other words, we need to consider only the segments $S_i$ for which $\Delta{y_i} \neq 0$.
We can parameterize each segment $S_i$ as shown below: $${\bf r}_i(s) = \left( \tilde{x}_i(s),\tilde{y}_i(s) \right) = (x_i + s\Delta{x_i}, y_i + s\Delta{y_i}),\quad s \in [0,1] \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_param_segm}$$ Let ${\bf r}_i'(s)$ be the derivative of ${\bf r}_i(s)$ with respect to $s$. The length of ${\bf r}_i'(s)$ is: $$\big\|{\bf r}'_i(s)\big\| = \big\|(\Delta{x_i},\Delta{y_i})\big\| = \sqrt{\Delta{x_i}^2 + \Delta{y_i}^2} = L_i$$ We can now use the parameterization from equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_param_segm} as well as the definition of line integral to compute the right-hand side of equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_int_qdp1}: $$\begin{eqnarray} \int_{\Omega} p_d(x,y) dxdy &=& \sum_{i=1}^{n} n^1_i \int_{S_i} q_{d+1}(x,y) dl \nonumber\\[5pt] &=& \sum_{i=1}^{n} n^1_i \int_{0}^{1} q_{d+1}(\tilde{x}_i(s),\tilde{y}_i(s)) \big\|{\bf r}'_i(s)\big\|ds \nonumber\\[5pt] &=& \sum_{i=1}^{n} n^1_i \int_{0}^{1} q_{d+1}(x_i + s\Delta{x_i}, y_i + s\Delta{y_i}) L_i ds \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_line_int_q} \end{eqnarray}$$ Consider now the following change of variable (from now on we will explicitly consider only segments with $\Delta{y_i} \neq 0$): $$y = y_i + s\Delta{y_i} \Longrightarrow dy = \Delta{y_i}ds \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_y_func_s}$$ As for the $x$ part, \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_y_func_s} implies: $$\displaystyle x_i + s\Delta{x_i} = x_i + (y - y_i)\frac{\Delta{x_i}}{\Delta{y_i}} = x_i + m_i(y - y_i), \quad m_i = \frac{\Delta{x_i}}{\Delta{y_i}} \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_x_func_s}$$ Using equations \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_y_func_s} and \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_x_func_s} on equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_line_int_q}, we obtain: $$\begin{eqnarray} \int_{\Omega} p_d(x,y) dxdy &=& \sum_{\underset{\Delta{y_i} \neq 0}{i=1}}^{n} n^1_i L_i \int_{y_i}^{y_{i+1}} q_{d+1}(x_i + m_i(y - y_i), y) \frac{1}{\Delta{y_i}} dy \nonumber\\[5pt] &=& \sum_{\underset{\Delta{y_i} \neq 0}{i=1}}^{n} \frac{n^1_i L_i}{\Delta{y_i}} \int_{y_i}^{y_{i+1}} q_{d+1}(x_i + m_i(y - y_i), y) dy \nonumber\\[5pt] &=& \sum_{\underset{\Delta{y_i} \neq 0}{i=1}}^{n} \int_{y_i}^{y_{i+1}} q_{d+1}(x_i - m_iy_i + m_i y, y) dy \nonumber\\[5pt] &=& \sum_{\underset{\Delta{y_i} \neq 0}{i=1}}^{n} \sum_{m,n=0}^{d} c_{mn} \int_{y_i}^{y_{i+1}} \frac{ (x_i - m_iy_i + m_i y)^{m+1} }{m+1} y^n dy \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_int_pd_interm} \end{eqnarray}$$ where equations \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_n_i} and \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_def_qdp1} were used. Since: $$(x_i - m_iy_i + m_iy)^{m+1} = \sum_{p=0}^{m+1} \binom{m+1}{p} (x_i - m_iy_i)^p m_i^{m+1-p} y^{m+1-p}$$ then equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_int_pd_interm} can be written as: $$\int_{\Omega} p_d(x,y) dxdy = \sum_{\underset{\Delta{y_i} \neq 0}{i=1}}^{n} \sum_{m,n=0}^{d} \sum_{p=0}^{m+1} \beta_{mnip}\int_{y_i}^{y_{i+1}} y^{m+1-p+n} dy \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_int_pd_interm2}$$ where (the reason for defining $\beta_{mnip} = 0$ when $\Delta{y_i} = 0$ will become clear soon): $$\beta_{mnip} = \begin{cases} c_{mn} \binom{m+1}{p} \frac{(x_i - m_iy_i)^p}{m+1} m_i^{m+1-p} & \textrm{ if } \Delta{y_i} \neq 0\\[5pt] 0 & \textrm{ otherwise } \end{cases} \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_beta}$$ where $m_i$, defined in equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_x_func_s}, is given by: $$\displaystyle m_i = \frac{\Delta{x_i}}{\Delta{y_i}}$$ Finally, computing the integrals over $y$ on equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_int_pd_interm2} gives us our final result: $$\boxed{ \displaystyle \int_{\Omega} p_d(x,y) dxdy = \sum_{i=1}^{n} \sum_{m,n=0}^{d} \sum_{p=0}^{m+1} \beta_{mnip}\frac{\left(y_{i+1}^{m+2-p+n} - y_i^{m+2-p+n}\right)}{m+2-p+n} } \label{post_f8cf6364586db30c14e1742f378bbc1c_eq_symb_integ_poly_over_domain}$$ On the equation above, the terms corresponding to segments with $\Delta{y_i} = 0$ vanish automatically since for those terms we have $\beta_{mnip} = 0$. This allows us to remove the restriction $\Delta{y_i} \neq 0$ from the sum over segments.
Unfortunately equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_symb_integ_poly_over_domain} is not very efficient for numerical computations (although it can be used as a last resource). Whenever possible, it is better to compute the integrals explicitly and hard-code them. For example, if for a given problem all polyomials that must be integrated have degree $d=2$ (as in equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_polyn_deg_2}), then one can use equation \eqref{post_f8cf6364586db30c14e1742f378bbc1c_eq_symb_integ_poly_over_domain} to obtain a formula which is easy to hard-code: $$\begin{eqnarray} \int_{\Omega} p_2(x,y) dxdy = \sum_{i=1}^{n} \Delta{y_i} \bigg[ \; & & c_{00} \left( \frac{\Delta{x_i}}{2} + x_i \right) \nonumber\\[5pt] \; &+ & c_{10} \left( \frac{\Delta{x_i}^2}{6} + \frac{\Delta{x_i} x_i}{2} + \frac{x_i^2}{2} \right) \nonumber\\[5pt] \; &+ & c_{01} \left( \frac{\Delta{x_i}\Delta{y_i}}{3} + \frac{\Delta{x_i} y_i + \Delta{y_i} x_i}{2} + x_iy_i \right) \nonumber\\[5pt] \; &+ & c_{20}\left( \frac{\Delta{x_i}^3}{12} + \frac{x_i\Delta{x_i}^2}{3} + \frac{x_i^2\Delta{x_i}}{2} + \frac{x_i^3}{3} \right) \nonumber\\[5pt] \; &+ & c_{11}\bigg( \frac{\Delta{x_i}^2\Delta{y_i}}{8} + \frac{\Delta{x_i}\Delta{y_i} x_i}{3} + \frac{x_i^2 \Delta{y_i}}{4} \nonumber\\[5pt] \; & & \quad \quad + \frac{\Delta{x_i}^2 y_i}{6} + \frac{\Delta{x_i} x_i y_i}{2} + \frac{x_i^2 y_i}{2} \bigg) \nonumber\\[5pt] \; &+ & c_{02} \bigg( \frac{\Delta{x_i}\Delta{y_i}^2}{4} + \frac{2\Delta{x_i}\Delta{y_i} y_i}{3} + \frac{\Delta{x_i} y_i^2}{2} \nonumber\\[5pt] \; & & \quad \quad + \frac{x_i\Delta{y_i}^2}{3} + \Delta{y_i} y_i x_i + x_i y_i^2 \bigg) \bigg] \end{eqnarray}$$
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9 teachers like this lesson
Print Lesson
## Objective
SWBAT explain radian measure as the length of an arc on the unit circle. SWBAT convert from degree to radian measures.
#### Big Idea
Why should we divide the circle into 360 parts? Isn't there a more natural unit? Yes - the radius!
## Extending the Domain
10 minutes
Based on the practice problems from the previous lesson, the class is now prepared to extend the domain of the trigonometric functions to include all real numbers. All we need to do is use the unit circle to show that it is reasonable to include angle values greater than one full rotation and to assign a meaning to negative angle measures.
I will ask the class to begin by taking a couple of minutes to check their solutions to the practice problems with one another. Together they should be able to catch some of the simpler mistakes.
After a few minutes, I'll begin calling on students to read their answers one-by-one. (On this set of problems, I'll accept decimal approximations for the radical values, but I'll make a point of asking for the exact values, as well (MP 6).) If the entire class agrees (and they're correct), then we'll move quickly to the next problem. There are a few things I'll emphasize along the way:
• How to sketch the reference triangle once we leave the first Quadrant.
• The repetition of the two special right triangles.
• The correct interpretation of an angle greater than 360 degrees. (Can anyone think of a real-life situation in which these angles might arise?)
• The reasonable interpretation of negative angles.
On this last point, I think it's very important to make the analogy with negative numbers explicit. How did mathematicians come to accept the existence of negative numbers? By interpreting positive & negative as different directions on the number line. In exactly the same way, we interpret positive & negative angle measures as rotations in different directions from the positive real axis (MP 2).
15 minutes
With the unit circle firmly established as the key to a new understanding of the trigonometric functions (they aren't ratios anymore), it's time to take the next step: introducing radian measure. I'll begin with the simple question, "We've been talking about degrees quite a bit. Can anyone tell me what a degree is?" My follow-up to this question is "Why are there 360 of them in a full rotation?" (MP 2)
This opening should be enough to launch a very brief overview of the history of degrees, minutes, and seconds. Along the way, I'll emphasize the fact that there isn't anything about the degree that makes it a natural way to measure angles. With this little prelude, I'll introduce the concept of radian measure as the length of the arc on the unit circle subtended by the central angle. The idea is simple: We could just as easily measure how far "around the circle" based on the arc length as on the central angle.
There's a pretty good explanation of this notion here. The main points are:
• Degree measure is used for historical/cultural reasons
• Radians were developed for mathematical reasons
• The radius, diameter, and circumference are the "natural" candidates for a new unit.
• 1 radian is defined as the angle that is subtended by an arc equal to 1 radius
• Example of how to convert from degrees to radians and vice versa
My students should already be very comfortable with other kinds of unit conversions, and this is really no different. The essential fact is that 360 degrees equals 2 pi radians.
20 minutes
Now that students know what radians are, the only thing left is to gain some familiarity. I'll handout the problem set, Radian Practice, and ask everyone to begin by working individually. As they begin, I'll move around the room checking for understanding and making sure that no one's simply letting their calculator do the conversions for them. If students are giving the radian measure in decimals, I'll ask them to also provide the measure in terms of pi.
Students will continue working on their own until I'm satisfied that everyone knows how to do the conversions in both directions. At that point, I'll let them begin working in small groups if they'd like. These problems are so straightforward, however, that they really should only be using their classmates to check their work.
On problem 3, watch to see who notices that parts b and d are negative. I will not warn the class about these ones; it's more instructive to make the mistake and have to correct it.
Problems 4 - 11 are intended to draw out some of the advantages of radian measure over degree measure. They should see that both arc lengths sector areas are easier to calculate when the angle is given in radians. Problems 10 and 11 lead to the conclusion that when x is close to zero, we can say that sin(x) is approximately equal to x but only when x is measured in radians. This will be vitally important in calculus, but for now it's merely a curiosity. We will discuss all of this tomorrow.
## Wrapping Up
5 minutes
Before class ends, I will review the solutions to problems 1 - 3 of Radian Practice with the class and we will discuss any issues that arise (especially the negative solutions in 3b and 3d). Depending on how far the class has progressed, we may also review the numerical solutions to problems 4, 5, 8, and 9. The rest will be homework and fodder for discussion tomorrow.
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# One Variable Statistics
## • Create dot plots, histograms, and box plots. • Use available classroom technology to create histograms and box plots and calculate measures of center and spread. • Use terms such as “flat,” “skewed,” “bell-shaped,” and “symmetric” to describe data distributions. • Analyze and compare data sets. • Understand relationships between mean and median for symmetrical and skewed data distributions. • Recognize outliers when they exist, and know to investigate their source. • Know that outliers affect the mean, but not the median of a data set. • Describe variability by calculating deviations from the mean. • Compare two data sets with the same means but different variabilities, and contrast them by calculating the deviation of each data point from the mean. • Understand that IQR is a description of variability better suited to a skewed distribution. • Work with two-way tables.
Instead of creating representations of data, the emphasis in high school is on interpreting representations and judiciously interpreting measures of center and spread. Students describe the shape of a data distribution in more detail (symmetric, skewed, flat, or bell-shaped). Students develop a more precise understanding of measures of center and understand relationships between mean and median for symmetrical and skewed data distributions. They learn that outliers affect the mean of a data set but not the median. They recognize outliers when they exist and learn to investigate their source. Students learn that standard deviation is a measure of spread, that a larger standard deviation means the data are more spread out, and to understand standard deviation as “typical distance from the mean” for a symmetrical distribution. They also understand that interquartile range is a description of variability better-suited to a skewed distribution. Finally, students are introduced to two-way frequency tables and understand how to interpret relative frequencies in the context of the data represented in the tables.
In unit M1.3 on bivariate statistics (which could take place either before or after this unit), students also build their statistics foundation by learning ways to determine whether two sets of data are correlated, and how strongly. Students identify linear association and interpret slope and intercept in the context of the data. Given different visual representations of data (linear models) students draw and justify significant and meaningful conclusions about the given situation. Students begin to use technology as a means to plot data and generate correlation coefficients. In unit M2.1 on probability, students revisit two-way frequency tables from a probability standpoint, and use them as a tool for conceptualizing and finding conditional probabilities. In unit M3.6 on inferences, students combine the ideas of distributions and probability. They learn about normal distributions and use them to solve problems, and use the distributions of probability models to find the likelihood of a particular outcome. In doing so, students build on their experience with standard deviations from S1, calculating standard deviations using technology, and interpreting the results. Every high school statistics and probability standard is a modeling standard, hence modeling pervades the four units.
## Sections
M1.1.0 Pre-unit diagnostic assessment
#### Summary
Assess students’ recall of middle grades statistics, specifically their ability to
• recognize a statistical question;
• describe the distribution of data collected to answer a statistical question by its center, spread, and overall shape;
• interpret statistical plots;
• summarize numerical data sets in relation to their context;
• informally assess the degree of visual overlap of two numerical data distributions with similar variabilities.
View Full Details
M1.1.1 How can data be represented and summarized meaningfully?
#### Summary
• Revisit various ways to plot data: dot plots, histograms, and box plots.
• Interpret plots of data within the context of the data.
• Use the terms “symmetric” and “skewed” as descriptors of distributions.
View Full Details
M1.1.8 Summative Assessment
#### Summary
Assess students’ ability to
• calculate mean, median, and mode;
• create box plots given data;
• compare and contrast two frequency distributions;
• articulate reasons to choose mean or use median as a measure of center;
• read and interpret relative frequencies.
View Full Details
M1.1.2 Analyze data distributions
#### Summary
• Create dot plots, histograms and box plots.
• Use available classroom technology to create histograms and box plots and calculate measures of center and spread.
• Use terms such as “flat,” “skewed,” “bell-shaped,” and “symmetric” to describe data distributions.
• Analyze and compare data sets.
View Full Details
M1.1.3 Measures of center
#### Summary
• Recall how to calculate mean and median.
• Understand mean and median as a “typical value” that can answer a statistical question.
• Know that mean and median are equal for a symmetrical data distribution.
• Explain why mean and median are unequal for a skewed data distribution.
• Select mean as the better measure for symmetrical distributions, and median as the better measure for skewed distributions.
• Make generalization what kinds of distributions have means larger than medians, and what kinds have medians larger than means.
• Recognize outliers when they exist, and know to investigate their source—that data point is way out there, why is that? Is there something weird about it that means we should disregard it?
• Know that outliers affect the mean, but not the median of a data set.
View Full Details
M1.1.4 Mid-unit assessment
#### Summary
Assess students’ ability to
• describe a set of data given a graph or table;
• identify and calculate spread, center, shape, outliers, quartiles, mean, median, mode;
• construct and interpret a box plot;
• compare, contrast, and draw conclusions when given two data sets.
View Full Details
M1.1.5 Standard deviation
#### Summary
• Describe variability by calculating deviations from the mean.
• Compare two data sets with the same means but different variabilities, and contrast them by calculating the deviation of each data point from the mean.
• Interpret sets with greater deviations as having greater variability.
• Calculate a standard deviation by hand for a small data set, and understand standard deviation as an indicator of a typical deviation from the mean of an element of the data set.
View Full Details
M1.1.6 Bringing it all together
#### Summary
• Represent a data set in different ways and decide which way is most appropriate.
• Select measures of center and spread appropriate to the shape of the distribution.
• Compare and contrast two or more distributions by using appropriate measures to describe center, variability, and shape.
View Full Details
M1.1.7 Two-way frequency tables
#### Summary
• Interpret a two-way table.
• Understand that the choices made when organizing data can lead to different conclusions.
View Full Details
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# Understanding Odds and Betting Lines in Sports
## The Basics of Betting Odds
When it comes to sports betting, odds play a crucial role in determining the potential payout. Understanding how odds work can help you make more informed betting decisions and increase your chances of winning. So, let’s start with the basics of betting odds.
Betting odds can be expressed in different formats, but the most common ones are decimal odds, fractional odds, and American odds. In this article, we will focus on American odds, also known as moneyline odds.
## What are American Odds?
American odds are represented by a positive or negative number. The positive number indicates the potential profit you can make from a \$100 bet, while the negative number indicates the amount you need to bet in order to win \$100.
For example, if the odds are +150, it means that a \$100 bet can yield a profit of \$150. On the other hand, if the odds are -200, it means that you need to bet \$200 to win \$100.
## The Favorite vs. the Underdog
In sports betting, you will often come across a favorite and an underdog, indicated by the plus and minus signs in the odds. The favorite is the team or player expected to win the game, while the underdog is the one expected to lose.
When the odds are positive, it indicates that the team or player is the underdog. In this case, a winning bet will yield a higher profit because they are considered less likely to win. On the other hand, when the odds are negative, it indicates that the team or player is the favorite. In this case, a higher bet is required to win a smaller profit.
## Calculating Payouts with American Odds
To calculate the potential payout using American odds, you can use the following formulas:
For positive odds:
Potential profit = (odds / 100) x bet amount
For negative odds:
Potential profit = (100 / odds) x bet amount
Let’s say you want to bet \$50 on a team with odds of +200. Using the formula for positive odds, the potential profit would be:
Potential profit = (200 / 100) x 50 = \$100
On the other hand, if you want to bet \$75 on a team with odds of -150, the potential profit would be:
Potential profit = (100 / 150) x 75 = \$50
## The Role of Point Spreads
In addition to moneyline odds, point spreads are another important aspect of sports betting. Point spreads are used to level the playing field between the favorite and the underdog.
Unlike moneyline bets, where you only need to predict the winner, point spread bets require you to consider the margin of victory. The favorite needs to win by a certain number of points, while the underdog needs to lose by fewer points or win the game outright.
The point spread is indicated by a plus or minus sign next to the odds. For example, if the odds are -4.5 for the favorite, it means they need to win by at least 5 points for the bet to be successful.
## Understanding Over/Under Bets
Another popular type of bet in sports betting is the over/under bet, also known as totals. This bet allows you to wager on whether the total combined points scored by both teams will be over or under a certain number.
For example, if the over/under for a basketball game is set at 200, you can bet on whether the total points scored by both teams will be over or under that number. The odds for over/under bets are typically represented as -110, meaning you need to bet \$110 to win \$100.
## Conclusion
Understanding odds and betting lines in sports is essential for anyone looking to engage in sports betting. Knowing how to interpret odds can give you a better understanding of the potential payout and the likelihood of an outcome.
Remember to always do your research, analyze the odds, and make informed decisions before placing your bets. And most importantly, only bet what you can afford to lose. Happy betting! Interested in further exploring the topic discussed in this article? Explore this detailed research, packed with supplementary and useful information to enhance your reading.
Delve deeper into the theme with the selected related links:
Examine this useful document
Learn from this detailed text
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# Download 3.4: Parallel and Perpendicular Lines 3.5: Parallel Lines and Triangles
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Transcript
```3.4: Parallel and Perpendicular Lines
Theorem: If two lines are parallel to the same line, then they are parallel to each other.
Theorem: In a plane, if two lines are perpendicular to the same transversal, then the two lines are parallel to each other.
Theorem: In a plane, if a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.
a ll b and a t
a
b
t
3.5: Parallel Lines and Triangles
Triangle: A figure formed by three segments joining three noncollinear points.
B
Triangle ABC ( ABC)
Vertices of C
A
Sides of Angles of
Classify Triangles by the number of congruent sides
Classify Triangles by their angles
Auxiliary Line: A line added to a diagram to help in a proof.
Triangle Sum Theorem
Example:
The sum of the measure of the angles of a triangle is 180.
Find the value of x. Then give the measure of each angle.
The measure of on angle of a triangle is 5 more than the measure of the smallest angle. The measure of the third angle is three times the measure of the smallest angle. What are the measures of the 3 angles?
Use the given information to ind the unknown angle measures in the triangle.
100
53
12
w
x
2x + y
y
5x + y
5x The perimeter of my triangle is 60 ft. The sides are in the ratio 3:4:5. What are the sides of my triangle?
```
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Student Instructions
### Commutative Property
Let's decompose (break) the word commutative word. Commutative includes the word "commute" which means to MOVE AROUND. Now we can discuss what commutative property has to do with our numbers. Commutative property means to move around the factors and continue to have the same multiple (product). Example: 4 x. 8= 32 same as 8 x 4 = 32 So we are basically moving around the numbers (factors) and continue to obtain the same multiple(product).
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# My answers (not including long response) (1 Viewer)
#### M@ster P
##### Member
umm 24b?
I got m between 1 and 3, because directly proportional means it is a straight line, and a straight line occurs between 1 and 3
Edit: when i think about it know directly proportional could mean anything, y = x^3, x^4 not just y =x, so not sure now
Last edited:
#### Macdwg
I'm going to say that the magnet question is open to interpretation, it really depends on the distance between the magnet and the wire, which we aren't given, and hence from this you are unable to assume that the field lines are acting horizontally like moke is suggesting, nor are you able to assume that its far enough for the field lines to pull back in the direction of the south pole. My guess is that you probably should assume they are acting horizontally to the right, however it's arguable.
Hence why this question sucks penis, the end.
#### Ruptured
##### New Member
umm 24b?
I got m between 1 and 3, because directly proportional means it is a straight line, and a straight line occurs between 1 and 3
Edit: when i think about it know directly proportional could mean anything, y = x^3, x^4 not just y =x, so not sure now
Directly propertional is where the ratio between the two values is constant. So just a straight line of any gradient
#### darkchild69
##### Nanotechnologist
umm 24b?
I got m between 1 and 3, because directly proportional means it is a straight line, and a straight line occurs between 1 and 3
Edit: when i think about it know directly proportional could mean anything, y = x^3, x^4 not just y =x, so not sure now
Directly proportional means y = kx where k is a constant of proportionality, hence plotting y vs x will be a straight line.
not y = kx^2 (this means y is directly proportional to x^2, not x. If you plot y vs x in this form, it will not be a straight line, hence not directly proportional)
mass should be between around 1.3-3
It is definately not 1-3, it begins after 1
#### M@ster P
##### Member
Directly proportional means y = kx where k is a constant of proportionality, hence plotting y vs x will be a straight line.
not y = kx^2 (this means y is directly proportional to x^2, not x. If you plot y vs x in this form, it will not be a straight line, hence not directly proportional)
mass should be between around 1.3-3
It is definately not 1-3, it begins after 1
sigh, wasn't accurate enough
#### youngminii
##### Banned
Directly proportional means y = kx where k is a constant of proportionality, hence plotting y vs x will be a straight line.
not y = kx^2 (this means y is directly proportional to x^2, not x. If you plot y vs x in this form, it will not be a straight line, hence not directly proportional)
mass should be between around 1.3-3
It is definately not 1-3, it begins after 1
1.3 - 3?
It was straight from 0 - 1.3, then became all wavy from 1.3 - 3.. Unless I remember wrong? Damn
#### darkchild69
##### Nanotechnologist
1.3 - 3?
It was straight from 0 - 1.3, then became all wavy from 1.3 - 3.. Unless I remember wrong? Damn
Shit yeah, i think you are right actually!
Damn paper is upstairs now, cbf checking, but im pretty sure you are right
#### Ruptured
##### New Member
Shit yeah, i think you are right actually!
Damn paper is upstairs now, cbf checking, but im pretty sure you are right
Glad to hear that because I was thinking a minute ago that I could've sworn it was 0-1.3
#### darkchild69
##### Nanotechnologist
LoL
I'm meant to know this stuff!
I just checked the paper and yep, definately 0-1.3
#### fallingstar
##### Member
Also, anyone who got A=34.5 is wrong and made this mistake.
I x I does not = 2I, it equals I squared. Therefore your answers were 69/2 = 34.5A when they should have been root 69 = 8.3A
dammit i did this! I'm an idiot, why can't I do basic math?!?!?
argh
#### tashisthebest
##### Member
this is what i rekon happen to the coil in the magnetic field
#### tommowhit
##### New Member
this is what i rekon happen to the coil in the magnetic field
um current flows from positive to negative
#### tashisthebest
##### Member
um current flows from positive to negative
thas a conventinal current
n normal electric current (electron flow) they travel from negative to positive
#### tommowhit
##### New Member
thas a conventinal current
n normal electric current (electron flow) they travel from negative to positive
u use the right hand in the right hand palm rule for convenionl current if ur going 2 use electron flow u still use direction of positive charge or use left hand
#### whitnall8
##### Member
We always deal in conventional current though, not in the flow of electrons.
#### tashisthebest
##### Member
umm no you still uses your right hand rule for a electric current
all the past papers n question i have done i have used the right hand rule n i have been right.
i have been taught this by ma teacher n my tutor.
#### tashisthebest
##### Member
i was rong
it seems like i have fuked up exam.
#### Mileyjoe
##### New Member
i was rong
it seems like i have fuked up exam.
LOLOLOLOLOLOL!!!!! funniest set of posts ever! i also was tossing that up in the exam but i got it right thankfully.
#### random-1005
##### Banned
Hmmmm, why was the weight on the asteroid given?
for a trick, you didnt need it, it was just to trick people who dont know that weight varies with location
#### random-1005
##### Banned
i did both to be sure!
I was almost ready to write on the paper: Thanks for giving us engineering students an unfair advantage!
But they look noting like conventional stress/strain curves!
they practically held your hand with that rubber band question.
lol, the questions people are asking, you are all shit, why did u even do physics, just roll over and die noobs
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This notebook was prepared by Donne Martin. Source and license info is on GitHub.
# Solution Notebook¶
## Constraints¶
• Is the array sorted?
• Yes
• Are the elements in the array distinct?
• No
• Does a magic index always exist?
• No
• If there is no magic index, do we just return -1?
• Yes
• Are negative values allowed in the array?
• Yes
• If there are multiple magic values, what do we return?
• Return the left-most one
• Can we assume this fits memory?
• Yes
## Test Cases¶
• None input -> -1
• Empty array -> -1
a[i] -4 -2 2 6 6 6 6 10
i 0 1 2 3 4 5 6 7
Result: 2
a[i] -4 -2 1 6 6 6 6 10
i 0 1 2 3 4 5 6 7
Result: 6
a[i] -4 -2 1 6 6 6 7 10
i 0 1 2 3 4 5 6 7
Result: -1
## Algorithm¶
We'll use a binary search to split the search space in half on each iteration. To obtain more efficiency, we can do a little better than a naive left and half split.
In the example below, we see that i == 5 cannot be the magic index, otherwise a[5] would have to equal 5 (note a[4] == 6).
a[i] -4 -2 2 6 6 6 6 10
i 0 1 1 3 4 5 6 7
mid
Similarly, in the example below we can further trim the left search space.
a[i] -4 -2 2 2 2 6 6 10
i 0 1 2 3 4 5 6 7
mid
• Calculate mid
• If mid == array[mid], return mid
• Recurse on the left side of the array
• start: 0
• end: min(mid-1, array[mid]
• Recurse on the right side of the array
• start: max(mid+1, array[mid]
• end: end
Complexity:
• Time: O(log(n))
• Space: O(log(n))
## Code¶
In [1]:
from __future__ import division
class MagicIndex(object):
def find_magic_index(self, array):
if array is None or not array:
return -1
return self._find_magic_index(array, 0, len(array) - 1)
def _find_magic_index(self, array, start, end):
if end < start or start < 0 or end >= len(array):
return -1
mid = (start + end) // 2
if mid == array[mid]:
return mid
left_end = min(mid - 1, array[mid])
left_result = self._find_magic_index(array, start, end=left_end)
if left_result != -1:
return left_result
right_start = max(mid + 1, array[mid])
right_result = self._find_magic_index(array, start=right_start, end=end)
if right_result != -1:
return right_result
return -1
## Unit Test¶
In [2]:
%%writefile test_find_magic_index.py
import unittest
class TestFindMagicIndex(unittest.TestCase):
def test_find_magic_index(self):
magic_index = MagicIndex()
self.assertEqual(magic_index.find_magic_index(None), -1)
self.assertEqual(magic_index.find_magic_index([]), -1)
array = [-4, -2, 2, 6, 6, 6, 6, 10]
self.assertEqual(magic_index.find_magic_index(array), 2)
array = [-4, -2, 1, 6, 6, 6, 6, 10]
self.assertEqual(magic_index.find_magic_index(array), 6)
array = [-4, -2, 1, 6, 6, 6, 7, 10]
self.assertEqual(magic_index.find_magic_index(array), -1)
print('Success: test_find_magic')
def main():
test = TestFindMagicIndex()
test.test_find_magic_index()
if __name__ == '__main__':
main()
Overwriting test_find_magic_index.py
In [3]:
%run -i test_find_magic_index.py
Success: test_find_magic
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# Math
A theater group made appearances in two cities. The hotel charge before tax in the second city was \$1000 higher than in the first. The tax in the first city was 4% , and the tax in the second city was 7% . The total hotel tax paid for the two cities was \$455 . How much was the hotel charge in each city before tax?
Please help me follow the steps. I understand that you have to put the two cities in equations and solve but the numbers confuse me. the total of the two cities and the number earned.
1. 👍 1
2. 👎 0
3. 👁 2,531
1. first hotel charge ---- \$x
2nd hotel charger ---- \$x+1000
.04x + .07(x+1000) = 455
how about multiplying each term by 100 if you don't like decimals
4x + 7(x+1000) = 45500
11x + 7000 = 45500
11x = 38500
x = 3500
first hotel cost \$3500
2nd hotel cost \$4500
check:
.04(3500) + .07(4500) = 455
1. 👍 0
2. 👎 0
2. A theater group made appearances in two cities. The hotel charge before tax in the second city was \$500 higher than the first. The tax in the first city was 8.5% and the tax in the second city was 4%. The total hotel tax paid for the two cities was \$613.75. How much was the hotel charge in each city before tax?
1. 👍 0
2. 👎 0
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In an affine space there's the concept of affine combination which states that any point in space can be represented as a affine combination in the form
$a + \sum_{i \in I} \lambda_i \mathbf{aa_i} \quad \quad \text{if \sum_{i \in I} \lambda_i = 1}$
We can add an additional restriction on the values of $$\lambda_i$$ to define a triangle built out of three points, if $$\lambda_1 = \beta, \lambda_2 = \gamma$$, $$\beta + \gamma = 1$$ and $$\beta, \gamma \in [0,1]$$ then a triangle is defined as the affine combination
$a + \beta \mathbf{ab} + \gamma \mathbf{ac}$
One geometric property of the scalar values is that they're the signed scaled distance from the lines that pass through the triangle sides, to compute the scalar values $$\beta$$ and $$\gamma$$ we can use the fact that when the implicit equation of the line that pass through a side is evaluated with points that don't lie on the line the result is equal to
$f(x,y) = d_{(x,y)} \cdot \sqrt{A^2 + B^2}$
Where $$d_{(x,y)}$$ is the distance from the point $$(x,y)$$ to the line, $$A$$ and $$B$$ are the coefficients of $$x$$ and $$y$$ of the general equation of the line that passes through $$a$$ and $$c$$
$Ax + Bx + C = 0$
To find the value of $$\beta$$ we can use the value of the implicit equation of the line to map the distance between any point to the line in the range $$[f_{ac}(x_a, y_a), f_{ac}(x_b, y_b)] = [0, f_{ac}(x_b, y_b)]$$, we can use a simple division to find the value of $$\beta$$
$\beta = \frac{f_{ac}(x,y)}{f_{ac}(x_b, y_b)} = \frac{d_{(x,y)}}{d_{(x_b, y_b)}}$
In a similar fashion the value of $$\gamma$$ is
$\gamma = \frac{f_{ab}(x,y)}{f_{ab}(x_c, y_c)} = \frac{d_{(x,y)}}{d_{(x_c, y_c)}}$
References
• Shirley, P. and Ashikhmin, M. (2005). Fundamentals of computer graphics. Wellesley, Mass.: AK Peters.
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# Physics
posted by .
A drowsy cat spots a flowerpot that saild first up and then down past an open window. The pot was in view for a total of .43 s, and then top-to-bottom height of the window is 2m. How high above the window top did the flowerpot go?
• Physics -
Let V2 be the velocity as the pot goes by the bottom of the windowm and V1 be the velocity as it goes by the bottom. The time-averaged velocity as it passes by is (V1 + V2)/2 = 2/0.43 = 4.65 m/s
The change in velocity as it goes by is (V1-V2)/2 = gt = 9.8 m/s^2*0.43 = 0.42 m/s
You now have two equations that let you solve for both V1 and V2
V1 + V2 = 9.30 m/s
V1 - V2 - 0.42 m/s
2 V2 = 8.88 m/s
V2 = 4.44 m/s
The height H above the window that the pot travels can be obtained by setting the kinetic energy at V2 equal to the gain of potential energy at the highest point:
M g H = (1/2) M V2^2
H = V2^2/2g = 1.01 meter
• Physics -
thank you SOOOO much, i spent a long time trying to figure this out
• Physics -
I made a mistake typing one sentence and equation. A "1/2" factor should not have been in the velocity charge equation. It should have read
<<The change in velocity as it goes by is (V1-V2)= gt = 9.8 m/s^2*0.43 = 0.42 m/s>>
It does not affect the answer because I ignored the "/2" when doing the numbers
• Physics -
YOUREWRONG
• Physics -
I don't understand how you get .42 from 9.8*.43
Every time I do this I get 4.2 m/s and I would like to know what I am doing wrong or if you are making a mistake.
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A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot was in view for a total of .43 s, and the top-to-bottom height of the window is 2 m. How high aboe the window top did the flowerpot go?
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## Letters of Euler to a German Princess, Vol. II, Letter VIII
This is the second of eleven Letters of Euler I will rewrite and post on the subject of infinitesimals (the infinitely small), an idea that is fundamental to a good understanding of calculus. Click here to read the previous letter.
### Letter VIII. Divisibility of Extension in Infinitum
The controversy between modern philosophies and geometricians to which I have alluded, turns on the divisibility of body. This property is undoubtedly founded on extension, and it is only in so far as bodies are extended that they are divisible, and capable of being reduced to parts.
You will recollect that in geometry it is always possible to divide a line, however small, into two equal parts. We are likewise, by that science, instructed in the method of dividing a small line, ai, into any number of equal parts at pleasure, and the construction of this division is there demonstrated beyond the possibility of doubting its accuracy.
You have only to draw a line AI (plate II. fig. 23) parallel to ai of any length, and at any distance you please, and to divide it into as many equal parts AB, BC, CD, DE, etc. as the small line given is to have divisions, say eight. Draw afterwards, through the extremities A, a, and I, i the straight lines AaO, IiO, till they meet in the point O: and from O draw toward the points of division B, C, D, E, etc. the straight lines OB, OC, OD, OE, etc., which shall likewise cut the small line ai into eight equal parts.
This operation may be performed, however small the given line ai, and however great the number of parts into which you propose to divide it. True it is, that in execution we are not permitted to go too far; the lines which we draw always have some breadth, whereby they are at length confounded, as may be seen in the figure near point O; but the question is not what may be possible for us to execute, but what is possible in itself. Now in geometry lines have no breadth*, and consequently can never be confounded. hence it follows that such division is illimitable.
###### *In Shormann Math, a line is defined as a widthless length, which is the same thing Euler is describing. In fact, all normal geometry courses define a line this way. The idea is that we are not concerned with how thick, or wide the line is. When you draw a line though, it has to have some thickness to it in order to be able to see it.
If it is once admitted that a line may be divided into a thousand parts, by dividing each part into two it will be divisible into two thousand parts, and for the same reason into four thousand, and into eight thousand, without ever arriving at parts indivisible. However small a line may be supposed, it is still divisible into halves, and each half again into two, and each of these again in like manner, and so on to infinity.
What I have said of a line is easily applicable to a surface, and, with greater strength of reasoning, to a solid endowed with three dimensions, length, breadth, and thickness. Hence is is affirmed that all extension is divisible to infinity, and this property is denominated divisibility in infinitum.
Whoever is disposed to deny this property of extension, is under the necessity of maintaining, that it is possible to arrive at last at parts so minute as to be unsusceptible of any farther division, because they ceased to have any extension. Nevertheless all these particles taken together must reproduce the whole, by the division of which you acquired them; and as the quantity of each would be a nothing, or cypher (0), a combination of cyphers would produce quantity, which is manifestly absurd. For you know perfectly well, that in arithmetic, two or more cyphers joined never produce any thing.
This opinion that in division of extension, or of any quantity whatever, we may come at last to particles so minute as to be no longer divisible, because they are so small, or because quantity no longer exists, is, therefore, a position absolutely untenable.
In order to render the absurdity of it more sensible, let us suppose a line of an inch long, divided into a thousand parts, and that these parts are so small as to admit of no farther division; each part, then, would no longer have any length, for if it had any, it would be still divisible. Each particle, then, would of consequence be a nothing. But if these thousand particles together constituted the length of an inch, the thousandth part of an inch would, of consequence, be a nothing; which is equally absurd with maintaining, that the half of any quantity whatever is nothing. And if it be absurd to affirm, that the half of any quantity is nothing, it is equally so to affirm, that the half of a half, or that the fourth part of the same quantity, is nothing; and what must be granted as to the fourth, must likewise be granted with respect to the thousandth, and the millionth part. Finally, however far you may have already carried, in imagination, the division of an inch, it is always possible to carry it still farther; and never will you be able to carry on your subdivision so far, as that the last parts shall be absolutely indivisible. These parts will undoubtedly always become smaller, and their magnitude will approach nearer and nearer to 0, but can never reach it.
The geometrician, therefore, is warranted in affirming, that every magnitude is divisible to infinity; and that you cannot proceed so far in your division, as that all farther division shall be impossible. But it is always necessary to distinguish between what is possible in itself, and what we are in a condition to perform. Our execution is indeed extremely limited. After having, for example, divided an inch into a thousand parts, these parts are so small as to escape our senses, and a farther division would to us, no doubt, be impossible.
But you have only to look at this thousandth part of an inch through a good microscope, which magnifies, for example, a thousand times, and each particle will appear as large as an inch to the naked eye; and you will be convinced of the possibility of dividing each of these particles again into a thousand parts: the same reasoning may always be carried forward, without limit and without end.
It is therefore an indubitable truth, that all magnitude is divisible in infinitum, and that this takes place not only with respect to extension, which is the object of geometry, but likewise with respect to every other species of quantity, such as time and number.
28th April, 1761.
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# The Folding Challenge - how low can your ratio go?
1. Nov 22, 2008
### qbit
I had a typical A4, 80 gsm sheet of paper and I had determined that its dimensions were:
210 x 297 x 0.11 mm.
The last dimension was obtained using an old micrometer I possess and is close to the average of four measurements taken in various places on that piece of paper.
Thus the volume of this sheet of paper is approximately 6861 mm cubed and its surface area slightly more than 124740 mm squared. This gives a surface area to volume ratio of about 18.2 to 1.
The smallest surface to volume ratio value would be to form a sphere with the material that makes up the sheet of paper. This is trivial to calculate: the radius is close to 11.8 mm, the surface area is 1746 mm squared and so the surface to volume ratio is a little over 0.25 to 1.
To the best of my knowledge, it is not possible to fold a sheet of paper into sphere in the conventional meaning of 'fold'.
The challenge is this: What is the smallest surface area to volume ratio a sheet of A4 80 gsm can be folded into such that it can be readily unfolded back into its original shape (not including creases) and critically, when in its folded shape, no objects like hefty books are used to confine it to its largest three dimensions. What do I mean by the largest three dimensions? Well, basically the measurement of its length by width by breadth so that nothing sticks out. I don't see why a clever designer couldn't use the paper's own weight to help compress itself when sitting on a desk as this aspect is, to some extent, inescapable in practice. But I'm thinking a folded sheet of paper should be able to maintain it's folded shape or rather, maximum surface area as defined by its largest three dimensions, for, say, five minutes. No tape. Obviously I'm not including surface area inside the folds. That would defeat the purpose of the challenge as the surface area of a folded sheet of paper does not, in fact, change.
As you will have already deduced, a criteria where the surface area is determined by the largest three dimensions, would mean that if a sphere with a radius of 11.8 mm could actually be constructed by folding, it would be then be determined to have a surface area of about 3342 mm squared and a volume of about 13144 mm cubed. Nearly twice the actual surface area and volume! But therefore, a surface area to volume ratio not dissimilar to that of the sphere of a little over 0.25. I just can't think of a more convenient way of determining a surface area of a folded object. Most of the more low ratio structures I've constructed have been cuboid anyway. So using the 'volume as defined by the largest three dimensions' rule, the lowest ratio achievable would be 2166:6861 or about 0.32.
My best efforts to date is getting close to a ratio of 2 to 1. I have, in some attempts, used a weighty book to hold my folding in place, but made my measurements of its three largest dimensions after removing the book and then after five minutes to observe if my compressed structure was inclined to expand.
This challenge offers no prizes except the gratification in the knowledge that you've broken the 2 to 1 ratio barrier that has eluded me. I'd be interested in knowing the sequence and types of folds used to accomplish it.
If anyone's interested, I'll post the folds that got me close to the 2 to 1 ratio.
And yes, I do realise there are better things to do with one's life than fold paper. But you never know where these sort of things can lead. Perhaps people used to say similar things about the quest for determining large prime numbers.
Last edited: Nov 22, 2008
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# clamp a 2D coordinate to fit within an ellipse
I need to clamp a 2D coordinate to fit within an ellipse.
Call of Duty: Modern Warfare 2 does something similar where capture points are translated from a 3D vector in the world to a 2D screen coordinate and then the 2D coordinates are clamped within an ellipse.
When the capture points are in view they're within the bounds of the ellipse.
When they're behind you they are clamped to be within the bounds of the ellipse.
Given a 2D coordinate that could be off screen, etc, what is the math behind clamping it within an ellipse?
• The ellipse is 2D too, right? If not, what plane is it in? – Josh Aug 10 '16 at 18:44
• Yes, the ellipse is 2D. – Rocky Breslow Aug 10 '16 at 18:48
• Can you scale the points and the ellipse to make the ellipse into a circle, clamp to the circle (relatively easy), then scale the points back? – amitp Aug 10 '16 at 19:17
• @amitp: Why? Just use the well known equation for an ellipse at the origin and just use a simple convolution by the transformation that takes it to the origin. – Pieter Geerkens Aug 10 '16 at 22:24
• simple: change coordinates so the ellipse is at 0,0 and round; limit vector length; change coordinates back. – Sarge Borsch Aug 11 '16 at 10:25
## 2 Answers
Here's a general method to clamp a point $$\A\$$ to the ellipse of center $$\C\$$ and half-dimensions $$\d\$$ (i.e. the ellipse is $$\2d_x\$$ wide and $$\2d_y\$$ high):
• $$\B = \frac{(A - C)}{d}\$$ gives you the coordinates of $$\A\$$ in the frame in which the ellipse is the unit circle. The fraction is memberwise division ($$\ \frac{a}{b} = (\frac{a_x}{b_x} ; \frac{a_y}{b_y})\$$)
• $$\B' = \frac{B}{|B|}\$$ normalizes $$\B\$$: $$\B'\$$ is now on the unit circle.
• $$\B'' = C + B' × d\$$, where $$\×\$$ is the memberwise product. This brings our clamped point back into the original frame : $$\B''\$$ is $$\A\$$, clamped onto the ellipse.
This method can be greatly simplified if the ellipse is centered ($$\C\$$ disappears), and/or if your frame has the same aspect ratio as your ellipse ($$\d_x = d_y\$$, so the scaling isn't necessary).
The equation for an ellipse centered at the origin is just:
$$\bigg( \frac{x ^ 2}{Rx ^ 2}\bigg) + \bigg(\frac{y ^ 2}{Ry ^ 2} \bigg) = 1$$
where:
• Rx is the X-dimension radius; and
• Ry is the y-dimension radius
Combining this with a transformation that maps the relevant ellipse to the origin is straightforward.
Then just replace the = sign with a <= to get a test for (inclusive) containment in the ellipse.
• right, I've gotten that far but all that lets us know is if the point is inclusive or exclusive in the ellipse, not where the closest point on the ellipse – Rocky Breslow Aug 10 '16 at 23:48
• extrude it: take the normal vector of the point to the center of the ellipse and then extrude it to the intersection point on the edge of the ellipse – CobaltHex Aug 11 '16 at 1:50
• @RockyBreslow: That's just going to be the intersection of the line segment from origin to point and the equation of the ellipse. – Pieter Geerkens Aug 11 '16 at 4:34
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### Home > GC > Chapter 7 > Lesson 7.3.2 > Problem7-114
7-114.
Examine the diagram at right.
1. Are the triangles in this diagram similar? Explain.
Do the marks help prove they are similar? If so, how? What makes these triangles similar? Are there any angles that are congruent?
Yes, by $\text{SAS}\sim$ because Segment $FI$ is proportional to segment $FG$, $∠F≅∠F$, and segment $FJ$ is proportional to segment $FH$.
2. Name all the pairs of congruent angles in this diagram you can.
Look at the triangles, what angles are corresponding to each other?
3. Are $\overline{GH}$ and $\overline{IJ}$ parallel? Explain how you know.
How can knowing sets of congruent angles, help you prove the sides are parallel?
Yes they are parallel, because corresponding angles are congruent.
4. If $GH=4x-3$ and $IJ=3x+14$, find $x$. Then find the length of $\overline{GH}$.
How are the lengths of $\overline{GH}$ and $\overline{IJ}$ related? How will this help you solve for $x$?
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# Chapter 5 Matrices and rasters
Last updated: 2024-06-24 11:04:57
## Aims
Our aims in this chapter are:
• Start working with spatial data (rasters)
• Install and use packages beyond “base R”
• Introduce the basic matrix and array data structures, and their analogous spatial data structure (class stars) for single band and multi-band rasters, respectively
• Learn to access the cell values and other properties of rasters
• Learn to read and write raster data
We will use the following R packages:
• stars
• mapview
## 5.1 Matrices
### 5.1.1 What is a matrix?
A matrix is:
• a two-dimensional (like a data.frame) collection of values
• of the same type (like a vector, and unlike a data.frame).
The number of values in all columns of a matrix is equal, and the same can be said about the rows. It is important to know how to work with matrices because it is a commonly used data structure, with many uses in data processing and analysis, including spatial data. Many R functions accept a matrix as an argument, or return a matrix as the returned object. For example, the st_distance function returns a distance matrix between all pairs of features in two vector layers (Section 8.3.2.3). Matrices are also used for data storage in more complex data structures, e.g., to store single-band raster values in stars raster objects (as we’ll see later, in Section 5.3.8.2).
### 5.1.2 Creating a matrix
A matrix can be created with the matrix function. The matrix function accepts the following arguments:
• data—A vector of the values to fill into the matrix
• nrow—The number of rows
• ncol—The number of columns
• byrow—Whether the matrix is filled by column (FALSE, the default) or by row (TRUE)
For example:
x = matrix(1:6, nrow = 2, ncol = 3)
x
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 2 4 6
Note that the class of matrix objects is a vector of length two, with the values 'matrix' and 'array':
class(x)
## [1] "matrix" "array"
This implies the fact that the matrix class inherits (Section 1.1.5) from the more general array class (Section 5.2).
The nrow and ncol parameters determine the number of rows and number of columns, respectively. When only one of them is specified, the other is automatically determined based on the length of the data vector:
matrix(1:6, nrow = 2)
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 2 4 6
matrix(1:6, ncol = 2)
## [,1] [,2]
## [1,] 1 4
## [2,] 2 5
## [3,] 3 6
What do you think will happen when we try to create a matrix with less, or more, data values than matrix size nrow*ncol? Run the following expressions to find out.
matrix(12:1, ncol = 4, nrow = 2)
matrix(12:1, ncol = 4, nrow = 4)
Create a $$3\times3$$ matrix where all values are $$1/9$$.
Finally, the byrow parameter determines the direction of filling the matrix with data values. In both cases the filling starts from the top-left corner (i.e., row 1, column 1), however with byrow=FALSE the matrix is filled one column at a time (the default), while with byrow=TRUE the matrix is filled one row at a time. For example:
matrix(1:12, nrow = 3) ## the default (byrow=FALSE)
## [,1] [,2] [,3] [,4]
## [1,] 1 4 7 10
## [2,] 2 5 8 11
## [3,] 3 6 9 12
matrix(1:12, nrow = 3, byrow = TRUE) ## byrow=TRUE
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] 5 6 7 8
## [3,] 9 10 11 12
### 5.1.3matrix properties
#### 5.1.3.1 Dimensions
To demonstrate several functions for working with matrices, let’s define a matrix named x as shown above (Section 5.1.2):
x = matrix(1:6, nrow = 2)
x
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 2 4 6
The length function returns the number of values in a matrix:
length(x)
## [1] 6
Just like with a data.frame (Section 4.1.4.1), the nrow and ncol functions return the number of rows and columns in a matrix, respectively:
nrow(x)
## [1] 2
ncol(x)
## [1] 3
Also like with a data.frame, the dim function gives both dimensions of the matrix as a vector of length 2, i.e., number of rows and columns, respectively:
dim(x)
## [1] 2 3
For example, R has a built-in dataset named volcano, which is a matrix of surface elevation. The sample script volcano.R, used in Section 2.1.3 to demontrate working with R code files, creates a 3D image of elevation based on that matrix (Figure 2.2).
Find out what are the number of elements, rows and columns in the built-in matrix named volcano.
#### 5.1.3.2 Row and column names
Like a data.frame (Section 4.1.4.2), matrix objects also have row and column names which can be accessed or modified using the rownames and colnames functions, respectively. Unlike data.frame row and column names, which are mandatory, matrix row and column names are optional. For example, matrices created with matrix as shown above (Section 5.1.2) initially do not have row and column names:
rownames(x)
## NULL
colnames(x)
## NULL
The matrix row and column names can be initialized, modified, or removed, by assignment to the rownames and colnames properties:
rownames(x) = c('a', 'b')
colnames(x) = c('var1', 'var2', 'var3')
x
## var1 var2 var3
## a 1 3 5
## b 2 4 6
rownames(x) = NULL
colnames(x) = NULL
x
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 2 4 6
### 5.1.4matrix conversions
#### 5.1.4.1matrix → vector
The as.vector function converts a matrix to a vector:
x = matrix(1:6, ncol = 3, byrow = TRUE)
x
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 4 5 6
as.vector(x)
## [1] 1 4 2 5 3 6
Note that the matrix values are always arranged by column in the resulting vector!
Does the volcano matrix contain any NA values? How can we check?
#### 5.1.4.2matrix → data.frame
The as.data.frame function converts a matrix to a data.frame:
as.data.frame(x)
## V1 V2 V3
## 1 1 2 3
## 2 4 5 6
Note that row and column names are automatically generated (if they do not exist) as part of the conversion, since they are mandatory in a data.frame (Section 5.1.3.2).
### 5.1.5 Transposing a matrix
The t function transposes a matrix. In other words, the matrix rows and columns are “switched”—rows become columns, and columns become rows:
x = matrix(1:6, ncol = 3)
x
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 2 4 6
t(x)
## [,1] [,2]
## [1,] 1 2
## [2,] 3 4
## [3,] 5 6
What will be the result of t(t(x))?
### 5.1.6 Image with contours
Using the image and contour functions we can graphically display matrix values in the form of a heatmap—where values are encoded to colors—and a contour layer, respectively.
The color scale in image can be set with col. R has several built-in color scale functions—such as terrain.colors and heat.colors—that return a specified number of color codes from the given color scale. The hcl.colors function offers a wider range of color scales, including many scales useful for mapping, such as the well-known ColorBrewer scales. The hcl.colors function requires the number of colors and the name of the color scale. For example, the following expression returns 11 colors from the 'Spectral' color scale:
hcl.colors(11, 'Spectral')
## [1] "#A71B4B" "#D44D35" "#ED820A" "#F7B347" "#FCDE85" "#FEFDBE" "#BAEEAE"
## [8] "#61D4AF" "#00B1B5" "#0084B3" "#584B9F"
The x/y aspect ratio, in either image or contour, can be set with asp.
Finally, add=TRUE can be used so that the contour layer is added on top of an existing plot, such as the one created with image, rather than initiated in a new plot.
Combining all of the above, the following expressions display a contour layer on top of a colored image of the volcano matrix. The resulting image is shown in Figure 5.1.
image(volcano, col = terrain.colors(30), asp = ncol(volcano)/nrow(volcano))
contour(volcano, add = TRUE)
Try replacing terrain.colors(30) with heat.colors(30), or with hcl.colors(11,'Spectral'), to see an image of the volcano matrix with other color scales. You can run hcl.pals() to get a list of possible scale types passed to hcl.colors (other than 'Spectal') and try some of those too.
Note that the argument asp=ncol(volcano)/nrow(volcano) makes sure the ratio between the x and y axes is 1:1, i.e., that the image cells are square17.
Also note that image does not display a matrix in the same orientation as its printout (rows from top to bottom, etc.). Here is the solution to obtain a matrix image in the exact same orientation (Figure 5.2):
volcano1 = t(apply(volcano, 2, rev))
image(volcano1, col = terrain.colors(30), asp = ncol(volcano1)/nrow(volcano1))
contour(volcano1, add = TRUE)
### 5.1.7 Matrix subsetting
#### 5.1.7.1 Individual rows and columns
Similarly to what we learned about data.frame (Section 4.1.5), matrix indices are two-dimensional. The first value refers to rows and the second value refers to columns. For example:
x
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 2 4 6
x[2, 1] # Row 2, column 1
## [1] 2
x[1, 3] # Row 1, column 3
## [1] 5
The following examples subset the volcano matrix:
volcano[20, 40] # Row 20, column 40
## [1] 190
volcano[81, 61] # Row 81, column 61
## [1] 95
How can we find out the top-right corner value of volcano?
Complete rows or columns can be accessed by leaving a blank space instead of the row or column index. By default, a subset that comes from a single row or a single column is simplified to a vector:
x
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 2 4 6
x[2, ] # Row 2
## [1] 2 4 6
x[, 2] # Column 2
## [1] 3 4
To “suppress” the simplification of individual rows/columns to a vector, we can use the drop=FALSE argument (Section 4.1.5.3):
x
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 2 4 6
x[2, , drop = FALSE] # Row 2
## [,1] [,2] [,3]
## [1,] 2 4 6
x[, 2, drop = FALSE] # Column 2
## [,1]
## [1,] 3
## [2,] 4
When referring to an elevation matrix, such as volcano, any given row or column subset is actually an elevation profile. For example, the following expressions extract two elevation profiles from volcano:
r30 = volcano[30, ] # Row 30
r70 = volcano[70, ] # Row 70
Figure 5.3 graphically displays those profiles (Section 3.2.1):
plot(r30, type = 'o', col = 'blue', ylim = range(c(r30, r70)), ylab = 'Elevation (m)')
lines(r70, type = 'o', col = 'red')
Figure 5.4 shows the location of the two profiles in a 3D image of volcano.
#### 5.1.7.2 Subsets with >1 row or column
We can also use vectors of length >1 when subsetting:
x
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 2 4 6
x[, 1:2]
## [,1] [,2]
## [1,] 1 3
## [2,] 2 4
x[2, c(1,3)]
## [1] 2 6
x[2, c(1,3), drop = FALSE]
## [,1] [,2]
## [1,] 2 6
We can assign new values to subsets of a matrix:
m = matrix(NA, ncol = 3, nrow = 3)
m
## [,1] [,2] [,3]
## [1,] NA NA NA
## [2,] NA NA NA
## [3,] NA NA NA
m[2:3, 1:2] = 1
m
## [,1] [,2] [,3]
## [1,] NA NA NA
## [2,] 1 1 NA
## [3,] 1 1 NA
m[1:2, 2:3] = 100
m
## [,1] [,2] [,3]
## [1,] NA 100 100
## [2,] 1 100 100
## [3,] 1 1 NA
### 5.1.8 Summarizing rows and columns
How can we calculate the row or column means of a matrix? One way is to use a for loop (Section 4.2.3), as follows:
result = rep(NA, nrow(volcano))
for(i in 1:nrow(volcano)) {
result[i] = mean(volcano[i, ])
}
The resulting vector of row means can be visualized as follows (Figure 5.5):
plot(result, type = 'b', xlab = 'Row', ylab = 'Elevation (m)')
What changes do we need to make in the for loop code to calculate column means?
We can use the apply function (Section 4.5) to do the same, using much shorter code:
result = apply(volcano, 1, mean)
What changes do we need to make in the apply expression to calculate column means?
Moreover, for the special case of mean there are further shortcuts, named rowMeans and colMeans18:
result = rowMeans(volcano)
Note: in both cases we can use na.rm to determine whether NA values are included in the calculation (default is FALSE).
How can we check whether the above two expressions give exactly the same result?
## 5.2 Arrays
### 5.2.1 Creating an array
An array is a data structure that contains values of the same type and can have any number of dimensions. We may therefore consider a vector (1 dimension) and a matrix (2 dimensions) as special cases of an array. In this scope of this book, we will be dealing only with three-dimensional arrays, which—as we will see later on (Section 5.3.8.2)—are used to store multi-band raster values in stars raster objects.
An array can be created with the array function, specifying the values (data) and the required dimensions (dim). Note that creating arrays, with the array function, is very similar to the way we create matrices, with the matrix function—just more general. For example, the following expression creates an array with the values 1:24 and three dimensions—2 rows, 3 columns, and 4 “layers” (Figure 5.6):
y = array(data = 1:24, dim = c(2, 3, 4))
y
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 2 4 6
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 7 9 11
## [2,] 8 10 12
##
## , , 3
##
## [,1] [,2] [,3]
## [1,] 13 15 17
## [2,] 14 16 18
##
## , , 4
##
## [,1] [,2] [,3]
## [1,] 19 21 23
## [2,] 20 22 24
### 5.2.2 Array subsetting
Subsetting an array works similarly to matrix subsetting (Section 5.1.7), only that we can have any number of indices—corresponding to the number of array dimensions—rather than two. Specifically, when subsetting a three-dimensional array we need to provide three indices. For example, here is how we can extract a particular row, column or “layer” from a three-dimensional array (Figure 5.7):
y[1, , , drop = FALSE] # Row 1
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 3 5
##
## , , 2
##
## [,1] [,2] [,3]
## [1,] 7 9 11
##
## , , 3
##
## [,1] [,2] [,3]
## [1,] 13 15 17
##
## , , 4
##
## [,1] [,2] [,3]
## [1,] 19 21 23
y[, 1, , drop = FALSE] # Column 1
## , , 1
##
## [,1]
## [1,] 1
## [2,] 2
##
## , , 2
##
## [,1]
## [1,] 7
## [2,] 8
##
## , , 3
##
## [,1]
## [1,] 13
## [2,] 14
##
## , , 4
##
## [,1]
## [1,] 19
## [2,] 20
y[, , 1, drop = FALSE] # "Layer" 1
## , , 1
##
## [,1] [,2] [,3]
## [1,] 1 3 5
## [2,] 2 4 6
We can also subset two or three dimensions at a time, to get an individual row/column, row/layer, column/layer or row/column/layer combination (Figure 5.8):
y[1, 1, ] # Row 1, column 1
## [1] 1 7 13 19
y[1, , 1] # Row 1, "layer" 1
## [1] 1 3 5
y[, 1, 1] # Column 1, "layer" 1
## [1] 1 2
y[1, 1, 1] # Row 1, column 1, "layer" 1
## [1] 1
### 5.2.3 Using apply with arrays
When using apply on a 3-dimensional array, we can apply a function:
• On one of the dimensions: rows (1), columns (2), or “layers” (3)
• On a combinations of any two dimensions: row/column combinations (c(1,2)), row/“layer” combinations (c(1,3)), or column/“layer” combinations (c(2,3))
For example:
apply(y, 1, mean) # Row means
## [1] 12 13
apply(y, 2, mean) # Column means
## [1] 10.5 12.5 14.5
For arrays representing spatial data, i.e., multi-band rasters, the two most useful dimension combinations are summarizing layer or pixel properties (Section 5.3). For example:
apply(y, 3, mean) # "Layer" means
## [1] 3.5 9.5 15.5 21.5
apply(y, 1:2, mean) # Row/Column combination ("pixel") means
## [,1] [,2] [,3]
## [1,] 10 12 14
## [2,] 11 13 15
### 5.2.4 Basic data structures in R
So far, we met four out of five of the basic data structures in R (Table 5.1). It is time for a short summary. We can classify the basic data structures in R based on:
• Number of dimensions
• One-dimensional19
• 2-dimensional
• N-demensional
• Homogeneity
• Homogeneous (values of the same type)
• Heterogeneous (values of different types)
Table 5.1: Five basic data structures in R
Dimensions Homogeneous Heterogeneous
One-dimensional vector (numeric, character, logical, Date) (Chapter 2) list (Chapter 11)
Two-dimensional matrix (Chapter 5) data.frame (Chapter 4)
N-dimensional array (Chapter 5)
Most of the data structures in R are combinations of these basic five ones.
## 5.3 Rasters
### 5.3.1 What is a raster?
A raster (Figure 5.9) is basically a matrix or an array representing a rectangular area on the surface of the earth. To associate a matrix, or an array, with the particular area it represents, the raster has some additional spatial properties:
• Origin ($$x_{min}$$, $$y_{max}$$) and resolution ($$delta_{x}$$, $$delta_{y}$$)
• Coordinate Reference System (CRS)
on top of the non-spatial properties that any ordinary matrix or array has:
• Values
• Dimensions (rows, columns, layers)
Let’s now clarify the spatial property terms listed above: origin, resolution, and CRS.
The purpose of the the origin and resolution is to convey the coordinates of all raster pixels. Since the raster is regular (i.e., its rows and columns are parallel to the axes, and step size in the x and y direction is fixed), there is no need to actually store all coordinate values: just four numbers are sufficient to calculate them. Raster origin is the location of the top-left corner of the raster, i.e., the top-left corner of the top-left raster pixel ($$x_{min}$$, $$y_{max}$$). Raster resolution is the step size between one pixel to the next, in the x and y directions ($$delta_{x}$$, $$delta_{y}$$). Knowing the origin and resolution, the x- and y-coordinates of each pixel at ($$row$$, $$col$$) can be calculated using Equations (5.1) and (5.2), respectively.
$$$x = x_{min} + (delta_{x} \times col) - (\frac{1}{2} \times delta_{x}) \tag{5.1}$$$
$$$y = y_{max} + (delta_{y} \times row) - (\frac{1}{2} \times delta_{y}) \tag{5.2}$$$
In plain language, to get the pixel coordinates we start from the origin, then go the specified number of steps along the axis, whereas step size is the resolution. Additionally, we subtract half of the step size, to get the coordinates of the pixel centroid, rather than the coordinates of its edge. By convention, the y-axis origin is specified as the maximum value (i.e., $$y_{max}$$), and, accordingly, the y-axis resolution is negative (we elaborate on this in Section 6.3.3).
Another commonly used raster property is the extent, the range of x- and y-axis coordinates that the raster occupies (Figure 5.10), that is, a set of four numbers: $$x_{min}$$, $$y_{min}$$, $$x_{max}$$, $$y_{max}$$. Raster extent holds the same information as the origin+resolution, and the two are interchangeable. Namely, given the origin, resolution, and the dimensions (number of rows and columns), we can always calculate the extent, and vice versa. For example, the x-axis resolution is equal to the x-axis range difference (i.e., $$x_{max}-x_{min}$$) divided by the number of columns. Therefore, in the leftmost panel in Figure 5.10, the x-axis and y-axis resolutions are both equal to $$\frac{8-0}{12}=\frac{8}{12}\approx 0.67$$.
As shown above, given the origin+resolution, we can determine the coordinates of all raster pixels. However, bare coordinates are just numbers. It is the Coordinate Reference System (CRS) definition that ties the coordinates to specific locations on the surface of the Earth. The CRS is the particular system that “associates” the raster coordinates (which are just pairs of x/y values) to geographic locations. For example, in the WGS84 geographic CRS, the point (0,0) (known as Null Island20) is located in the Gulf of Guinea, in the Atlantic Ocean. In a different CRS, the (0,0) pair of coordinates may refer to a completely different location, or even a location supposedly beyond the surface of the Earth. Therefore, we must know the CRS to “make sense” of numeric coordinates.
### 5.3.2 Raster file formats
Commonly used raster file formats (Table 5.2) can be divided in two groups. “Simple” raster file formats, such as GeoTIFF, are single-band or multi-band rasters (Figure 5.11) with a geo-referenced extent, as discussed above (Section 5.3.1). “Complex” raster file formats, such as HDF, contain additional complexity, such as more than three dimensions (Figure 5.12), and/or metadata, such as band names, time stamps, units of measurement, and so on.
Table 5.2: Common raster file formats
Type Format File extension
“Simple” GeoTIFF .tif
Erdas Imagine Image .img
“Complex” (>3D and/or metadata) HDF e.g., .hdf or he5
NetCDF .nc
### 5.3.3 Using R packages
Functions for working with rasters are not included in the “base” R installation, but in external packages, such as the stars package, which we are going to learn about in a moment (Section 5.3.5). Before that, let’s say a few words about what exactly are R packages, and how to use them.
An R package is a collection of code files used to load objects—mostly functions—into memory. All object definitions in R are contained in packages. To use a particular package, we need to:
• Install the package with the install.packages function (once, for every installation of R21)
• Load the package using the library function (every time we start a new R session)
Installing a package with install.packages downloads the package code from a repository and stores it on our computer. Loading a package with library basically means that all of its code files are executed, loading all objects the package defined into the RAM.
Note that all function calls we used until now did not require installing or loading any package. Why is that? The answer is that several packages are installed along with R and loaded on R start-up (known as “base” packages). There are several more packages which are installed by default but not loaded on start-up (known as “recommended packages”). The number of packages in both of these groups is 29, as of May 2024 (Table 5.3).
Table 5.3: Packages included with the installaton of R as of May 2024, including “base” packages (loaded by default), and “recommended” packages (installed but not loaded default)
Package Version Priority
base 4.4.0 base
compiler 4.4.0 base
datasets 4.4.0 base
graphics 4.4.0 base
grDevices 4.4.0 base
grid 4.4.0 base
methods 4.4.0 base
parallel 4.4.0 base
splines 4.4.0 base
stats 4.4.0 base
stats4 4.4.0 base
tcltk 4.4.0 base
tools 4.4.0 base
utils 4.4.0 base
boot 1.3-30 recommended
class 7.3-22 recommended
cluster 2.1.6 recommended
codetools 0.2-19 recommended
foreign 0.8-86 recommended
KernSmooth 2.23-22 recommended
lattice 0.22-5 recommended
MASS 7.3-60 recommended
Matrix 1.6-5 recommended
mgcv 1.9-1 recommended
nlme 3.1-163 recommended
nnet 7.3-19 recommended
rpart 4.1.23 recommended
spatial 7.3-15 recommended
survival 3.5-8 recommended
Most of the ~20,775 official R packages (as of May 2024) are not installed by default. To use one of these packages we first need to install it on the computer. Installing a package is a one-time operation using the install.packages function. After the package is installed, each time we want to use it we need to load it using library.
In the following examples, we are going use a package called stars to work with rasters. The stars package is not installed with R, therefore we need to install it ourselves with:
install.packages('stars')
If the package is already installed, running install.packages overwrites the old installation. This is done intentionally if you you want to update the package, i.e., to install a newer version of the package.
Once the package is installed, we need to use the library function to load it into memory. Note how the package name can be passed to library (with or without quotes):
library(stars)
Other than stars, in this book we are going to use the mapview, sf, units, data.table, dplyr, gstat, and automap packages. You can now take the time and install all of them, using a series of install.packages expressions, or all of them at once with:
install.packages(c('stars','mapview','sf','units','data.table','dplyr','gstat','automap'))
### 5.3.4 The raster and terra packages
Before moving on to stars package, the raster and terra packages deserve to be mentioned. The raster package is a powerful and well-established (2010-) package for working with rasters in R (Table 0.1). terra (2020-) is the new and improved package by the same authors, providing very similar capabilities, only easier to use, and faster. Each of these packages contains several classes, and a very extensive collection of functions, for working with rasters in R.
The raster and terra packages have several limitations compared to stars, most notably that they are limited to three dimensions, and cannot hold raster metadata such as measurement units. Moreover, they are less well-integrated with the sf package—which is the leading package for working with vector layers in R (Chapters 78). On the other hand, raster and terra provide some “classical” GIS operations that are unavialble in stars, such as built-in focal filter functions (Section 9.4). Otherwise, however, there is a very large overlap between stars and raster/terra in terms of capabilities, and the choice is ultimately a matter of preference. In this book we will be working with the stars package for rasters (Section 5.3.5), which we now move on to introduce.
### 5.3.5 The stars package
The stars package is a relatively new (2018-) R package for working with rasters (Table 0.1). The stars package is comprehensive, flexible, and tightly integrated with the leading package for vector layers named sf, which we learn about later on (Section 7.1.4).
The stars package defines the stars class, used for representing all types of rasters, and numerous functions for working with rasters in R. A stars object is basically a list of matrices or arrays, along with metadata describing their dimensions. Don’t worry if this is not clear; we will elaborate on this shortly (Section 6.3).
Notably, the stars class is very flexible, and can represent complex rasters:
In this book we are going to work with “simple” regular rasters, such as those typically stored in GeoTIFF files (Section 5.3.2). Nevertheless, it is important to be aware that stars can handle a wide variety of raster complexity which you may encounter in more specialized use cases, which are beyond the scope of this book, such as when working with climatic datasets (Figure 5.12).
### 5.3.6 Reading raster from file
The most common and most useful method for creating a raster object in R (and elsewhere) is reading from a file, such as a GeoTIFF file. We can use the read_stars function to read a GeoTIFF file and create a stars object in R. The first parameter (.x) is the file path to the file we want to read, or just the file name—when the file is located in the working directory.
As an example, let’s read the dem.tif raster file, which contains a coarse Digital Elevation Model (DEM) of the area around Haifa. First, we have to load the stars package:
library(stars)
Then, we can use the read_stars function to read the file into a stars object named s22:
s = read_stars('data/dem.tif')
The file MOD13A3_2000_2019.tif is a multi-band raster with monthly Normalized Difference Vegetation Index (NDVI) values in Israel, for the period between February 2000 and June 2019 from the MODIS instrument on the Terra satellite. This is a multi-band raster with 233 bands, where each band corresponds to an average monthly NDVI image (233 months in total). Uncertain NDVI values, such as in clouded areas, were replaced with NA. Let’s try reading this file as well, to create another stars object, named r, in the R environment:
r = read_stars('data/MOD13A3_2000_2019.tif')
Two stars objects, named r and s, are now available in our R session.
### 5.3.7 Visualization with plot and mapview
#### 5.3.7.1 Raster images with plot
The simplest way to visualize a stars object is to use the plot function. This produces a static image of the raster, such as the ones shown in Figures 5.13 and 5.14:
plot(s)
plot(r)
## downsample set to 2
Useful additional parameters when running plot on stars objects include:
• text_values—Logical, whether to display text labels (default is FALSE)
• axes—Logical, whether to display axes (default is FALSE)
• col—A vector of color codes or names
For example (Figure 5.15):
plot(s, text_values = TRUE, axes = TRUE, col = terrain.colors(10))
The expression terrain.colors(10) uses one of the built-in color palette functions in R to generate a vector of length 10 with terrain color codes, as discussed above (Section 5.1.6). The default color breaks are calculated using quantiles (breaks='quantile'). We can use other break types (e.g., breaks='equal'), or pass our own vector of custom breaks (Figure 5.16):
plot(s, text_values = TRUE, col = terrain.colors(10), breaks = 'equal')
plot(s, text_values = TRUE, col = terrain.colors(3), breaks = c(0, 100, 300, 500))
Note that the number of colors in the second expression (e.g., 4) needs to match the number of breaks minus one (e.g., 3) (why?).
#### 5.3.7.2 Interactive maps with mapview
The mapview function from package mapview lets us visually examine spatial objects—vector layers or rasters—in an interactive map on top of various background layers, such as OpenStreetMap, satellite images, etc. We can use the mapview function, after loading the mapview package, as follows. In this case, we are displaying the second layer of r (we learn about layer subsetting in Section 6.2.1):
library(mapview)
mapview(r[,,,2])
### 5.3.8 Raster values and properties
#### 5.3.8.1 Class and structure
Before moving on to cover the structure and properties of stars objects (Section 5.3.8.25.3.8.3), let us get a sense of the way their structure is reflected in the R console through the print, class, and str commands.
The print method for raster objects gives a summary of their properties:
s
## stars object with 2 dimensions and 1 attribute
## attribute(s):
## Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
## dem.tif 3 7 39 110.4906 179 448 17
## dimension(s):
## from to offset delta refsys point x/y
## x 1 7 679624 2880 UTM Zone 36, Northern Hem... FALSE [x]
## y 1 10 3644759 -2880 UTM Zone 36, Northern Hem... FALSE [y]
The class function returns the class name, which is stars in this case:
class(s)
## [1] "stars"
As discussed in Section 1.1.4, a class is a “template” with pre-defined properties that each object of that class has. For example, a stars object is actially a collection (namely, a list) of matrix or array objects, along with additional properties of the dimensions, such as dimension names, Coordinate Reference Systems (CRS), etc. When we read the dem.tif file with read_stars, the information regarding all of the properties was transferred from the file and into the stars “template”. Now, the stars object named s is in the RAM, filled with the specific values from dem.tif.
We can display the structure of the stars object with the specific values with str:
str(s)
## List of 1
## $dem.tif: num [1:7, 1:10] NA NA NA NA NA 3 3 NA NA NA ... ## - attr(*, "dimensions")=List of 2 ## ..$ x:List of 7
## .. ..$from : num 1 ## .. ..$ to : num 7
## .. ..$offset: num 679624 ## .. ..$ delta : num 2880
## .. ..$refsys:List of 2 ## .. .. ..$ input: chr "UTM Zone 36, Northern Hemisphere"
## .. .. ..$wkt : chr "PROJCRS[\"UTM Zone 36, Northern Hemisphere\",\n BASEGEOGCRS[\"WGS 84\",\n DATUM[\"unknown\",\n "| __truncated__ ## .. .. ..- attr(*, "class")= chr "crs" ## .. ..$ point : logi FALSE
## .. ..$values: NULL ## .. ..- attr(*, "class")= chr "dimension" ## ..$ y:List of 7
## .. ..$from : num 1 ## .. ..$ to : num 10
## .. ..$offset: num 3644759 ## .. ..$ delta : num -2880
## .. ..$refsys:List of 2 ## .. .. ..$ input: chr "UTM Zone 36, Northern Hemisphere"
## .. .. ..$wkt : chr "PROJCRS[\"UTM Zone 36, Northern Hemisphere\",\n BASEGEOGCRS[\"WGS 84\",\n DATUM[\"unknown\",\n "| __truncated__ ## .. .. ..- attr(*, "class")= chr "crs" ## .. ..$ point : logi FALSE
## .. ..$values: NULL ## .. ..- attr(*, "class")= chr "dimension" ## ..- attr(*, "raster")=List of 4 ## .. ..$ affine : num [1:2] 0 0
## .. ..$dimensions : chr [1:2] "x" "y" ## .. ..$ curvilinear: logi FALSE
## .. ..$blocksizes : int [1, 1:2] 7 10 ## .. .. ..- attr(*, "dimnames")=List of 2 ## .. .. .. ..$ : NULL
## .. .. .. ..$: chr [1:2] "x" "y" ## .. ..- attr(*, "class")= chr "stars_raster" ## ..- attr(*, "class")= chr "dimensions" ## - attr(*, "class")= chr "stars" The 1st line in the str printout, tells us that the s raster object is a list of length 1, namely that the raster has one attribute. The 2nd line in the str printout specifies the name and contents of the first (and only) item in the list, namely its type, dimensions, and a sample of the first few values. In this case, the list item is named 'dem.tif', and it is a matrix with 7 rows and 10 columns. We elaborate on raster values and methods to access them in Sections 5.3.8.2, 5.3.8.4, and 6.2. The remaining lines in the str printout comprise the 'dimensions' component of the raster object, which contains the spatial properties of the raster. We elaborate on the 'dimensions' component of rasters in Sections 5.3.8.3 and 6.3. Recall that “simple” rasters (Section 5.3.2), such as the ones we are working with in this book, always contain just one attribute. #### 5.3.8.2 Raster attributes and values stars objects are collections of matrices or arrays. Each matrix or array is known as an attribute and is associated with a name. A GeoTIFF file, being a simple raster format (Section 5.3.2), can contain just one attribute. Attribute names are not specified as part of a GeoTIFF file, and therefore automatically given default values based on the file name. We can get the attribute name(s) using the names function: names(s) ## [1] "dem.tif" names(r) ## [1] "MOD13A3_2000_2019.tif" We can change the attribute names through assignment. For example, it makes sense to name the attribute after the physical property or measurement it represents: names(s) = 'elevation' names(s) ## [1] "elevation" names(r) = 'NDVI' names(r) ## [1] "NDVI" Accessing an attrubute, by name or by numeric index, returns the matrix (if single-band raster) or array (if multi-band raster) object with the values of that attribute. Accessing attributes uses the list access methods (Section 11.1.2), namely: • list$name
• list[[index]]
• list[['name']]
where list is the list object, name is the element name, and index is the element’s numeric index.
For example, in the following expressions we access the (only) attribute of the s raster by name:
s$elevation ## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] ## [1,] NA NA NA NA NA NA NA 3 5 7 ## [2,] NA NA NA 61 106 47 31 39 32 49 ## [3,] NA NA NA 9 132 254 233 253 199 179 ## [4,] NA NA NA 4 11 146 340 383 357 307 ## [5,] NA 4 6 9 6 6 163 448 414 403 ## [6,] 3 6 9 10 6 6 13 152 360 370 ## [7,] 3 4 7 16 27 12 64 39 48 55 Here we demonstrate that this is a matrix with 7 rows and 10 columns: class(s$elevation)
## [1] "matrix" "array"
dim(s$elevation) ## x y ## 7 10 r$NDVI is an array that is too big to print here, but we can print the first five rows and columns from the 1st layer:
r$NDVI[1:5, 1:5, 1] ## [,1] [,2] [,3] [,4] [,5] ## [1,] NA NA NA NA NA ## [2,] NA NA NA NA NA ## [3,] NA NA NA NA NA ## [4,] NA NA NA NA NA ## [5,] NA NA NA NA NA Here we demonstrate that this is an array with 167 rows, 485 columns, and 233 “layers”: class(r$NDVI)
## [1] "array"
dim(r$NDVI) ## x y band ## 167 485 233 Here, we do the same using a numeric index and the [[ operator: class(s[[1]]) ## [1] "matrix" "array" dim(s[[1]]) ## x y ## 7 10 class(r[[1]]) ## [1] "array" dim(r[[1]]) ## x y band ## 167 485 233 The $ and [[ operators actually select individual elements from a list; this works because a stars object is, internally, a list of matrices or arrays (i.e., the “attributes”). We will elaborate on lists and their subsetting methods, namely the $ and [[ operators, in Section 11.1. By now we met all three subset operators ([, $, and [[) which we are going to use in this book, so let’s review and summarize the operators and their use cases (Table 5.4). The fourth subset operator @ concerns objects of classes known as S4, which is more rarely encountered and we will not use it in the book.
Table 5.4: Subset operators in R
Syntax Objects Returns
x[i] vector, list Subset i
x[i, j] data.frame, matrix Subset i,j
x[i, j, k] array Subset i,j,k
x[[i]] vector, list Single element i
x$i data.frame, list Single column or element i x@n S4 objects Slot n #### 5.3.8.3 Dimensions and spatial properties The nrow, ncol, and dim functions return the number or rows, column or all available dimensions of a stars object, respectively. These functions return a named numeric vector, where the names correspond to dimension names (Section 6.3.2). For example, here are the outputs of nrow, ncol, and dim for the single-band raster s: nrow(s) ## Number of rows ## x ## 7 ncol(s) ## Number of columns ## y ## 10 dim(s) ## All dimensions ## x y ## 7 10 and here are the outputs of the same functions for the multi-band raster r: nrow(r) ## Number of rows ## x ## 167 ncol(r) ## Number of columns ## y ## 485 dim(r) ## All dimensions ## x y band ## 167 485 233 As mentioned above (Section 5.3.1), the spatial properties, determining raster placement in geographical space, include the extent, or the origin and resolution, and the Coordinate Reference System (CRS). The extent can be accessed using function st_bbox (“bounding box”). Here are the extents of s and r: st_bbox(s) ## xmin ymin xmax ymax ## 679624 3615959 699784 3644759 st_bbox(r) ## xmin ymin xmax ymax ## 3239946 3267745 3394692 3717158 The extent is returned as an object of class bbox. The object is basically a numeric vector of length 4, including the xmin, ymin, xmax and ymax values (in that order). How can we get the bottom y-coordinate (i.e., $$y_{min}$$) out of st_bbox(s)? Write an expression that returns it. The origin and resolution, as well as other properties of a stars dimensions, can be accessed using the st_dimensions function. We will elaborate on this in Section 6.3.1. In the meanwhile, for completeness, here are the origins ('offset') and resolutions ('delta') of the x- and y-axes of s: st_dimensions(s)$x$offset ## x-axis origin ## [1] 679624 st_dimensions(s)$y$offset ## y-axis origin ## [1] 3644759 st_dimensions(s)$x$delta ## x-axis resolution ## [1] 2880 st_dimensions(s)$y$delta ## y-axis resolution ## [1] -2880 and of r: st_dimensions(r)$x$offset ## x-axis origin ## [1] 3239946 st_dimensions(r)$y$offset ## y-axis origin ## [1] 3717158 st_dimensions(r)$x$delta ## x-axis resolution ## [1] 926.6254 st_dimensions(r)$y\$delta ## y-axis resolution
## [1] -926.6254
Note that the resolution is separate for the 'x' and 'y' dimensions. The (absolute) resolution is usually equal for both, in which case raster pixels are square. However, the 'x' and 'y' resolutions can also be unequal, in which case raster pixels are non-square rectangles.
The CRS definition of a stars object can be accessed with the st_crs function. For example, the rasters s and r have different CRS definitions. Here is the CRS definitions of raster s, known as “UTM Zone 36N”:
st_crs(s)
## Coordinate Reference System:
## User input: UTM Zone 36, Northern Hemisphere
## wkt:
## PROJCRS["UTM Zone 36, Northern Hemisphere",
## BASEGEOGCRS["WGS 84",
## DATUM["unknown",
## ELLIPSOID["WGS84",6378137,298.257223563,
## LENGTHUNIT["metre",1,
## ID["EPSG",9001]]]],
## PRIMEM["Greenwich",0,
## ANGLEUNIT["degree",0.0174532925199433,
## ID["EPSG",9122]]]],
## CONVERSION["Transverse Mercator",
## METHOD["Transverse Mercator",
## ID["EPSG",9807]],
## PARAMETER["Latitude of natural origin",0,
## ANGLEUNIT["degree",0.0174532925199433],
## ID["EPSG",8801]],
## PARAMETER["Longitude of natural origin",33,
## ANGLEUNIT["degree",0.0174532925199433],
## ID["EPSG",8802]],
## PARAMETER["Scale factor at natural origin",0.9996,
## SCALEUNIT["unity",1],
## ID["EPSG",8805]],
## PARAMETER["False easting",500000,
## LENGTHUNIT["metre",1],
## ID["EPSG",8806]],
## PARAMETER["False northing",0,
## LENGTHUNIT["metre",1],
## ID["EPSG",8807]]],
## CS[Cartesian,2],
## AXIS["easting",east,
## ORDER[1],
## LENGTHUNIT["metre",1,
## ID["EPSG",9001]]],
## AXIS["northing",north,
## ORDER[2],
## LENGTHUNIT["metre",1,
## ID["EPSG",9001]]]]
and here is the CRS definition of the raster r, known as “Sinusoidal”:
st_crs(r)
## Coordinate Reference System:
## User input: unnamed
## wkt:
## PROJCRS["unnamed",
## BASEGEOGCRS["unnamed ellipse",
## DATUM["unknown",
## ELLIPSOID["unnamed",6371007.181,0,
## LENGTHUNIT["metre",1,
## ID["EPSG",9001]]]],
## PRIMEM["Greenwich",0,
## ANGLEUNIT["degree",0.0174532925199433,
## ID["EPSG",9122]]]],
## CONVERSION["Sinusoidal",
## METHOD["Sinusoidal"],
## PARAMETER["Longitude of natural origin",0,
## ANGLEUNIT["degree",0.0174532925199433],
## ID["EPSG",8802]],
## PARAMETER["False easting",0,
## LENGTHUNIT["metre",1],
## ID["EPSG",8806]],
## PARAMETER["False northing",0,
## LENGTHUNIT["metre",1],
## ID["EPSG",8807]]],
## CS[Cartesian,2],
## AXIS["easting",east,
## ORDER[1],
## LENGTHUNIT["metre",1,
## ID["EPSG",9001]]],
## AXIS["northing",north,
## ORDER[2],
## LENGTHUNIT["metre",1,
## ID["EPSG",9001]]]]
The CRS definition is an object of class crs, which contains the textual definition of the CRS, in the WKT format. As we will see later on, the crs of a spatial layer can be initialized using an EPSG code (Section 7.9.2), or by importing it from an existing layer. Therefore, in practice, we never have to manually type WKT definitions such as the ones shown above.
#### 5.3.8.4 Accessing raster values
As shown above (Section 5.3.8.2), raster values can be accessed directly, as a matrix or an array, by selecting a raster attribute, either by name or by index. For example:
s[[1]]
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,] NA NA NA NA NA NA NA 3 5 7
## [2,] NA NA NA 61 106 47 31 39 32 49
## [3,] NA NA NA 9 132 254 233 253 199 179
## [4,] NA NA NA 4 11 146 340 383 357 307
## [5,] NA 4 6 9 6 6 163 448 414 403
## [6,] 3 6 9 10 6 6 13 152 360 370
## [7,] 3 4 7 16 27 12 64 39 48 55
A histogram can give a first impression of the raster values distribution (Figure 5.17), using the hist function:
hist(s[[1]])
For example, the above histogram tells us that the overall range of elevation values in the raster s is roughly 0-450 ($$m$$), but most pixels are in the 0-50 $$m$$ elevation range.
Note that we can pass the matrix (or array) of raster values to many other functions that accept a numeric vector, such as mean or range.
Calculate the mean, minimum and maximum of the cell values in the raster s (excluding NA).
The matrix or array rows and columns are swapped, compared to the visual arrangement of the raster, because in a stars object the first dimension (matrix rows) refers to x (raster columns) and the second dimension (matrix columns) refers to y (raster rows)! For example, the following expression returns the 7th column of the matrix that holds the raster values, which actually refers to the 7th row in terms of the spatial arrangement of the raster (Figure 5.15):
s[[1]][, 7]
## [1] NA 31 233 340 163 13 64
We can modify (a subset of) raster values using assignment. For example, the following code creates a copy of the raster s, named u, then replaces the values in the 7th row (of the raster) with the value -1:
u = s
u[[1]][, 7] = -1
The resulting raster u is shown in Figure 5.18:
plot(u, text_values = TRUE, axes = TRUE, col = terrain.colors(8))
We can even replace the entire matrix or array of values with a custom one. This can be done using assignment to an “empty” subset, which implies selecting all cells, as in r[[1]][]. For example, the following code section creates another copy named u, then replaces all values with a consecutive vector:
u = s
u[[1]][] = 1:length(u[[1]])
u[[1]]
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,] 1 8 15 22 29 36 43 50 57 64
## [2,] 2 9 16 23 30 37 44 51 58 65
## [3,] 3 10 17 24 31 38 45 52 59 66
## [4,] 4 11 18 25 32 39 46 53 60 67
## [5,] 5 12 19 26 33 40 47 54 61 68
## [6,] 6 13 20 27 34 41 48 55 62 69
## [7,] 7 14 21 28 35 42 49 56 63 70
The resulting raster u is shown in Figure 5.19:
plot(u, text_values = TRUE, axes = TRUE, breaks = 'equal', col = terrain.colors(30))
Sometimes it is useful to assign the same value to all cells, to create a uniform raster. This can be done using assignment of a single value, which is replicated, to the subset of all raster cells:
u = s
u[[1]][] = -3
u[[1]]
## [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
## [1,] -3 -3 -3 -3 -3 -3 -3 -3 -3 -3
## [2,] -3 -3 -3 -3 -3 -3 -3 -3 -3 -3
## [3,] -3 -3 -3 -3 -3 -3 -3 -3 -3 -3
## [4,] -3 -3 -3 -3 -3 -3 -3 -3 -3 -3
## [5,] -3 -3 -3 -3 -3 -3 -3 -3 -3 -3
## [6,] -3 -3 -3 -3 -3 -3 -3 -3 -3 -3
## [7,] -3 -3 -3 -3 -3 -3 -3 -3 -3 -3
The resulting raster u is shown in Figure 5.20:
plot(u, text_values = TRUE, axes = TRUE, breaks = 'equal', col = terrain.colors(1))
### 5.3.9 Writing raster to file
We will not need to export rasters (or vector layers) in this book, since we will be working exclusively in the R environment. In practice, however, one often needs to export spatial objects from R to a file, to share them with other colleagues, further edit or process them in GIS software such as ArcGIS or QGIS, and so on.
Writing a stars raster object to a file on disk is done using write_stars. To run write_stars, we need to specify:
• obj—The stars object to write
• dsn—The file name (or file path) of the raster file being created
The function can automatically detect the required file format based on the file extension. For example, the following expression exports the stars object named s to a GeoTIFF file named dem_copy.tif in the current working directory:
write_stars(s, 'dem_copy.tif')
### References
Wickham, Hadley et al. 2011. “The Split-Apply-Combine Strategy for Data Analysis.” Journal of Statistical Software 40 (1): 1–29.
1. Due to the conventions of image, a matrix image is actually reversed in 90 degrees compared to the textual representation of a matrix. To get the same orientation as in the textual representation, use image(t(apply(volcano,2,rev)))↩︎
2. There are similar functions named rowSums and colSums for calculating row and column sums, respectively.↩︎
3. Note that there are no data structure for zero-dimensional data structures (i.e., scalars) in R.↩︎
4. Unless a new version of a package was released and we want to update it, in which case we need to re-install the package.↩︎
5. GeoTIFF files can come with both *.tif and *.tiff file extension, so if one of them does not work you should try the other.↩︎
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# Percent
Percent
1. Zoomer Bix plans to buy a microcomputer that costs \$690, excluding sales tax. If the sales tax rate is 5 1/2%, how much tax will Zoomer have to pay?
Solution
Multiply 690 by 0.055 = 37.95
So the tax is \$37.95
1. Wheel World sold 875 cars last year. If sales this year are 120% of sales last year, how many cars will Wheel World sell this year?
Solution
Multiply 875 by 1.20 = 1050
So Wheel World will sell 1050 cars this year.
1. Klutz Schlump borrowed \$500. He agreed to pay back the entire loan, plus interest, at the end of the year. If the interest rate is 18% per year, what is the total amount Klutz agreed to pay back?
Solution
Multiply 500 by 1.18 = 590
So Klutz will pay back 500 in principal and 90 in interest.
1. The “suggested retail’ price of a certain camera is \$340, but a discount store sells the camera at a 30% discount. What is the price of the camera at the discount store?
Solution
If there is a 30% discount then 70% of the price is kept.
Multiply 340 by .70 = 238
So the camera at the discounted price is \$238
1. Mr. J. Doe has a taxable income of \$9000. If the income tax rate is 16% on the first \$4000 of income, 19% on the next \$4000, and 23% on the next \$1000, how much is Mr. Doe’s tax?
Solution
Multiply 4000 by .16 = 640
Multiply 4000 by .19 = 760
Multiply 1000 by .23 = 230
640 + 760 + 230 = 1630
So Mr. J Doe will pay \$1630 tax on \$9000 in income.
1. After taking his girl friend out to dinner, Osgood decides to leave a tip of 15% of the bill. If the bill is \$38, how much should the tip be?
Solution
Multiply 38 by .15 = 5.70
So Osgood will leave a tip of \$5.70
1. Profits of Calculess Corporation this year were 140% of profits last year. If profits last year were \$5200, what were profits this year?
Solution
Multiply 5200 by 1.40 = 7280
So Calculess Corp will have profits of \$7280 this year.
1. Klutz got 10 out of 16 problems on an algebra test correct. What percent were correct?
Solution
10 divided by 16 = .625
So Klutz had 62.5% of the problems correct.
1. A team won 12 games, lost 15 games, and tied 3 games. What percent of its games did the team win?
Solution
30 games were played. (12 + 15 + 3)
12 divided by 30 = .4
So the team won 40% of its games.
1. In a magazine drive a school keeps 40% of all sales dollars. How many dollars worth of magazines must be sold for the school to earn \$5000?
Solution
5000 = 40% of some number
.40x = 5000
x = 12500
So \$12500 worth of magazines must be sold for the school to earn \$5000.
1. A real estate broker earns 2 ½% of her sales as a commission. How many dollars in sales does she need in order to earn a commission of \$1000?
Solution
1000 = 2.5% of some number
.025x = 1000
x = 40,000
So the real estate broker must sell \$40,000 to earn \$1000 when the sales commission is 2.5%
1. A steel cable expands 0.2% of its length when its temperature is increased 100°C. How much longer will a 750 meter cable become with this increase in temperature?
Solution
Multiply 750 by 1.2 = 900
So the 750 meter cable will become 900 meters
1. Elmo Buckets made 54 out of 80 free throws. What percent did he miss?
Solution
If Elmo made 54 of 80 then he missed 26..
Divide 26 by 80 = .325 = 32.5%
So Elmo missed 32.5% of his free throws.
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# In its most recent approach, the comet Crommelin passed the
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In its most recent approach, the comet Crommelin passed the [#permalink]
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29 Mar 2012, 02:02
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In its most recent approach, the comet Crommelin passed the Earth at about the same distance and in about the same position, some 25 degrees above the horizon, that Halley’s comet will pass the next time it appears.
(A) that Halley’s comet will pass
(B) that Halley’s comet is to be passing
(C) as Halley’s comet
(D) as will Halley’s comet
(E) as Halley’s comet will do
What's wrong with "that" in a?
Is it impossible that the clause after "that" modifies position?
I might be wrong, but I think that I've seen "the same X that Y is Z" <- this structure somewhere.
[Reveal] Spoiler: OA
If you have any questions
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Re: PT #15 SC 22 [#permalink]
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29 Mar 2012, 04:04
eybrj2 wrote:
In its most recent approach, the comet Crommelin passed the Earth at about the same distance and in about the same position, some 25 degrees above the horizon, that Halley’s comet will pass the next time it appears.
(A) that Halley’s comet will pass
(B) that Halley’s comet is to be passing
(C) as Halley’s comet
(D) as will Halley’s comet
(E) as Halley’s comet will do
What's wrong with "that" in a?
Is it impossible that the clause after "that" modifies position?
I might be wrong, but I think that I've seen "the same X that Y is Z" <- this structure somewhere.
that is relative pronoun use to refer noun
In this sentence we are comparing the way---the comet Crommelin passed as will Halley's comet
D & E
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Re: PT #15 SC 22 [#permalink]
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30 Mar 2012, 03:26
1
This post was
BOOKMARKED
In its most recent approach, the comet Crommelin passed the Earth at about the same distance and in about the same position, some 25 degrees above the horizon, that Halley’s comet will pass the next time it appears.
(A) that Halley’s comet will pass
(B) that Halley’s comet is to be passing
(C) as Halley’s comet--> changes the meaning
(D) as will Halley’s comet
(E) as Halley’s comet will do--> use of "do" is incorrect here use "will" is sufficient for the implied comparison.
"That" is used without comma because the clause it introduces is essential one, hence A and B are wrong.
hth
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Re: PT #15 SC 22 [#permalink]
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01 Jul 2012, 23:00
'that' is wrong because comparison is being done by two things - distance and position .If we use 'that' , then it modifies only position and not distance;furthermore , 'some 25 degrees above the horizon' is an appositive modifies 'position' and so it is wrong to say 'that' is not correct after comma.for example;
I like to read the book that is blue--> correct
I like to read the book , "bible" , that is blue --> is also correct.
D is correct since 'As' is correct here because it compares action 'passed' of Crommelin and Halley’s comet .
C - > Wrong because 'As' is used in wrong comparison :- verb with noun
E-> Wrong because 'will do' redundant.
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Re: In its most recent approach, the comet Crommelin passed the [#permalink]
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Re: In its most recent approach, the comet Crommelin passed the [#permalink]
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17 Sep 2015, 03:59
we should look at "same " in dictionary
same has two meaning: the same thing and the two things alike.
same can go with "that" and "as"
when going with "that" , the object of the following clause is "the same"
when going with "as" there can be one of two meaning above
after "as" we can use a noun which is the same as previous noun
or
we can use only subject of the second clause to show the second context of the comparion, not showing the second noun
apply inhere
we can use only the subject of the second clause. D and E are correct
hard for me
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Re: In its most recent approach, the comet Crommelin passed the [#permalink]
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17 Sep 2015, 04:06
same has 2 meaning,
the two things similar
in this meaning
we use a noun after the "as" this noun is similar to the first noun
your pen is the same as mine
second meaning is : one thing
in this meaning we use the subject after "as" to provide the second context because there is one thing, not two things.
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Re: In its most recent approach, the comet Crommelin passed the [#permalink]
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23 Feb 2016, 07:39
eybrj2 wrote:
In its most recent approach, the comet Crommelin passed the Earth at about the same distance and in about the same position, some 25 degrees above the horizon, that Halley’s comet will pass the next time it appears.
(A) that Halley’s comet will pass
(B) that Halley’s comet is to be passing
(C) as Halley’s comet
(D) as will Halley’s comet
(E) as Halley’s comet will do
What's wrong with "that" in a?
Is it impossible that the clause after "that" modifies position?
I might be wrong, but I think that I've seen "the same X that Y is Z" <- this structure somewhere.
the same that
this phrase is easier to use . that-clause is a relative pronoun. I speak the same language that you speak.
but the phrase "the same+noun+as+ second clause " is harder. in the phrase, only the different thing which has the same +noun is retain. I used to make mistake by repeating the noun again. this is not idomatic because the noun need to appear one time
I speak the same english as you
here there is one kind of english and this english appear one time in the first clause of comparison. in the second clause we need to show the different factor of comparision
we can not say
I speak the same english as the english you do. this is redundant and not idiomatic.
come back to our question.
in d
as will harley commet
in here the position and distance do not appear in the second clause of comparison.
this pattern is shown in dictionary but we misuse it often . I make mistake also
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In its most recent approach, the comet Crommelin passed the [#permalink]
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23 Feb 2016, 13:25
Whats wrong with E?I fail to understand.
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Re: In its most recent approach, the comet Crommelin passed the [#permalink]
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23 Feb 2016, 13:38
1
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Expert's post
goforgmat wrote:
Whats wrong with E?I fail to understand.
Option E has nothing wrong grammatically. Option D is just more economical than option E in expressing the idea - you would recollect that we are supposed to choose the best option from the given choices.
Re: In its most recent approach, the comet Crommelin passed the [#permalink] 23 Feb 2016, 13:38
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# DIGITAL SAT MATH PROBLEMS AND SOLUTIONS(Part - 25)
Problem 1 :
y = 2x2 – 21x + 64
y = 3x + a
In the given system of equations, a is a constant. The graphs of the equations in the given system intersect at exactly one point (x, y) in the xy-plane. What is the value of x?
A) -8
B) -6
C) 6
D) 8
Solution :
Problem 2 :
The table gives the coordinates of two points on a line in the xy-plane. The y-intercept of the line is (k - 5, b), where k and b are constants. What is the value of b?
Solution :
Problem 3 :
A certain park has an area of 11,863,808 square yards. What is the area in square miles of this park?
(1 mile = 1,760 yards)
A) 1.96
B) 3.83
C) 3,444.39
D) 6,740.8
Solution :
Problem 4 :
In the given system of equations, r is a constant. If the system has no solution, what is the value of r?
Solution :
Problem 5 :
The scatterplot above shows the federal-mandated minimum wage every 10 years between 1940 and 2010. A line of best fit is shown, and its equation is y = 0.096x − 0.488. What does the line of best fit predict about the increase in the minimum wage over the 70-year period?
A. Each year between 1940 and 2010, the average increase in minimum wage was 0.096 dollars.
B. Each year between 1940 and 2010, the average increase in minimum wage was 0.49 dollars.
C. Every 10 years between 1940 and 2010, the average increase in minimum wage was 0.096 dollars.
D. Every 10 years between 1940 and 2010, the average increase in minimum wage was 0.488 dollars.
Solution :
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## Boolean financial advisors
Alex and Bob work as financial advisors for the same company. They draw equal salaries from the company. They behave well at the office. Both work on similar assignments. Each assignment requires a yes-no decision. The company uses the decisions made by them to make profits.
After the recession hit the company very badly, one of them has to be fired. Both Alex and Bob have worked on almost the same number of assignments in the last ten years. Alex has been consistently taking about 80% decisions correctly every year. Bob, on the other hand, has been taking only about 5% correct decisions every year.
Assuming that the performances of Alex and Bob would remain the same in future, who should the company fire to maximize its profits in the years to come? Why?
[SOLVED]
### 5 comments
#### Abhishek Sethi solved this puzzle:
Alex should be fired. Since a yes-no decision has to be taken, Bob being 5% right is equivalent to Bob being 95% wrong. If every decision of Bob is inverted, we get 95% accuracy.
#### Rino Raj solved this puzzle:
Consistently 95% wrong means the company could reverse each decision and be 95% right. Hence Bob should be retained. Alex should be fired.
#### Indhu Bharathi solved this puzzle:
Fire Alex and inverse Bob's decisions to get a success rate of 95%.
#### Gnanasenthil G solved this puzzle:
Alex should be fired. Bob takes only 5% correct decisions. The company can do the exact opposite of what Bob recommends and take 95% right decisions. However, Alex can help the company take only 80% right decisions.
#### Reiner Rückwald solved this puzzle:
If Bob keeps on performing so badly that 95% of his decisions are taken wrongly every year, then the company should better keep him and fire Alex instead, provided that the company always inverts Bobs decisions and decides upon the opposite. This ensures that 95% of those decisions are taken correctly to maximize the company's profit.
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Solve the following
Question:
If $S_{2}$ and $S_{4}$ denote respectively the sum of the squares and the sum of the fourth powers of first $n$ natural numbers, then $\frac{S_{4}}{S_{2}}=$ __________________ .
Solution:
S2 : Sum of the squares of first n natural numbers.
S: Sum of the fourth powers of first n natural numbers.
To find :- $\frac{S_{4}}{S_{2}}$
Since
$S_{4}=\frac{n(n+1)(2 n+1)\left(3 n^{2}+3 n-1\right)}{30}$
$S_{2}=\frac{n(n+1)(2 n+1)}{6}$
Hence,$\frac{S_{4}}{S_{2}}=\frac{n(n+1)(2 n+1)\left(3 n^{2}+3 n-1\right)}{30 \times n(n+1)(2 n+1)} \times 6$
Hence,$\quad \frac{S_{4}}{S_{2}}=\frac{3 n^{2}+3 n-1}{5}$
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https://nrich.maths.org/public/topic.php?code=131&cl=2&cldcmpid=2644
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| 602,142,210 | 7,449 |
# Resources tagged with: Reflections
Filter by: Content type:
Age range:
Challenge level:
### There are 33 results
Broad Topics > Transformations and constructions > Reflections
### Hexpentas
##### Age 5 to 11 Challenge Level:
How many different ways can you find of fitting five hexagons together? How will you know you have found all the ways?
### Building with Longer Rods
##### Age 7 to 14 Challenge Level:
A challenging activity focusing on finding all possible ways of stacking rods.
### Penta Play
##### Age 7 to 11 Challenge Level:
A shape and space game for 2,3 or 4 players. Be the last person to be able to place a pentomino piece on the playing board. Play with card, or on the computer.
##### Age 11 to 14 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Transformation Game
##### Age 11 to 14 Challenge Level:
Why not challenge a friend to play this transformation game?
### Coordinating Classroom Coordinates
##### Age 7 to 11
This article describes a practical approach to enhance the teaching and learning of coordinates.
### Making Maths: Indian Window Screen
##### Age 7 to 11 Challenge Level:
Can you recreate this Indian screen pattern? Can you make up similar patterns of your own?
### Matching Frieze Patterns
##### Age 11 to 14 Challenge Level:
Sort the frieze patterns into seven pairs according to the way in which the motif is repeated.
### Transformation Tease
##### Age 7 to 11 Challenge Level:
What are the coordinates of this shape after it has been transformed in the ways described? Compare these with the original coordinates. What do you notice about the numbers?
### National Flags
##### Age 7 to 11 Challenge Level:
This problem explores the shapes and symmetries in some national flags.
### Let Us Reflect
##### Age 7 to 11 Challenge Level:
Where can you put the mirror across the square so that you can still "see" the whole square? How many different positions are possible?
### Transforming the Letters
##### Age 7 to 11 Challenge Level:
What happens to these capital letters when they are rotated through one half turn, or flipped sideways and from top to bottom?
### Reflector ! Rotcelfer
##### Age 7 to 11 Challenge Level:
Can you place the blocks so that you see the reflection in the picture?
### Decoding Transformations
##### Age 11 to 14 Challenge Level:
See the effects of some combined transformations on a shape. Can you describe what the individual transformations do?
### So It's Times!
##### Age 7 to 14 Challenge Level:
How will you decide which way of flipping over and/or turning the grid will give you the highest total?
### Building with Rods
##### Age 7 to 11 Challenge Level:
In how many ways can you stack these rods, following the rules?
### Hidden Meaning
##### Age 7 to 11 Challenge Level:
What is the missing symbol? Can you decode this in a similar way?
### 2010: A Year of Investigations
##### Age 5 to 14
This article for teachers suggests ideas for activities built around 10 and 2010.
### It's Times Again
##### Age 7 to 14 Challenge Level:
Which way of flipping over and/or turning this grid will give you the highest total? You'll need to imagine where the numbers will go in this tricky task!
### What Am I?
##### Age 7 to 11 Challenge Level:
Can you draw the shape that is being described by these cards?
### Frieze Patterns in Cast Iron
##### Age 11 to 16
A gallery of beautiful photos of cast ironwork friezes in Australia with a mathematical discussion of the classification of frieze patterns.
### 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?
### ...on the Wall
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
### Friezes
##### Age 11 to 14
Some local pupils lost a geometric opportunity recently as they surveyed the cars in the car park. Did you know that car tyres, and the wheels that they on, are a rich source of geometry?
### Times
##### Age 7 to 11 Challenge Level:
Which times on a digital clock have a line of symmetry? Which look the same upside-down? You might like to try this investigation and find out!
### Clocks
##### Age 7 to 11 Challenge Level:
These clocks have been reflected in a mirror. What times do they say?
### Combining Transformations
##### Age 11 to 14 Challenge Level:
Does changing the order of transformations always/sometimes/never produce the same transformation?
### Simplifying Transformations
##### Age 11 to 14 Challenge Level:
How many different transformations can you find made up from combinations of R, S and their inverses? Can you be sure that you have found them all?
### Mirror, Mirror...
##### Age 11 to 14 Challenge Level:
Explore the effect of reflecting in two parallel mirror lines.
### The Frieze Tree
##### Age 11 to 16
Patterns that repeat in a line are strangely interesting. How many types are there and how do you tell one type from another?
### Paint Rollers for Frieze Patterns.
##### Age 11 to 16
Proofs that there are only seven frieze patterns involve complicated group theory. The symmetries of a cylinder provide an easier approach.
### Shaping up with Tessellations
##### Age 7 to 14
This article describes the scope for practical exploration of tessellations both in and out of the classroom. It seems a golden opportunity to link art with maths, allowing the creative side of your. . . .
### Reflecting Lines
##### Age 11 to 14 Challenge Level:
Investigate what happens to the equations of different lines when you reflect them in one of the axes. Try to predict what will happen. Explain your findings.
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https://ask.truemaths.com/question/in-fig-9-32-abcd-is-a-parallelogram-and-bc-is-produced-to-a-point-q-such-that-ad-cq-if-aq-intersect-dc-at-p-show-that-ar-bpc-ar-dpq-hint-join-ac-q-4/
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• 0
Guru
# In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). [Hint : Join AC.] Q.4
• 0
Hello sir i want to know the best solution of the question from exercise 9.4 of math of Areas of Parallelograms and Triangles chapter of class 9th give me the best and easy for solving this question how i solve it of question no. 4 In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). [Hint : Join AC.
Share
1. Given:
ABCD is a parallelogram
To prove:
ar (â–³BPC) = ar (â–³DPQ)
Proof:
∠APD = ∠QPC [Vertically Opposite Angles]
∠ADP = ∠QCP [Alternate Angles]
, △ABO ≅ △ACD [AAS congruency]
, DP = CP [CPCT]
In â–³CDQ, QP is median. [Since, DP = CP]
Since, median of a triangle divides it into two parts of equal areas.
, ar(△DPQ) = ar(△QPC) —(i)
In △PBQ, PC is median. [Since, AD = CQ and AD = BC ⇒ BC = QC]
Since, median of a triangle divides it into two parts of equal areas.
, ar(△QPC) = ar(△BPC) —(ii)
From the equation (i) and (ii), we get
ar(â–³BPC) = ar(â–³DPQ)
• 0
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https://convertertoolz.com/km-to-miles/146-km-to-miles/
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# 146 km to miles
## Heading 2: Step 5: Understand the Final Conversion
The final step in the conversion process is to understand the final conversion. Once you have completed all the necessary calculations and conversions, you need to make sure you have the correct units for the final answer. Whether you are converting distances, weights, or temperatures, it is important to have the proper unit to ensure accurate understanding and application.
For example, if you are converting kilometers to miles, it is crucial to know which unit you are using for the final result. In this case, the final conversion would be in miles. Understanding this will help you communicate the converted value accurately and effectively. It is also important to double-check your calculations to avoid any errors in the final conversion. Taking the extra time to review your work can save you from potentially embarrassing or costly mistakes. So, always make sure you are clear about the final conversion and that it is in the correct unit before applying it in any practical or real-life situations.
## Heading 1: Practical Applications and Examples
Practical Applications and Examples
To truly understand the practical applications of converting kilometers to miles, let’s consider a few examples. Imagine you’re planning a road trip and need to know the distance between two cities. In most countries, the distance is measured in kilometers, but you are more familiar with miles. By converting the kilometers to miles, you can gauge the approximate amount of time your journey will take and plan your stops accordingly. This simple conversion allows you to have a better understanding of the distances you will cover and make your trip planning smoother.
Another practical application of converting kilometers to miles can be found in fitness and exercise. Many people track their running or jogging distance to monitor their progress. If your running app displays the distance in kilometers, but you are more comfortable with miles, converting the values allows you to easily compare and track your performance over time. It’s a handy tool for staying motivated and pushing yourself towards achieving your fitness goals.
Remember, these are just a couple of examples of how converting kilometers to miles can be useful in everyday life. The ability to convert between these two units of measurement opens up opportunities for better planning, communication, and understanding, making our lives a little bit easier.
## Heading 2: Converting 146 Kilometers to Miles: An Example Calculation
To understand how to convert 146 kilometers to miles, we need to know the conversion rate between the two units of measurement. The most commonly used conversion rate is 1 kilometer equals 0.62137119 miles. Using this conversion rate, we can calculate the equivalent distance in miles for 146 kilometers.
With 146 kilometers, all we have to do is multiply it by the conversion rate of 0.62137119. So, when we do the math, 146 kilometers is approximately equal to 90.7205 miles. It may not be a precise whole number, but it gives us a good estimate of the equivalent distance in miles.
Now that we have successfully converted 146 kilometers to miles, we can apply this knowledge in various practical situations. Whether it’s for understanding a road trip distance or analyzing the fuel consumption per mile, knowing how to convert kilometers to miles can come in handy. By mastering this conversion, we can easily navigate the world of everyday measurements and calculations, making our lives a little bit easier.
## Heading 2: Converting Kilometers to Miles in Everyday Life
Converting kilometers to miles is not just a mathematical exercise, but a practical skill that can be used in everyday life. Whether you’re estimating the distance between two locations or trying to make sense of international units, understanding this conversion can come in handy.
One common scenario where converting kilometers to miles is useful is when reading the specifications of a car’s fuel efficiency. In many countries, the fuel consumption of vehicles is measured in liters per kilometer. However, if you’re more familiar with miles per gallon, it’s important to know how to convert these measurements. This knowledge allows you to accurately compare the fuel efficiency of different vehicles and make informed decisions when purchasing a car. Additionally, when traveling abroad, having a basic understanding of how kilometers relate to miles can help you navigate and estimate distances more effectively. So, whether you’re planning a road trip or just trying to make sense of your car’s specifications, having a grasp of this conversion can be truly practical in everyday life.
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https://acemyhw.com/projects/30961/Mathematics/mat101
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# Project #30961 - MAT101
I will need assistance with completing the following progect. All work must be shown. Thank you.
Name:
Date:
Instructor:
Use this template to insert your answers for the assignment. Please use one of the four methods for showing your work (EE, Math Type,ALT keys, or neatly typed). Remember that your work should be clear and legible.
1. Identify the coefficients, variable terms (with exponents), and constants in the following expression.
2x3+5y2-3z+1
Coefficients: _____________________
Variable terms: _______________________
Constants: ______________________
2. Identify the coefficients, variable terms (with exponents), and constants in the following expression.
4z5-8x2-6
Coefficients: _____________________
Variable terms: _______________________
Constants: ______________________
3. Combine like terms in the following expression. (Hint: You can color code the like terms.)
8x2+3x+9-x2+7x-2+y
Calculations:
4. Distribute and combine like terms in the following expression.
3(6y2-9+7x-2x2-3x-6)
Calculations:
5. Write and simplify an expression that applies the distributive property. Include at least 3 different terms.
Calculations:
6. Simplify the expression using the order of operations. (Note: * stands for multiplication)
(6*2-4) – 3(8-5) * 7
2
Calculations:
7. Simplify the expression using the order of operations.
(3-5) * -| -22 - 52 * 4|
Calculations:
8. Translate the following statement.
The product of 3 more than a number and 3 less than the same number.
Calculations:
9. Translate and solve the following statement.
The quotient of 2x and 4 is the same as the product of 6 and 3.
Calculations:
10. Write and translate your own statement using at least two different operations (i.e. - add, subtract, multiply, divide).
Calculations:
11. Simplify the expression. (Hint: Careful with the signs)
-6(-42-7)
Calculations:
12. Simplify the expression.
(-10)2 * -|23-7+12|
Calculations:
For problems 13-14, evaluate the expressions using the following values.
x= -3 y= 8 z= -12
13. 2y+3z
4x
Calculations:
14. 4x2-2z2
Calculations:
For problems 15-16, evaluate the expressions using the following values.
a= -1 b= 11 c= -7
15. 14a + (7- 6b)
c
Calculations:
16. (a2+b2)(b2-c2)
Calculations:
For problems 17-20, solve the equation. Check your answer by plugging it back into the equation.
17. 10x = 9x-15
Calculations:
18. 4x-9 = 7x+3
Calculations:
19. -3(8x-2x) = 72
Calculations:
20. 9(4y-3)-12y = 4(27+5y)
Calculations:
Subject Mathematics Due By (Pacific Time) 05/14/2014 12:00 am
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CC-MAIN-2017-39
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https://learn.careers360.com/school/question-need-solution-for-rd-sharma-maths-class-12-chapter-17-maxima-and-minima-exercise-17-point-1-question-7/?question_number=7.0
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#### Need solution for RD Sharma maths class 12 chapter 17 Maxima and Minima exercise 17 point 1 question 7
Maximum value is 3 and minimum value does not exist.
Hint:
f(x) have max value in [a, b] such that f(x) ≤ f(c) for all x belongs to [a ,b] and if f(x) ≥ f(c) then f(x) has minimum value.
Given:
$f(x)=-|x+1|+3 \text { on } R$
Explanation:
We have,
$f(x)=-|x+1|+3 \text { on } R$
We have,
\begin{aligned} &-|x+1| \leq 0 \text { for every } x \in R\\ &-|x+1|+3 \leq 3 \text { for every } x \in R\\ \end{aligned}
Thus maximum value of f is attained when |x+1|=0
X = -1
So maximum value of f(x) = f(-1) = - |-1+1| + 3
So its maximum value is 3 and it does not have minimum value.
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https://texasgateway.org/resource/determining-slopes-equations-graphs-and-tables
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| 534,350,462 | 34,279 |
# Introduction
Slope is a constant rate of change for a linear function. You can use equations, graphs, and tables to represent the slope of a linear function, and you can use the slope formula to calculate the slope between two points.
In this resource, you will examine equations, tables, and graphs that all represent linear functions. You will use these representations to determine the slope of the linear function, and then use the slopes to make comparisons among linear functions.
Remember, slope is a rate of change and can appear in a variety of ways.
# Determine the Slope Given an Equation
Compare the two sets of equations below. One set is presented in slope-intercept form, and the other set is not.
How can you tell if an equation is written in slope-intercept form?
One way to determine the slope of a line, given its equation, is to change the equation to slope-intercept form, and then identify the coefficient of the x term. The coefficient of the x term is the slope of the line.
To write an equation in slope-intercept form, you must solve the equation for y.
Directions:
1. Copy the table below into your notes.
2. Solve the equation for y, determine the slope, and then write your solutions in the table.
3. Use the Equation Calculator and Solver below to check your answers. Click on the image to access the Equation Calculator and Solver.
4. After you use the Equation Calculator and Solver, click on the link below to see a completed table.
Equation Equation in Slope-Intercept Form Slope
-5x + y = 10
3y + 15x = 8
-4 + 2y = 10
6y + 3x = 18
### Equation Calculator and Solver
Need help using the Equation Solver? Click to see an image with additional directions.
Pause and Reflect
1. List the slopes from your table in order from least to greatest.
2. Which of the slopes are greater than −1?
Practice
Determine the slopes of the lines represented by each of the equations below.
1. y = 0.45x – 18
2. y = $\frac{15}{6}$x – 18
3. 5x + 8y = 38
4. 3.2x + 1.6y = 72
# Determine the Slope Given a Table
Let’s consider a situation where the Williams family is coming home from visiting relatives. Mrs. Williams has to drive 350 miles in order to get home. Even though there is road construction, the family still drives at a constant rate; however, it takes them longer to drive home than it did to get to their relatives' house. Shonda Williams was bored, so she kept track of how many miles, y, remained until they got home after a certain amount of time, x, had passed. Shonda’s data appears in the table below.
Return Trip Home Amount of Time Since Leaving Relatives (in hours), x Number of Miles Remaining, y 0.5 325 1 300 1½ 275 2.5 225 4 150
Use the interactive below to help you determine the slope of the function representing the Williams family’s return trip home.
What is the slope of the linear function representing the relationship between the amount of time since leaving the Williams family’s relatives’ house, x, and the number of miles remaining, y, until the Williams family is back home?
Remember, slope can be found using the following formula:
$\text{Slope =§#xA0;}\frac{\mathrm{§#x394;}y}{\mathrm{§#x394;}x}\text{§#xA0;=§#xA0;}\frac{\text{Change in y}}{\text{Change in x}}$
Pause and Reflect
1. If you select a different pair of numbers from the table, do you get a different value for the slope? Why do you think that is the case?
2. How is this method related to the slope formula?
# Determine the Slope Given a Graph
There are several methods you can use to determine the slope of a line when you are given its graph. In this section, you will study two different methods.
Method 1
Pause and Reflect
If you are using the rise over run method for determining the slope from a graph, how can you tell from looking at the line if the slope will be positive or negative?
Method 2
Choose two points from the line of the graph. Given the coordinates of those two points, use the slope formula $m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-x1}$ to determine the slope.
Pause and Reflect
How does using the slope formula compare to using the rise over run method?
Practice
Click on the image below to open a new website and practice determining the slope given a graph. Complete problems 1-10. Enter your answers into the box on the left. Check your answers once you are done.
Source: Practice Problems- Determining Slope From a Graph, algebrahelp.com
# Defining Zero Slope, No Slope, and Undefined Slope
The slope of a line can be positive, negative, zero, or undefined. Below are graphical representations of a positive, negative, zero, and undefined slope.
Positive Slope Negative Slope Zero Slope Undefined Slope
Deciphering the difference between zero, undefined, and no slopes lines can be difficult. Remember, mathematics is like learning another language.
Click on the chart below next to the given slope to translate each phrase. Use the slope and the translation to decide which is the correct real world example in the third column.
# Summary
In this resource, you investigated different ways to determine the slope of a linear function. One of the most helpful ways to determine the slope is to use the slope formula.
$\text{Slope of a line =§#xA0;}m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-x1},$
$\text{where§#xA0;}m\text{§#xA0;represents the slope of the line.}$
If you are given a table or graph, the slope formula can indirectly be used by finding the ratio of the change in the y-values and the change in the x-values.
$\text{Slope =§#xA0;}\frac{\mathrm{§#x394;}y}{\mathrm{§#x394;}x}\text{§#xA0;=§#xA0;}\frac{\text{Change in y}}{\text{Change in x}}$
In a graph, the ratio of the change in y to the change in x is called rise over run. In a table, we can use the notation $\frac{\mathrm{§#x394;}y}{\mathrm{§#x394;}x}$ to represent this ratio, where Δ is the Greek letter delta, which is used to mean "change in" a quantity.
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# 1058.9 hours in hours and minutes
## Result
1058.9 hours equals 1058 hours and 54 minutes
You can also convert 1058.9 hours to minutes.
## How to convert 1058.9 hours to hours and minutes?
In order to convert 1058.9 hours to hours and minutes we can take the decimal part of 1058.9 hours and convert it into minutes. In this case we need to convert 0.9 hours to minutes.
We know that 1 hours equals 60 minutes, therefore to convert 0.9 hours to minutes we simply multiply 0.9 hours by 60 minutes:
0.9 hours × 60 minutes = 54 minutes
Finally, we can say that 1058.9 hours in hours and minutes is equivalent to 1058 hours and 54 minutes:
1058.9 hours = 1058 hours and 54 minutes
One thousand fifty-eight point nine hours is equal to one thousand fifty-eight hours and fifty-four minutes.
## Conversion table
For quick reference purposes, below is the hours and hours to minutes conversion table:
hours(hr) hours(hr) minutes(min)
1059.9 hours 1059 hours 54 minutes
1060.9 hours 1060 hours 54 minutes
1061.9 hours 1061 hours 54 minutes
1062.9 hours 1062 hours 54 minutes
1063.9 hours 1063 hours 54 minutes
1064.9 hours 1064 hours 54 minutes
1065.9 hours 1065 hours 54 minutes
1066.9 hours 1066 hours 54 minutes
1067.9 hours 1067 hours 54 minutes
1068.9 hours 1068 hours 54 minutes
## Units definitions
The units involved in this conversion are hours and minutes. This is how they are defined:
### Hours
An hour (symbol: h, also abbreviated hr) is a unit of time conventionally reckoned as 1⁄24 of a day and scientifically reckoned between 3,599 and 3,601 seconds. In the modern metric system, hours are an accepted unit of time defined as 3,600 atomic seconds. There are 60 minutes in an hour, and 24 hours in a day.
### Minutes
The minute is a unit of time usually equal to 1⁄60 (the first sexagesimal fraction) of an hour, or 60 seconds. In the UTC time standard, a minute on rare occasions has 61 seconds, a consequence of leap seconds (there is a provision to insert a negative leap second, which would result in a 59-second minute, but this has never happened in more than 40 years under this system). Although not an SI unit, the minute is accepted for use with SI units. The SI symbol for minute or minutes is min (without a dot).
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# Motion Graphs Physics Worksheet
Posted on
Motion Graphs Physics Worksheet. For instance, consider the addition of the same three vectors in a unique order. Energy is required to cause a optimistic test charge to move towards the electric field between the adverse and the optimistic terminal. Determine the acceleration of the bullet . Fill each grid area with an appropriately concise reply.
The results of including 11 km, north plus eleven km, east is a vector with a magnitude of 15.6 km. Later, the tactic of figuring out the direction of the vector might be mentioned. This problem asks to find out the result of adding two displacement vectors which might be at proper angles to each other.
• Uniform round motion is when an object travels on a round path at a relentless velocity.
• Find how far the practice will go before it is brought to relaxation.
• This larger slope is indicative of a larger velocity.
• Connect with us around-the-clock for any orders or pressing questions.
An object has moved by way of a distance. If sure, help your Solution with an instance. A dragster accelerates to a pace of 112 m/s over a distance of 398 m. Determine the acceleration of the dragster. A aircraft has a takeoff pace of 88.three m/s and requires 1365 m to achieve that speed. Determine the acceleration of the aircraft and the time required to reach this speed.
## Components Affecting Wave Speed
We know that the displacement is 304 m north and the time is a hundred and eighty s. We can use the method for average velocity to unravel the issue. For displacement as an alternative of d, as used on this text. Explain the usage of small arrows over variables is a common approach to denote vectors in higher-level physics courses.
Compare the centripetal acceleration for this fairly gentle curve taken at highway pace with acceleration because of gravity . For the second Worked Example, we’ll calculate the pressure required to make a automotive round a curve. When added together in this different order, these similar three vectors nonetheless produce a resultant with the identical magnitude and course as earlier than (20. m, 312 degrees).
Which of the next portions is equal to the gradient of the sort of graph shown? The second methodology makes use of the graph and an equation of movement. Worksheet 26 r kinematic equations 1.
The graph exhibits continuous accelerated motion. Graphing movement kinematics worksheet goal students will apply determining relationships between the shapes and slopes of graphs and the. Solve worded issues with the assist of diagrams. Activity will use a worksheet and speed vs.
### Part Key Phrases
When velocity is positive, the displacement-time graph ought to have a positive slope. When velocity is adverse, the displacement-time graph should have a negative slope. When velocity is zero, the displacement-time graph should be horizontal. The magnitude of common velocity of an object is equal to its common speed, solely when an object is moving in a straight line.
### Sources
Galileo additional noticed that regardless of the angle at which the planes have been oriented, the ultimate peak was almost always equal to the preliminary height. If the slope of the alternative incline have been reduced, then the ball would roll an additional distance to be able to attain that original top. Ballpark Estimate Math WorksheetsActual worksheet, first grade rounding worksheets and math worksheet rounding to nearest 10 are t… Determine unknown parameters from graphs of motion; The graph beneath describes the movement of a fly that starts out flying left. The graph under exhibits how the movement of a car is altering.
Complete the three small velocity-time graphs from the data offered under every graph. The worksheet for this exercise consists of three small and one large displacement-time graph. Complete the three small displacement-time graphs from the data supplied under every graph. Two cars are adjacent to each other on a four-lane freeway. The first automotive accelerates uniformly from rest the second the light modifications to green.
Like we noticed with displacement and distance in the last section, changes in course over a time interval have an even bigger effect on speed and velocity. The battery establishes an electrical potential difference across the 2 ends of the external circuit and thus causes the charge to move. The battery voltage is the numerical value of this electrical potential distinction. In an identical method, it is the distinction in water stress between the highest of the water slide and the underside of the water slide that the water pump creates. This difference in water pressure causes water to move down the slide.
Equations can simply comprise the information equal of several sentences. Galileo’s description of an object moving with fixed velocity required one definition, four axioms, and six theorems. All of those relationships can now be written in a single equation. Activity for training plotting graph data factors, drawing a line of finest match, calculating a gradient and analysing the line so as to calculate quantities. The first page consists of the activity for college students to finish and the second page includes solutions to the questions. Suitable for AQA and OCR Physics A AS-Level and A-Level specs.
### Electrical Potential Distinction And Simple Circuits
The first is to only agree with what the text description says. Free fall acceleration on Earth is only a quantity — a number that you need to memorize in case you have an expert purpose for studying physics. Displacement is the product of velocity and time. To discover displacement, calculate the world under each interval. The practice will cover a distance of 625 m before it involves relaxation.
## Related posts of "Motion Graphs Physics Worksheet"
#### Text Structure Worksheet Pdf
Text Structure Worksheet Pdf. This will assist them make connections between textual content and visual representations of text. K5 Learning presents free worksheets, flashcardsand inexpensiveworkbooksfor youngsters in kindergarten to grade 5. There are a number of completely different graphic organizers included (link to obtain within the section titled “Download the Free Text Structure Activities Here”)....
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Question
# A triangle and a parallelogram are on the same base and between the same parallel lines they have:
A
Equal area
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B
Equal perimeter
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C
Both
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D
None of these
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Solution
## The correct option is B None of these⇒ Let △ABP and a parallelogram ABCD be on the base AB and between the same parallels AB abd PC.⇒ Draw BQ∥AP to obtain another parallelogram ABQP and ABCD are on the same base AB and between the same parallel AB and PC.∴ ar(ABQP)=ar(ABCD) ---------- ( 1 )⇒ But △PAB ≅ △BQP [Diagonals PB divides parallelogram ABQP into two congruent triangles.]⇒ So ar(PAB)=ar(BQP) ----------- ( 2 )∴ $ar(PAB)=ar(ABQP) ----------- ( 3 ) [From ( 2 )]This gives ar(PAB)=12ar(ABCD) [From ( 1 ) and ( 3 )]⇒ So, we have prove that a triangle and a parallelogram are on same base and between the same parallel lines they don't have equal area or equal perimeter.⇒ So, correct answer is option D. Suggest Corrections 0 Join BYJU'S Learning Program Related Videos Triangles on the Same Base (Or Equal Bases) and Equal Areas Will Lie between the Same Parallels_$008
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# How do you figure out the angle of a roof?
Contents
The angle, or pitch, of a roof is calculated by the number of inches it rises vertically for every 12 inches it extends horizontally. For example, a roof that rises 6 inches for every 12 inches of horizontal run has a 6-in-12 pitch.
## How do you convert roof pitch to degrees?
Conversion Formula for Standard Roof Pitch to Degrees
1. To convert a roof slope expressed as “X-in-12″ to a roof slope expressed in degrees, find the arctangent of (rise/run).
2. Divide the rise (that’s the “X”, yours will vary) by the run (always 12).
3. Using a scientific calculator, find the arctangent of the result.
## What is the formula for a roof?
Multiply your house length by your house width to get the area. (For example, 40 feet x 30 feet = 1,200 square feet.) Next, multiply the area by your roof’s pitch. (1,200 x 1.05 = 1,260 square feet.)
## What is the most common angle of a roof?
The most commonly used roof pitches fall in a range between 4/12 and 9/12. Pitches lower than 4/12 have a slight angle, and they are defined as low-slope roofs. Pitches of less than 2/12 are considered flat roofs, even though they may be very slightly angled.
## How do you find an angle?
The formula for finding the total measure of all interior angles in a polygon is: (n – 2) x 180. In this case, n is the number of sides the polygon has. Some common polygon total angle measures are as follows: The angles in a triangle (a 3-sided polygon) total 180 degrees.
INTERESTING: What is the purpose of a roof fan?
## How do you find the slope of a roof angle?
If you know the roof’s angle in degrees, you can find the roof pitch by converting the angle in degrees to a slope, then finding the rise by multiplying the slope by 12. First, find the slope by finding the tangent of the degrees, e.g. slope = tan(degrees). Then multiply the slope by 12 to get the rise.
## How do you find the slope of a roof area?
Determine the area protected by the sloping roof by multiplying the length of the roof by the width. The result is the area covered by the sloping roof. For example, a single sloping roof covering a house 30 feet long and 20 feet wide covers an area of 600 square feet — 30 x 20 = 600.
## How do you determine the slope?
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 is a good slope for a roof?
For metal panel roof systems, NRCA recommends slopes of 1/2:12 or more for structural panel systems and 3:12 or more for architectural panel systems. For asphalt shingle, clay and concrete tile, metal shingle, slate and wood shake and shingle roof systems, NRCA recommends slopes of 4:12 or more.
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# Solve using the elimination method. Show your work. If the system has no solution or an infinite number of solutions, state this. 2x + 10y = 58 -14x + 8y = 140
91,330 questions, page 17
1. ## Rational Inequality with Absolute Value
Here's a question I'm having trouble with: |x|/|x+2|=0 and x+2>0, x must be greater than or equal to zero. When I solved for x in this case, I got x>-4. Reconciling the constraints leaves me back at x>=0. Did I do that correctly? Case 2: x is greater than
asked by Randy on March 26, 2011
2. ## 8th G-Math
Write an equation and show all work. One Monday, 405 students went on a trip to the zoo. All 7 buses were filled and 6 students had to travel in cars. How many students were in each bus? (My teacher taught the class but no one even understand, can you
asked by Cassidy on September 22, 2013
3. ## chemistry
How many g of NaOH is needed to make 300.0 mL of a 20 % (w/v) solution? please please show all work I'm very lost
asked by Thomas Ruger on April 2, 2017
4. ## math
Determine the solution of the following systems of linear equations using elimination. 2x + 3y = 7 x – 2y – 4 = 0
asked by jordan on December 20, 2015
5. ## math
Determine the solution of the following systems of linear equations using elimination. 2x + 3y = 7 x – 2y – 4 = 0
asked by jordan on December 20, 2015
6. ## algebra
Solve each of the following systems by using either addition or substitution. If a unique solution does not exist, state whether the system is dependent or inconsistent. 10x+2y=7 y==5x+3
asked by Becky on December 27, 2007
7. ## A number thoery question
Please help me! Thank you very much. Prove Fermat's Last theorem for n=3 : X^3 + Y^3 = Z^3 where X, Y, Z are rational integers, then X, Y, or Z is 0. Hint: * Show that if X^3 + Y^3 = Epsilon* Z^3, where X, Y, Z are quadratic integers in Q[sqrt(-3)], and
asked by kate on November 15, 2006
8. ## algebra
Please check my system of linear equations using any algebraic method. Show the check,if possible. 8x-6y=12 / 12x-9y=18 8x-6y=12 (eq.1) 12x-9y=18 (eq.2) 4x-3y=6 (eq.1) 4x-3y=6 (eq.2) y=4x-6 / 3 Check: Thank you for your help.
asked by Anneliese on December 26, 2016
9. ## Algebra
Find two consecutive positive integers such that the sum of the square is 85..(show your work) If X is the first integer, then X+1 is the next; therefore, the sum of those squares will be 85. X^2 + (X+1)^2 = 85 Solve for X, then add 1 to it for the second
asked by Tomika on June 13, 2007
Calculate the new molarity if each of the following dilutions is made. Assume the volumes are additive. (a) 52.3 mL of water is added to 27.1 mL of 0.124 M KOH solution (b) 110. mL of water is added to 54.2 mL of 0.808 M NaCl solution ((please show work,
asked by Jasmin on November 10, 2010
11. ## Math
Mrs.Perry hires a landscaper that charges \$15 per hr. The landscaper says the total charge for the work done is \$120.00 Write an inequality to show h, the number of hours it took to do the work and solve the inequality.
asked by Amy on March 16, 2016
12. ## calculus
prove that 5x - 7 - sin3x = 0 has at least one zero and prove that it has exactly one real zero. How am I supposed to show my work for this? I don't really understand how to show my work for the IVT. Without that I can't do any work for the MVT.
asked by Rob on December 16, 2011
13. ## Algebra Help
Please solve and check all proposed solutions. Show work for solving and checking: (x-2)/(x-1) + (2)/(x^2-1) = 0
asked by Jessica on August 3, 2009
14. ## Algebra
I can't figure out this problem... 2x-2-(x-1) = 4-1+x-(-3x) solve for x Will you also show your work and check step Thanks, :)Juliet
asked by Juliet on December 6, 2011
15. ## Chemistry
HA(aq) H+(aq) + A-(aq) K = 5.0 x 10 -9 The initial concentration of HA is 0.30 M What is the final equilibrium concentrations of HA, H+, A-? Show work, solve if you can
asked by Kay on April 12, 2018
16. ## Algebra 2
4Xsquared+7x+3 find the quadratic equation by factoring to solve. Show work
asked by Andres on November 9, 2009
17. ## Math
Solve and check. ( 3 marks) 5/(x-1) + 2/(x+1)=-6 Here's my work on it: 5/(x-1)+2/(x+1)=? 5/(-3/2-1)+2/(-3/2+1)=-6 Can you show how I would have found x in this situation and also if I went wrong anywhere.
asked by Alex on January 9, 2015
18. ## algebra
Simplify using long division.Show all work. (-72-4x^2+8x^3-36x)/(x-3) Also,please solve (8b^3-6)/(2b-1) I greatly appreciate your help.
asked by Jack on January 12, 2017
19. ## Algebra
Please solve the Linear Equation of P = 200(10)+6000 Please show work in all steps
asked by Anna on December 3, 2008
20. ## Math- Problem3
Solve the following inequality and write your answer using interval notation. Please show all of your work. (1 / (x+8) ) ≤ ((8-5x)/43)
asked by Greg - PV on June 9, 2011
21. ## Geometry
solve for x and AC if C is between A and B, AC = 3x + 55, CB = 2x and AB = 65. Draw figure and show algebra work. Can someone help me please I don't quiet get it?
asked by SiSi on August 26, 2015
22. ## Algebra
how do you solve this quadratic equation? tickets= -0.3x^2+12x+10 Please show the work so I can understand.
asked by CathyT on August 11, 2011
23. ## algebra
Solve and check all proposed solutions. Show work for solving and for checking. x+1/x-1 - 2/x^2-x=0
asked by math help on December 5, 2010
24. ## algebra
d=(1/2)at^2; solve for a. i got a=d/([1/5])t^2) but i don't see that in any of the possible responses: a. a= (2d)/t^2 b. a= 2dt^2 c. a= (dt^2)/2 d. a= d/2t^2 What did I do wrong? & can you please show detailed work?
asked by Ana-x00 on October 14, 2010
25. ## Algebra
Solve using set builder notation, show all work. 12x-25 is greater than or = to 3x-5(x+6)
asked by Ellen on July 28, 2012
26. ## chem/math
I need to fid the activation energy but i forgot how to solve for Ea 4.58*10^-15=e^(Ea/2537.02) can you show the work thank you
asked by help on October 28, 2010
27. ## Algebra 1
I know the answers, I just don't know how to solve it and show my work. 1. (3√6 + 5√10)^2 - 286 + 60√15 2. (2 + √10)(2 - √10) - -6 3. y = √6x-18 - x (greater then or equal to) 3 Thank you
asked by Wes on March 13, 2017
28. ## math
solve each of the following equations by factoring and show your work 12x² + 15x + 10= -10x -2
asked by Derek on September 7, 2008
29. ## algebra
which method do you use to solve this equation.(x-1)^2=7. Is it the factoring method or the sqare root method I would take the square root of each side. (x-1)= +- sqrt7 Now for the factoring method: (x-1)^2=7 (x-1)^2-7=0 you have a difference of two
asked by Tony on May 17, 2007
30. ## Number Systems of Mathematics
Solve using elimination and express answer as ordered pair: 2x + 3y = 7 x + 5y = 7
asked by James on January 1, 2016
31. ## Algebra 2
maximize p=6x+10y minimize p=14x+9y use elimination to solve: 4x+5y=41 7x+5y=53 -4x-y=-16 -4x-5y=-32 2x-4=8 x+2y=9 9x-5x=13 4x-6y=2
asked by Roman on April 16, 2013
32. ## math
Solve the systems of equations by elimination. 1. -2x+y=7 2. 2x+y=-9 6x+12y=24 4x+3y=1 3. 7x+3y=-1.5 4.2x-3y=-0.27 2x-5y=-30.3 5x2y=0.04
asked by pam on February 4, 2011
33. ## algebra 1
3a+b=52a+5=10use elimination to solve each systems of equations.
asked by Anonymous on March 20, 2013
Solve using elimination.......answers must be in ordered pair form 0.3x-0.2y=4 0.4x+0.5y=-55/17
asked by Michelle on July 13, 2010
35. ## math
Solve using elimination.......answers must be in ordered pair form 1) 5x+5y=-11 7x-2y=17
asked by Michelle on July 13, 2010
reed's bookstore is open 9 hours each day. it opens at 9:30a.m. when does it close? i know the answer but i don't know how to show my work. please help. Two ways to do it. 12:00 noon - 9:30 = 2:30 hours in the morning. That leaves 6:00 hours - 2:30 minutes
asked by zeke on April 24, 2007
37. ## Math
1.) Can a system of linear equations have exactly two solutions? A. True B. False 2.) The solution of a system of equations with no solution describes what kind of lines? a.) parallel b.) perpendicular c.) skew d.) coincidental
asked by Josh on December 7, 2013
38. ## medicine
how can i solve this?prepare 150ml of a 7.5%dextrose sol.you have D5%W and D10%W. how much of each sol. is needed? I assume the 7.5 dextrose is going to be on a w/w basis also. This is really simple, no algebra needed. Since 7.5 is halfway between 5 and
asked by rimple on December 2, 2006
39. ## math
Solve the system of equations: x - y - z = 2 x + 2y - 2z = 3 3x - 2y - 4z = 5 I know how to do the whole process, but sometimes I make careless errors that I never catch when I go over my work. This is the answer I worked out: x = 5 y = 1 z = 2 could
asked by Jennifer on May 5, 2008
40. ## maths
the mid-point of the line FG is (6,10). if the coordinates of G is (8,14) find the coordinates of F. There are a number of ways to solve this but I think the simplest is to use the definiton of midpoint. We know mdpt(FG) = 1/2(F+G) where FG is the segement
asked by Jen on September 7, 2006
41. ## algebra
i have 2 solve by elimination an i keep gettin fractions when the answer is (-3,-3) my teacher gave me the answer an problem but she wants me 2 work it out idk how i keep gettin fractions the problem is: -3x-4y=21 8x-5y=-9 please help
asked by angel on January 31, 2011
42. ## calculus
Solve the system by the method of your choice. Identify systems with no solutions and systems with infinitely many solutions, using set notation to express their solutions: (0 ,-5) (0 , 0) (-2 ,-1) (-2,-3)
asked by kelly on December 2, 2007
43. ## calculus
Solve the system by the method of your choice. Identify systems with no solutions and systems with infinitely many solutions, using set notation to express their solutions: x + y = 3 -2 + y = -3 (0 ,-5) (0 , 0) (-2 ,-1) (-2,-3)
asked by kelly on December 2, 2007
44. ## statistics
Researchers conducted a survey of parents of 66 kindergarten children. The parents were asked whether they played games with their children. The parents were divided into two groups: working class and middle class. The researchers wanted to know if there
asked by dd on August 23, 2013
45. ## Algebra
Use substitution to determine whether the ordered pair (–2, –3) is a solution of the equation x2 – y = 7. Show some work. Be sure to state your conclusion about whether the ordered pair is a solution or not.
asked by Jen on June 9, 2009
46. ## Algebra
Use substitution to determine whether the ordered pair (– 3, – 2) is a solution of the equation 5x + y2 = –11. Show some work. Be sure to state your conclusion about whether the ordered pair is a solution or not.
asked by Math Loser :( on October 29, 2010
47. ## Physics
A system undergoes a two-step process. In the first step, the internal energy of the system increases by 368 J when 144 J of work is done on the system. In the second step, the internal energy of the system increases by 24 J when 248 J of work is done on
asked by Sally on April 27, 2008
48. ## physics/chemistry(Thermodynamics)!!!!!helppppp!
In a certain chemical process,a lab technician supplies 254 j of heat to a system. At the same time,73 j of work are done on the system by iys surroundings.What is the increase in the internal energy of the system?
asked by duke on May 15, 2013
49. ## Geometry
Which ordered pair is the solution to the system of equations below? {3x-7y=-10 {5x-y=-6 A system of equations is shown below 2x+y=7 3x-2y=0 What is the solution to the system?
asked by Phillip on May 8, 2016
50. ## physics urgent!!!
A canoe of mass 70 kg with a paddler of mass 55 kg are in a river. If the river's current exerts a force of 15 N [E] while the paddler is paddling with an average force of 22 N [N38degreesW], find the acceleration of the canoe and paddler. Use both a
asked by Jess on April 7, 2011
51. ## calculus
Use Newton's method to solve the equation sec x = 4 in the interval x in (0, pi/2). In other words, use Newton's Method to compute arcsec(4). (You need to make a good initial guess for the root otherwise Newton's method will probably fail. Please justify
asked by Anonymous on April 13, 2011
52. ## Chem
I need help please! You have been given 40ml of a 0.6M solution of NaCl and 40ml of a 0.1M solution of KCl. You are also given as much diluent as you need. Show all your work and indicate how you would make 40ml of a single solution that contains 0.005M
asked by Jen on May 7, 2012
53. ## Algebra 1
What is an equation of a line perpendicular to the line with equation y = 1 - 3x a.y=-3x + 5 b. y=3x + 5 c.y=-1/3x + 5 d. y=1/3x + 5 To solve the linear system below, which substitution of unkowns is proper ? x + 4y= -6 5x - 3y = -16 a. substitute -4y-6
asked by Ley on April 22, 2013
54. ## Algebra
2. Tell whether the system has one solution infinitely many solutions or no solution. x=-7y+34 x+7y=32 a. one solution b. infinitely many solutions c. no solution*** can you please solve this thank you
asked by Leslie on December 12, 2016
Find the value fir each expression. Show your work. (13.28 - 7.8) divided by 4. I have the answer in the back of my book and when I try to work the problem out I keep getting the wrong answer. The answer is supposed to be 1.345 I can get the 1.34 but I get
56. ## Algebra
CEO Salaries In 2002, hourly wages for the top three CEOs (chief executive officers) of U.S. corporations were calculated based on their working 14 hours per day for 365 days. Together the three CEOs made \$44,000 per hour. The top CEO earned \$2,000 more
asked by genie on November 8, 2011
57. ## Chemistry
How many grams of magnesium sulfate would you need to make 0.500 L of a 10.0% M/V solution in water? Please show work.
asked by Margaret on February 18, 2013
58. ## Chemistry
What is the molarity of a potassium chloride solution that has a volumeof 400.0 mL and contains 85.0 g KCl? Please show work.
asked by MEGAN!!!! on April 30, 2010
59. ## Chemistry
Express the concentration of a commercial hydrogen peroxide solution, which is 3.0 (w/w), as a molarity. Show all work
asked by Amber on March 13, 2014
60. ## Chemistry
(Paper Chromatography) 1. What carrier solution might work well to separate newspaper inks? Why? 2. What carrier solution might work to separate different salts? Why? 3. What carrier solution might work to separate different fatty acids? Why? I really have
asked by Mikaela on January 26, 2017
61. ## Science
A repair shop tested two methods(a & B) of replacing a broken roller. In testing method A, these times in minutes were measured (5 trials), 2,4,9,3,2. In testing methods B, these times in minutes were measured (6 trials), 3,7,5,8,4,3. Use the 0.10
asked by Debra on December 19, 2011
62. ## Algebra
Solve. Show your work. A high school athletic department bought 40 soccer uniforms at a cost of \$3,000. After soccer season, they returned some of the uniforms but only received \$40 per uniform. What was the difference between what they paid for each
asked by anon on August 23, 2015
63. ## Quantitative analysis
I am trying to solve this question but I am lost can anyone help me pls Suppose your supervisor asks you to use the transportation method to rearrange the desks of everyone in the office. How would you do it? What other factors are important but will not
asked by amy on June 2, 2010
64. ## was not in class bio/ math
Preparing a Solution: A. Calculate the amount of methylene blue chloride (MW = 319.8 g/mol) needed to make 50 ml of a 10 mM stock solution. (Show calculations below) B. If you wanted to then dilute the solution from Part A to a final concentration of 0.1mM
asked by emily on September 20, 2016
65. ## Math
Which graph represents the solution of the given system? -2x+5y=-10 and -3x+5y=-20 I cannot show the graphs, but if anyone did the Systems of Equations and Inequalities Unit Test, in 7th grade Connexus, you might know what i'm talking about. My answer is
asked by Me on January 11, 2019
66. ## Math
Solve the following system of equations by graphing. If the system is inconsistent or the equations are dependent, say so 5x-2y=3 10x-4y=6 I know the equations are dependent but I am confused on what is the solution set I'm thinking its (10x-4y=6) but I'm
asked by Lataya on November 22, 2015
67. ## Math
Simplify the expressions in each pair. (2 marks each) (Show your work) i) (x^2-x-2)/(x+1),g(x)=(x-2) My work on simplifying the functions: (x-1)(x+2)/(1) (x-1)(x+2) ii) (x^2-x-12)/(x+3), g(x)=(x^2-6x+8)/(x-2) My work on simplifying the functions:
asked by Alex on January 10, 2015
68. ## chemistry
Assume that it is necessary to determine the [FE^3+ in a solution that may be 0.0001 to 0.0005 M range . Outline the spectrophotometric method that would accurately give the required data. Be spectific about the solutions that you would use. I know I have
asked by jxv on November 13, 2016
69. ## Chemistry
The Russian Mir space station used a chemical oxygen generator system to make oxygen for the crew. The system ignited a tube of solid lithium perchlorate (LiClO4) to make oxygen and lithium chloride (LiCl): LiClO4 (s) ---> 2O2 (g) + LiCl (s) If you have
asked by O on March 6, 2015
70. ## chemistry
The Russian Mir space station used a chemical oxygen generator system to make oxygen for the crew. The system ignited a tube of solid lithium perchlorate (LiClO4) to make oxygen and lithium chloride (LiCl): LiClO4 (s) 2O2 (g) + LiCl (s) If you have 500
asked by O on February 20, 2015
71. ## physics
A block and tackle system of 5 pulleys is used to raise a load of 50kg through a height of 20m the work done against friction is 200J.Calculate the work done by effort,the efficiency of the system and the effort applied.
asked by Mohamedy on June 25, 2016
72. ## College Math II
x/x – 1 - 3/x = 1/2 I am not sure how to solve this equation and show my work. My problem just says to solve the equation. Thanks.
asked by Lori on November 21, 2009
73. ## math
Pick a country of your choice that is experiencing population growth. Using the Library, web resources, and/or other materials to find the most recent population count of the country you have chosen and the population growth rate of that country. Use that
asked by Anonymous on September 17, 2008
74. ## Math
Convert 22 miles to yards. Use a proportion to solve. Show your work. Where do I start with this problem?
asked by Steve on March 26, 2015
75. ## Algebra
Use factoring and the zero-product property to solve the following problems: z(z - 1)(z + 3) = 0 x^2 - x - 10 = 2 4a^ - 11a + 6 = 0 9r^r - 30r + 21 = -4 Please show work. Thank you so much!
asked by Kathy on February 20, 2014
76. ## Math - Inverse Functions
Find the inverses of the following functions. y = 3(x - 1)^2, x >= 1 Work: x = 3(y - 1)^2 x = 3(y - 1)(y - 1) x = 3(y^2 - y - y + 1) x = 3y^2 - 6y + 3 And now what do I do!? Please explain and show me how to solve this inverse function! ... Thank you
asked by Anonymous on January 10, 2008
77. ## Math
Solve the equation: 4/3-x + 1/x+9 = 5/3-x Check all proposed solutions. Show work in solving and in checking.
asked by Crystal on July 7, 2009
78. ## Algebra
Write using set builder notation.Show work....Solve..-2(10x-9)-17x
asked by sherry on March 16, 2012
79. ## Maths
Solve the differential equation: d2y/dx2 - 2 dy/dx + y = 3sinhx The answer should be: y(x) = e^x (Ax+ B) + ( 3^8 )(2[x^2][e^x] - [e^-x] ) Can someone please show me how to work it out?
asked by Claire on November 30, 2009
80. ## algebra
Solve the following inequality and write your answer using interval notation . Please show all your work 10^3 + 6x^2 - 90x - 54 < 0
asked by anon on December 10, 2011
81. ## math
solve the inequality, write answer using interval notation pleas show the work 50/x+4+1
asked by mz tee on November 11, 2011
82. ## Algebra 2
Solve the equation. Round your answers to four decimal places. Show all of your work! a. 54x = 90
asked by denise on April 30, 2014
83. ## pre calculus
Solve the exponential equation Express you solutions in exact form only Please show your work e^2x - 9e^x - 20 = 2
asked by anon on December 20, 2011
84. ## Math
I was having trouble solving these equations: 7a+10=2a 11x=24+8x 9g-14=2g m-18=3m. I have to show my work and solve the answer to get the variable. Thank you!
asked by Karissa on April 3, 2018
85. ## algebra II
Solve the quadratic equation 12x^2 + 36a^2= 43ax by factorization Show your work
86. ## Math
3. Factor completely and solve. Check your solutions. Show all work. 6x^3 – 15x^2 – 36x = 0
asked by Anonymous on November 1, 2012
87. ## college algebra
Solve the exponential equation. Express your solutions in exact form only. Please show all of your work. 3^x-70*3^-x=9
asked by Sunny on January 4, 2012
88. ## Math
Use inverse operations to solve the equation, and show your work: 19 = h / 3 - 8 Please help ! This is the last question on my quiz, and I've been stuck on it for a while..
asked by Vixen - Connexus on October 26, 2018
89. ## algrebra
Solve by applying the quadratic Formula: all radicals should be simplified as far as possible. Show your work. a^2 -12a + 35 =0
asked by dan on April 25, 2010
90. ## geometry
what do I do to solve this problem? I don't know how to work it out, so could somone show me hoe to work it out. Please help! the lenght of a rectangular playing field is 5 ft less than twice its width.the perimeter is 230 ft,so what is the lenght and
asked by keshia on March 21, 2007
91. ## MATH
se a graphing calculator or Excel to find the solution of the system of equations. (If the system is dependent, enter DEPENDENT. If there is no solution, enter NO SOLUTION.) 5x + 3y = 2 3x + 7y = −4
asked by RICH on March 2, 2017
92. ## algebra
If a system has an infinite number of solutions, use set-builder notation to write the solution set. If a system has no solution, state this. m - 2n = 16, 4m + 4 = 1
asked by Anita on December 5, 2010
93. ## Intermediate Algebra
If a system has an infinite number of solutions, use set−builder notation to write the solution set. If a system has no solution, state this. y = x + 2, 3y − 2x = 4
asked by Donna on November 11, 2010
94. ## Math
If a system has an infinite number of solutions, use set-builder notation to write the solution set. If a system has no solution, state this. 6x − 2y = 2, 9x − 3y = 1
asked by Anonymous on August 5, 2011
95. ## Intermediate Algebra
if a system has an infinite number of solutions, use set-builder notation to write the solution set. If a system has no solution, state this. m - 2n = 16, 4m + n = 1
asked by Donna on November 11, 2010
96. ## Math
An area is bounded by the x-axis and the parabola y = 16 - x^2. Use four rectangles of equal width and the midpoint approximation method to estimate the bounded area. Could you please show me how to work out this problem? Thanks!
asked by Johnny on May 21, 2016
97. ## Completing the square
Please show the procedure for answering this quadratic equation using the completing the square method: x^2+14x+98 I think i understand the basics of the method, but in this particular question I'm finding it hard to use the appropriate factors of 98 - 7
asked by Barbie on May 20, 2011
98. ## Math
asked by ALLISON on September 23, 2011
99. ## math
I need help on this math problem Solve the systems of equations by elimination. -2x+y=7 6x+12y=24
asked by pdub on February 6, 2011
100. ## math
solve gauss jordan elimination 6w-2x-4y+2z=2 3w-3x-6y+z=-4 -12w+8x+21y-8z=8 -6w-10y+7z=43
asked by rehuel on January 24, 2019
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ALGEBRIDGE Making the Deck: The deck consists of two parts
```ALGEBRIDGE
From "MATH NOTS, A Newsletter for Math Teachers," Ann Arbor, Michigan, Volume I, Number 2.
Making the Deck: The deck consists of two parts, playing cards and replacement
cards. The 52 playing cards can be 4 x 2.5 inch pieces of posterboard. The suits are
represented by 4 colors of felt pens. On each card print one of the following
algebraic expressions: (x2+3x), (x3-2x), (x2), (x3), (2x2-2x+2), (x4-x2+1), (x3+2), (7xx2), (x3+x2+1), (4x-x2), (x4-1), (1/2 x3), and (x3+7). These same expressions (or any
preferred ones) are repeated in each suit {color). The second part of the deck is a
set of 17 cards.
These are used to determine replacement value for x during play -- make 4 with numeral
4, 4 with 3, 4 with 2, 4 with 1 and 1 with 5. Shuffle well and keep separate from
playing cards.
Using the deck: The game is played like bridge. There are four players. The
replacement set is shuffled and placed face down in the center of the table. The
playing cards are dealt so that each player receives 13. The dealer turns over the top
replacement card -- this is the value by which to replace the variable for that trick.
The dealer now leads a card from any suit. All players must follow suit if they can or
if they cannot, they discard any card from their hand. (A card from the wrong suit
cannot win the trick.) The winner of the trick is the player whose card has the
highest value when replaced by the replacement value showing. It is now his turn to
turn over the next replacement card and lead. The winning team is the one that wins
the most tricks.
```
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# Physics Standing Waves Lab
Topics: Wave, Sound, Wavelength Pages: 5 (713 words) Published: March 18, 2013
SCHOOL OF PHYSICS
Physics 1001: Laboratory 3.
Standing Waves.
Date:_________ Marker’s signature:______________ Mark:______
Pre-lab Question 1:
What is the wavelength of a sound wave of frequency 500 Hz in air (you will need to look up the speed of sound in air)? ________ m.
Pre-lab Question 2:
If a person inhales helium gas the sound of their voice changes quite dramatically. Why?
Pre-lab Question 3:
A shower cubicle measures 0.86x0.86x2.1 m. If you were to sing in this shower what frequencies will be amplified? (The amplified frequencies will correspond to the resonant frequencies of the cubicle, assume it acts like a pipe closed at both ends so there must be a node at both ends. Take the range of the human voice to be 130 to 2000 Hz.)
Part 1. Resonant frequencies of open and closed pipes.
Following the instructions in the lab measure the resonant frequencies of the tube open at both ends. You may not be able to fill all the spaces in the table, but should find at least five resonant frequencies. The frequency difference column should be filled with the differences between the resonant frequency you have just measured and the next lower resonant frequency.
Open tube resonant frequencies.
Divisions
Time / Division
Period (T)
Frequency (1 / T)
Frequency difference
By considering the differences between the resonant frequencies, construct a linear equation describing the resonant frequencies.
______________________________.
Again following the directions in the lab manual close one end of the tube and measure the resonant frequencies.
Closed tube resonant frequencies.
Divisions
Time / Division
Period (T)
Frequency (1 / T)
Frequency Difference
As you did for the open tube, construct a linear equation describing the resonant frequencies of the closed tube.
______________________________.
Compare the equations you have determined for the open and closed tubes. What differences do you notice?
The resonant wavelengths are determined by the length of the tube, if it is open λ = 2 Lopen (n+1)-1 or if it is closed λ = 4 Lclosed (2n+1)-1 in both cases taking the fundamental as n=0. Being careful of what value you use for n, work out the wavelengths of the resonant modes you have observed and thus the speed of sound from each observation.
Table of the measured sound velocities.
Open Tube
Frequency
Wavelength
Velocity
Closed Tube
Frequency
Wavelength
Velocity
What is your average speed of sound, and how does it compare to the standard value of 340 m s-1?
What are some of the factors that could cause the speed to be different to the standard value?
Part 2. Geometry of standing waves.
Choose the resonant frequency for the open tube closest to 600 Hz and measure the positions of the extrema of the standing wave using the microphone.
Positions of maxima and minima of the standing wave in an open tube. Frequency
Max 1
Min 1
Max 2
Min 2
Max 3
Min 3
Max 4
Min 4
Repeat the measurement of one of these positions five times as described in the lab manual. Extremum
1
2
3
4
5
Since they are being measured with a microphone the positions found are extrema of the pressure in the tube. Is the pressure wave you have just measured in phase or out of phase with the displacement wave usually described when discussing waves? (Have a look at figure 4 in the lab manual, and the accompanying explanation for a hint.)
Make a sketch of what is going on inside the tube, including both pressure and displacement waves.
Reconstruct the closed tube and repeat your measurements of the positions of the maxima and minima of the resonant mode with frequency nearest to 600 Hz.
Positions of maxima and minima of the standing wave in a closed tube. Frequency
Max 1
Min 1
Max 2
Min 2
Max 3
Min 3
Max 4
Min 4
Make another sketch of what...
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# Homework Help: Geometric probability question
1. Sep 13, 2010
### kazuchan
A dog is running around in a fenced-off rectangular field with dimensions of 40ft by 50ft. If the position of the dog is uniformly random throughout the field, what is the probability that the dog is 10 feet or more away from the fence at any given time?
In know that since the dog is uniformly random throughout the field that the probability that the dog will be in any region of the field is equal to the ratio of the area of that region to the area of the entire field.
So the area of the entire field is 50x40=2000 ft
and the area of the marked off field is (50-10)x(40-10)=1200 ft
Now, I find the ratio (1200/2000)=3/5=.6 which gives me a probability of the dog being 10 feet away from the fence 60% of the time.
When I went to check my answer, it said that the dog was away 30% of the time. How could this be?
Thank you for your help!
2. Sep 13, 2010
### Office_Shredder
Staff Emeritus
Re: Probability
If you're ten feet away from the top.
AND you're ten feet away from the bottom
That's twenty less feet of room you have to move in
3. Sep 13, 2010
### kazuchan
Re: Probability
got it! thanks for pointing that out!
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# Probability and normal distribution : key information
This content was STOLEN from BrainMass.com - View the original, and get the already-completed solution here!
1. At a computer manufacturing company, the actual size of computer chips is normally distributed with a mean of 1 centimeter and a standard deviation of 0.1 centimeter. A random sample of 12 computer chips is taken. What is the standard error for the sample mean?
a) 0.029
b) 0.050
c) 0.091
d) 0.120
2. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 25 fish yields a mean of 3.6 pounds, what is the Z-score for this observation?
a) 18.750
b) 2.500
c) 1.875
d) 0.750
3. The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 16 fish is taken, what would the standard error of the mean weight equal?
a) 0.003
b) 0.050
c) 0.200
d) 0.800
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#### Solution Summary
The solution solves the questions on probability and normal distributions.
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## Probability Using Normal Distribution: Key Information
The mean salary offered to students who are graduating from Coastal State University this year is \$24,230, with a standard deviation of \$3,712. A random sample of 80 Coastal State students graduating this year has been selected. What is the probability that the mean salary offer for these 80 students is 24,250 or less?
Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.
Can you explain this how to do this?
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# Propositional Formula
In propositional logic, a propositional formula is a type of syntactic formula which is well formed and has a truth value. If the values of all variables in a propositional formula are given, it determines a unique truth value. A propositional formula may also be called a propositional expression, a sentence, or a sentential formula.
A propositional formula is constructed from simple propositions, such as "x is greater than three" or propositional variables such as P and Q, using connectives such as NOT, AND, OR, and IMPLIES; for example:
(x = 2 AND y = 4) IMPLIES x + y = 6.
In mathematics, a propositional formula is often more briefly referred to as a "proposition", but, more precisely, a propositional formula is not a proposition but a formal expression that denotes a proposition, a formal object under discussion, just like an expression such as "x + y" is not a value, but denotes a value. In some contexts, maintaining the distinction may be of importance.
### Other articles related to "formulas, propositional, propositional formula":
Substitution (logic) - Definition
... Where Ψ and Φ represent formulas of propositional logic, Ψ is a substitution instance of Φ if and only if Ψ may be obtained from Φ by substituting formulas for symbols in Φ, always replacing ... A) is a substitution instance of (A A) In some deduction systems for propositional logic, a new expression (a proposition) may be entered on a line of a derivation if it is a substitution instance of a previous ... In first-order logic, every closed propositional formula that can be derived from an open propositional formula by substitution is said to be a substitution instance of ...
Boolean Algebra (logic) - Propositional Logic
... Propositional logic is a logical system that is intimately connected to Boolean algebra ... Many syntactic concepts of Boolean algebra carry over to propositional logic with only minor changes in notation and terminology, while the semantics of propositional ... Syntactically, every Boolean term corresponds to a propositional formula of propositional logic ...
Propositional Formula - Historical Development
... Although a propositional calculus originated with Aristotle, the notion of an algebra applied to propositions had to wait until the early 19th century ... It is here that what we consider "modern" propositional logic first appeared ... connective, the "stroke"
### Famous quotes containing the word formula:
Beauty, like all other qualities presented to human experience, is relative; and the definition of it becomes unmeaning and useless in proportion to its abstractness. To define beauty not in the most abstract, but in the most concrete terms possible, not to find a universal formula for it, but the formula which expresses most adequately this or that special manifestation of it, is the aim of the true student of aesthetics.
Walter Pater (1839–1894)
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# How to convert Kelvin to fahrenheit
2
by vish1407
2015-06-23T07:26:13+05:30
We can convert kelvin to fahrenheit by the following formula
k-starting point in thermometer/number of divisions=f-starting point in thermometer/number of divisions.
starting point in kelvin thermometer is 273
the number of divisions is 100
starting point in fahrenheit thermometer is 32
the number of divisions =180
......k-273/100=f-32/180
if we cancel 180 and 100 the formula
.....
k-273/5=f-32/9
2015-06-23T08:11:52+05:30
It is simple if it is in Fahrenheit add273 if it is in kelvin scale subtract 273. and at 0 Kelvin we assume that all gases will occupies 0 volume
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You are viewing an older version of this Concept. Go to the latest version.
# Solving Trigonometric Equations
## Identities and solving equations on an interval or with no solutions.
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Practice Solving Trigonometric Equations
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Trigonometric Equations
Solving a trigonometric equation is just like solving a regular equation. You will use factoring and other algebraic techniques to get the variable on one side. The biggest difference with trigonometric equations is the opportunity for there to be an infinite number of solutions that must be described with a pattern. The equation $\cos x=1$ has many solutions including 0 and $2 \pi$ . How would you describe all of them?
#### Watch This
http://www.youtube.com/watch?v=26EWKD2Xha4 James Sousa: Solve Trigonometric Equations I
http://www.youtube.com/watch?v=ABKO3ta_Azw James Sousa: Solve Trigonometric Equations II
http://www.youtube.com/watch?v=7thuFLqC7z0 James Sousa: Solve Trigonometric Equations III
#### Guidance
The identities you have learned are helpful in solving trigonometric equations. The goal of solving an equation hasn’t changed. Do whatever it takes to get the variable alone on one side of the equation. Factoring, especially with the Pythagorean identity, is critical.
When solving trigonometric equations, try to give exact (non-rounded) answers. If you are working with a calculator, keep in mind that while some newer calculators can provide exact answers like $\frac{\sqrt{3}}{2}$ , most calculators will produce a decimal of 0.866... If you see a decimal like 0.866..., try squaring it. The result might be a nice fraction like $\frac{3}{4}$ . Then you can logically conclude that the original decimal must be the square root of $\frac{3}{4}$ or $\frac{\sqrt{3}}{2}$
When solving, if the two sides of the equation are always equal, then the equation is an identity. If the two sides of an equation are never equal, as with $\sin x=3$ , then the equation has no solution.
Example A
Solve the following equation algebraically and confirm graphically on the interval $[-2 \pi, 2 \pi]$ .
$\cos 2x=\sin x$
Solution:
$\cos 2x &= \sin x\\1-2 \sin^2 x &= \sin x\\0 &= 2 \sin^2 x+\sin x-1\\0 &= (2 \sin x-1)(\sin x+1)$
Solving the first part set equal to zero within the interval yields:
$0 &= 2 \sin x-1\\\frac{1}{2} &= \sin x\\x &= \frac{\pi}{6}, \frac{5 \pi}{6}, - \frac{11 \pi}{6}, -\frac{7 \pi}{6}$
Solving the second part set equal to zero yields:
$0 &= \sin x+1\\-1 &= \sin x\\x &= -\frac{\pi}{2}, \frac{3 \pi}{2}$
These are the six solutions that will appear as intersections of the two graphs $f(x)=\cos 2x$ and $g(x)=\sin x$
Example B
Determine the general solution to the following equation.
$\cot x-1=0$
Solution:
$\cot x-1 &= 0\\\cot x &= 1$
One solution is $x=\frac{\pi}{4}$ . However, since this question asks for the general solution, you need to find every possible solution. You have to know that cotangent has a period of $\pi$ which means if you add or subtract $\pi$ from $\frac{\pi}{4}$ then it will also yield a height of 1. To capture all these other possible $x$ values you should use this notation.
$x=\frac{\pi}{4} \pm n \cdot \pi$ where $n$ is a integer
Example C
Solve the following equation.
$4 \cos^2 x-1=3-4 \sin^2 x$
Solution:
$4 \cos^2 x-1 &= 3-4 \sin^2 x\\4 \cos^2 x+4 \sin^2 x &= 3+1\\4 (\cos^2 x+\sin^2 x) &= 4\\4 &= 4$
This equation is always true which means the right side is always equal to the left side. This is an identity.
Concept Problem Revisited
The equation $\cos x=1$ has many solutions. When you type $\cos^{-1} 1$ on your calculator, it will yield only one solution which is 0. In order to describe all the solutions you must use logic and the graph to figure out that cosine also has a height of 1 at $-2 \pi, 2 \pi, -4 \pi, 4 \pi \ldots$ Luckily all these values are sequences in a clear pattern so you can describe them all in general with the following notation:
$x=0 \pm n \cdot 2 \pi$ where $n$ is an integer, or $x=\pm n \cdot 2 \pi$ where $n$ is an integer
#### Vocabulary
The terms “general solution,” “completely solve” , and “solve exactly” mean you must find solutions to an equation without the use of a calculator. In addition, trigonometric equations may have an infinite number of solutions that repeat in a certain pattern because they are periodic functions. When you see these directions remember to find all the solutions by using notation like in Example B.
#### Guided Practice
1. Solve the following equation on the interval $(2 \pi, 4 \pi)$ .
$2 \sin x+1=0$
2. Solve the following equation exactly.
$2 \cos^2 x+3 \cos x-2=0$
3. Create an equation that has the solutions:
$\frac{\pi}{4} \pm n \cdot 2 \pi$ where $n$ is an integer
1. First, solve for the solutions within one period and then use logic to find the solutions in the correct interval.
$2 \sin x+1 &= 0\\\sin x &= -\frac{1}{2}\\x &= \frac{7 \pi}{6}, \frac{11 \pi}{6}$
You must add $2 \pi$ to each of these solutions to get solutions that are in the interval.
$x=\frac{19 \pi}{6}, \frac{23 \pi}{6}$
2. Start by factoring:
$2 \cos^2 x+3 \cos x-2 &= 0\\(2 \cos x-1)(\cos x+2) &= 0$
Note that $\cos x \neq -2$ which means only one equation needs to be solved for solutions.
$2 \cos x-1 &= 0\\\cos x &= \frac{1}{2}\\x&=\frac{\pi}{3}, -\frac{\pi}{3}$
These are the solutions within the interval $-\pi$ to $\pi$ . Since this represents one full period of cosine, the rest of the solutions are just multiples of $2 \pi$ added and subtracted to these two values.
$x=\pm \frac{\pi}{3} \pm n \cdot 2 \pi$ where $n$ is an integer
3. There are an infinite number of possible equations that will work. When you see the $\frac{\pi}{4}$ you should think either of where tangent is equal to one or where sine/cosine is equal to $\frac{\sqrt{2}}{2}$ . The problem with both of these initial guesses is that tangent repeats every $\pi$ not every $2 \pi$ , and sine/cosine have a second place where they reach a height of 1. An option that works is:
$\tan \frac{x}{2}=1$
This equation works because the period of $\tan \frac{x}{2}$ is $2 \pi$
#### Practice
Solve each equation on the interval $[0, 2 \pi)$ .
1. $3 \cos^2 \frac{x}{2}=3$
2. $4 \sin^2 x=8 \sin^2 \frac{x}{2}$
Find approximate solutions to each equation on the interval $[0, 2 \pi)$ .
3. $3 \cos^2 x+10 \cos x+2=0$
4. $\sin^2 x+3 \sin x=5$
5. $\tan^2 x+\tan x=3$
6. $\cot^2 x+5 \tan x+14=0$
7. $\sin^2 x+\cos^2 x=1$
Solve each equation on the interval $[0,360^\circ)$ .
8. $2 \sin \left(x-\frac{\pi}{2}\right)=1$
9. $4 \cos (x-\pi)=4$
Solve each equation on the interval $[2 \pi, 4 \pi)$ .
10. $\cos^2 x+2 \cos x+1=0$
11. $3 \sin x=2 \cos^2 x$
12. $\tan x \sin^2 x=\tan x$
13. $\sin^2 x+1=2 \sin x$
14. $\sec^2 x=4$
15. $\sin^2 x-4=\cos^2 x-\cos 2x-4$
### Vocabulary Language: English
general solution
general solution
The term general solution refers to all solutions to an equation. Remember that trigonometric equations may have an infinite number of solutions that repeat in a certain pattern because they are periodic functions.
Pythagorean Identity
Pythagorean Identity
The Pythagorean identity is a relationship showing that the sine of an angle squared plus the cosine of an angle squared is equal to one.
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You are Here: Home >< Maths
# Integral of 4Tan3x, and AQA Formula booklet Watch
1. Basically i was integrating Tan3x however in the AQA formula booklet it states that the integral of Tanx = ln|secx| which is fine but when i integrated Tan3x i made it equal to ln|sec3x| not 1/3ln|sec3x| which is what somebody said from a thread a while back.
Usually in the formula booklet it writes the integral of say sec^2kx as 1/k tankx, however it does not do this with the integral of Tanx which is why i did not stick a 1/3 at the front of my integral of Tan3x. I know i should divide by 3 when i integrate the function but i don't understand why the formula booklet has not indicated the 1/k part.
so basically is the Integral of 4Tan3x = 4(1/3)ln|sec3x| + c ?
I am sorry if this sounds confusing btw!
2. EDIT: ignore this sorry i found further down that the integral of tankx is something else, sorry!
3. (Original post by raveen789)
Basically i was integrating Tan3x however in the AQA formula booklet it states that the integral of Tanx = ln|secx| which is fine but when i integrated Tan3x i made it equal to ln|sec3x| not 1/3ln|sec3x| which is what somebody said from a thread a while back.
Usually in the formula booklet it writes the integral of say sec^2kx as 1/k tankx, however it does not do this with the integral of Tanx which is why i did not stick a 1/3 at the front of my integral of Tan3x. I know i should divide by 3 when i integrate the function but i don't understand why the formula booklet has not indicated the 1/k part.
so basically is the Integral of 4Tan3x = 4(1/3)ln|sec3x| + c ?
I am sorry if this sounds confusing btw!
You are doing correct.
4. (Original post by raheem94)
You are doing correct.
oh it says that the integral of tanKx = ln|coskx| :/ which is quite confusing!
5. (Original post by raveen789)
oh it says that the integral of tanKx = ln|coskx| :/ which is quite confusing!
It would probably be
6. (Original post by raheem94)
It would probably be
Why the minus sign?
7. (Original post by ghostwalker)
Why the minus sign?
Please correct me if i am wrong.
8. (Original post by raheem94)
Please correct me if i am wrong.
Barring the typo, the second line is correct, however the derivative of sech(x) is minus sech(x)tanh(x).
Ignore: Irrelevant.
9. (Original post by ghostwalker)
Barring the typo, the second line is correct, however the derivative of sech(x) is minus sech(x)tanh(x).
Maybe I'm missing something but I don't see the relevance?
10. (Original post by ghostwalker)
Barring the typo, the second line is correct, however the derivative of sech(x) is minus sech(x)tanh(x).
These aren't hyperbolic functions.
It is a 'k' not a 'h'.
Where is the typo, i don't see it?
11. (Original post by hassi94)
Maybe I'm missing something but I don't see the relevance?
Is the formula i have written correct?
Sorry for missing the 'dx', i forgot to write it.
12. (Original post by raheem94)
Please correct me if i am wrong.
(Original post by hassi94)
Maybe I'm missing something but I don't see the relevance?
I'm being a moron!
Reading k as h, and thinking it's hyperbolics. Duh!
Ignore me: I'll just go and shoot myself.
13. (Original post by raheem94)
These aren't hyperbolic functions.
It is a 'k' not a 'h'.
Where is the typo, i don't see it?
The typo was you missing 'ln' on the last bit of the second line
Otherwise I think it's correct.
14. (Original post by ghostwalker)
I'm being a moron!
Reading k as h, and thinking it's hyperbolics. Duh!
Ignore me.
Aha okay that makes sense now, don't worry about it
15. (Original post by ghostwalker)
I'm being a moron!
((((hugs)))) for ghostwalker
we have all been there
16. (Original post by ghostwalker)
I'm being a moron!
Reading k as h, and thinking it's hyperbolics. Duh!
Ignore me.
Never knew you also make mistakes!
(Original post by hassi94)
The typo was you missing 'ln' on the last bit of the second line
Otherwise I think it's correct.
Thanks i will edit my post, its too difficult to spot your own mistakes.
17. Thanks for the help guys, its just the formula booklet does not indicate the 1/k ln|seckx| like it does for other formulas :/
18. (Original post by raveen789)
Thanks for the help guys, its just the formula booklet does not indicate the 1/k ln|seckx| like it does for other formulas :/
Yeah it seems fairly hit and miss. Just remember you must as it's 'the reverse chain rule'.
Updated: May 11, 2012
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# 8.1: Atomic and Molecular Calculations are Expressed in Atomic Units
##### Learning Objectives
• Demonstrate how solving electron structure problems are less cluttered by switching to atomic units instead of SI units.
Atomic units (au or a.u.) form a system of natural units which is especially convenient for atomic physics calculations. Atomic units, like SI units, have a unit of mass, a unit of length, and so on. However, the use and notation is somewhat different from SI. Suppose a particle with a mass of m has 3.4 times the mass of electron. The value of mass $$m$$ can be written in three ways:
• $$m=3.4\; m_e$$: This is the clearest notation (but least common), where the atomic unit is included explicitly as a symbol.
• $$m=3.4\; a.u.$$: This notation is ambiguous, but is common. Here, it means that the mass $$m$$ is 3.4 times the atomic unit of mass. If considering a length $$L$$ of 3.4 times the atomic unit of length, the equation would look the same, $$L= 3.4 \;a.u.$$ The dimension needs to be inferred from context, which is sloppy.
• $$m = 3.4$$: This notation is similar to the previous one, and has the same dimensional ambiguity. It comes from formally setting the atomic units to 1 (Table 8.1.1 ).
This article deals with "Hartree type" of atomic units, where the numerical values of the following four fundamental physical constants are all unity by definition:
Dimension Name Symbol/Definition Value in SI units Value in Atomic Units
Table 8.1.1 : Fundamental atomic units
mass electron rest mass $$m_e$$ 9.109×10−31 kg 1
charge elementary charge $$e$$ 1.602×10−19 C 1
action reduced Planck's constant $$\hbar = \dfrac{h}{2\pi}$$ 1.054×10−34 J·s 1
electric constant−1 Coulomb force constant $$\displaystyle k_e = \frac{1}{4 \pi \epsilon_o}$$ 8.987 x 109 kg·m3·s−2·C−2 1
##### Example 8.1.1 : Simplifying the Hamiltonian
Use the atomic units definitions in Table 8.1.1 to contrast the Hamiltonian for a Helium atom in Si units and in atomic units.
Solution
In SI units, the Hamiltonian for a Helium atom is
$\hat {H} = -\dfrac {\hbar ^2}{2m_e} (\nabla ^2_1 + \nabla ^2_2) -\dfrac {2e^2}{4 \pi \epsilon _0 r_1} - \dfrac {2e^2}{4 \pi \epsilon _0 r_2} + \dfrac {e^2}{4 \pi \epsilon _0 r_{12}} \nonumber$
In atomic units, the same Hamiltonian
$\hat {H} = -\dfrac {1}{2} (\nabla ^2_1 + \nabla ^2_2) - \dfrac {2}{r_1} - \dfrac {2}{r_2} + \dfrac {1}{r_{12}} \nonumber$
All the units that make the SI version of the Hamiltonian disappear to emphasize the key aspects of the operator.
Atomic units are derived from certain fundamental properties of the physical world, and are free of anthropocentric considerations. It should be kept in mind that atomic units were designed for atomic-scale calculations in the present-day universe, with units normalize the reduced Planck constant and also mass and charge of the electron are set to 1, and, as a result, the speed of light in atomic units is a large value, $$1/\alpha \approx 137$$. For example, the orbital velocity of an electron around a small atom is of the order of 1 in atomic units. Table 8.1.2 give a few derived units. Some of them have proper names and symbols assigned, as indicated in the table.
Table 8.1.2 : Derived atomic units
Dimension Name Symbol Expression Value in SI units Value in more common units
length bohr $$a_o$$ $$4\pi \epsilon_0 \hbar^2 / (m_\mathrm{e} e^2) = \hbar / (m_\mathrm{e} c \alpha)$$ 5.291×10−11 m 0.052 nm = 0.529 Å
energy hartree $$E_h$$ $$m_\mathrm{e} e^4/(4\pi\epsilon_0\hbar)^2 = \alpha^2 m_\mathrm{e} c^2$$ 4.359×10−18 J 27.2 eV = 627.5 kcal·mol−1
time $$\hbar / E_\mathrm{h}$$ 2.418×10−17 s
velocity $$a_0 E_\mathrm{h} / \hbar = \alpha c$$ 2.187×106 m·s−1
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## Temple University Department of Economics
### Hypothesis Testing Homework
1. Consider a multiple regression;
, with 17 observations given in the table:
Y x2 x3 15 12 360 13 8 372 17 10 355 18 12 340 24 16 326 23 14 304 28 18 284 32 21 271 32 21 253 37 26 239 39 29 205 42 31 195 44 33 172 46 34 158 49 37 137 53 41 112 55 43 98
Based on computer estimation (attach a printout), find:
(a) The least squares estimates b1 , b2 , and b3 for the model coefficients.
From EVIEWS
Dependent Variable: Y Method: Least Squares Date: 03/26/11 Time: 08:26 Sample: 1 17 Included observations: 17 Variable Coefficient Std. Error t-Statistic Prob. C 44.09476 14.11416 3.124150 0.0075 X2 0.481075 0.258989 1.857511 0.0844 X3 -0.090392 0.032315 -2.797211 0.0143 R-squared 0.991784 Mean dependent var 33.35294 Adjusted R-squared 0.990611 S.D. dependent var 13.63333 S.E. of regression 1.321059 Akaike info criterion 3.553529 Sum squared resid 24.43275 Schwarz criterion 3.700567 Log likelihood -27.20500 Hannan-Quinn criter. 3.568145 F-statistic 845.0196 Durbin-Watson stat 1.387993 Prob(F-statistic) 0.000000
(b) An estimate for the error variance (sigma hat squared).
sigma hat squared = SSR/df = 24.43275/(17-3) = 1.74519 Note that this is also the square of the " S.E. of regression" in the above table.
(c) An estimate for the variance for b2 .
Var(b2) = .2589892 = .06707
(d) R2, SSE, SST, and SSR.
R2 = .991
SSR = 24.43275
R2 = 1-SSR/SST ==> .991784 = 1 - (24.43275/SST) ==> SST = 2973.795
SST = SSR + SSE ==> 2973.79503 = 24.43275 + SSE ==> SSE = 2949.362
2. For this problem you will need the data in rdchem.wf1 for 32 firms in the chemical industry. This is an EVIEWS workfile. You will need the EVIEWS software to do the homework. The variable rdintens is is expenditures on research and development as a percent of company sales. Sales and R&D expenditures are both measured in millions of dollars. The variable profmarg is profits as a percent of sales; both in millions of dollars. From the data use EVIEWS to construct estimates of the model parameters of
A. Report your estimates of the coefficients and their standard errors.
Dependent Variable: RDINTENS Method: Least Squares Date: 03/26/11 Time: 08:53 Sample: 1 32 Included observations: 32 Variable Coefficient Std. Error t-Statistic Prob. C 0.469548 1.676242 0.280120 0.7814 LSALES 0.321472 0.215592 1.491111 0.1467 PROFMARG 0.050167 0.045780 1.095830 0.2822 R-squared 0.098652 Mean dependent var 3.265625 Adjusted R-squared 0.036490 S.D. dependent var 1.874079 S.E. of regression 1.839569 Akaike info criterion 4.146000 Sum squared resid 98.13642 Schwarz criterion 4.283412 Log likelihood -63.33599 Hannan-Quinn criter. 4.191548 F-statistic 1.587016 Durbin-Watson stat 1.652507 Prob(F-statistic) 0.221790
B. How much of the variation in rdintens is explained by the two independent variables?
You are being asked to report the R-sq = .098.
C. Interpret your estimate of the coefficient on the log of sales. In particular, if sales increases by 10%, what is the estimated percentage increase in rdintens? Is this and economically large effect?
From Chapter 2 of the text (Wooldridge, Introduction to Econometrics) we know that the marginal effect a change in sales is in the third column of this table and the elasticity of rdintens with respect to sales is in the fourth column.
Level - Log
Our estimate of beta1 is .32. The variable rdintens is the ratio of R&D spending to sales. Its mean is 3.26. Therefore the elasticity of RDINTENS with respect to sales is .32*3.26 = 1.04. So, if sales rise 10% then we expect RDINTENS to rise by 10%. If the denominator of RDINTENS rises by 10% and RDINTENS has risen by 10% then R&D spending must have risen 100%, i.e. it doubled! This is an economically meaningful magnitude.
D. At the 5% level, test the hypothesis that sales has no impact on rdintens.
This is just a t-test that we can read out of the table of estimation results. The observed t is 1.49 which is smaller than the critical t of 2.045. Also, the p-value is much greater than the stipulated 5% in the two tails.
Do not reject the null that sales don't matter.
E. Are sales and profmarg jointly significant in explaining rdintens?
The F-statistic and p-value reported in the table are 1.587 and .221 respectively. The critical F(29,.05) = 3.328. We do not reject the null that sales and profmarg do not matter.
3. Some cities are economically dominated by the universities that they host. Examples include University of Illinois in Champaign-Urbana, University of Wisconsin in Madison, University of MIchigan in Ann Arbor, or Penn State in Happy Valley. A perennial complaint in these towns is that the student population drives the monthly rent fo apartments. The data for 127 college towns is in the EVIEWS file rental .wf1. Let rent be average monthly rent paid on apartments in the town, pop denote the total city population, avginc the average city income, and pctstu the student population as a percent of the total population. A model of rental costs in such towns is
A. What do you expect for the sign of the coefficient on log(avginc)?
As income in the community rises we expect rents to rise as well.
B. Use the data to fit the stated model.
Dependent Variable: LRENT Method: Least Squares Date: 03/26/11 Time: 09:44 Sample (adjusted): 1 127 Included observations: 127 after adjustments Variable Coefficient Std. Error t-Statistic Prob. C -3.355510 0.468364 -7.164325 0.0000 LPOP 0.031448 0.027184 1.156853 0.2496 LAVGINC 0.875790 0.041813 20.94546 0.0000 PCTSTU 0.006569 0.001209 5.434882 0.0000 R-squared 0.794053 Mean dependent var 5.749841 Adjusted R-squared 0.789030 S.D. dependent var 0.331448 S.E. of regression 0.152239 Akaike info criterion -0.895742 Sum squared resid 2.850731 Schwarz criterion -0.806162 Log likelihood 60.87964 Hannan-Quinn criter. -0.859347 F-statistic 158.0806 Durbin-Watson stat 1.795774 Prob(F-statistic) 0.000000
C. Was your supposition in part A. confirmed? (Note that your supposition is a testable hypothesis.)
Yes, the supposition is confirmed and the coefficient on LAVGINC is statistically significant (t=20.9 and p = 0.0).
D. Test the hypothesis, at the 1% level, that the effect of log(pop) is five times as great as the effect of pctstu on log(rent).
This is a test of significance of a linear combination of random variables.
At the 1% level the crtical t(123,.005) = +/- 2.616.
The statement of hypothesis is
The test statistic is
with
So the observed t is
This is such a small t that we cannot reject the null.
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# AP Statistics Curriculum 2007 EDA Shape
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## General Advance-Placement (AP) Statistics Curriculum - Measures of Shape
### Definitions
• A distribution is unimodal if it has one mode. Unimodal distributions include:
• Bell shaped distributions (symmetric, Normal)
• Skewed right or skewed left
• We can use the mean and median to help interpret the shape of a distribution. For a unimodal distribution we have these properties:
• If mean = median, then the distribution is symmetric
• If mean > median, then the distribution is right skewed
• If mean < median, then the distribution is left skewed
• Multimodal distributions have two or more than one modes. Examples of multimodal distributions are:
### Examples
What seems like a logical choice for the shape of the hot dog calorie data? Try looking at the histogram of the calories for the Hot-dogs dataset.
### Activities
Collect data, draw the sample histogram or dot-plot and classify the shape of the distribution accordingly. Also, if unimodal, classify symmetry (symmetric, skewed right or skewed left).
• Data collected on height of randomly sampled college students.
• Data collected on height of randomly sampled female college students.
• The salaries of all persons employed by a large university.
• The amount of time spent by students on a difficult exam.
• The grade distribution on a difficult exam.
### Problems
Translate this page:
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1. ## optimization with constraint
question: $z=100x^{.25}y^{.75}$ where $x=48-4y$ Use chain rule to find $\frac{dz}{dy}$. Solve $\frac{dz}{dy}=0$ for y in terms of x and use the constraint to find x and y to maximize z.
NOTE: these are normal derivatives, not partial
work:
$z=100(48-4y)^{.25}y^{.75}$
I take the derivative and subsitute x back in to get $\frac{dz}{dy}=-x^{-.75}y^{.75}+\frac{3}{4}y^{-.25}x^{.25}$
Then I set it equal to 0 to get $\frac{3}{4}y^{-.25}x^{.25}=x^{-.75}y^{.75}$
2. Looks good so far. To solve for y, use power properties. $y^a*y^b=y^{a+b}$. You can multiply both sides by y^a, where a will both cancel one y term and make the other y^1, which is very convenient.
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## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
$P(2)=P(4)=P(1)=P(3)=P(5)=0.2=\dfrac{1}{5}$
$P(2)=P(4)=P(6)=P(1)=P(3)=P(5)=x.$ and $P(6)=0.$ We know that $P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1\\x+x+x+x+x+0=1\\5x=1\\x=0.2$ Thus, $P(2)=P(4)=P(1)=P(3)=P(5)=0.2=\dfrac{1}{5}$
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## sonja_lee2 3 years ago Find the slope of the line passing through the given points. Then tell whether the line falls, rises, is horizontal, or is vertical. (14,-3),(4,11)
1. Hero
Are you familiar with y = mx + b?
2. sonja_lee2
yes
3. sonja_lee2
i know how to find the slope i just dont know how to find out whether if it falls, rises that stuff
4. Hero
Okay, good, now, I'm going to simply plug each point into that formula: -3 = 14m + b 11 = 4m + b Are you with me so far?
5. sonja_lee2
yes
6. Hero
Okay, m is the slope, so I'm going to isolate b in both equations: b = -3 - 14m b = 11 - 4m Do you see how I got that?
7. sonja_lee2
kinda
8. Hero
It takes a little practice to understand it, but now I'm going to set b = b to get: -3 - 14m = 11 - 4m
9. Hero
Now all I have to do is solve for m: -3 - 11 = 14m - 4m -14 = 10m -14/10 = m - 7/5 = m
10. Hero
There we go
11. sonja_lee2
ok i was wait thats not right
12. Hero
You entered the answer as soon as I posted it, lol
13. sonja_lee2
kk i got that too
14. sonja_lee2
haha yeah
15. Hero
You shouldn't enter the answer so fast :P I guess I can take my own advice in that regard.
16. sonja_lee2
okay
17. Hero
Sometimes, I make typos, so you should wait at least one minute after someone posts their answer.
18. sonja_lee2
um ok
19. Hero
My fingers are not as bright as my brain
20. Hero
Sometimes, my fingers hit keys they shouldn't.
21. sonja_lee2
oh okay its a
22. sonja_lee2
all G
23. Hero
By the way, If the slope is NEGATIVE, it FALLS If the slope is POSITIVE, it RISES
24. sonja_lee2
always?
25. Hero
If something falls, it's always going down, right? If a plane rises from the ground, it is always going up, right?
26. sonja_lee2
right
27. Hero
|dw:1348449900048:dw|
28. sonja_lee2
ohh i see
29. Hero
Yeah just think about a plane either taking a nose dive or lift off
30. sonja_lee2
ohh okay thank u!
31. Hero
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# Number Combinations :: 25 Random Numbers (5 Digit Combinations)
Dec 15, 2008
I have 25 random numbers and I would like to get a possible 5 digit combinations of these numbers. Can anybody help me with the possible formula?
## Number Combinations :: Between 7 To 15 (6 Digit Combinations)
Jun 30, 2008
What I am looking for is to select between 7 and 15 numbers in total, I want all the possible 6 digit combinations for this.
EG: if I choose 2,9,11,13,15,17&26, it would look something like this
2,9,11,13,15,17
2,9,11,13,15,26
9,11,13,15,17,26
And so on.
If I chose more numbers (10) 1,2,3,4,3,6,7,8,9,10 it would start something like this
1,2,3,4,5,6
1,2,3,4,5,7
1,2,3,4,5,8
1,2,3,4,5,9
1,2,3,4,5,10
And so on.
Please remenber I would like to be able to secelt between 7 and 15 number and be given all the possible combinations.
I would like it to be in one sheet but if that can not be done on as many as it takes.
It would be good if I could just type the required number into A1,B1,C1 and so on and they just gave the combinations required.
## Generate All Combinations 3 Digit Number Produces?
Aug 3, 2013
How can I generate all the combinations a 3 digit number produces? Example 123 can be written as the following:
123
321
213
312
132
231
I THINK this is all the combinations it will produce.
## Combinations From Set Of Numbers
Jan 4, 2007
I want to do all columns combination that is (C=4^N/(3*N+1); N=4) for numbers 1,2,3,4.
as like this,
1111....2222....3333.....4444.....
1111....2222....3333.....4444.....
1111....2222....3333.....4444.....
1234....1234....1234.....1234....
## All Possible Combinations When Adding 2 Numbers
Feb 16, 2014
I have stamps of 2 denominations: .32 and .33
I want to know how I would write/format a spreadsheet that would tell me what are all the possible outcomes between the two numbers and what the components of these two numbers would be.
For example.
(8 * .32) + (3 * .33) = 3.55
So somewhere I would be able to see that in order to achieve 3.55 I need (8) .32 stamps and (3).33 stamps
## All Combinations Of A Set Of Numbers Using Excel?
Aug 22, 2012
Lets say I have 10 numbers (1,2,4,5,6,8,9,12,19,13). Now, I am trying to get a list of all possible combinations (single digit, two digit, 3 digit, 4 digit... 13 digits).
Assume that the 10 numbers (this is a variable, it can sometimes be 9 or 11 etc) are in column A and I want all the possible combinations in column B.
## Identifying Certain Combinations Of Numbers?
Mar 15, 2013
Is there a way in Excel to identifying certain combinations of numbers? If tried every IF statement I know.
I have an Excel with two columns of numbers and I need to identify their combinations.
18
18
18
19
18
19
18
20
18
18
18
20
The combinations 18 and 18 would return a "LOW" message in the third column, the combination 18 and 19 would return a message "HIGH" etc.
## Permutations Or Combinations Of A Set Of Numbers
Mar 11, 2009
I want to output every combination of a set of numbers. These permutations must include combinations that use only a few of the numbers as well as all eight...ie.
1,
1,2
1,2,3
as well as 1,2,3,4,5,6,7,8
The macro below is from
http://www.j-walk.com/ss/excel/tips/tip46.htm
This only produces combinations using every number. I'm not sure how this macro works but hopefully someone with better know how could run with it or break it down for me!
I only need to achieve this once but am pretty sure doing it manually will cause error and or madness
Dim CurrentRowSub GetString() Dim InString As String InString = InputBox("Enter text to permute:") If Len(InString) < 2 Then Exit Sub If Len(InString) >= 8 Then MsgBox "Too many permutations!" Exit Sub Else ActiveSheet.Columns(1).Clear CurrentRow = 1 Call GetPermutation("", InString) End IfEnd SubSub GetPermutation(x As String, y As String)' The source of this algorithm is unknown Dim i As Integer, j As Integer j = Len(y) If j < 2 Then Cells(CurrentRow, 1) = x & y CurrentRow = CurrentRow + 1 Else For i = 1 To j Call GetPermutation(x + Mid(y, i, 1), _ Left(y, i - 1) + Right(y, j - i)) Next End IfEnd Sub
## All Permutation And Combinations Of 4 Letters And 4 Numbers?
Jan 18, 2013
getting All permutation and combinations of 4 letters and 4 numbers?
## Given Number Combinations
Mar 16, 2008
I want to have a sheet with all the previous drawings for the florida lottery.
Then I want excel to ask me how many numbers do I wish to play.
Then when I say 12 (for example) it would pick 12 numbers from 1 to 53 and match all possible comibinations of thoses numbers to the the previous drawings and tell me wich 12 number combination would have won more times then others. So it would need to compare each possible 12 nmber combination to each other and see wich one won more 6outof6, 5outof5, 4outof4, and 3outof3. Then give me the one with the highest winings
## Number Combinations
Oct 28, 2008
Here is what my table kind of looks like
A B
4 0.25
7 0.3
2 0.1
3 0.15
4 0.30
1 0.4
8 0.05
2 0.1
Now there is the number and the number(A) that it value it represents (B). This is what I want. I want to see all possible number combinations between the upper and lower A columns listed veritcally under the whole set, with out duplicating any numbers and with the upper numbers only being in the 1st position and the lower numbers being in the second position Ex( 4-1, 4-8, 4-2, 7-4, 7-1, ect, ect (listed vertically). Now in column B under, right beside each possible combination I want to see the result of the two values multiplied (ex. beside 4-1 I would like to see 0.1)
So below that mess listed above I want to see something like this
A B
4-1 0.1
4-8 0.0125
ect
ect
ect
I want to be able to swith the values around and have it automatically make the correction.
## Output All Combinations Of List Of Given Numbers With Repetition?
May 10, 2014
I need to be able to create a list of all combinations (where position doesn't matter, therefore its not a permutation) of a list of numbers with repetition enabled. I need to provide anywhere up to at least 10-15 numbers if possible and they are integers normally between 1 and 72. Example: (for numbers 1, 2, 3)
VB:
1
2
3
[Code]....
## Produce A List Of Combinations From A Range Of Numbers?
May 9, 2014
Am looking for an easy way to produce a list of combinations (maximum 6 numbers) from a range of numbers listed in 6 different columns:
Example
Column A contains : 1,2,3
Column B contains : 7,8,11,15
Column C contains : 12,16,18,19
Column D contains : 17,30,31
Column E contains : 30,31,32,33,34
Column F contains : 37,39,40
The rules are:
In each combination of 6 numbers, numbers should always be taken from ALL 6 columns. In each combination, numbers cannot repeat.
show me a formula to arrive the results.
## Number Of Combinations Across Fields
Jan 12, 2007
I wish to calculate the total number of combinations using the number of entries in 6 fields regardless of what the actual entries are. In other words only the number of entries are relevant. They look as follows for example:
Field A1=4; Field B1=3,5,12; Field C1=9,2,11; Field D1=14,1,5; Field E1=18,2; Field F1=3,7
Calculating it using a calculator is too tedious since I have several rows of data with similar entries. I just need to know what the total number of combinations are eg in the above example 1X3x3X3X2X2=108 using some excel formula.
## List All The Combinations Of A Group Of Cells Containing Letters, But Not Numbers
Feb 9, 2008
I have 7 cells containing strings but not numbers on a row.
Now I want to list out all the combinations of drawing out 3 cells out of these 7 cells while the remaining cells that haven't been drawn out could also be listed out one column next to the drawn cells.
For example, I got 7 cells like this.
A B C D E F G (each letter in ONE cell)
And I want to list out all the combinations like this:
ABC DEFG
ABD CEFG
ABE CDFG
ABF CDEG
etc.
For more details, please refer to the attached sample (an .xls file being zipped).
## Listing All Part Number Combinations
Mar 28, 2007
im looking to piece together all possible part number combinations with a one condition:
The values in Column A must come before the values in Column B. The values in Column B must come before the values in Column C. And So forth...
Example: If Column A had the values of A, B, C; Column B had the values of -1, -2, -3; and Column C had the values of -01, -02, -03
The outcome(s) would be: A-1-01, A-1-02, A-1-03 and so forth...
How can I get it to do that? But I need the flexibility of adding on more "options" should the part number be bigger (for future part number combinations).
## Identify Matches Of Number Combinations
Jan 14, 2008
After looking at other similar post titles I could not find a formula to work exactly proper.
I have attatched a sample of my problem. What I need is for cells E4:J4 to look for a match in cells B4:B16 and give a value of " Yes or No " in cell K4 if a match is found. Then copy down formula through E5:J5 to B4:B16 - E9:J9 to B4:B16. As my examples indicate Cells K4 would = yes, K5 = yes, K6 = yes, K7 = no, K8 = no, K9 = yes. Column D is the results, Columns E-J represents all combinations of column D.
I have tried the following formulas but they did not work entirely correct.
= NOT( ISNA( MATCH(\$b\$4:\$b\$16, e5:j9, 0 ) ) )
= COUNTIF(e9:j9,"="&b4:b16)>0
## Match Number Combinations To Equal Set Value
Aug 1, 2008
I am looking to do the following:
Say I enter the following values into a spreadsheet:
5
7
4
24
32
Is there a way to enter the value "36" and have Excel find the two values of 32 and 4 which add up to 36 for me? I will obviously be dealing with much larger spreadsheets and would like to somehow save time by not having to hunt down possible combinations of the total value I'm searching for.
## Search And Delete Combinations Of 4 Specific Numbers Stored In 5 Columns?
Aug 28, 2012
I attached partial file so you can see what i mean. I pasted only a few combinations cause the file was to big and i wasnt able to attach it...there are 142506 combinations . But you can see what i mean. A VBA or a macro on the worksheet will do the job?
## Formula Or Macro To Show All Number Combinations?
Nov 19, 2007
as an example i will use the national lottery. numbers 1 to 49 inclusive. i need a formula that will list all the possibile 6 number combinations not repeating any.
## Find All Number Combinations In Dataset That Add Up To Specific Number?
Jun 3, 2014
I have attached an example. If I have a set of numbers such as the one attached, is it possible to create a formula that will show me all the combinations of numbers that add up to 55.52? In the attached I have highlighted in different colours all the number combinations that add up to 55.52. The numbers highlighted in blue appear within more then one combination. Is there a formula that can do this for me, instead of randomly adding numbers hoping they add up to 55.52.
## Find All Number Combinations In Dataset That Add Up To Specific Number
Jun 4, 2014
I have attached an example. If I have a set of numbers such as the one attached, is it possible to create a formula that will show me all the combinations of numbers that add up to 55.52? In the attached I have highlighted in different colours all the number combinations that add up to 55.52. The numbers highlighted in blue appear within more then one combination. Is there a formula that can do this for me, instead of randomly adding numbers hoping they add up to 55.52.
examples.xlsx
## Calculate Number Of Hours Between Two Date/time Combinations
Jan 21, 2010
I am trying to find a way to calculate the number of hours between date/times found in separate rows. The attached data set will help to envision what I am talking about.
For each couple of rows, I need to find a way to calculate the number of hours elapsed from row 1 to row 2. In the first example, to calculate the number of hours between 12/2/2009 8:56:51 and 12/4/2009 6:35:27.
## Count Number Of Times Text Combinations Appear In Same Column
Oct 6, 2013
A1:A10 contains text (say colors) and B1:B10 also text (say vegetables). I need a formula to count the number of times a certain combination of numbers and vegetables appear in the same column, so if "red" and "carrot" appeared in A4 and B4 and also in A6 and B6, the result would be 2.
## All Number Combinations From Entries In 7 Columns And 3 Rows With No Duplicates?
Dec 2, 2013
I have 7 columns x 3 rows with unique numbers. I wish to determine every combination of the above numbers without repeats.
5
11
17
24
29
35
40
[Code] ........
Column 1 would only use column 1 numbers.
Column 2 would only use column 2 numbers...etc.
Also have the same for 5 rows of unique numbers.
I've been working on this for 10 years and this is the last step to finalizing the WINNING numbers.
Back checking using the 5 row technique I would have spent 10,000,000.00 playing over 8 years with a winners purse of 33,000,000.00
With the 3 row I would have spent close to 200k and won 2.5 mil.
Only problem is that I don't have the 10 to start with, the 200k is a maybe .
I need to be able to generate the combinations to play the numbers.
## VBA To Generate Random 5 Digit Serial Numbers?
May 31, 2014
I have a userform for keeping records and would love to incorporate a new feature. I would want to generate a random serial number for each entry made with the userform.
## Seperate 2 Digit Number Into 2 Single Digit Numbers
Apr 3, 2008
I know you can take a number from one cell and combine it with number from another cell and make it one number. What I need to do is the reverse. Take a two digit number in a cell and separate it into single digits in two cells. If you have the number 50 in a cell, then is there a formula that will take the 5 and put it in cell and take the 0 and put it in the cell beside it?
## Generate A Random 9 Digit Number
Jan 7, 2009
I have a spreadsheet with approx 11,000 rows and I would like to generate a unique 9 digit number for each line.
I know I could just put 100 000 000 and then increase that by one all the way down but I was wondering if there was any formula or code
## Excel 2010 :: Generate 6 Digit Unique Random Number For ID Column A
Jul 18, 2012
How do I create a 6 digit unique random number for use as an ID in column A. Once created the rows with preexisting 6 digit unique random ID numbers must not change every time new rows are added.
## All Possible Combinations That Sum Up To X
Oct 19, 2007
I have a column of transactions lets say 500 or so, of irregular numbers such as 257,273.80 (not something ou can add up in your head...). I know that of those 500 transactions a handful of them have a sum of lets say 2,877,000.00, which I know. Now I need to find all possible combinations of values in this column that equal a sum I enter.
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# Scott Grissom, copyright 2004 Chapter 5 Slide 1 Analysis of Algorithms (Ch 5) Chapter 5 focuses on: algorithm analysis searching algorithms sorting algorithms.
## Presentation on theme: "Scott Grissom, copyright 2004 Chapter 5 Slide 1 Analysis of Algorithms (Ch 5) Chapter 5 focuses on: algorithm analysis searching algorithms sorting algorithms."— Presentation transcript:
Scott Grissom, copyright 2004 Chapter 5 Slide 1 Analysis of Algorithms (Ch 5) Chapter 5 focuses on: algorithm analysis searching algorithms sorting algorithms recursion Comparing Algorithms (5.1) Given two algorithms that do the same thing, which is better? Is either good enough? What about special cases hardware differences software differences small data sets
Scott Grissom, copyright 2004 Chapter 5 Slide 2 Runtime Time Working code is not enough Running time is an important issue How to improve How to compare Space is not as important
Scott Grissom, copyright 2004 Chapter 5 Slide 3 Efficiency (5.2) Running Time based on algorithm AND input data Each statement takes constant time Determine total operations int total = 3; total = total + 3 * 6 - 4; What about this? int total = 0; for (int I=0; I<=10; I++) total = total+ I * I * I;
Scott Grissom, copyright 2004 Chapter 5 Slide 4 Data Set Size of input data N? Predict time t = f(n) Using a stopwatch (benchmarking) is not sufficient. What about this? int Total = 0 for (int i=0; i <=N; i++) Total = Total + i * i * i;
Scott Grissom, copyright 2004 Chapter 5 Slide 5 Big O Notation Predicting exact counts is a nuisance Instead, use Big Oh or Order t = O [f(n)] t n o We can round off t = 6n + 14 t = O(n) 6n 4 + 3n 3 + 12n + 6453 2n 2 + 6 6 NlogN + 2n 2 + 13 12,345 The Most Common n 3 n logN n 2 NlogN 1 2 n place in correct order
Scott Grissom, copyright 2004 Chapter 5 Slide 6 Linear Search Start at the beginning and compare with every item until it is found (or not) Code position = -1; for (int I=0; I<array.length; I++) if (target == array[I]) position = I; Consider best case average case worst case Benchmarking see Demo/LinearSearch.java Consider Enhancements Figure 5.4 on page 217
Scott Grissom, copyright 2004 Chapter 5 Slide 7 Binary Search Items must be sorted Guess middle item If target is less than item then look in first half If target is greater than item then look in second half Else, you found it Code position = -1; while (start <= end) middle = (start + end) / 2 if target < array[middle] end = middle - 1 if target > array[middle] start = middle + 1 if target == array[middle] position = middle Benchmarking see Demo/BinarySearch.java Consider Enhancements Figure 5.5 on page 219
Scott Grissom, copyright 2004 Chapter 5 Slide 8 Asymptotic Analysis 5.3 Nested Loops work inside out for (i=0; i< N; i++) for (j=0; j< N; j++) S; What about this? for (int i=1; i<1000; i++) S; And this one? for (int i=1; i<=N; i++) method (i);
Scott Grissom, copyright 2004 Chapter 5 Slide 9 Sorting Place items in correct order (ascending) Be aware of these terms swap comparison pass Many algorithms have been developed Selection Sort Bubble Sort Merge Sort Quick Sort
Scott Grissom, copyright 2004 Chapter 5 Slide 10 Selection Sort For each pass find smallest remaining item place in correct location Repeat N-1 times Benchmark see Demos/SelectionSort.java
Scott Grissom, copyright 2004 Chapter 5 Slide 11 Bubble Sort For each pass swap neighbors out of order Repeat N-1 times Consider Enhancements Figure 5.6 on page 220
Scott Grissom, copyright 2004 Chapter 5 Slide 12 Sorting Analysis Goal of sorting is to remove inversions How many inversions are there worst case? Consider algorithms that compare and swap neighbors. How many inversions are removed with a single swap? Therefore, what is the expected runtime for ALL sorting algorithms that compare and swap neighbors?
Scott Grissom, copyright 2004 Chapter 5 Slide 13 Other Complexity Measures (5.4) Big Omega best case t >= f(n) Big Theta best and worst case t == f(n) Consider Linear Search Binary Search Bubble Sort in book
Scott Grissom, copyright 2004 Chapter 5 Slide 14 Recursion -- 5.5 Recursion is a fundamental programming technique that can provide an elegant solution to certain kinds of problems Book coverage is limited to the analysis of recursive algorithms A recursive definition is one which uses the word or concept being defined in the definition itself All recursive definitions have to have a non-recursive part called the base case If they didn't, there would be no way to terminate the recursive path called infinite recursion
Scott Grissom, copyright 2004 Chapter 5 Slide 15 Recursive Definitions N!, for any positive integer N, is defined to be the product of all integers between 1 and N inclusive This definition can be expressed recursively as: 1! = 1 N! = N * (N-1)! Eventually, the base case of 1! is reached
Scott Grissom, copyright 2004 Chapter 5 Slide 16 Recursive Programming A method in Java can invoke itself a recursive method The code of a recursive method must be structured to handle both the base case and the recursive case Each call to the method sets up a new execution environment, with new parameters and local variables
Scott Grissom, copyright 2004 Chapter 5 Slide 17 Summation Consider the problem of computing the sum of all the numbers between 1 and any positive integer N This problem can be recursively defined as: Summation Solution int Sum (int N){ if (N < = 1) return 1 else return N + sum(N - 1) } Draw the recursive call stack Too much recursion can cause a Stack Overflow error See Demos/Recursion.java
Scott Grissom, copyright 2004 Chapter 5 Slide 18 Practice Recursion mystery (N){ if (N <= 1) print something else mystery (N - 1); Do it again with (N - 2) Do it again with (N / 2) big difference!
Scott Grissom, copyright 2004 Chapter 5 Slide 19 Recursive Binary Search The binary search algorithm can be implemented recursively It has the same general strategy as the iterative version Each recursive call cuts the search space in half Study code in Figure 5.10 Use the substitution method to solve T(n) = T(n/2) + 3
Scott Grissom, copyright 2004 Chapter 5 Slide 20 Group Projects Write code to solve the following problems How many methods are called for each solution with respect to N? Print the values 1 to N 1 2 3 4 5 6 Print the values N to 1 6 5 4 3 2 1 Fibonacci sequence 1 1 2 3 5 8 13 21 fibonacci (10) prints the first ten numbers in the sequence Factorial value = factorial (8); Power result = power (base, exponent)
Scott Grissom, copyright 2004 Chapter 5 Slide 21 Elegant Solutions? Note that just because we can use recursion to solve a problem, doesn't mean we should For instance, we usually would not use recursion to solve the sum of 1 to N problem, because the iterative version is easier to understand However, for some problems, recursion provides an elegant solution, often cleaner than an iterative version
Scott Grissom, copyright 2004 Chapter 5 Slide 22 Towers of Hanoi A classic problem! Three spindles and N disks of different sizes Move all disks from spindle 1 to 3 The world will end when the monks successfully move 64 disks How long will it take?
Scott Grissom, copyright 2004 Chapter 5 Slide 23 Merge Sort The approach: Divide the list into two equal parts Recursively sort both parts Merge the two parts into a single list again The merge process takes O(N) The runtime is O(n log n) T(n) = 2T(n/2) + n Use the substitution method to solve
Scott Grissom, copyright 2004 Chapter 5 Slide 24 Quick Sort The approach of Quick Sort: Select a pivot Divide the list into two lists based on the pivot all items less than the pivot all items greater than or equal to the pivot Recursively sort both lists Piece together the first list, pivot, and second list into the original list It is considered the fastest known sorting algorithm The runtime is O (n log n)
Scott Grissom, copyright 2004 Chapter 5 Slide 25 Summary Given a computer that performs one billion operations per second on input of one million log n would take almost no time n would take.001 seconds n log n would take _______ n 2 would take ________ n 3 would take ________ 2 n would take ________ 2 n - exponential growth is very slow if computer performs 1 bip n = 40 takes 18 minutes n = 50 13 days n = 60 310 years
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Longitudinal Stability Longitudinal Stability (Pitching)
In designing an airplane a great deal of effort is spent in developing the desired degree of stability around all three axes. But longitudinal stability about the lateral axis is considered to be the most affected by certain variables in various flight conditions.
As we learned earlier, longitudinal stability is the quality which makes an airplane stable about its lateral axis. It involves the pitching motion as the airplane's nose moves up and down in flight. A longitudinally unstable airplane has a tendency to dive or climb progressively into a very steep dive or climb, or even a stall. Thus, an airplane with longitudinal instability becomes difficult and sometimes dangerous to fly.
Static longitudinal stability or instability in an airplane, is dependent upon three factors:
1. Location of the wing with respect to the center of gravity;
2. Location of the horizontal tail surfaces with respect to the center of gravity; and
3. The area or size of the tail surfaces.
In analyzing stability it should be recalled that a body that is free to rotate will always turn about its center of gravity.
To obtain static longitudinal stability, the relation of the wing and tail moments must be such that, if the moments are initially balanced and the airplane is suddenly nosed up, the wing moments and tail moments will change so that the sum of their forces will provide an unbalanced but restoring moments which in turn, will bring the nose down again. Similarly, if the airplane is nosed down, the resulting change in moments will bring the nose back up.
We have spoken of the airplane's center of gravity and the airfoil's center of lift in preceding sections. Now let us reexamine the center of lift or as it is sometimes called, the center of pressure. As previously pointed out, the center of pressure in most unsymmetrical airfoils has a tendency to change its fore and aft position with a change in the angle of attack. The center of pressure tends to move forward with an increase in angle of attack and to move aft with a decrease in angle of attack. This means that when the angle of attack of an airfoil is increased, the center of pressure (lift) by moving forward, tends to lift the leading edge of the wing still more. This tendency gives the wing an inherent quality of instability. Figure 17-24 shows an airplane in straight and level flight. The line CG-CL-T represents the airplane's longitudinal axis from the center of gravity (CG) to a point T on the horizontal stabilizer. The center of lift (or center of pressure) is represented by the point CL.
Most airplanes are designed so that the wing's center of lift (CL) is to the rear of the center of gravity. This makes the airplane "nose heavy" and requires that there be a slight downward force on the horizontal stabilizer in order to balance the airplane and keep the nose from continually pitching downward. Compensation for this nose heaviness is provided by setting the horizontal stabilizer at a slight negative angle of attack. The downward force thus produced, holds the tail down, counterbalancing the "heavy" nose. It is as if the line CG-CL-T was a lever with an upward force at CL and two downward forces balancing each other, one a strong force at the CG point and the other, a much lesser force, at point T (downward air pressure on the stabilizer). Applying simple physics principles, it can be seen that if an iron bar were suspended at point CL with a heavy weight hanging on it at the CG, it would take some downward pressure at point T to keep the "lever in balance.
Even though the horizontal stabilizer may be level when the airplane is in level flight, there is a downwash of air from the wings. This downwash strikes the top of the stabilizer and produces a downward pressure which, at a certain speed, will be just enough to balance the "lever." The faster the airplane is flying, the greater this downwash and the greater the downward force on the horizontal stabilizer (except "T" tails) (Fig. 17-25). In airplanes with fixed position horizontal stabilizers, the airplane manufacturer sets the stabilizer at an angle that will provide the best stability (or balance) during flight at the design cruising speed and power setting (Fig. 17-26).
If the airplane's speed decreases, the speed of the airflow over the wing is decreased. As a result of this decreased flow of air over the wing, the downwash is reduced, causing a lesser downward force on the horizontal stabilizer. In turn, the characteristic nose heaviness is accentuated, causing the airplane's nose to pitch down more. This places the airplane in a nose low attitude, lessening the wing's angle of attack and drag and allowing the airspeed to increase. As the airplane continues in the nose low attitude and its speed increases, the downward force on the horizontal stabilizer is once again increased. Consequently, the tail is again pushed downward and the nose rises into a climbing attitude.
As this climb continues, the airspeed again decreases, causing the downward force on the tail to decrease until the nose lowers once more. However, because the airplane is dynamically stable, the nose does not lower as far this time as it did before. The airplane will acquire enough speed in this more gradual dive to start it into another climb, but the climb is not so steep as the preceding one.
After several of these diminishing oscillations, in which the nose alternately rises and lowers, the airplane will finally settle down to a speed at which the downward force on the tail exactly counteracts the tendency of the airplane to dive. When this condition is attained the airplane will once again be in balanced flight and will continue in stabilized flight as long as this attitude and airspeed are not changed.
A similar effect will be noted upon closing the throttle. The downwash of the wings is reduced and the force at T in Fig. 17-24 is not enough to hold the horizontal stabilizer down. It is as if the force at T on the lever were allowing the force of gravity to pull the nose down. This, of course, is a desirable characteristic because the airplane is inherently trying to regain airspeed and reestablish the proper balance.
Power or thrust can also have a destabilizing effect in that an increase of power may tend to make the nose rise. The airplane designer can offset this by establishing a "high thrustline" wherein the line of thrust passes above the center of gravity (Figs. 17-27, 17-28). In this case, as power or thrust is increased a moment is produced to counteract the down load on the tail. On the other hand, a very "low thrust line" would tend to add to the nose up effect of the horizontal tail surface.
It can be concluded, then, that with the center of gravity forward of the center of lift, and with an aerodynamic tail down force, the result is that the airplane always tries to return to a safe flying attitude.
A simple demonstration of longitudinal stability may be made as follows: Trim the airplane for "hands off" control in level flight. Then momentarily give the controls a slight push to nose the airplane down. If, within a brief period, the nose rises to the original position and then stops, the airplane is statically stable. Ordinarily, the nose will pass the original position (that of level flight) and a series of slow pitching oscillations will follow. If the oscillations gradually cease, the airplane has positive stability; if they continue unevenly the airplane has neutral stability; if they increase the airplane is unstable.
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analytic geometry question
• Mar 5th 2008, 10:01 PM
bluebirdsfly
analytic geometry question
The perimeter of a triangle is 30, and the points (0,-5) and (0,5) are two fo the vertices. Find the equation of the ellipse of the third vertex.
Any idea as to how to begin solving this problem...?
• Mar 6th 2008, 05:17 AM
earboth
Quote:
Originally Posted by bluebirdsfly
The perimeter of a triangle is 30, and the points (0,-5) and (0,5) are two fo the vertices. Find the equation of the ellipse of the third vertex.
Any idea as to how to begin solving this problem...?
An ellipse is the locus of all points which have the constant sum of distances from 2 fixed points (=focus).
Therefore:
1. (0, -5), (0, 5) are the focii.
2. The sum of distances is 20. (The distance between the focii is 10!)
3. The length of the main semiaxis must be 10.
4. The center of the ellipse must be the origin.
5. The length of the minor semi axis must be:$\displaystyle b = \sqrt{10^2-5^2} = 5 \cdot \sqrt{3}$
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# Synthetic division problem solver
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Kinetic Energy
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In addition to the revision notes for Kinetic Energy on this page, you can also access the following Work, Energy and Power learning resources for Kinetic Energy
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5.2Kinetic Energy
In these revision notes for Kinetic Energy, we cover the following key points:
• What is Kinetic Energy?
• What are the factors affecting the Kinetic Energy of an object?
• What is the relationship between Work and the change in Kinetic Energy of an object?
• How to calculate the Kinetic Energy of moving objects?
Kinetic Energy Revision Notes
Kinetic Energy is the energy of a moving object. This means when an object is at rest it does not possess Kinetic Energy.
Kinetic Energy is denoted in equations as KE. As all the other types of energy, it is also measured in Joules [J].
Kinetic Energy of a moving object depends on two factors:
1. The Mass m of object, and
2. The Velocity v of the object.
The equation that relates Kinetic Energy with the abovementioned factors is
KE = 1/2 m × v2
The Work-Kinetic Energy Theorem, states that
"The work done by the sum of all forces acting on an object equals the change in the kinetic energy of the object itself."
Mathematically, the Work-Kinetic Energy Theorem is written as:
F × ∆x = 1/2 m × v2f - 1/2 m × v2i
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Maths Class 3 Contents
# Maths Class 3 Contents
NCERT (National Council of Educational Research and Training) designed the syllabus of all subjects of CBSE board. CBSE (Central Board of Secondary Education) is responsible for providing the necessary guidelines to all the schools which are affiliated to CBSE. It improves the standard of education in the country as it modifies the syllabus accordingly so that students can learn and grab the topics in a very easy way.
## Chapter 1: Large Numbers
Ø Ø Number Names
Ø Ø Expanded Form and Standard Form
Ø Ø Successor and Predecessor
Ø Ø Even and Odd Numbers
## Chapter 3: Multiplication
Ø Multiplication Tables Ø Multiplication on the Number Line
## Chapter 4: Division
Ø Division Using the Number Line Ø Division as Repeated Subtraction
Ø Division Using Multiplication Tables Ø Properties of Division
Ø Dividing 2-digit Number by a 1-digit Number
Ø Dividing 3-digit Number by a 1-digit Number
Ø Dividing 4-digit Number by a 1-digit Number
Ø Verification of Division Using Multiplication Tables
## Chapter 5: Fractions
Ø Properties of Fractions Ø Fractions as a Collection of Objects
Ø Ø Tangram
## Chapter 7: Measurement
Ø Ø Measuring Length
Ø Ø Capacity and Volume
Ø Ø Calendar
Ø Ø Timeline
Ø Ø Bills
## Chapter 10: Data Handling
Ø Ø Pictograph
## Chapter 11: Patterns and Symmetry
Ø Ø Types of Patterns
Ø Ø Symmetry
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# ROOTS OF INTEGERS
Roots of integers :
To find roots of integers, first we have to notice the index of the given root. According to the index of the given radical, we have to split the number and simplify.
## What is index of the root ?
We have to read the first term as cube root of 27 and second term as 5th root of 125.
## How to find roots of integers
Step 1 :
Let "x" be the root of the given radical term
Step 2 :
According to the index, we have to raise power on both sides.For example, if we have square root, then we have to take squares on both sides, in order to remove the square root.
Step 3 :
After cancelling the power and the radical sign in right hand side, we have to express the number in the exponential form.
Step 4 :
Check whether we have same powers on either sides of equal sign.
Step 5 :
Since the power are equal, we can decide that bases are also equal.
From the above steps, we can find the roots of the integers.
Let us look into some examples based on the above concept.
Example 1 :
Find the real number root of √64
Solution :
Step 1 :
Index of the given square root = 2
Let "x" be the required root
x = √64
Step 2 :
In order to remove the square root, take squares on both sides
x² = (√64)²
x² = 64
Step 3 :
Express 64 as the product of two same numbers
That is, 64 = 8 x 8
64 = 8²
x² = 8²
Step 4 :
Since the powers are equal, then their base are also equal.
x = 8
Hence the required root is 8.
Example 2 :
Find the real number root of ∛512
Solution :
Step 1 :
Index of the given square root = 3
Let "x" be the required root
x = ∛512
Step 2 :
In order to remove the square root, raise power 3 on both sides
x³ = (∛512)³
x³ = 512
Step 3 :
Express 512 as the product of two same numbers
That is, 512 = 8 x 8 x 8
512 = 8³
x³ = 8³
Step 4 :
Since the powers are equal, then their base are also equal.
x = 8
Hence the required root is 8.
Example 3 :
Find the real number root of ∛-8000
Solution :
Step 1 :
Index of the given square root = 3
Let "x" be the required root
x = ∛-8000
Step 2 :
In order to remove the square root, raise power 3 on both sides
x³ = (∛-8000)³
x³ = -8000
Step 3 :
Express -8000 as the product of two same numbers
That is, -8000 = (-20) x (-20) x (-20)
-8000 = (-20)³
x³ = (-20)³
Step 4 :
Since the powers are equal, then their base are also equal.
x = -20
Hence the required root is -20.
Example 4 :
Find the real number root of ∜16
Solution :
Step 1 :
Index of the given square root = 4
Let "x" be the required root
x = ∜16
Step 2 :
In order to remove the square root, raise power 3 on both sides
x = (∜16)
x = 16
Step 3 :
Express 16 as the product of two same numbers
That is, 16 = 2 x 2 x 2 x 2
16 = 2
x = 2
Step 4 :
Since the powers are equal, then their base are also equal.
x = 2
Hence the required root is 2.
After having gone through the stuff given above, we hope that the students would have understood "Roots of integers".
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
WORD PROBLEMS
HCF and LCM word problems
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
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Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
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Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
| 1,446 | 5,400 |
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| 844,001,311 | 9,680 |
# CALCULUS Curriculum Guide
PROFICIENCY 1: THE LEARNER WILL DEVELOP AN UNDERSTANDING OF
FUNCTIONS AND GRAPHS
1.1 Define functions from problematic situations
1.2 Determine whether a given function is even or odd
1.3 Determine the zeros of a function
1.4 Graph trigonometric functions and determine their periods, amplitudes, and phase shifts
1.5 Find the slope of a linear function
1.6 Use the standard forms to write an equation of a line that fits given conditions
1.7 Graph exponential and logarithmic functions and use their inverse relationships
1.8 Find the asymptotes of the graph of a function
1.9 Describe the symmetry of the graph of a function
PROFICIENCY 2: THE LEARNER WILL DEVELOP AN UNDERSTANDING OF LIMITS
AND CONTINUITY
2.1 Find or approximate limits intuitively using a calculator
2.2 Find limits of functions by substitution
2.3 Find limits using the constant, sum, difference , product , and quotient rules
2.4 Find the limit of a rational function that has an indeterminate form
2.5 Find one-sided limits
2.6 Find limits at infinity
2.7 Determine when a limit is infinite
2.8 Use the definition of continuity to determine whether a function is continuous at a point
2.9 Use the intermediate value theorem on a function over a closed interval
2.10 Determine the types of discontinuities of a function
2.11 Apply the theorem, “If f (x) is continuous over a closed interval, then f has a maximum
and a minimum value on the interval”
PROFICIENCY 3: THE LEARNER WILL DEVELOP AN UNDERSTANDING OF THE
CONCEPTS OF DIFFERENTIAL CACLULUS
3.1 State and apply the definition of derivative
3.2 Find the derivatives of elementary functions including algebraic , trigonometric,
exponential, and logarithmic
3.3 Find the derivatives of sums, products, and quotients
3.4 Determine the derivative of a composite function ( chain rule )
3.5 Find the derivatives of implicitly defined functions
3.6 Find derivatives of higher order
3.7 Find the derivative of the inverse of a function
3.8 Find derivatives using logarithmic differentiation
3.9 Use the relation between differentiability and continuity
3.10 Apply the Mean Value Theorem
3.11 Use Rule
PROFICIENCY 4: THE LEARNER WILL DEVELOP AN UNDERSTANDING OF THE
APPLICATIONS OF THE CONCEPTS OF A DERIVATIVE
4.1 Find the slope of a curve
4.2 Find the tangent and normal lines
4.3 Determine where a function is increasing and where it is decreasing
4.4 Find critical points, relative (local), and absolute maximum and minimum points
4.5 Determine the concavity and points of inflection of a function
4.6 Use first and second derivatives to help sketch a curve
4.7 Use differentials to approximate change
4.8 Use Newton’s Method to approximate the zeros of a function or the intersection of two
functions
4.9 Solve optimization problems
4.10 Find average and instantaneous rates of change
4.11 Find the velocity and acceleration of a particle moving in a straight line
4.12 Find related rates of change
PROFICIENCY 5: THE LEARNER WILL DEVELOP AN UNDERSTANDING OF THE
CONCEPTS OF INTEGRAL CALCULUS
5.1 Use rectangle approximation techniques to find approximate value of integrals
5.2 Calculate the values of Riemann Sums
5.3 Recognize and write definite integrals as limits of Riemann Sums and vice versa
5.4 Use the Fundamental Theorem of Calculus
5.5 Use properties of antiderivatives and the Fundamental Theorem of Calculus to evaluate
definite and indefinite integrals
5.6 Use properties of definite integrals
5.7 Use the technique of integration by substitution ( change of variables ) to find values of
integrals
5.8 Use the technique of integration by parts
5.9 Use the technique of integration by trigonometric substitution
5.10 Use numeric techniques such as the Trapezoidal Rule, Simpson’s Rule, or technology to
approximate definite integrals
PROFICIENCY 6: THE LEARNER WILL DEVELOP AN UNDERSTANDING OF THE
APPLICATIONS OF INTEGRAL CALCULUS
6.1 Derive velocity functions from acceleration functions and/or position functions from
velocity functions given the necessary initial conditions
6.2 Solve separable differential equations of the form f (x)dx = g (y)dy
6.3 Solve differential equations of the form y’ = k y as applied growth and decay problems
6.4 Use definite integrals to find the area under a curve and above the x-axis
6.5 Use definite integrals to find the area between two curves
6.6 Use definite integrals to find the average value of a function over a closed interval
6.7 Use definite integrals to find the volume of a solid with known cross-sectional areas
6.8 Use definite integrals to find the volume of a solid obtained by revolving an area about the
x-axis, y-axis, or a line parallel to either axis (Disc, washer, and shell methods)
6.9 Use definite integrals to calculate the surface area of a solid obtained by revolving a region
about one of the coordinate axes
6.10 Write improper integrals as limits of definite integrals to determine their nature (convergent
or divergent) and find the values of those that converge
Prev Next
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# 5402 (number)
5,402 (five thousand four hundred two) is an even four-digits composite number following 5401 and preceding 5403. In scientific notation, it is written as 5.402 × 103. The sum of its digits is 11. It has a total of 3 prime factors and 8 positive divisors. There are 2,592 positive integers (up to 5402) that are relatively prime to 5402.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 11
• Digital Root 2
## Name
Short name 5 thousand 402 five thousand four hundred two
## Notation
Scientific notation 5.402 × 103 5.402 × 103
## Prime Factorization of 5402
Prime Factorization 2 × 37 × 73
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 5402 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) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 5,402 is 2 × 37 × 73. Since it has a total of 3 prime factors, 5,402 is a composite number.
## Divisors of 5402
1, 2, 37, 73, 74, 146, 2701, 5402
8 divisors
Even divisors 4 4 4 0
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 8436 Sum of all the positive divisors of n s(n) 3034 Sum of the proper positive divisors of n A(n) 1054.5 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 73.4983 Returns the nth root of the product of n divisors H(n) 5.12281 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 5,402 can be divided by 8 positive divisors (out of which 4 are even, and 4 are odd). The sum of these divisors (counting 5,402) is 8,436, the average is 105,4.5.
## Other Arithmetic Functions (n = 5402)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 2592 Total number of positive integers not greater than n that are coprime to n λ(n) 72 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 719 Total number of primes less than or equal to n r2(n) 16 The number of ways n can be represented as the sum of 2 squares
There are 2,592 positive integers (less than 5,402) that are coprime with 5,402. And there are approximately 719 prime numbers less than or equal to 5,402.
## Divisibility of 5402
m n mod m 2 3 4 5 6 7 8 9 0 2 2 2 2 5 2 2
The number 5,402 is divisible by 2.
• Deficient
• Polite
• Square Free
• Pronic
• Sphenic
## Base conversion (5402)
Base System Value
2 Binary 1010100011010
3 Ternary 21102002
4 Quaternary 1110122
5 Quinary 133102
6 Senary 41002
8 Octal 12432
10 Decimal 5402
12 Duodecimal 3162
20 Vigesimal da2
36 Base36 462
## Basic calculations (n = 5402)
### Multiplication
n×y
n×2 10804 16206 21608 27010
### Division
n÷y
n÷2 2701 1800.67 1350.5 1080.4
### Exponentiation
ny
n2 29181604 157639024808 851566012012816 4600159596893232032
### Nth Root
y√n
2√n 73.4983 17.5463 8.57311 5.57842
## 5402 as geometric shapes
### Circle
Diameter 10804 33941.8 9.16767e+07
### Sphere
Volume 6.60317e+11 3.66707e+08 33941.8
### Square
Length = n
Perimeter 21608 2.91816e+07 7639.58
### Cube
Length = n
Surface area 1.7509e+08 1.57639e+11 9356.54
### Equilateral Triangle
Length = n
Perimeter 16206 1.2636e+07 4678.27
### Triangular Pyramid
Length = n
Surface area 5.0544e+07 1.85779e+10 4410.71
## Cryptographic Hash Functions
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| 1,441 | 4,003 |
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# Surface area of a rectangular prism example
### Surface Area and Volume of Box Calculator
Right Prisms And Cylinders Measurements Siyavula. A right rectangular prism has Schläfli The surface area of a right prism whose base is a regular n-sided polygon with Example: Pentagonal prism, {5, A right rectangular prism has Schläfli The surface area of a right prism whose base is a regular n-sided polygon with Example: Pentagonal prism, {5.
### Surface Area of a Rectangular Prism AAAKnow
What Are Some Real-Life Examples of Triangular Prisms. Area of a rectangular prism, area of a cuboid, examples and solved problems., The surface area of a triangular prism can be found in the A triangular prism has three rectangular sides and Example 1: Find the Surface Area of the Right.
Area of a rectangular prism, area of a cuboid, examples and solved problems. Surface Area Rectangular Prism lateral surfaces and its rectangular base is the total surface area of the Example 2. Find the surface area of the
... Finding the surface area of a rectangular prism. Worked example 3: Finding the surface area of The volume of right prisms and cylinders is simply Total Surface Area and Volume of Box Calculator. A box is nothing but a right rectangular prism. It can also be called as the rectangular parallelepiped.
A right rectangular prism has Schläfli The surface area of a right prism whose base is a regular n-sided polygon with Example: Pentagonal prism, {5 7/07/2015 · Quickly master how to find the surface area of a rectangular prism. Watch more lessons like this and try our practice at https://www.studypug.com/geometry
... Finding the surface area of a rectangular prism. Worked example 3: Finding the surface area of The volume of right prisms and cylinders is simply Calculate the area of each side and find the sum of the areas of all sides to find the surface area of a prism. Example: the surface area of a rectangular prism of
Surface area of trapezoidal prism could be of two types lateral surface area of trapezoidal prism and total surface area of trapezoidal prism. Calculate the area of each side and find the sum of the areas of all sides to find the surface area of a prism. Example: the surface area of a rectangular prism of
For example, if you consider an 8 m-by-9 m face a base for the right rectangular prism in this problem, The surface area of the right rectangular prism is 348 m 2. What Are Some Real-Life Examples of Triangular Prisms? The Flatiron Building in New York City is an example of a triangular prism in real-life. This building is a
What Are Some Real-Life Examples of Triangular Prisms? The Flatiron Building in New York City is an example of a triangular prism in real-life. This building is a The figure is the net for a rectangular prism Example #1: Find the surface area of a rectangular Amira is painting a rectangular banner 2 1/4 yards
The lateral surface area of a prism is the sum of the areas of its lateral faces. Menu. About Academic Tutoring Test Prep Example 2: Find ... Finding the surface area of a rectangular prism. Worked example 3: Finding the surface area of The volume of right prisms and cylinders is simply
Learn how to calculate the surface area, volume, and perimeter for geometric shapes, including cylinders, Surface Area and Volume of a Rectangular Prism . ... Finding the surface area of a rectangular prism. Worked example 3: Finding the surface area of The volume of right prisms and cylinders is simply
How can the surface area of a triangular prism be The surface area of a right triangular prism with rectangular faces of equal dimensions and equilateral Lateral surface area is finding the area Lateral Area: Definition, Formula & Examples. Lateral Surface Area of a Rectangular Prism. Let's try another example,
A rectangular prism consists of three different dimensions. The prism's length, height and width create its volume and surface area, which are its internal and After watching this video lesson, you will be able to use the formula for the surface area of a rectangular prism. Definition & Examples 4:17
How can the surface area of a triangular prism be The surface area of a right triangular prism with rectangular faces of equal dimensions and equilateral The surface area of a triangular prism can be found in the A triangular prism has three rectangular sides and Example 1: Find the Surface Area of the Right
What Are Some Real-Life Examples of Triangular Prisms? The Flatiron Building in New York City is an example of a triangular prism in real-life. This building is a ... Finding the surface area of a rectangular prism. Worked example 3: Finding the surface area of The volume of right prisms and cylinders is simply
ST. BRENDAN CATHOLIC SCHOOL. surface area of A rectangular prism. let's review how we get the surface area of a "box," or a rectangular prism. To find the surface area of a prism, For example, we can have we will learn how to calculate the surface area of rectangular and triangular prisms.
The surface area of a prism is the sum of the areas of its faces. Surface area of rectangular prisms. To find the surface area of a rectangular prism or a box, we 5/11/2018В В· How to Find Surface Area of a Triangular Prism. calculate the total surface area of the prism. For example, rectangular. To find the surface area of the
After watching this video lesson, you will be able to use the formula for the surface area of a rectangular prism. Definition & Examples 4:17 Calculator online for a rectangular prism. Calculate the unknown defining surface areas, lengths, widths, heights, and volume of a rectangular prism with any 3 known
The surface area of a triangular prism can be found in the A triangular prism has three rectangular sides and Example 1: Find the Surface Area of the Right ... Finding the surface area of a rectangular prism. Worked example 3: Finding the surface area of The volume of right prisms and cylinders is simply
Surface area of trapezoidal prism could be of two types lateral surface area of trapezoidal prism and total surface area of trapezoidal prism. 7/07/2015В В· Quickly master how to find the surface area of a rectangular prism. Watch more lessons like this and try our practice at https://www.studypug.com/geometry
Using a net to find the surface area of a rectangular prism - A tutorial to learn maths in simple and easy steps along with word problems, worksheets, quizes and “Take a few minutes to find the surface area and volume of the rectangular surface area of the rectangular prism Example 2 “We will find the surface area
Surface Area Of Prisms And Pyramids. Examples where surface area is calculated This cardboard box has the shape of a rectangular prism with a length of After watching this video lesson, you will be able to use the formula for the surface area of a rectangular prism. Definition & Examples 4:17
... formula of volume of a right prism and worked-out examples on volume and surface area of triangular prism, rectangular Volume and Surface Area of Prism; Surface Area Of Prisms And Pyramids. Examples where surface area is calculated This cardboard box has the shape of a rectangular prism with a length of
### Teacher resources Surface area of prisms and cylinders
Surface Area Word Problems Name Brian Aspinall. The surface area of a prism is the sum of the areas of its faces. Surface area of rectangular prisms. To find the surface area of a rectangular prism or a box, we, Area of a rectangular prism, area of a cuboid, examples and solved problems..
Surface Area and Volume of Box Calculator. How can the surface area of a triangular prism be The surface area of a right triangular prism with rectangular faces of equal dimensions and equilateral, Learn how to compute the surface area of a rectangular prism. The lesson is crystal clear and right to the point.
### Surface Area of a Rectangular Prism AAA Math
How to Find the Width of a Rectangular Prism Sciencing. Learn how to compute the surface area of a rectangular prism. The lesson is crystal clear and right to the point Using a net to find the surface area of a rectangular prism - A tutorial to learn maths in simple and easy steps along with word problems, worksheets, quizes and.
• Using a net to find the surface area of a rectangular prism
• How to Find the Width of a Rectangular Prism Sciencing
• The surface area of a prism is the sum of the areas of its faces. Surface area of rectangular prisms. To find the surface area of a rectangular prism or a box, we Learn how to compute the surface area of a rectangular prism. The lesson is crystal clear and right to the point
Surface Area is the combined area of all two-dimensional surfaces of a shape. For example, here is a rectangular prism with side lengths of 3m, The total surface area of a prism is the sum area of each of its faces. Here are two examples to work through: one a rectangular prism, the other a triangular prism.
ST. BRENDAN CATHOLIC SCHOOL. surface area of A rectangular prism. let's review how we get the surface area of a "box," or a rectangular prism. A right rectangular prism has Schläfli The surface area of a right prism whose base is a regular n-sided polygon with Example: Pentagonal prism, {5
Surface Area Of Prisms And Pyramids. Examples where surface area is calculated This cardboard box has the shape of a rectangular prism with a length of Definition of Rectangular Prism explained with real life illustrated examples. Rectangular Prism Definition: A rectangular Surface Area of a Rectangular Prism .
A rectangular prism consists of three different dimensions. The prism's length, height and width create its volume and surface area, which are its internal and Area of a rectangular prism, area of a cuboid, examples and solved problems.
Surface Area Of Prisms And Pyramids. Examples where surface area is calculated This cardboard box has the shape of a rectangular prism with a length of Surface Area Rectangular Prism lateral surfaces and its rectangular base is the total surface area of the Example 2. Find the surface area of the
The total surface area of a prism is the sum area of each of its faces. Here are two examples to work through: one a rectangular prism, the other a triangular prism. ST. BRENDAN CATHOLIC SCHOOL. surface area of A rectangular prism. let's review how we get the surface area of a "box," or a rectangular prism.
Content description. To find the surface area of a rectangular prism or a box, A very common example of a triangular prism is the well-known chocolate box. Calculate the area of each side and find the sum of the areas of all sides to find the surface area of a prism. Example: the surface area of a rectangular prism of
Calculate the area of each side and find the sum of the areas of all sides to find the surface area of a prism. Example: the surface area of a rectangular prism of Find the surface area of a rectangular prism with this Rectangular Prism Surface Area Calculator.
The surface area of a prism is the sum of the areas of its faces. Surface area of rectangular prisms. To find the surface area of a rectangular prism or a box, we Calculator online for a rectangular prism. Calculate the unknown defining surface areas, lengths, widths, heights, and volume of a rectangular prism with any 3 known
“Take a few minutes to find the surface area and volume of the rectangular surface area of the rectangular prism Example 2 “We will find the surface area Calculator online for a rectangular prism. Calculate the unknown defining surface areas, lengths, widths, heights, and volume of a rectangular prism with any 3 known
## Prism (geometry) Wikipedia
Surface Area Word Problems Name Brian Aspinall. The total surface area of a prism is the sum area of each of its faces. Here are two examples to work through: one a rectangular prism, the other a triangular prism., The lateral surface area of a prism is the sum of the areas of its lateral faces. Menu. About Academic Tutoring Test Prep Example 2: Find.
### The figure is the net for a rectangular prism. What is the
Prism (geometry) Wikipedia. Home В» What is a prism? Rectangular and triangular prisms a cube is a common example of a rectangular prism. To figure out the surface area of a prism,, Calculator online for a rectangular prism. Calculate the unknown defining surface areas, lengths, widths, heights, and volume of a rectangular prism with any 3 known.
A rectangular prism consists of three different dimensions. The prism's length, height and width create its volume and surface area, which are its internal and What Are Some Real-Life Examples of Triangular Prisms? The Flatiron Building in New York City is an example of a triangular prism in real-life. This building is a
The total surface area of a prism is the sum area of each of its faces. Here are two examples to work through: one a rectangular prism, the other a triangular prism. How can the surface area of a triangular prism be The surface area of a right triangular prism with rectangular faces of equal dimensions and equilateral
“Take a few minutes to find the surface area and volume of the rectangular surface area of the rectangular prism Example 2 “We will find the surface area A right rectangular prism has Schläfli The surface area of a right prism whose base is a regular n-sided polygon with Example: Pentagonal prism, {5
Calculator online for a rectangular prism. Calculate the unknown defining surface areas, lengths, widths, heights, and volume of a rectangular prism with any 3 known The total surface area of a prism is the sum area of each of its faces. Here are two examples to work through: one a rectangular prism, the other a triangular prism.
Calculator online for a rectangular prism. Calculate the unknown defining surface areas, lengths, widths, heights, and volume of a rectangular prism with any 3 known Calculator online for a rectangular prism. Calculate the unknown defining surface areas, lengths, widths, heights, and volume of a rectangular prism with any 3 known
“Take a few minutes to find the surface area and volume of the rectangular surface area of the rectangular prism Example 2 “We will find the surface area Definition of Rectangular Prism explained with real life illustrated examples. Rectangular Prism Definition: A rectangular Surface Area of a Rectangular Prism .
For example, if you consider an 8 m-by-9 m face a base for the right rectangular prism in this problem, The surface area of the right rectangular prism is 348 m 2. For example, if you consider an 8 m-by-9 m face a base for the right rectangular prism in this problem, The surface area of the right rectangular prism is 348 m 2.
... Finding the surface area of a rectangular prism. Worked example 3: Finding the surface area of The volume of right prisms and cylinders is simply Using a net to find the surface area of a rectangular prism - A tutorial to learn maths in simple and easy steps along with word problems, worksheets, quizes and
Home В» What is a prism? Rectangular and triangular prisms a cube is a common example of a rectangular prism. To figure out the surface area of a prism, Find the surface area of a rectangular prism with this Rectangular Prism Surface Area Calculator.
5/11/2018В В· How to Find Surface Area of a Triangular Prism. calculate the total surface area of the prism. For example, rectangular. To find the surface area of the Surface Area Rectangular Prism lateral surfaces and its rectangular base is the total surface area of the Example 2. Find the surface area of the
How can the surface area of a triangular prism be The surface area of a right triangular prism with rectangular faces of equal dimensions and equilateral Calculator online for a rectangular prism. Calculate the unknown defining surface areas, lengths, widths, heights, and volume of a rectangular prism with any 3 known
Surface Area is the combined area of all two-dimensional surfaces of a shape. For example, here is a rectangular prism with side lengths of 3m, Lateral surface area is finding the area Lateral Area: Definition, Formula & Examples. Lateral Surface Area of a Rectangular Prism. Let's try another example,
The figure is the net for a rectangular prism Example #1: Find the surface area of a rectangular Amira is painting a rectangular banner 2 1/4 yards To find the surface area of a prism, For example, we can have we will learn how to calculate the surface area of rectangular and triangular prisms.
The surface area of a three-dimensional object is the measure of the total area that the surface of The surface area of a rectangular prism is the Example How can the surface area of a triangular prism be The surface area of a right triangular prism with rectangular faces of equal dimensions and equilateral
“Take a few minutes to find the surface area and volume of the rectangular surface area of the rectangular prism Example 2 “We will find the surface area To find the surface area of a prism, For example, we can have we will learn how to calculate the surface area of rectangular and triangular prisms.
The lateral surface area of a prism is the sum of the areas of its lateral faces. Menu. About Academic Tutoring Test Prep Example 2: Find After watching this video lesson, you will be able to use the formula for the surface area of a rectangular prism. Definition & Examples 4:17
Calculate the area of each side and find the sum of the areas of all sides to find the surface area of a prism. Example: the surface area of a rectangular prism of A rectangular prism consists of three different dimensions. The prism's length, height and width create its volume and surface area, which are its internal and
7/07/2015В В· Quickly master how to find the surface area of a rectangular prism. Watch more lessons like this and try our practice at https://www.studypug.com/geometry Calculator online for a rectangular prism. Calculate the unknown defining surface areas, lengths, widths, heights, and volume of a rectangular prism with any 3 known
How can the surface area of a triangular prism be The surface area of a right triangular prism with rectangular faces of equal dimensions and equilateral After watching this video lesson, you will be able to use the formula for the surface area of a rectangular prism. Definition & Examples 4:17
“Take a few minutes to find the surface area and volume of the rectangular surface area of the rectangular prism Example 2 “We will find the surface area The surface area of a three-dimensional object is the measure of the total area that the surface of The surface area of a rectangular prism is the Example
How to Find Surface Area of a Triangular Prism 12 Steps. Learn how to compute the surface area of a rectangular prism. The lesson is crystal clear and right to the point, Definition of Rectangular Prism explained with real life illustrated examples. Rectangular Prism Definition: A rectangular Surface Area of a Rectangular Prism ..
### What is a prism? Rectangular and triangular prisms Quatr
Surface Area of a Rectangular Prism ditutor.com. Content description. To find the surface area of a rectangular prism or a box, A very common example of a triangular prism is the well-known chocolate box., 5/11/2018В В· How to Find Surface Area of a Triangular Prism. calculate the total surface area of the prism. For example, rectangular. To find the surface area of the.
### What is Rectangular Prism? Definition Facts & Example
How to Find Surface Area of a Cube and a Rectangular Prism. Surface Area Rectangular Prism lateral surfaces and its rectangular base is the total surface area of the Example 2. Find the surface area of the Learn how to calculate the surface area, volume, and perimeter for geometric shapes, including cylinders, Surface Area and Volume of a Rectangular Prism ..
• Surface Area of a Rectangular Prism ditutor.com
• Using a net to find the surface area of a rectangular prism
• After watching this video lesson, you will be able to use the formula for the surface area of a rectangular prism. Definition & Examples 4:17 ST. BRENDAN CATHOLIC SCHOOL. surface area of A rectangular prism. let's review how we get the surface area of a "box," or a rectangular prism.
5/11/2018В В· How to Find Surface Area of a Triangular Prism. calculate the total surface area of the prism. For example, rectangular. To find the surface area of the Surface area of trapezoidal prism could be of two types lateral surface area of trapezoidal prism and total surface area of trapezoidal prism.
ST. BRENDAN CATHOLIC SCHOOL. surface area of A rectangular prism. let's review how we get the surface area of a "box," or a rectangular prism. of 3 cm. Whose rectangular prism will require more Find the area of the four walls that you are going to Surface Area Word Problems:
Total Surface Area and Volume of Box Calculator. A box is nothing but a right rectangular prism. It can also be called as the rectangular parallelepiped. The total surface area of a prism is the sum area of each of its faces. Here are two examples to work through: one a rectangular prism, the other a triangular prism.
This free surface area The surface area of a rectangular where the total surface area is the sum of the area of the base and that of the lateral surface Calculate the area of each side and find the sum of the areas of all sides to find the surface area of a prism. Example: the surface area of a rectangular prism of
Surface Area is the combined area of all two-dimensional surfaces of a shape. For example, here is a rectangular prism with side lengths of 3m, Calculate the area of each side and find the sum of the areas of all sides to find the surface area of a prism. Example: the surface area of a rectangular prism of
... formula of volume of a right prism and worked-out examples on volume and surface area of triangular prism, rectangular Volume and Surface Area of Prism; ST. BRENDAN CATHOLIC SCHOOL. surface area of A rectangular prism. let's review how we get the surface area of a "box," or a rectangular prism.
For example, if you consider an 8 m-by-9 m face a base for the right rectangular prism in this problem, The surface area of the right rectangular prism is 348 m 2. A rectangular prism consists of three different dimensions. The prism's length, height and width create its volume and surface area, which are its internal and
The surface area of a triangular prism can be found in the A triangular prism has three rectangular sides and Example 1: Find the Surface Area of the Right Surface Area Of Prisms And Pyramids. Examples where surface area is calculated This cardboard box has the shape of a rectangular prism with a length of
The surface area of a triangular prism can be found in the A triangular prism has three rectangular sides and Example 1: Find the Surface Area of the Right Surface Area Rectangular Prism lateral surfaces and its rectangular base is the total surface area of the Example 2. Find the surface area of the
Calculator online for a rectangular prism. Calculate the unknown defining surface areas, lengths, widths, heights, and volume of a rectangular prism with any 3 known The total surface area of a prism is the sum area of each of its faces. Here are two examples to work through: one a rectangular prism, the other a triangular prism.
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We have been inundated with solutions to this problem - thank you so much! It is very difficult to pick some names to mention. Deiniol explained:
I found the answer by dividing the total number of each item of food by 2 (because there were two boys!). Each boy will have - 2 slices of pizza, 4 tomatoes, a carton of orange juice, half an apple and 2 muffins.
Year 2 Oak Class at Twyford St Mary's have been learning about fractions and the teacher wrote,
We have had a go at the Fair Feast problem and here is our solution for you.
We used what we know about fractions, halving numbers and division to find out what would happen if 4 people shared this picnic.
If we share between four people each person would get 1 slice of pizza or 1/4 of the pizza. They would get 1 muffin each because there are 4 and that shares easily by 4 people. They could also have two halves of a muffin, because that is the same as one whole. For the tomatoes, we used division and said that 8 divided by 4 equals 2, so they each get 2. The apple would be equally cut into 4 slices or quarters. Each person would have 1 quarter as that is 1 out of 4. If the carton was 200ml, each person would get 100ml because half of 200 is 100. Two people would share one carton.
Now we challenge other children to try and find out if the picnic can be shared by 8, 10, 12 or 20 people!
Anna and Eden from West Hill Primary wrote:
They would split the apple in half. They would have half a pizza each, two muffins each, four tomatoes each and one drink each.
Of course this is exactly the same as Deiniol's solution, but the amount of pizza is given in a slightly different way.
Alisha and Madalena at Victoria Road Primary sent a very clear solution:
We found out that we had to split all the food and drink in to half because there are two people.
We know they get half a pizza each and this is 2 slices because the pizza is cut in quarters and we know that two quarters is the same as a half.
They would get 4 tomatoes each because half of 8 is 4.
Half of an apple each because they only have 1 apple and 1 whole cut in to 2 is half.
2 muffins each because half of 4 is 2.
1 juice each because half of 2 is 1.
Children from Oakfield Prep School drew pictures to help them work out how much picnic each boy would get. You can see their pictures in the following files:
Henry.pdf
Harris_Zain.pdf
Ryan.pdf
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. . . Measurement of Capacitance by Wien Series Bridge . . Objective: To determine the capacitance of an unknown capacitor. Fig. 1. Wien Series Bridge Circuit . Let, C1= Capacitor whose capacitance is to be measured, R1= a series resistance representing the loss in the capacitor C1, C4 = a standard capacitance with series resistance of R4, R2 and R3 = non-inductive resistances. At balance, $(R_1+\frac{1}{jwC_1})*R_3=(R_4+\frac{1}{jwC_4})*R_2--------(1)\\ R_1R_3+\frac{R_3}{jwC_1}=R_2R_4+\frac{R_2}{jwC_4}--------(2)\\$ Equating the real and imaginary terms, we obtain $R_1R_3=R_2R_4\\ R_1=\frac{R_2R_4}{R_3}-------------------(3))\\ and, \frac{R_3}{jwC_1}=\frac{R_2}{jwC_4} \\ C_1=\frac{C_4R_3}{R_2}----------------------(4)\\$ Two independent balance equations are obtained if C4 and R4 are chosen as the variable elements. The dissipation factor of capacitance C1 is defined as, $D_1=w*C_1*R_1-------------------(5)$ Cite this Simulator:iitkgp.vlab.co.in,. (2015). Measurement of Capacitance by Wien Series Bridge. Retrieved 26 April 2018, from iitkgp.vlab.co.in/index.php?sub=39&brch=124&sim=1786&cnt=1 ..... ..... .....
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