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https://codermetrics.org/page/3/
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# Codermetrics.org
Sharing information on codermetrics
## Who Is +1? Coder VAR
leave a comment »
If you follow baseball statistics or read baseball sportswriters like those at ESPN.com, over the last few years you may have heard about VORP or WAR. These intense-sounding acronyms stand for Value Over Replacement Player (VORP) and Wins Above Replacement (WAR). While the method of how these are calculated for baseball players is somewhat arcane (and in fact there are at least 3 different methods that people use to calculate WAR) the concept itself is simple and persuasive, which has led to the increased popularity in these statistics.
The general concept behind VORP and WAR is that a good way to identify the value of a player is to rate them relative to an average replacement player. For example, let’s say you want to rate the 3rd baseman for your favorite baseball team. You could look at his offensive and defensive statistics, and compare the details to 3rd basemen on other teams. But VORP and WAR attempt to create a single number that rates your 3rd baseman against an “average” replacement 3rd baseman. An above-average VORP/WAR means that your player contributes more than an average replacement.
This is based on common statistical techniques (various comparisons to average). A similar idea can be useful as a way to categorize coders and their skills. I call this Value Above Replacement, or VAR. Unlike baseball’s VORP or WAR, which seek to create a single metric to rate baseball players, the concept of VAR can be applied to any metric. The formula is based on using standard deviation, as follows:
For any given metric X –
1. Calculate X for every coder
2. Calculate the average (mean) of X across all coders
3. Calculate the population standard deviation of X across for all coders
4. Calculate VAR X for each coder as VAR X = ((X of coder) – (average X)) / (standard deviation X) and then truncate
For any given metric, this shows you how many “standard deviations” each coder is from the average. If you are familiar with normal distributions or bell curves, this applies the same concept. Someone who is in the top 3% for a given area (measured by metric X) might be a +2 for example, meaning that the person is 2 standard deviations above average (and, conversely, someone who is in the bottom 3% might be a -2).
Those of you whose teachers “graded on the curve” in school might be recoiling at the thought of applying this to coders. But as I’ve mentioned in other writing on metrics, this isn’t a grading system. Metrics are best used as a categorization system useful to help you more objectively identify or confirm the strengths and weaknesses of individuals and teams. In this regard, VAR can be extremely helpful as a way to focus on the most meaningful data, namely the distribution and categorization of contributions and skills.
For example, let’s say you have 7 coders on a software team, and you measure their productivity by looking at the number of tasks each coder completes and the complexity of each task (this metric is called Points in Codermetrics). For a one month period, you might have data like the following:
• Coder A Productivity = 24 tasks completed x average task complexity 2 = 48
• Coder B Productivity = 20 tasks completed x average task complexity 2 = 40
• Coder C Productivity = 26 tasks completed x average task complexity 1 = 26
• Coder D Productivity = 38 tasks completed x average task complexity 1 = 38
• Coder E Productivity = 17 tasks completed x average task complexity 3 = 51
• Coder F Productivity = 22 tasks completed x average task complexity 2 = 44
• Coder G Productivity = 15 tasks completed x average task complexity 3 = 45
For this group, then, the average productivity for the month (rounded) is 42, and the standard deviation is 8. With these values, you can calculate the Productivity VAR for each coder. I’ve created an example Google Docs spreadsheet which you can access here and it has also been posted in Shared Resources. Below is a screenshot showing the calculated values for this group.
This provides a good example of how Coder VAR helps. If you just look at the Productivity metrics, as highlighted in the pie chart, it isn’t that easy to see how the values are grouped, and which values stand out as separate from the others. With Productivity VAR, you can easily see that there are three groups, one high (Coder E at +1), one low (Coder C at -2), and one in the middle (everyone else at 0).
In studying codermetrics for your software team, this is often the kind of information that can be extremely useful. How many (if any) coders are above-average or below-average in a specific area? What specific areas of strength might be lost if someone left and was replaced by an average coder? Do areas of strength or weakness correlate with coders’ level of experience, and what areas of weakness might be improved?
Coder VAR also provides a useful way to discuss and convey key findings from your metrics. This is part of what has driven the popularity of WAR in baseball. It’s easier to understand if you say “Coder E is plus one for Productivity” or “Coder B’s Productivity is average,” than if you say “Coder E’s Productivity was fifty-one” or “Coder B’s Productivity is forty.” Or if you are looking to hire a new coder, it might be useful to know and discuss that you are looking for “a coder whose Productivity is plus one.”
There are some limitations to be aware of when using VAR. For example, VAR draws a line that separates values that may in fact be close, such as Coder A (Productivity 48 and Producivity VAR 0) and Coder E (Productivity 51 and Productivity VAR 1) in the data above. This points to the usefulness of VAR, which is as a method of categorization, not a method of detailed analysis and certainly not a method of grading. Also, you should be careful about VAR analysis of coders who are known to have different levels of experience or who have very different roles on your team. Coder VAR is best used as an analysis of coders who are generally similar in experience and roles, and therefore you might want to analyze VAR separately for your senior and junior coders, for example.
The biggest limitation of Coder VAR is that it is clearly relative to your population, so it is limited if your population (the number of coders analyzed) is small and isolated. For example, if you have a team with three senior coders, then it can be somewhat useful to look at their Productivity VAR. But it would also be helpful to know how the coders’ Productivity compares to other senior coders, either on other teams in your organization, or in other organizations. Maybe your senior coders are all similarly productive (Productivity VAR of 0 within the team) but maybe they are highly productive compared to other senior coders (Productivity VAR +1 or +2 when compared across teams). This is a general problem with codermetrics, namely the lack of normalized data we all have and share, and something I hope to address more in the future. For now, those in larger organizations would be able to address this by measuring across teams and applying techniques to establish normalized baselines (something also discussed in my book).
As with other metrics, however, if you are aware of the limitations then Coder VAR can still be very useful. It can help you to increase your understanding of team dynamics, to identify and analyze the characteristics of successful teams, and to plan ways to improve your software team.
Advertisement
Written by Jonathan Alexander
December 4, 2011 at 10:07 am
Posted in New Ideas
## New Article on O’Reilly Radar
leave a comment »
O’Reilly Radar published a new article today Moneyball for Software Engineering. This is a quick introduction to concepts about how Moneyball-like metrics can be applied to software development teams.
Written by Jonathan Alexander
October 10, 2011 at 1:49 pm
Posted in Publications
## Upcoming Event: O’Reilly Webcast- Moneyball for Software Teams?
leave a comment »
Time: Sept 16, 2011 10am PT Location: Online Cost: Free
The new movie Moneyball starring Brad Pitt will be released in the US on Sept 23rd. Based on the bestselling book of the same name, the movie explores the use of “sabermetrics” to build winning baseball teams. In this webcast, Jonathan Alexander, author of Codermetrics, suggests that these same ideas can be applied to software teams. Jonathan will discuss how you can apply similar ideas to improve your software teams, giving examples of specific metrics and techniques to help make the team a more cohesive and productive unit.
For more information and free registration click here.
Written by Jonathan Alexander
September 3, 2011 at 9:00 am
Posted in Upcoming Events
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# Physics for Engineers | Kinematics
Kinematics: The Geometry of Motion
Kinematics is a branch of mechanics that describes the motion of points, bodies (objects), and systems of bodies without considering the forces that cause them to move. It focuses on the geometric aspects of motion.
1. Displacement, Velocity, and Acceleration
• Displacement is a vector quantity that represents the change in position of an object. It has both magnitude and direction.
• Velocity is also a vector quantity that describes the rate of change of displacement. It tells us how fast an object is moving and in which direction.
• Acceleration is the rate of change of velocity over time. Like velocity and displacement, acceleration is a vector quantity.
2. Equations of Motion
The equations of motion, also known as the kinematic equations, describe the relationship between displacement, velocity, acceleration, and time under uniform acceleration. They are:
1. v = u + at
2. s = ut + 1/2 at^2
3. v^2 = u^2 + 2as
where:
• u is the initial velocity,
• v is the final velocity,
• a is the acceleration,
• s is the displacement,
• t is the time.
3. Projectile Motion
Projectile motion is a form of motion experienced by an object or particle that is thrown near the Earth’s surface and moves along a curved path under the action of gravity only. The key points about projectile motion are:
• The horizontal and vertical motions are independent of each other.
• The only acceleration acting on the projectile is the acceleration due to gravity, which acts downward.
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1. ## Integration by substitution
$
\int \frac{2x-1}{(4x^2-4x)^{2}} dx\
$
In this case what would u be? that the integral would be
$
\frac{1}{16} \, \int u^{-2} du
$
At first glance i thought it would just be
$
u=4x^2-4x
$
but obviously thats not it.
2. Originally Posted by simsima_1
$
\int \frac{2x-1}{(4x^2-4x)^{2}} dx\
$
In this case what would u be? that the integral would be
.
$(4x^2-4x)^2 = 16(x^2 - x)$.
Let $u=x^2-x$.
3. Originally Posted by ThePerfectHacker
$(4x^2-4x)^2 = 16(x^2 - x)$.
Let $u=x^2-x$.
....dumb question ... but how did u get that? the 16(x^2-x)??
4. Originally Posted by simsima_1
....dumb question ... but how did u get that? the 16(x^2-x)??
$(4x^2-4x)^2 = (4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2$
EDIT: Mistake fixed.
5. $(4x^2-4x)^2=(4(x^2-x))^2$
if $x^2-x = u$ then $(2x-1)dx=du$ so you have
$\int \frac{du}{(4u)^2}\ = \frac{1}{16} \int \frac{1}{u^2} du\ = \frac{1}{16} \int u^{-2} du\ = -\frac {1}{16u} + C = -\frac{1}{16(x^2-x)} + C$
The way he wrote it made it seem that $(4x^2-4x)^2=16(x^2-x)$ <-- NOT TRUE
Below are the correct calulations
$(4x^2-4x)^2=(4(x^2-x))^2=16(x^2-x)^2$ we used that middle one for our substitution $(4u)^2 = 16u^2$
Edit : Mistake fixed
6. Originally Posted by ThePerfectHacker
$(4x^2-4x)^2 = (4(x^2-x))^2 = 4^2(x^2-x) = 16(x^2-x)$
Isn't it $(4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2$ ?
Originally Posted by Lopared
Below are the correct calulations
$(4x^2-4x)^2=(4(x^2-x))^2=16(x^4-x^2)$
$(x^2 - x)^2 \neq x^4 - x^2$.
7. Originally Posted by wingless
Isn't it $(4(x^2-x))^2 = 4^2(x^2-x)^2 = 16(x^2-x)^2$ ?
$(x^2 - x)^2 \neq x^4 - x^2$.
yea your right. my mistake (fixed now)
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Vous êtes sur la page 1sur 16
# Considering the
entire area
By offsets to the
baes line
Based on field
measurements
By latitudes
&departures
Area
Based on
measurements
scaled from a
map
By coordinates
Instrumental
method
Usually by a
planimeter
## Based on field measurements considering the entire area
In this method, the area is divided into number of geometrical figures such as triangles,
rectangles, squares and trapeziums and then the area can be calculated according to one of
following method. Computing the area can be achieved in two steps.
1 st step
Area of a triangle:
Area= s ( sa ) ( sb )( sc )
Where a, b and c are sides and s = (a + b + c) / 2
Area of rectangle
Area=a b
Where a and b are sides.
Area of square
Area=a2
Where a is the side of the square.
Area of the trapezium
1
Area= ( a+b ) d
2
Where a and b are the parallel sides and d is the perpendicular distance between them.
2 n d step
The area along the boundaries is calculated as follows.
X1
X22
01 , 02=ordinates
x 1 , x 2=chainages
01+ 02
( x 2x 1 )
2
## Areas from offsets to a baseline offsets at regular intervals
This method is suitable for long narrow strips of land. The offsets are measured from the
boundary to the base line or a survey line at regular intervals. This method can also be
applied to a plotted plan from which the offsets to a line can be scaled off. There are 4
methods in order to calculate the area.
Mid-ordinate rule
Average-ordinate rule
Trapezoidal rule
Simpsons rule
Mid-ordinate rule
In order to apply this method, we assume that the boundaries between the extremities of the
ordinates are straight lines.
The base line is divided into a number of divisions and the ordinates are measured at the mid
points of each division.
Area=common distance*sum of mid ordinates
Mid ordinate=o1+o2/2=h1
Area=d (h1+h2+)
Average-ordinate rule
In this method also we assume that the boundaries between the extremities of the ordinates
are straight lines. The offsets are measured to each of the points of the divisions of the base
line.
Area=sum of ordinate/no.of ordinates *length of base line
Area=o1+o2++on/o (n+1)*l
Trapezoidal rule
This rule is based on the assumption that the figures are trapeziums. This rule is more
accurate than the previous two rules which are approximate versions of this rule. This rule
can be applied for any number of ordinates.
Area enclosed by one trapezium=o0+o1/2*d
Total area is the sum of each separate trapezium.
Total area= o0+o1/2*d +
Total area=d/2*(o1+2o2++2on-1+on)
Total area=common distance/2*(1st ordinate + last ordinate+2[sum of other ordinate]
Add the average of the end offsets to the sum of the intermediate offsets. Multiply the total
sum thus obtained by the common distance between the ordinates to get the required area.
Simpsons rule
In this rule, we assume that the short lengths of boundary between the ordinates are parabolic
arcs. This method is more useful when the boundary line departs considerably from the
straight line.
The area is equal to the sum of the two end ordinates plus four times the sum of the even
intermediate ordinates plus twice the sum of the odd intermediate ordinates, the whole
multiplied by one-third the common interval between them.
Even though this method gives more accurate results out of other three methods, this is only
applicable when the number of divisions is even.
## Areas from offsets to a baseline offsets at irregular
intervals
There are two methods to find the area when the ordinates distances are irregular.
1st method
In this method, the area of each trapezoid is calculated separately and then added together to
compute the total area.
2nd method
Area by coordinates
## Latitudes and departures double meridian distances
This method is one of the most frequently used for computing the area of a closed traverse.
In order to calculate the area by this method, the latitudes and departures of each line of the
traverse are calculated. Then the traverse is balanced.
A reference meridian is then assumed to pass through the most westerly station of the traverse
& the double meridian distances of the lines are computed.
Meridian distances
The meridian distance of any point in a traverse is the distance of that point to the reference
meridian, measured at right angles to the meridian.
The meridian distance of any line is equal to the meridian distance of the preceding line plus
half the departure of the preceding line plus half the departure of the line itself.
## Area by latitudes and meridian distances
East-west lines drawn from each station to the reference meridian, thus getting triangles and
trapeziums.
One side of each triangle or trapezium will be one of the lines, the base of the triangle or
trapezium will be the latitude of the line, and the height of the triangle or trapezium will be
the meridian distance of that line.
Area of triangle or trapezium=latitude of the line*meridian distance of the line
## Double meridian distance
The double meridian distance of a line is equal to the sum of the meridian distances of the
two lines.
The D.M.D. of any line is equal to the D.M.D. of the preceding line plus the departure of the
preceding line plus the departure of the line itself.
Area by latitudes and double meridian distances
A=area of dDcC+area of CcbB-area of dDA-area of ABb
Area from departures and total latitudes
## Area by double parallel distances and departures
A parallel distance of any line of a traverse is the perpendicular distance from the middle
point of that line to a reference line (chosen to pass through most southerly station) at right
angles to the meridian.
The DPD of any line is the sum of the parallel distances of its ends.
The principle of finding area by D.M.D method & D.P.D. method are identical.
Area by coordinates
This method can be applied when the offset intervals are irregular.
The procedure for this method is that from the given distances & offsets, a point is selected as
the origin.
Then the coordinated of all other points are arranged with reference to the origin.
Instrumental method
The instrument used for computation of area from a plotted map is the planimeter. The area
obtained from this instrument is more accurate than other graphical methods.
The two types of planimeter are:
Amsler polar planimeter
Roller planimeter
Procedure of finding the area with a planimeter
The Vernier of the index mark is set to the exact graduation marked on the tracer arm
corresponding to the scale as obtained from the table.
The anchor point is fixed firmly in the paper outside or inside the figure. It should be ensured
that the tracing point is easily able to reach every point on the boundary line.
A good starting point is marked on the boundary line.
By observing the disc, wheel and Vernier the initial reading (IR) is recorded.
The tracing point is moved gently in a clockwise direction along the boundary of the area.
The number of times the zero mark of the dial passes the index mark in a clockwise or
anticlockwise direction should be observed.
Finally, observing the disc, wheel and Vernier the final reading (FR) is recorded.
Then, the area of the figure may be obtained from the following expression.
Are=M (FR-IR+- 10N +C)
M=multiplier given in the table
N=number of times the 0 mark of the dial passes the index mark
C=constant given in the table
Level
section
Two-level
section
From
cross
sections
Side hill
two-level
section
Three level
section
Volume
From spot
levels
Multi-level
section
By cross
sections
From
contours
By equal
depth
contours
By
horizontal
planes
## Volume calculation from cross sections
This is the most widely used method.
The total volume is divided into a series of solids by the planes of cross sections.
Cross-sections are established at some convenient intervals along a center line of the works.
Volumes are calculated by relating the cross-sectional areas to the distances between them.
In order to compute the volume it is first necessary to evaluate the cross-sectional areas,
which may be obtained by the following methods:
By calculating from the formula or from first principles the standard cross-sections of
constant formation widths and side slopes.
By measuring graphically from plotted cross-sections drawn to scale, areas being obtained by
planimeter or division into triangles or square.
The various cross sections may be categorized as,
Level section
Two level sections
Side hill two level sections
## Three level sections
Multi- level section
Level section
In this case the ground is level transversely.
## Depth of centre line or height of embankment = h
Formation width = b
Side width = w
Area = h(b + mh)
## Area = 1/2m [(b/2 + mh)(w1 + w2) b2/2]
Side hill two level sections
In this case, the ground slope crosses the formation level so that one portion of the area is in
cutting and the other in filling.
## Area of fill = [(b/2 + kh)2/(k-m)]
Area of cut = [(b/2 - kh)2/(k-n)]
## Area = m [(w1+w2) (mh+b/2) b2/2)
The volumes of the prismoids between successive cross-sections are obtained either by
trapezoidal formula or by prismoidal formula.
The prismoidal formula
A prismoid is defined as a solid whose end faces lie in parallel and consist of any two
polygons, not necessarily of the same number of sides, the longitudinal faces being surface
extended between the end planes.
The longitudinal faces take the form of triangles, parallelograms or trapezium.
The total volume of the pyramid can be stated as follows.
Where h is the length of the prismoid measured perpendicular to the two end parallel planes
and.
## A1=area of cross-section of one end plane.
A2= area of cross-section of other end plane.
A=the mid area
In order to calculate the volume of earth work between a number of sections having area a1,
a2, a3, , an spaced at a constant distance h apart.
Total voume=h/3[(a1+an)+4(a2+a4an-1)+2(a3+a5an-2)]
This is also known as Simpsons rule for volume.
In order to apply this formula, it is necessary to have an odd number of cross-sections.
If there are even numbers of sections, the end strip must be calculated separately, and the
volume between the remaining sections may be calculated by prismoidal formula.
The trapezoidal formula (Average End Area Method)
It is no more accurate to use the prismoidal formula where the mid-sectional areas have not
been directly measured than it is to use the end areas formula, particularly as the earth solid is
not exactly represented by a prismoid.
Hence this method is based on the assumption that the mid-area is the mean of the end areas.
In that case the volume of the above prismoid is given as,
v = h [ ( A1 + A2) / 2 ]
but this is true only if the prismoid is composed of prisms and wedged only and not of
pyramids.
In some cases the volume is calculated and then a correction is applied the correcton being
equal to the difference between the volume as calculated and that which could be obtained by
the use of the prismoidal formula. the correction is known as the prismoidal correction.
In order to calculate the volume of earth work between a number of sections having area a1,
a2, a3, , an spaced at a constant distance h apart.
V = d{[( A1 + A2) / 2 ] + A2 + A3 + An-1}
level section
prismoidal
correct
Cp=
two
sections
level sections
sections
level
Ds
h1h2 )2
(
6
ion
## Volume from spot levels
In this method, the field work consists in dividing the area into a number fo squares,
rectangles or triangles and measuring the levels of their corners before and after the
construction.
Thus, the depth of excavation or height of filling at every corner is known.
## V = Area of the figure average depth.
Therefore for a rectangle with corner depth ha, hb, hc and hd,
## Volume of a group of rectangles or squares having the same area
h1 = some of depths used once
h2 = sum of depths used twice
h3 = sum of depths used thrice
h4 = sum of depths used four times.
A=horizontal area of the cross-section of the prism
## Volume from contour plan
Contour lines may be used for volume calculations and theoretically this is the most accurate
method.
However, as the small contour interval necessary for accurate work is seldom provided due to
cost, high accuracy is not often obtained.
Unless the contour interval is less than 1m or 2m at the most, the assumption that there is an
even slope between the contours is incorrect and volume calculation from contours become
unreliable.
There are four distinct methods depending upon the type of work.
By cross-sections
On the same cross-section used to draw the ground surface the grade line of the proposed
work can be drawn and the area of the section can be estimated either by ordinary methods or
with the help of a planimeter.
The area of the cut and fill can be found from the cross-section.
The volumes of earth work between adjacent cross-sections may be calculated by the use of
average end areas.
## By equal depth contours
In this method, the contours of the finished or graded surface are drawn on the contour map at
the same interval as that of contours.
At every point, where the contours of the finished surface intersect a contour of the existing
surface the cut or fill can be found by simply subtracting the difference elevation between the
two contours.
By joining the points of equal cut or fill, a set of lines is obtained.
These lines are the horizontal projections of the lines cut from the existing surface by planes
parallel to the finished surface.
The volume between any two successive areas is determined by multiplying the average of
the two areas by the depth between them, or by the prismoidal formula.
H=contour interval
V-total volume
V=sigmah/2 (a1+A2) by trapezoidal
V=Sigma h/3(A1+4A2+A3) by prismoidal formula
By horizontal planes
This method consists in determining the volumes of earth to be moved between the horizontal
planes marked by successive contours.
Chainag
e along
the
survey
line
Ordinate
(m)
0.0
20.0
40.0
60.0
80.0
100.0
130.0
160.0
190.0
8.4
9.5
7.34
6.23
7.89
7.3
9.81
6.65
4.5
According to the given data above we can plot a graph to determine the enclosed area of the
given survey line.
12
10
9.81
9.5
8.4
8
7.89
7.34
7.3
6.63
6.23
4.5
4
2
0
0
20
40
60
80
100
120
140
160
180
200
In this particular survey line there are two distinct Chainage intervals of 20m and 30m.
Therefore when calculating the area it has to be separated into two parts.
The method used to calculate the area is the trapezoidal rule since it is more accurate than the
mid-ordinate rule and average-ordinate rule and Simpsons rule cannot be applied because the
number of divisions is not even in the plotted plan.
Y-Values
12
10
8
Axis Title
Y-Values
6
4
2
0
20
40
60
80
Axis Title
100
130
160
190
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# Why there's a whirl when you drain the bathtub?
At first I thought it's because of Coriolis, but then someone told me that at the bathtub scale that's not the predominant force in this phenomenon.
• Related (perhaps duplicate even): physics.stackexchange.com/q/32 – Marek Mar 29 '11 at 9:23
• @Marek: partial duplicate. I was expecting a "definitive" answer as to what DOES cause the whirl. We can agree Coriolis is out... – Dan Mar 29 '11 at 9:44
The whirl is due to the net angular momentum the water has before it starts draining, which is pretty much random.
If the circulation were due to Coriolis forces, the water would always drain in the same direction, but I did the experiment with my sink just now and observed the water to spin different directions on different trials.
The Coriolis force is proportional to the velocity of the water and the angular velocity of Earth. Earth's angular velocity is $2\pi/24\ {\rm hours}$, or about $10^{-4}\ s^{-1}$. If water's velocity as it drains is $v$ the Coriolis acceleration is about $10^{-4} v\ s^{-1}$.
The water moves about a meter while draining, which takes a time $1\ m/v$, so the total velocity imparted by Coriolis forces could be at most $10^{-4} v\ s^{-1} * 1\ m/v = 10^{-4} \ m/s$.
So the Coriolis effect is quite a small effect. But this first-order Coriolis effect does not cause the water to rotate.
The direction of Coriolis force depends on your direction of motion. All the water in your tub is moving the same direction, so the Coriolis force pushes it all the same direction. The effect is that if the bathtub starts out perfectly flat and begins draining (and it points north), all the water will get pushed east. The two edges of the tub will have very slightly different depths of water, because the Coriolis force is pushing sideways.
The Coriolis force could create "spinning" on uniformly-moving water, but only as a second-order effect. As you move away from the equator, the Coriolis force changes. This change in the Coriolis force is because the angle between "north" and the angular velocity vector of Earth changes as you move around; as you go further north (in the Northern Hemisphere) the "north" direction gets closer and closer to making a right angle with the angular velocity vector, so the Coriolis force increases in strength. The size of this effect would be proportional to the ratio of the size of your tub to the radius of Earth. That ratio is $10^{-7}$, so this effect is completely negligible.
The Coriolis force could also create some "spinning" if different parts of the water are moving different speeds. If the tub is draining to the north in the northern hemisphere, and water near the drain is moving faster than water far away, then the water near the drain would be pushed east more than water far away is. If you subtracted out the average effect of the Coriolis force, what remained would be an easterly push near the drain and a westerly push far away. This gives a clockwise spin as viewed from above.
We've already estimated the typical velocities as $\omega L$, so the angular momentum per unit mass induced this way would be on the order of $\omega L^2$ (but maybe smaller by a factor of 10). That's only $10^{-4}\ m^2/s$. To get an equivalent effect, in a tub of $100\ L$, you could give just one liter of water on the edge of the pool a velocity of a few cm/s, something you surely do many time over when removing your body from the tub.
This effect is too small to affect your bathtub, but it's still observable under the right conditions. According to Wikipedia, Otto Tumlirz conducted several experiments in the early 20th century that demonstrated the effects of the Coriolis forces on a draining tub of water. The tub was allowed to settle for 24 hours in a controlled environment before the experiment began. This was enough to damp out the residual angular momentum left over from filling the tub up to the point where Coriolis effects were dominant.
• If I keep the water still for a very long time, so that it has very little angular momentum to begin with, would the water swirl? – Bernhard Heijstek Apr 26 '11 at 17:11
• @phycker According to the reports of experiments, yes, eventually you would see a swirl. – Mark Eichenlaub Apr 26 '11 at 18:47
• Coriolis effect is not limited to that caused by the earth's rotation. It is just another consequence of conservation of angular momentum, regardless of the scale. – Mike Dunlavey May 28 '12 at 3:25
• >The whirl is due to the net angular momentum the water has before it starts draining< This can't be true, as I explain in this question: physics.stackexchange.com/q/365171 – Adrian May Oct 26 '17 at 14:16
• I don't see why the speed of the vortex should be proportional to the initial angular momentum in the water, and you don't seem to have supplied any argument as to why this should be so. – Mark Eichenlaub Oct 26 '17 at 15:30
Since you want to explain it to your daughter, take a plastic bottle, cut the bottom open, turn it upside town, hold the top closed and fill it with water. Give her that bottle and have her release the top (which is on the bottom now, sorry for the bad phrasing). The water will whirl in different orientations whenever you repeat this (if it whirls at all) and she can influence it by accelerating the bottle in a circular motion to understand that an initial disturbance is responsible for the whirl orientation.
• I am not sure if it is an honest answer. What is going to happen during the release is that the potential energy of the water will be moved to two componentes: the aceleration of the mass center, and the internal energy. The later will appear as a whirl around the mass center. – arivero Mar 31 '11 at 20:54
• @arivero: one could of course use a spoon to introduce a small turbulence instead, I was just focusing on using as little resources as possible... – Tobias Kienzler Apr 1 '11 at 8:27
• Tobias, my point is that even if you dont whirl by hand, it will appear. If it doesn't, it should mean that all the energy is going to energy of the mass center, which is unlikely. Of course I agree thatyou can influence it. – arivero Apr 2 '11 at 19:15
The whirl happens in the draining tube, whose optimal solution to drain the bathtub is a laminar flow allowing for some rotation in the tube. What you see in the surface is the match between the solution of flow in the tube and the solution of flow in the surface.
Angular momentum of the flow gets modified a lot as the tube twists and twists, sometimes even siphoning up and down.
Consider this picture of water running from a faucet:
The stream narrows as it falls. The shape of the stream has a definite mathematical shape, which follows from (1) conservation of energy, (2) conservation of mass, and (3) the approximation that the horizontal velocity of the water makes a negligible contribution to its kinetic energy. By conservation of energy, the water lower down is moving faster. If the stream didn't become narrower, then more water would be coming out at the bottom than went in at the top.
A draining bathtub is somewhat different because the water flows out at a rate that is approximately determined by the diameter of the drain, and there is also an open surface at the top, which can develop a conical indentation when there's rotation.
The open surface complicates things, so let's start by considering the case where there is no open surface. Then I think the same considerations apply, at least qualitatively, as in the case of the stream of water from the faucet. The bathtub doesn't have a shape that constricts strongly at the bottom. It has a nearly constant horizontal cross-sectional area. Presumably there is some very special shape for a bathtub that would allow the water to flow freely downward while satisfying constraints 1-3. For any other shape, the water has to violate #3 by rotating rapidly. This would suggest that the vortex forms spontaneously in all cases, and that it has a fixed speed of rotation regardless of the initial conditions. Feynman claims (Feynman 1964, section 40-4) that the flow is irrotational (in the sense that $\nabla\times\mathbf{v}=0$), so the speed of the rotation is proportional to $1/r$ (the same as the magnetic field around a wire). A possible loophole in all this is that according to the experiments and calculations in Andersen 2003, there is a layer of water near the drain hole that goes up and then comes back down. This allows us to have more kinetic energy for a fixed rate of flow.
In the case where there's an open surface, this gets even more complicated. Here we get an indentation, whose shape is shown by Feynman to be $z\propto 1/r^2$. The depth of this indentation is a variable that classifies different solutions. These different solutions have different rates of rotation, as predicted by Andersen. I don't know whether there is a solution for zero rotation. As discussed in the answers to another question, sufficiently careful experiments are able to see the tiny Coriolis effect, which suggests that there is a zero-rotation solution, but it's unstable.
In this experiment, they used two holes and found that only one whirlpool formed. This would seem consistent with my analysis in terms of conservation laws, since conservation laws are additive; only the total has to be conserved.
Andersen 2003 gives a complete mathematical analysis and comparison with experiment in the case where a cylindrical container with a hole at the bottom is intentionally rotated.
Andersen, Phys Rev Lett 91 (2003) 104502-1, http://sites.apam.columbia.edu/courses/apph4200x/prl-bathtub-vortex-2003.pdf
Feynman, The Feynman Lectures
• That experiments can see the Coriolis effect doesn't imply there's a zero-rotation solution. Experiments can claim to detect it with p > .05 only by using statistical tests on the results of many (at least 5) repeated experiments. – Phil Goetz May 8 '19 at 1:36
The main effect is angular momentum (rotational inertia) in the water set up by various movements before you start observing, such as getting out of your bath.
This results in the water level being lower near the centre of rotation than further away, setting up centripital forces which maintain the rotation. When the difference in levels is significant relative to the average water level, you notice the typical whirl effect.
There are other things happening too, including the Coriolis force.
You can think about it like this: It takes one day for the earth to perform a full rotation (about 86k seconds), on the other hand, it takes a few seconds for your sink to drain (lets say 10 seconds). So it takes 8600 times longer for the earth to do a full rotation than it takes the water to drain down the sink. It is not too hard to imagine that the earth's rotation can have no influence on the process of draining a sink.
However, if the sink was the size of lake Michigan and you were to drain it, Coriolis would play a role.
A discussion by 'The Straight Dope' website
references experimental work carried out but Ascher Schapiro in 1962, which concluded something like it all depends on the shape of container and how its stirred before being left to empty.
Here is Schapiro's paper but I feel you will need academic access via a university or library to read the full PDF:
http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=368912
• The "Straight Dope" article is a pretty muddy explanation, claiming "Coriolis forces in the Northern Hemisphere act in a counterclockwise direction." That's actually backwards. In the northern hemisphere, Coriolis forces are always making you turn to the right - clockwise. Hurricanes do turn counterclockwise, but that's because air is flowing in from all sides towards their low-pressure center. en.wikipedia.org/wiki/… – Mark Eichenlaub Mar 29 '11 at 10:57
• ""In the northern hemisphere, Coriolis forces are always making you turn to the right - clockwise."" Yes, to the right, but not clockwise. The wind moving to the center of low pressure moves to the right, thus missing the center and that starts the counterclockwise movement of the anticyclone. Fig 13 in the page You liked is about that. – Georg Mar 31 '11 at 21:32
The Coriolis effect works for big things like cyclones, but for a bathtub, the slightest asymmetry of the tub, air movement on top, convection current from uneven temperature etc. would have a far greater effect. Once there is the slightest sideways movement of the water any where around the hole, it will deflect the incoming water as in the diagram. That will cause the incoming water to push the water around causing it to rotate. This will deflect the incoming water more, increasing the rotation, & so on. It's like standing on a wheel. If you are right in the middle, where your weight is straight down towards the shaft, it will not rotate. But move slightly to the side, & it will start to turn, moving your feet to the side, which will make it accelerate more, moving your feet more to the side & so on.
It's because of the Coriolis effect. In the Northern hemisphere, it goes around in one direction, in the Southern hemisphere around the opposite direction, and goes straight down bang on the equator.
Black natives demostrate the effect to white tourists in the video clip Water flow at the equator, Coriolis effect.
Edit:
The above is nonsense, but I'll leave it on to prevent others doing the same. Videos like the above fake the effect by pouring in the water close to the edge to give the required initial momentum.
• It's already been explained quite clearly that it is not due to the Coriolis effect. Further, if it were, you still wouldn't be able to see it at the equator because the Coriolis effect is zero there. – Mark Eichenlaub Mar 29 '11 at 15:42
• Pff. The video is a show for tourists, not a scientific experiment. The vessels are not equally shaped and are not filled the same way. Just look at the math done in one of the answers. – Lagerbaer Mar 29 '11 at 15:49
• @user2146 Sorry, for a moment there I forgot that all YouTube videos are completely credible and that guys running tourist traps in Kenya know more about physics than the entire scientific establishment. – Mark Eichenlaub Mar 29 '11 at 15:53
• I wish I had more karma so I could downvote this further – Dan Mar 29 '11 at 16:00
• @Larry: It is trivial to fake this. All you have to do is pour the water into the tub from slightly to one side or the other. Then it has angular momentum and coriolis force entirely due to the way you poured it. To Mark, I am very shy of argument by popularity ("entire scientific establishment"). There is an Einstein anecdote on that. – Mike Dunlavey May 28 '12 at 14:01
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# Sampling Interval for Data: Calculating Omega and Number of Samples Per Period
• andrey21
In summary, a sampling interval is the time between each measurement in a data set and is important for determining the accuracy and precision of the data collected. It can be calculated by dividing the total time period by the number of samples taken. Omega (Ω) is a measure of signal frequency that is used to determine the maximum frequency that can be accurately represented in a data set. The number of samples per period also affects the accuracy of the data, with a higher number of samples per period resulting in a more accurate representation of the signal. Other factors such as the type of data, method of collection, and equipment used should also be considered when determining the sampling interval for data collection.
andrey21
I have been given a data and asked to state the sampling interval
Heres what I know
The data has a value generated for each month of the year, I have calculated a period of 11 years. Now I know the formula:
2PI/omega = number of samples per period
So will Omega = PI/66
not too sure what you're actually asking here? maybe need to be a little clearer with the question
Sorry I think I may have solved what I was asking, after reading it was very unclear:)
## 1. What is a sampling interval and why is it important in data collection?
A sampling interval is the time period between each measurement or sample taken in a data set. It is important because it determines the accuracy and precision of the data collected. A longer sampling interval can result in missing important data points, while a shorter interval can increase the amount of data collected and potentially improve the accuracy of the data.
## 2. How do you calculate the sampling interval for a data set?
The sampling interval can be calculated by dividing the total time period of the data collection by the number of samples taken. For example, if data is collected over a period of 10 hours and 100 samples are taken, the sampling interval would be 10 hours divided by 100, resulting in a sampling interval of 0.1 hours or 6 minutes.
## 3. What is the significance of Omega in determining the sampling interval?
Omega (Ω) is a measure of signal frequency in relation to the sampling interval. It is used to determine the maximum frequency that can be accurately represented in a data set. The sampling interval must be at least 2Ω (twice the signal frequency) in order to accurately capture the signal in the data.
## 4. How does the number of samples per period affect the accuracy of the data?
The number of samples per period refers to the number of times a complete cycle of the signal is measured. The more samples taken per period, the more accurately the signal can be represented in the data. This is especially important for signals with high frequencies, as a lower number of samples per period may result in missed or inaccurate data points.
## 5. Are there any other factors that need to be considered when determining the sampling interval for data collection?
Yes, there are other factors that can affect the sampling interval and accuracy of data. These include the type of data being collected, the method of data collection, and the equipment used. It is important to carefully consider these factors and make adjustments as needed to ensure accurate and reliable data collection.
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16200 in Words
16200 in words is written as “Sixteen Thousand Two Hundred”. For example, Rs.16200 is written as “Rupees Sixteen Thousand Two Hundred only” in a cheque. In Maths, 16200 is a cardinal number that expresses a quantity or a value. 16200 can represent any value, like the population of a town, the number of employees in a company, 16200 in dollars or any other currency. Learn more about Numbers In Words and writing the number names in English at BYJU’S.
16200 in Words Sixteen Thousand Two Hundred Sixteen Thousand Two Hundred in Numbers 16200
How to Write 16200 in Words?
The number 16200 in words can be written using a place value chart. Since 16200 is a five-digit number, thus, we can represent the digits using the table given below.
Ten-thousands Thousands Hundreds Tens Ones 1 6 2 0 0
From the above table,
1 → Ten Thousands
6 → Thousands
2 → Hundreds
0 → Tens
0 → Ones
Hence, when we read the number 16200 from right to left, it is Sixteen Thousand Two Hundred.
Expanded Form of 16200
We can write the expanded form as:
1 x Ten thousand + 6 x Thousand + 2 × Hundred + 0 × Ten + 0 × One
= 1 x 10000 + 6 x 1000 + 2 × 100 + 0 × 10 + 0 × 1
= 1 x 10000 + 6000 + 200 + 0 + 0
= Sixteen Thousand Two Hundred
16200 is a whole number that is succeeded by 16199 and preceded by 16201. Learn more about the number 16200 below:
• 16200 in Words – Sixteen Thousand Two Hundred
• Is 16200 an odd number? – No
• Is 16200 an even number? – Yes
• Is 16200 a perfect square number? – No
• Is 16200 a perfect cube number? – No
• Is 16200 a prime number? – No
• Is 16200 a composite number? – Yes
Frequently Asked Questions on 16200 in words
Q1
What is 16200 in words?
16200 in words is given by Sixteen Thousand Two Hundred.
Q2
What is the rule to write 16200 in words?
To write the 16200 in words, we should know the place value of all the digits.
1 → Ten Thousands
6 → Thousands
2 → Hundreds
0 → Tens
0 → Ones
Thus, the number is read as Sixteen Thousand Two Hundred .
Q3
What is the value of 16200 + 100 in words?
16200 + 100 = 16300, i.e., Sixteen thousand three hundred in words.
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# A piece of plywood 24 inches wide is cut into strips 2 1/2 inches wide. How many strips of this width can be cut?
1
by LouLo
2015-09-18T08:57:08-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
1. Given situation:
=> A piece of plywood is 24 inches wide
=> it will be cut into strips of 2 ½ inches wide.
Now, we need to find the numbers of strips in 24 inches plywood.
To do that, we simply need to divide 24 inches from 2 ½ inches.
=> Since 2 ½ is a mixed fraction (combination of whole number and fraction), we need to convert it into improper fraction.
=> 2 ½ = 5/2
=> 24 = 24/1
Now, we have 2 fractions. Let’s divide
=> 24 / 5
1 2
=>24 / 5 = 48 / 5
1 2 2 2
=> 9.6 or 9 3/5
Thus, there will be 9.6 or 9 3/5 of plywood strips.
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# General expression for entropy and free energy
The specific heat capacity of a solid at constant pressure and low temperatures is given by $$c_p=aT+bT^3$$ with $a$ and $b$ are constants.
How can I calculate a general expression for the entropy $S(T,p)$ and Gibbs free energy $G(T,p)$ as a function of $T$, assuming that $S$ and $G$ are zero at $T = 0~\mathrm{K}$?
I started with trying to calculate entropy by $$\mathrm{d}S = \left(\frac{\partial S}{\partial T}\right)_p \mathrm{d}T = \frac {c_p}{T} \mathrm{d}T$$ and $$S = \int_0^T \frac{c_p}{T}\mathrm{d}T = (aT + bT^3)\ln T$$
• $c_p$ is itself a function of $T$, so $c_p/T = a + bT^2$ and $\int c_p\mathrm{d}T/T = aT + bT^3/3$ - you cannot simply take it out of the integral and integrate $\mathrm{d}T/T = \ln T$ – orthocresol Jan 15 '16 at 17:46
• Understood, but is that approach correct, can I do it that way? For the free energy we have $G(T,p)=H-TS$, but I dont know how to determine H from either its total differential nor $H=U+pV$... – user149868 Jan 15 '16 at 18:22
• The entropy part seems sensible to me, if there's a problem with it, I don't know. Gibbs free energy part, I suspect that you may need to use the equation $(\partial G/\partial T)_p = -S$, but I'm not entirely sure either. Sorry - I'm not that familiar with doing absolute thermodynamic quantities. – orthocresol Jan 15 '16 at 19:27
• For $\mathrm{d}G=\left( \frac {\partial G}{\partial T}\right)_p \mathrm {d}T + \left(\frac {\partial G}{\partial p}\right)_T\mathrm {d}p = -S\mathrm{d}T+V\mathrm{d}p$ maybe the second part disappears because it's an isobaric process (because of the way the heat capacity is given) and because at $0$ K the free energy is zero ($\Delta G=G$) we would just have to integrate $S$ with respect to $T$? – user149868 Jan 15 '16 at 20:11
• Your equation is messed up but I get what you mean. I was thinking along those lines too - as with the process for $S$, it seems very reasonable to me. – orthocresol Jan 15 '16 at 20:16
Since G and S are taken to be zero at T = 0, H is also 0 at T = 0. So you can calculate H by integrating the heat capacity from 0 to T: $$H=\int_0^TC_pdT=\frac{a}{2}T^2+\frac{b}{4}T^4$$ Then you substitute H into the equation $$\frac{\partial (G/T)}{\partial T}=-\frac{H}{T^2}=-\frac{a}{2}-\frac{b}{4}T^2$$ and integrate between 0 and T. This will give you G. You can then solve for S from G = H - TS. This gives you the same result as integrating directly to get S.
• I calculated $G=-(aT)/2-(bT^3)/12$ and $S=a/2+(at)/2+(bT^2)/12-(bT^3)/4$, however I dont quite understand how you get to the expression $\frac{\partial (G/T)}{\partial T}$, how ddo you come up with $G/T$? – user149868 Jan 16 '16 at 8:35
• Your solution for G is correct, but your solution for S is not. Probably an algebra error. Try again. The expression your are asking about comes from the equation for dG at constant pressure: $$dG=-SdT=-\frac{(H-G)}{T}dT$$. You can arrive at the final equation using the quotient rule for differentiation on the terms involving G. This equation is very useful because you can get the temperature dependence of G without involving S, and only involving H (which is much more easily obtained). – Chet Miller Jan 16 '16 at 12:58
• If I differentiate $-\frac{(H-G)}{T}dT$ with respect to $T$ I get $\frac{(H-G)}{T^2}$, but I still dont understand how you get from there to partially differentiating $G/T$ with respect to $T$ – user149868 Jan 16 '16 at 15:42
• $\frac{dG}{dT}-\frac{G}{T}=-\frac{H}{T}$ so $\frac{T\frac{dG}{dT}-G}{T^2}=\frac{d(G/T)}{dT}=-\frac{H}{T^2}$ – Chet Miller Jan 16 '16 at 15:58
• Thats really neat. I calculated entropy to $S=-\frac {G-H}{T}=-a/2+bT^3/12-aT/2-bT^3/4$. Thanks man! – user149868 Jan 16 '16 at 18:54
| 1,233 | 3,732 |
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| 3.53125 | 4 |
CC-MAIN-2020-10
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latest
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https://www.calculatorbit.com/en/length/9-gigameter-to-meter
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text/html
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crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00810.warc.gz
| 641,734,047 | 7,820 |
9 Gigameter to Meter Calculator
Result:
9 Gigameter = 9000000000 Meter (m)
Rounded: ( Nearest 4 digits)
9 Gigameter is 9000000000 Meter (m)
9 Gigameter is 9000000km
How to Convert Gigameter to Meter (Explanation)
• 1 gigameter = 1000000000 m (Nearest 4 digits)
• 1 meter = 1e-9 Gm (Nearest 4 digits)
There are 1000000000 Meter in 1 Gigameter. To convert Gigameter to Meter all you need to do is multiple the Gigameter with 1000000000.
In formula distance is denoted with d
The distance d in Meter (m) is equal to 1000000000 times the distance in gigameter (Gm):
Equation
d (m) = d (Gm) × 1000000000
Formula for 9 Gigameter (Gm) to Meter (m) conversion:
d (m) = 9 Gm × 1000000000 => 9000000000 m
How many Meter in a Gigameter
One Gigameter is equal to 1000000000 Meter
1 Gm = 1 Gm × 1000000000 => 1000000000 m
How many Gigameter in a Meter
One Meter is equal to 1e-9 Gigameter
1 m = 1 m / 1000000000 => 1e-9 Gm
gigameter:
The gigameter (symbol: Gm) is a unit of length in the metric system equal to 1000000000 meters that is 10^9 meters. 1 Gigameter is equal to 1000000 kilometer or you can say 1 billion meters
meter:
The meter (symbol: m) is unit of length in the international System of units (SI), Meter is American spelling and metre is British spelling. Meter was first originally defined in 1793 as 1/10 millionth of the distance from the equator to the North Pole along a great circle. So the length of circle is 40075.017 km. The current definition of meter is described as the length of the path travelledby light in a vacuum in 1/299792458 of a second, later definition of meter is rephrased to include the definition of a second in terms of the caesium frequency (ΔνCs; 299792458 m/s).
Gigameter to Meter Calculations Table
Now by following above explained formulas we can prepare a Gigameter to Meter Chart.
Gigameter (Gm) Meter (m)
5 5000000000.000001
6 6000000000.000001
7 7000000000.000001
8 8000000000.000001
9 9000000000.000002
10 10000000000.000002
11 11000000000.000002
12 12000000000.000002
13 13000000000.000002
14 14000000000.000002
Nearest 4 digits
Convert from Gigameter to other units
Here are some quick links to convert 9 Gigameter to other length units.
Convert to Gigameter from other units
Here are some quick links to convert other length units to Gigameter.
More Meter to Gigameter Calculations
Converting from one Gigameter to Meter or Meter to Gigameter sometimes gets confusing.
Is 1000000000 Meter in 1 Gigameter?
Yes, 1 Gigameter have 1000000000 (Nearest 4 digits) Meter.
What is the symbol for Gigameter and Meter?
Symbol for Gigameter is Gm and symbol for Meter is m.
How many Gigameter makes 1 Meter?
1e-9 Gigameter is euqal to 1 Meter.
How many Meter in 9 Gigameter?
Gigameter have 9000000000 Meter.
How many Meter in a Gigameter?
Gigameter have 1000000000 (Nearest 4 digits) Meter.
| 902 | 2,867 |
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https://www.tutorialspoint.com/find-maximum-or-minimum-sum-of-a-subarray-of-size-k-in-cplusplus
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| 1,135,605,989 | 10,132 |
# Find maximum (or minimum) sum of a subarray of size k in C++
In this problem, we are given an array arr[] and a number k. Our task is to Find the maximum (or minimum) sum of a subarray of size k.
Let’s take an example to understand the problem,
Input: arr[] = {55, 43, 12, 76, 89, 25, 99} , k = 2
Output: 165
Explanation:
The subarray of size 2 has sum = 76 + 89 = 165
## Solution Approach
A simple approach to solve the problem is by finding all k sized subarrays and then return the sum with maximum value.
Another Approach is using the sliding window, we will find the sum of k sized subarrayes. For this, the next k sized subarray, we will subtract the last index element and add the next index element.
And then return the subarray sum with the maximum value.
## Example
Live Demo
#include <iostream>
using namespace std;
int findMaxSumSubarray(int arr[], int n, int k) {
if (n < k) {
cout << "Invalid";
return -1;
}
int maxSum = 0;
for (int i=0; i<k; i++)
maxSum += arr[i];
int curr_sum = maxSum;
for (int i=k; i<n; i++) {
curr_sum += arr[i] - arr[i-k];
maxSum = max(maxSum, curr_sum);
}
return maxSum;
}
int main() {
int arr[] = {55, 43, 12, 76, 89, 25, 99};
int n = sizeof(arr)/sizeof(arr[0]);
int k = 2;
cout<<"The sum of subarray with max sum of size "<<k<<" is "<<findMaxSumSubarray(arr, n, k);
return 0;
}
## Output
The sum of subarray with max sum of size 2 is 165
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| 4 | 4 |
CC-MAIN-2023-14
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latest
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https://gmatclub.com/forum/in-pqs-above-if-pq-3-and-ps-4-then-pr-66224.html
| 1,571,762,099,000,000,000 |
text/html
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crawl-data/CC-MAIN-2019-43/segments/1570987822458.91/warc/CC-MAIN-20191022155241-20191022182741-00063.warc.gz
| 517,052,902 | 159,877 |
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# In ΔPQS above, if PQ =3 and PS = 4, then PR =?
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In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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26 Jun 2008, 21:39
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In ΔPQS above, if PQ =3 and PS = 4, then PR =?
A. 9/4
B. 12/5
C. 16/5
D. 15/4
E. 20/3
Project PS Butler : Question #88
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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05 Mar 2012, 11:52
5
9
carcass wrote:
In the figure, if the length of PQ is 4 and the length of PS is 3, what is the length of line PR?
(A) 9/4
(B) 12/5
(C) 16/5
(D) 15/4
(E) 20/3
Hypotenuse $$QS=\sqrt{3^2+4^2}=5$$;
The area of the triangle PQS is $$area=\frac{1}{2}*PQ*PS=6$$;
But the are can be found in another way too: $$area=\frac{1}{2}*PR*QS=\frac{5}{2}*PR$$ --> $$\frac{5}{2}*PR=6$$ --> $$PR=\frac{12}{5}$$.
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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26 Jun 2008, 21:58
7
prasannar wrote:
In ΔPQS attached, if PQ =3 and PS = 4, then PR=?
(A) 9/4
(B) 12/5
(C) 16/5
(D) 15/4
(E) 20/3
its just equating the ares
1/2(pq*ps)=1/2(qs*pr)
1/2(3*4)=1/2 (5*pr)
pr=12/5
##### General Discussion
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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26 Jun 2008, 21:51
Let QR=x and PR=h
x^2 +h^2 = 9
5-x)^2 + h^2 = 16
Solving for x = 9/5 and h=12/5
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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26 Jun 2008, 21:53
1
B
3-4-5 triangle
let qr = x
then rs = 5-x
let pr = h
2 equations:
3^2 = h^2 + x^2
and
4^2 = h^2 + (5-x)^2
expand and substract them to get x = 9/5
substitute 9/5 into the first equation (or second, your choice)
to get
9 = h^2 + 81/25
solve the equation to get h = 12/5
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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27 Jun 2008, 07:39
4
1
This is quite a well-known diagram in mathematics: it can be used to prove the Pythagorean Theorem (using a and b for the two legs, and not 3 and 4). There are (at least) three entirely different ways to solve this problem, two of which were described above. Call the length we're looking for 'd':
-The area of the triangle must be the same no matter which base you choose. Thus 3*4/2 = 5*d/2 --> d = 12/5.
-Let QR = c; then QS = 5-c. We have two right angled triangles, and can use Pythagoras to set up two equations, two unknowns (c and d; this is the most time-consuming approach).
-The approach not mentioned above: notice that the two smaller triangles in the diagram are each similar to the 3-4-5 triangle. Because PQR is similar to QSP, we have d/3 = 4/5 --> d = 12/5.
It's the similarity of the triangles that can let you prove Pythagoras, incidentally.
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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09 Nov 2010, 16:36
4
PQ =3 and PS = 4,
So QS = 5
area of the triangle = 1/2 * PQ * PS = 1/2*QS*PR
OR, 1/2 * 3 * 4 = 1/2 * QS * PR
OR PR = 12/5
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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05 Mar 2012, 11:45
Attachment:
Triangle.jpg [ 21.24 KiB | Viewed 24868 times ]
In the figure, if the length of PQ is 4 and the length of PS is 3, what is the length of line PR?
(A) 9/4
(B) 12/5
(C) 16/5
(D) 15/4
(E) 20/3
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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05 Mar 2012, 12:18
Be patient. I have not fully understood
Why 5/2 * PQ...should be QS ??
Secondly why my reasoning do not lead me to the answer and is flawed, completely wrong ???
We have a right triangle PQS where angle P is 90. Also two sides PS=3 and PQ= 4 so from this we have a 30-60-90 triangle.
So, if we look at angle R is 90, from this we can see that angle P is shared between PQS and PRS so P for PRS should be 60 (likewise angle P for triangle PRQ should be 30, so angle P is 60+30=90).
So, we have: for triangle PRS angle R is 90 (PS is the hypotenuse), angle P is 60 (RS long leg) and angle S is 30 (short leg PR).
If PS is 3 (hypotenuse opposite 90 angle) PR should be 1.5 (short leg opposite angle S that is 30). This based on ratio 30:60:90.
I know that it does not hold anywater, but is useful to understand
Thanks
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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05 Mar 2012, 12:33
4
carcass wrote:
Be patient. I have not fully understood
Why 5/2 * PQ...should be QS ??
Secondly why my reasoning do not lead me to the answer and is flawed, completely wrong ???
We have a right triangle PQS where angle P is 90. Also two sides PS=3 and PQ= 4 so from this we have a 30-60-90 triangle.
So, if we look at angle R is 90, from this we can see that angle P is shared between PQS and PRS so P for PRS should be 60 (likewise angle P for triangle PRQ should be 30, so angle P is 60+30=90).
So, we have: for triangle PRS angle R is 90 (PS is the hypotenuse), angle P is 60 (RS long leg) and angle S is 30 (short leg PR).
If PS is 3 (hypotenuse opposite 90 angle) PR should be 1.5 (short leg opposite angle S that is 30). This based on ratio 30:60:90.
I know that it does not hold anywater, but is useful to understand
Thanks
First of all 5/2 * PQ does not equal to QS. We equate the areas, which can be found in two ways:
1. 1/2*Leg1*Leg2 --> $$area=\frac{1}{2}*PQ*PS=6$$;
2. 1/2*Perpendicular to hypotenuse*Hypotenuse --> $$area=\frac{1}{2}*PR*QS=\frac{5}{2}*PR$$ (since hypotenuse QS=5);
Now, equate the areas: $$6=\frac{5}{2}*PR$$ --> $$PR=\frac{12}{5}$$.
Next, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio $$1 : \sqrt{3}: 2$$. In PQS sides PQ and PS are NOT in the ratio $$1 : \sqrt{3}$$, so PQS is not a 30°, 60°, and 90° right triangle. I think you are mixing 3-4-5 Pythagorean Triples triangle with 30°-60°-90° triangle.
Hope it's clear.
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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05 Mar 2012, 14:19
It does make sense.
Is clear. Thanks.
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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07 Sep 2012, 01:39
5
As mentioned by Bunuel, there are other methods as well to solve this problem
One of the method is Similarity
Triangle QPS is similar to PRS (because one common side & angle is 3 & 90 degree)
So,
QS/QP = PS/PR
5/4 = 3/x
x = 12/5
Hope it helps.
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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24 Feb 2016, 04:26
prasannar wrote:
Attachment:
Diag-1.JPG
In ΔPQS above, if PQ =3 and PS = 4, then PR =?
A. 9/4
B. 12/5
C. 16/5
D. 15/4
E. 20/3
Two approaches. the area is 6. so area of pqr + area of prs = 6. solution will result in 12/5. B
second: testing answer choices. if assume that line(pr) is an answer choice and try to rearrange to get line(rs), then you discover line rs is squareroot of some negative numbers in all the options except B
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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09 Apr 2016, 10:19
prasannar wrote:
Attachment:
Diag-1.JPG
In ΔPQS above, if PQ =3 and PS = 4, then PR =?
A. 9/4
B. 12/5
C. 16/5
D. 15/4
E. 20/3
QS will be 5 being the hypotenuse of 90 degree triangle.
As we can calculate the area by two ways here:
1/2 * PS * PQ = 1/2 * PR * QS
4 * 3 = PR * 5
PR = 12/5
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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05 Oct 2016, 11:22
Very simple hypotenuse will be 5 as right angle triangle (3,4,5) . A perpendicular drawn to hypotenuse will bisect it , that means 5/2 or 2.5 ; 12/5 nearest option.
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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04 Aug 2017, 01:32
Let QR=x and RS=y.
Area of QPS= 1/2*4*3 = 6
We know, Area of QPS = Area of QRP + Area of PRQ --> (1)
Let PR=a.
From Pythagoras Theorem, QS=5. ==> x+y=5
In (1), 6 = 1/2*y*a +1/2*x*a ==> 1/2(ay + ax) ==> a(x+y) =12
As x+y=5, a*5=12 ==> a=12/5 = PR
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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21 Dec 2018, 10:56
1
PQS is a 3-4-5 triangle, so QS = 5. PR is an altitude of PQS. The altitude of a right triangle forms two more triangles that are similar to the original and each other. So pick any of the smaller triangles and draw it with the same orientation as PQS. Then we can set up a proportion since corresponding sides of similar triangles are proportional. From that we get RP/PS = QP/QS --> RP/4 = 3/5 --> RP =12/5 B)
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In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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22 Dec 2018, 03:55
Bunuel wrote:
carcass wrote:
In the figure, if the length of PQ is 4 and the length of PS is 3, what is the length of line PR?
(A) 9/4
(B) 12/5
(C) 16/5
(D) 15/4
(E) 20/3
Hypotenuse $$QS=\sqrt{3^2+4^2}=5$$;
The area of the triangle PQS is $$area=\frac{1}{2}*PQ*PS=6$$;
But the are can be found in another way too: $$area=\frac{1}{2}*PR*QS=\frac{5}{2}*PR$$ --> $$\frac{5}{2}*PR=6$$ --> $$PR=\frac{12}{5}$$.
Bunuel, Gladiator59, chetan2u i have questions
Area of which traigle do you find through this $$area=\frac{1}{2}*PR*QS=\frac{5}{2}*PR$$
I know hypotenuse of trangle QPS is 5, and 5/2 means you are dividing hypotenuse by 2 since PR bisects triangle QPS
can please help me to understand the logic when you equate length of shorter leg 2.5 to area of triangle QPS $$\frac{5}{2}*PR=6$$
Also does PR bisect triangle QPS ? i think it bisects, because point R is 90 degrees. If my statement is true Yes than shorter leg (RS) of traingle PSR will be (5/2) =2.5
So if hypotenuse (PS) of triangle PSR is 4 and shorter leg (RS) is 2.5 find longer leg PR
we can use a pythogorean formula
$$X^2+2.5^2=4^2$$
$$x^2+6.25= 16$$
$$x^2=16 - 6.25$$
$$x^2=9.75$$ on taking square root
$$x = 3.12$$(approximatif)
Whats wring with my reasoning
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Re: In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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22 Dec 2018, 04:05
Please find my responses in-line in red.
dave13 wrote:
Area of which traigle do you find through this $$area=\frac{1}{2}*PR*QS=\frac{5}{2}*PR$$ Area of the big triangle - as PR is perpendicular to the hypotenuse and the sides are perpendicular to each other - we can find the area by applying the formula to either PR and QS or PQ and PS
I know hypotenuse of trangle QPS is 5, and 5/2 means you are dividing hypotenuse by 2 since PR bisects triangle QPS - Just applying the formula for area
can please help me to understand the logic when you equate length of shorter leg 2.5 to area of triangle QPS $$\frac{5}{2}*PR=6$$ Equating the areas as it is the SAME TRIANGLE!!
Also does PR bisect triangle QPS ? i think it bisects, because point R is 90 degrees. If my statement is true Yes than shorter leg (RS) of traingle PSR will be (5/2) =2.5 Does not bisect
So if hypotenuse (PS) of triangle PSR is 4 and shorter leg (RS) is 2.5 find longer leg PR
we can use a pythogorean formula
$$X^2+2.5^2=4^2$$
$$x^2+6.25= 16$$
$$x^2=16 - 6.25$$
$$x^2=9.75$$ on taking square root
$$x = 3.12$$(approximatif)
LAST PART ( IN MAROON) IS WRONG AS PR DOESNT BISECT
Whats wring with my reasoning
Just equate the area by two different methods - this is the fastest solution. There is another solution by similar triangles - can you find it?
Regards,
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In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink]
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22 Dec 2018, 05:37
Just equate the area by two different methods - this is the fastest solution. There is another solution by similar triangles - can you find it?
Regards,
Gladiator59 thank you, so we are equating areas of triangles based on the rule below ? correct ?
If the measures of the corresponding sides of two triangles are proportional then the triangles are similar. Likewise if the measures of two sides in one triangle are proportional to the corresponding sides in another triangle and the including angles are congruent then the triangles are similar.
source: https://www.mathplanet.com/education/ge ... /triangles
but how can I know that PR is not a median ?
have a look here https://www.khanacademy.org/math/geomet ... equal-area
In ΔPQS above, if PQ =3 and PS = 4, then PR =? [#permalink] 22 Dec 2018, 05:37
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Timeline andydude Long Time Fellow Posts: 510 Threads: 44 Joined: Aug 2007 05/01/2009, 08:17 PM I made this timeline to put hyperopertions research in perspective, so let me know if I got anything wrong, or missed anything. I'll try to remember to add Maurer and Kneser in a later post. All of these are chronological. 1852 E. M. Lemeray writes Sur les fonctions iteratives er sur une nouvelle fonction (Association Francaise pour l'Avancement des Sciences, Congres Bordeaux 2) in which he calls tetration the "fourth natural algorithm" (according to Knoebel). 1898 H. Schubert writes Grundlagen der Arithmetik (Encyklopadie der Mathematischen Wissenschaften mit Einschluss ihrer Anwendungen) in which he discusses tetration, and calls this the "operation of the fourth kind" (according to Knoebel). 1915 A.A. Bennett writes Note on an Operation of the Third Grade (available here) where he mentions the idea of hyperoperations (specifically commutative hyperoperations). 1928 Wilhelm Ackermann writes Zum Hilbertschen Aufbau der reellen Zahlen (available here) in which he defined the hierarchy $ \phi(a, b, n) = \left\{ \begin{tabular}{ll} a + b &\text{if } n = 0 \\ 0 &\text{if } n = 1, z = 0 \\ 1 &\text{if } n = 2, z = 0 \\ a &\text{if } n > 2, z = 0 \\ \phi(a, \phi(a, b - 1, n), n - 1) &\text{if } n > 2, z > 0 \end{tabular}$ While at first glance this looks identical to hyperoperations, it is in fact not. $ \begin{tabular}{rl} \phi(a, b, 0) & = a + b \\ \phi(a, b, 1) & = a b \\ \phi(a, b, 2) & = a^b \\ \phi(a, b, 3) & = a \uparrow\uparrow (1 + b) \\ \phi(a, 0, 4) & = a \\ \phi(a, 1, 4) & = a \uparrow\uparrow (1 + a) \\ \phi(a, 2, 4) & = a \uparrow\uparrow (1 + a \uparrow\uparrow (1 + a)) \\ \phi(a, 3, 4) & = a \uparrow\uparrow (1 + a \uparrow\uparrow (1 + a \uparrow\uparrow (1 + a))) \\ \end{tabular}$ As you can see, $\phi(a, b, 4)$ is different from pentation in a nontrivial way. 1935 Rozsa Peter writes one of these (not sure which one, got this from Robert Munafo here). This is where the modern Ackermann function was formed, and where the base was assumed to be 2 (which then made the Ackermann function a bivariate function). However, Robert Munafo says that Peter's initial conditions make the "zeration" equal $2b + 1$, which would make this function produce very different values from the Ackermann function today. 1947 Reuben Louis Goodstein writes Transfinite Ordinals in Recursive Number Theory (available here) in which he defines the hierarchy $G(k, a, n) = a {\uparrow}^{k-2} n$. Apparently, this is the first time that addition was assigned $k=1$. Together with the counting of exponents (as opposed to counting operators), this makes it equivalent to hyperoperations as we know them today. Goodstein also coins the terminology tetration, pentation, etc. 1948 Raphael M. Robinson writes Primitive recursive functions (available here) in which he references Peter's work. Robert Munafo says that it is here that the "zeration" becomes $b + 1$ making the Ackermann function we know today. This means that the 2-argument Ackermann function would more correctly be called the Ackermann-Peter-Robinson function (which is quite a mouthful). 1976 Donald E. Knuth writes Coping with Finiteness (available here), in which he coins the up-arrow notation we use all the time. 1987 Nick Bromer writes Superexponentiation (available here), which is probably the source of the super-prefix (used for super-logarithm). He defines an arrow notation similar to Knuth's arrow, but offset by one such that $a \uparrow b = {}^{b}a$ and $a \downarrow b = a^{a^{b-1}}$. Bromer references Knuth, so he was certainly aware of his notation, but apparently he dropped an arrow somewhere... 1993 Markus Müller writes Reigenalgebra (available here) in which he defines both up-arrow and down-arrow notations exactly like Bromer's, but offset from Knuth's arrow notation. 1994 K. K. Nambiar writes Ackermann functions and transfinite ordinals (available here) defines yet another notation for hyperoperations. 1999 Marc Wirz writes Characterizing the Grzegorczyk hierarchy by safe recursion (available here). 2000 Stephen R. Wassell writes Superexponentiation and Fixed Points of Exponential and Logarithmic Functions (available here) which continues Bromer's terminology. Campagnolo and Moore and Costa write An analog characterization of the Grzegorczyk hierarchy (available here, later published in 2002) in which they consider various algorithmic complexity classes, which involve iterated exponentials. They use the notation $\exp^{[n]}(x)$ (base e) and $2^{[n]}(x)$ (base 2) for iterated exponentials. I believe this is the earliest reference to what has evolved into $\exp_b^n(x)$ notation. 2001 Harvey M. Friedman writes Long Finite Sequences (available here) Yunhi Cho and Kyunghwan Park write Inverse Functions of $y = x^{1/x}$ (available where?) uses MacDonnell's hyperpower terminology. Andrew Robbins bo198214 Administrator Posts: 1,616 Threads: 102 Joined: Aug 2007 05/01/2009, 09:40 PM G. Königs: Recherches sur les intégrales de certaines equations functionelles Schroeder, 1884 P. Lévy: Functions à croissants régulière et itération d'ordre fractionnaire, 1928 I think they found the regular iteration limit formulas for the Schröder and Abel functions in the non-parabolic and parabolic case, respectively. BenStandeven Junior Fellow Posts: 27 Threads: 3 Joined: Apr 2009 05/02/2009, 09:57 PM Here's a few more: 1777: Leonhard Euler: De formulis exponentialibus replicatis, Acta Academiae Scientarum Imperialis Petropolitinae, No. 1. (177. He also wrote a second paper, IIRC. I don't remember its name. 1927: Gabriel Sudan: Bull. Math. Soc. Roumaine Sci. 30 (1927), 11 - 30; Jbuch 53, 171. About a precursor to the Ackermann function. 1953, Andrzej Grzegorczyk: Some classes of recursive functions, Rozprawy Matematyczne, 4, 1953, 3-45. at http://matwbn.icm.edu.pl/ksiazki/rm/rm04/rm0401.pdf : Original definition of the Grzegorczyk hierarchy. Ivan Newbie Posts: 1 Threads: 0 Joined: Jul 2020 07/07/2020, 05:00 PM When Hilbert (1926, but in a footnote, it says “Vortrag, gehalten am 4. Juni 1925”) reported about Ackermann’s result, he used a slightly different function from the one that Ackermann used in his 1928 paper. Hilbert’s $\phi_n(a,b)$ is actually the same as $\mathrm{hyper}n(a,b)$. (He started at $n=1$.) Hermann Schubert (1899) already considered the general concept of hyperoperations (implied by “and so on”) and addition as a direct operation of first order (“Stufe”), anticipating Hilbert. Here is what he wrote (translated from German): Quote:Addition and subtraction are called basic arithmetic operations of first order, multiplication and division of second order. Furthermore, addition and multiplication are called direct, subtraction and division indirect basic arithmetic operations. […] In the same way as multiplication emerges from addition, exponentiation from multiplication, one could derive from exponentiation as the direct operation of third order a direct operation of fourth order, from it one of fifth order etc. But already the definition of a direct operation of fourth order is, although logically justified, unimportant to the progress of mathematics, because the commutative law already loses its validity at the third order. tommy1729 Ultimate Fellow Posts: 1,742 Threads: 382 Joined: Feb 2009 07/15/2020, 12:27 PM Much to discuss. I have not read everything yet, well maybe I have. I mean i have not looked at all links but things look very familiar. I looked at this link : http://www.mrob.com/pub/math/ln-2deep.html#ack And was reminded of some of my ideas and some of my friend mick. Ignoring zeration , exp^[n](ln^[n](a) + ln^[n](b) , and the alike mainly stuff like this : https://math.stackexchange.com/questions...eroperator Posted on mathstackexchange by mick where F is his and T is mine. ( I am that friend ) I had ideas of making it analytic , changing the base , growing faster or slower then tetration , more variables etc. actually what i originally had in mind was T(0,a,b,v,u) = a + b + v T(n,0,c,v,u) = n + c + v = T(n,c,0,v,u) otherwise T(n,a,b,v,u) = T(n-1,T(n,a-1,b,v,u),T(n,a,b-1,v,u),v+u,u) I believe this can express nice asymt to tetration for a given base or superfactorial type functions. more research is needed. Regards tommy1729 « Next Oldest | Next Newest »
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# A fast train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed ofthe train
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distance = 200 km
speed be x and x +10
200/x+10 -200/x = 1
solve it.....and you'll get an answer
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# Beat the Drum
Make some noise by counting! Help your preschoolers count from one to 10 while strengthening their fine motor and interpersonal skills.
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• Begin the lesson by explaining to your students that they will play a game about counting cubes, or determining how many cubes there are.
• Tell your students that they will use tongs to put one-inch cubes into a container. After putting all 10 cubes into the container, inform them that they can strike the drum with a drumstick.
• Inform your students that they will then switch positions with another student so that everyone gets a chance to count and strike the drum.
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# Chi Square Calculation in SPSS, Excel, or ACCESS
Does anyone know how if it's possible to do a Chi Square analysis in SPSS, Excel, or ACCESS. And do I need to have previous data to compare it to or can I just plug in the data I have to run the query?
###### Who is Participating?
ProfessorCommented:
Chi square in SPSS is done in the crosstabs tool. See: http://academic.uofs.edu/department/psych/methods/cannon99/level2d.html
0
Commented:
This gives a good overview of how to do a Chi-square test in Excel: http://www.gifted.uconn.edu/siegle/research/ChiSquare/chiexcel.htm
0
Commented:
In Excel 2007 there are three formula functions to calculate chi square.
Two of them make an independence test. In Spanish they are called PRUEBA.CHI (in English it may be TEST.CHI or CHITEST) and PRUEBA.CHI.INV.
The other one uses a chi square distribution. In Spanish it is called DISTR.CHI (in English it may be CHIDIST.
The best way to know how to use them is viewing the Excel help, but you can also find a lot of examples through Google.
0
Author Commented:
Calacuccia,
Thanks very much for the step by step guide on how to do the Chi Square. I did it with my data for male and female (see attached), and it seemed to work. I got 6%. Does this mean that my data has an error of about 6%?
See attached.
Lyndy
Chi-Test.xlsx
0
Commented:
It's been a while since I've done extensive statistics, but your approach is the right one: you test whether the result differt a lot from the expected result with the right test....
However, I would suggest to go through this lesson, which I found very clear & nice: http://mste.illinois.edu/patel/chisquare/intro.html
0
Commented:
Carlynne,
I believe this to be right but I'm open to be corrected.
Chi^2 is not a percentage. The value obtained in your example, using Yates' correction is 1.71. That value needs to be looked up in a Chi^2 chart using 1 degree of freedom.
Patrick
ChiSq-01.xls
0
Commented:
Hope this helps
Chi-sqrd-011.jpg
0
Commented:
As I recall, in order to do a proper Chi-Square, you must first know the distribution you are trying to evaluate, you must then establish the number of intervals you want to test, evaluate the number of expected occurances within each of those intervals, and then compute the difference between the observed and expected value, and test to determine whether that is within specific parameters.
if you are only testing against a uniform distribution, then this is pretty simple. If you are testing against a
different distribution (Bernoulli, binomial, Poisson, geometric, continuous uniform, normal (bell curve), exponential, gamma, ...) this becomes a little more challenging, because you will need the cumulative distribution tables so that you can compute your expected values.
You might try Total Access Statistics (FMS): http://www.fmsinc.com/products/statistics/
0
ProfessorCommented:
There are two types of chi-square tests typically used in this context: tests of independence and tests of goodness of fit. In this case, you are testing goodness of fit - matching data to a hypothetical model of randomness.
Your model, in this case, is 600 cases expected in each. Your observed data are 568 and 632.
When you run CHITEST, you are producing a p-value, which is a probability. In this case, you got 0.06. This is interpreted like this:
If we were to assume that in the population, 600 cases of each gender were to be expected, there is only a 6% chance that we'd see deviations from 600/600 of at least the magnitude observed in this sample.
In inferential statistics/hypothesis testing, we traditionally compare p-values to 0.05. Since 0.06 > 0.05, we would conclude from your data that the balance of men and women does not deviate from what we would expect from chance.
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Time Dilation
The aim here is to imagine asking two observers, who are in relative motion, to make measurements of the time between the same two events, using identical clocks. The two events will be: 1. a very brief pulse of light is produced and 2. that pulse of light hits a mirror. The observers will be called A and B, not very imaginative I know but what did you expect? (You might have already met them in the relativity of simultaneity page.) Observer A holds the flash-light and mirror (note that this means that A has zero velocity relative to the apparatus) A holds the apparatus such that the line between flash-light and mirror is at 90° to the direction of the relative motion of A and B. This diagram shows the situation from A's point of view. The speed of light (the same for all inertial observers) is c. If the time between the two events, as measured by A, is to then we have so We will now consider the same situation from B's point of view. This diagram shows the trajectory of the light pulse from B's point of view (in B's frame of reference, if you prefer). We will suppose that, at the instant when the light pulse is produced, B and A are very close together. Notice that we are here assuming that the speed of B relative to A is a significant fraction of the speed of light. As B sees things, the mirror moves (to his/her right) during the time that the light is in motion. So B concludes that the light moved further to get to the mirror. In the time taken for the light to reach the mirror, A and B have moved a distance x relative to each other. For B, the distance moved by the light to reach the mirror is so B will say that the time between the two events, t, is also, as the speed of A relative to B is v we have From the above we can write In this equation we see the relation between the two times as measured by A and B. Rearranging this, we obtain the more convenient expression If we imagine A to repeatedly send a new flash of light towards the mirror each time a reflected pulse returns, then we can say that A has a sort of “light beam clock”. We can further imagine that A uses this to drive ordinary mechanical clock hands which could be observed (probably with the assistance of a telescope!) by B as they race past each other. The conclusion from our thought experiment is that B would see A’s clock running slowly compared with any similar clock carried by (and therefore at rest relative to) him/her-self. However, note that A would, of course, think the same thing about a clock carried by B. This is where the term time dilation comes from. It refers to the “stretching out” of time intervals as indicated by clocks which are in motion relative to the observer.
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Theorem
If the order of the integer
$a$
modulo
$n$
is
$c$
then
$a^k \equiv 1 \; (mod \; n)$
if and only if
$k$
is a multiple of
$c$
. Also
$c$
divides the Euler totient function
$\phi (n)$
.
Proof
If
$k$
is a multiple of
$c$
,
$k=cr$
then
$a^k=a^{cr}=(a^c)^r \equiv (1)^r \; (mod \; n) \equiv 1 \; (mod \; n)$
Suppose
$a^k \equiv 1 \; (mod \; n)$
. Dividing
$k$
by
$c$
, the Division Algorithm gives
$k=qc+r$
for some integers
$q, \; 0 \le r \lt c$
. Then
$1 \equiv a^k \equiv a^{qc+r} \; (mod \; n) \equiv (a^c)^q a^r \; (mod \; n) \equiv (1^q)^r \; (mod \; n) \equiv a^r \; (mod \; n)$
This forces
$r=0$
$c$
$a^c \equiv 1 \; (mod \; n)$
$k=qc$
$a^{\phi (n)} \equiv 1 \; (mod \; n)$
$\phi (n)$
$c$
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HSC Commerce (Marketing and Salesmanship) 12th Board ExamMaharashtra State Board
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# Question Paper Solutions - Mathematics and Statistics 2014 - 2015 HSC Commerce (Marketing and Salesmanship) 12th Board Exam
SubjectMathematics and Statistics
Year2014 - 2015 (March)
Marks: 80
[12]1 | Attempt any SIX of the following
[2]1.1
Express the following statement in symbolic form and write its truth value.
"If 4 is an odd number, then 6 is divisible by 3 "
Chapter: [1] Mathematical Logic
Concept: Mathematical Logic - Statement Patterns and Logical Equivalence
[2]1.2
Find the values of x and y, if
2[[1,3],[0,x]]+[[y,0],[1,2]]=[[5,6],[1,8]]
Chapter: [2] Matrices
Concept: Solution of Equations
[2]1.3
Find the value of 'k' if the function
f(X)=(tan7x)/(2x) ,
=k, for x=0
is continuos at x=0
Chapter: [3] Continuity
Concept: Continuous Function of Point
[2]1.4
Find dy/dx if y=cos^-1(sqrt(x))
Chapter: [4] Differentiation
Concept: Derivative - Derivative of Inverse Function
[2]1.5
Price P for demand D is given as P = 183 +120D - 3D2 Find D for which the price is increasing
Chapter: [5] Applications of Derivative
Concept: Increasing and Decreasing Functions
[2]1.6
Evaluate : int1/(x(3+logx))dx
Chapter: [6] Indefinite Integration
Concept: Indefinite Integration - Methods of Integration
[2]1.7
If a=[[2,1],[1,1]] " then show taht " A^2-3A+I=0
Chapter: [2] Matrices
Concept: Solution of Equations
[2]1.8
evaluate : int xcosxdx
Chapter: [6] Indefinite Integration
Concept: Indefinite Integration - Integration by Parts
[14]2
[6]2.1 | Attempt any TWO of the following
[3]2.1.1
Prove that the following statement pattern is equivalent :
(p ∨ q) r and (p → r) ∧ (q → r)
Chapter: [1] Mathematical Logic
Concept: Mathematical Logic - Statement Patterns and Logical Equivalence
[3]2.1.2
Examine the continuity of the following function :
f(x)=x^2-x+9 , for x ≤ 3
=4x+3 for x > 3
Chapter: [3] Continuity
Concept: Continuous Function of Point
[3]2.1.3
find dy/dx if y=tan^-1((6x)/(1-5x^2))
Chapter: [4] Differentiation
Concept: Derivative - Derivative of Inverse Function
[8]2.2 | Attempt any TWO of the following :
[4]2.2.1
Find the inverse of the following matrix by elementary row transformations if it exists. A=[[1,2,-2],[0,-2,1],[-1,3,0]]
Chapter: [2] Matrices
Concept: Inverse of Matrix
[4]2.2.2
Find the area of elipse x^2/a^2+y^2/b^2=1
Chapter: [7] Definite Integrals
Concept: Applications of Definite Integrals
[4]2.2.3
The expenditure Ec of a person with income I is given by Ec = (0.000035) I2 + (0.045) I. Find marginal propensity to consume (MPC) and marginal propensity to save (MPS) when I = 5000. Also find A (average) PC and A (average)
PS.
Chapter: [5] Applications of Derivative
Concept: Maxima and Minima
[14]3
[6]3.1 | Attempt any TWO of the following :
[3]3.1.1
Express the truth of each of the following statements by Venn diagram:
(a) Some hardworking students are obedient.
(b) No circles are polygons.
(c) All teachers are scholars and scholars are teachers.
Chapter: [1] Mathematical Logic
Concept: Venn Diagrams
[3]3.1.2
If 'f' is continuous at x = 0, then find f(0).
f(x)=(15^x-3^x-5^x+1)/(xtanx) , x!=0
Chapter: [3] Continuity
Concept: Continuous Function of Point
[3]3.1.3
find dy/dx if x=e2t , y=e^sqrtt
Chapter: [4] Differentiation
Concept: Derivatives of Functions in Parametric Forms
[8]3.2 | Attempt any TWO of the following :
[4]3.2.1
Evaluate : int (1+logx)/(x(2+logx)(3+logx))dx
Chapter: [6] Indefinite Integration
Concept: Indefinite Integration - Methods of Integration
[4]3.2.2
Evaluate :int_0^(pi/2)dx/(1+cotx)
Chapter: [6] Indefinite Integration
Concept: Indefinite Integration - Integration by Parts
[4]3.2.3
A firm wants to maximize its profit. The total cost function is C = 370Q + 550 and revenue is R = 730Q-3Q2. Find the output for which profit is maximum and also find the profit amount at this output.
Chapter: [5] Applications of Derivative
Concept: Maxima and Minima
[12]4 | Attempt any SIX of the following :
[2]4.1
The ratio of number of boys and girls in a school is 3 : 2. If 20% of the boys and 30% of the girls are scholarship holders, find the percentage of students who are not scholarship holders.
Chapter: [8] Ratio, Proportion and Partnership
Concept: Ratio, Proportion and Partnership
[2]4.2
Obtain crude death rates (C. D. R.) for city A and city B from the data given below :
Age groups (in years) City A City B Population No. of deaths Population No. of deaths Below 15 800 32 900 12 15-25 3000 12 1500 8 25-65 4800 48 4500 38 65 and above 1400 42 600 30
Chapter: [11] Demography
Concept: Measurements of Mortality
[2]4.3
Coefficient of rank correlation between x and y is 0.5 and sumd_i^2= 42. Assuming that no ranks are repeated, find the number of pairs of observations.
Chapter: [12] Bivariate Data and Correlation
Concept: Rank Correlation
[2]4.4
An agent charges 12% commission on the sales. What does he earn if the total sale amounts to Rs. 36,000? What does the seller get?
Chapter: [9] Commission, Brokerage and Discount
Concept: Commission, Brokerage and Discount
[2]4.5
Find the age standard death rate (S. D. R.) for the following data:
Age groups (in years) Population (in '000) No. of deaths 0-10 11 240 10-20 12 150 20-60 9 125 60 and above 2 90
Chapter: [11] Demography
Concept: Life Tables
[2]4.6
Following table gives the age of husbands and wives
Age of wives (in years) Age of husbands (in years) 20-30 30-40 40-50 50-60 15-25 5 9 - — 25-35 — 10 25 2 35-45 — 1 12 2 45-55 — — 4 16 55-65 — — — 4
Find:
(a)The marginal frequency distribution of the age of husbands.
(b)The conditional distribution of the age of husbands
Chapter: [11] Demography
Concept: Uses of Vital Statistics in Demography
[2]4.7
The present worth of the sum of Rs. 5,830, due 9 months hence, is Rs. 5,500. Find the rate of interest.
Chapter: [9] Commission, Brokerage and Discount
Concept: Commission, Brokerage and Discount
[2]4.8
For a binomial distribution mean is 6 and variance is 2. Find n and p.
Chapter: [14] Random Variable and Probability Distribution
Concept: Binomial Theorem and Binomial Distribution
[14]5
[6]5.1 | Attempt any TWO of the following :
[3]5.1.1
For the following problem, find the sequence that minimizes total elapsed time (in hours) required to complete jobs on two machines M1 and M2 in the order M1-M2 Also find the minimum elapsed time T.
Jobs A B C D E Machine M1 5 1 9 3 10 Machine M2 2 6 7 8 4
Chapter: [15] Management Mathematics
Concept: Sequencing in Management Mathematics
[3]5.1.2
Natarajan and Mr. Gopalan are partners in the company having capitals in the ratio 4 : 5 and the profits received by them are in the ratio 5 :4. If Mr. Gopalan invested capital in the company for 16 months, how long was Mr. Natarajan’s investment in the company?
Chapter: [9] Commission, Brokerage and Discount
Concept: Commission, Brokerage and Discount
[3]5.1.3
From a lot of 25 bulbs of which 5 are defective a sample of 5 bulbs was drawn at random with replacement. Find the probability that the sample will contain -
(a) exactly 1 defective bulb.
(b) at least 1 defective bulb.
Chapter: [14] Random Variable and Probability Distribution
Concept: Random Variables and Its Probability Distributions
[8]5.2 | Attempt any TWO of the following :
[4]5.2.1
Given the following table which relates to the number of parrots at age x, complete the life table for parrots.
X 0 1 2 3 4 5 lx 1000 940 780 590 25 0
Chapter: [11] Demography
Concept: Life Tables
[4]5.2.2
You are given the following information about advertising expenditure and sales:
Advertisement Expenditure (Rs. in lakh) (X) Sales (Rs. in lakh) (Y) Arithmetic mean 10 90 Standard deviation 3 12
Correlation coefficient between X and Y = 0.8.
(a) Obtain the two regression equations.
(b) What is the likely sales when the advertising budget is ? 15 lakh?
(c) What should be the advertising budget if the company
Chapter: [13] Regression Analysis Introduction
Concept: Lines of Regression of X on Y and Y on X Or Equation of Line of Regression
[4]5.2.3
Electro Corp. Co. manufactures two electrical products :Air conditioners and Fans. The assembly process for each is similar in which both require a certain amount of wiring and drilling. Each air conditioner takes 4 hours for wiring and 2 hours for drilling. Each fan also takes 2 hours for wiring and 1 hour for drilling. During the next production period, 240 hours of wiring time are available and upto 100 hours of drilling time may be used. Each air-conditioner assembled may be sold for Rs. 2,000 profit and each fan assembled may be sold for Rs. 1,000 profit. Formulate this problem as an L.P.P. in order to maximize the profit.
Chapter: [9] Commission, Brokerage and Discount
Concept: Commission, Brokerage and Discount
[14]6
[6]6.1 | Attempt any TWO of the following :
[3]6.1.1
The equations given of the two regression lines are :
2x + 3y-6 = 0 and 5x + 7y-12 = 0.
Find :
(a) Correlation coefficient
(b) sigma_x/sigma_y`
Chapter: [13] Regression Analysis Introduction
Concept: Lines of Regression of X on Y and Y on X Or Equation of Line of Regression
[3]6.1.2
Find graphical solution for the following system of linear inequations:
2x + 3y ≥ 2, -x + y ≤ 3, x ≤ 4, y ≥ 3
Chapter: [15] Management Mathematics
Concept: Linear Programming Problem in Management Mathematics
[3]6.1.3
The number of complaints which a bank manager receives per day is a Poisson random variable with parameter m = 4. Find the probability that the manager will receive -
(a) only two complaints on any given day.
(b) at most two complaints on any given day
[Use e-4 =0.0183]
Chapter: [14] Random Variable and Probability Distribution
Concept: Poisson Distribution
[8]6.2 | Attempt any TWO of the following :
[4]6.2.1
A warehouse valued at Rs. 10,000 contained goods worth Rs. 60,000. The warehouse was insured against fire for Rs. 4,000 and the goods to the extent of 90% of their value. A fire broke out and goods worth Rs. 20,000 were completely destroyed, while the remainder was damaged and reduced to 80% of its value. The damage to the warehouse was to the extent of Rs. 2,000. Find the total amount that can be claimed.
Chapter: [10] Insurance and Annuity
Concept: Insurance and Annuity
[4]6.2.2
In the following data, one of the values of Y is missing. Arithmetic means of X and Y series are 6 and 8
X 6 2 10 4 8 Y 9 11 ? 8 7
(a) Estimate the missing observation.
(b) Calculate correlation coefficient.
Chapter: [12] Bivariate Data and Correlation
Concept: Statistics - Karl Pearson’s Coefficient of Correlation
[4]6.2.3
A job production unit has four jobs A, B, C, D which can be manufactured on each of the four machines P, Q, R and S. The processing cost of each job is given in the following table:
Jobs Machines P Q R S Processing Cost (Rs.) A 31 25 33 29 B 25 24 23 21 C 19 21 23 24 D 38 36 34 40
How should the jobs be assigned to the four machines so that the total processing cost is minimum?
Chapter: [15] Management Mathematics
Concept: Assignment Problem
S
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# How does escape velocity relate to energy and speed?
I have several confusions regarding escape velocity. I am sure I am missing something(s) obvious or maybe I am taught wrong.
1. Lets say we throw an object of any mass at exactly escape velocity of earth calculated from $v^2=\frac{2GM}{r}$ which is almost $11 \text{km s}^{-1}$ but I am talking about exact escape speed. That ball initially has $KE=\frac{1}{2}mv^2$ and $PE=mgh$. Wikipedia says and I quote
In physics, escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero.
How is that possible?
2. As $F=\frac{GmM}{r^2}$ no matter how much the particle travels away from earth's surface it will always be accelerated towards earth. With increasing $r$ the $F$ will decrease but it will never reach $0$. That means that there will be no point where the particle will stop and will continue to move with slower and slower speed will never reaches zero. Am I right?
3. A request: Please explain exactly what happens to particle's $KE$ and $PE$ at different points such as at $r=0$ and at $r=\infty$.
-
Comment on the first part: $U = mgh$ is a result coming from a first order approximation of the full Newtonian gravitational potential, with which the escape velocity is calculated. Dropping the approximation gets rid of (what I think is your) problem. – ACuriousMind Sep 1 '14 at 15:41
I think this perhaps should have been split into more than one question... but since it's been answered, I've just edited the title to improve it. – David Z Sep 2 '14 at 3:11
1. Gravitational potential energy is usually measured as a negative value. We do this because an object that is so far away from a gravity well that it practically is unaware of it shouldn't be considered as having any potential energy. So as $r\to\infty$, $PE\to0$. As an object falls into a gravity well, it loses potential energy, so gravitational PE is a negative value. Since energy is conserved, if you can sum the KE (a positive value) and the PE (a negative value) and the result is $0$, that means there is enough KE for the projectile to reach $r\to\infty$, where $PE=0$ just as $KE=0$. At that point, the object never returns. Thus that KE defines the escape speed. Also, $KE=\frac{1}{2}mv^2$ as you wrote, but $PE=-\frac{GMm}{r}$. The equation $PE=mgh$ is only valid near Earth's surface and it represents the change in potential energy from the surface, not the total potential energy (which is why it is positive, not negative).
2. If $KE+PE=0$ then at $r=\infty$, the speed will be zero. However, in a more practical sense, you are correct. An object at escape speed in a universe with just one gravity source and the object will never reach zero speed, it will simply move slower and slower forever.
3. At $r=\infty$, the PE goes to zero. The KE then becomes the only contributor to the total energy. so whatever the total energy of the object is, that's its KE. $r=0$ is a complicated case. For a point source of gravity, the PE would become infinite, but (other than a black hole, which isn't covered by Newtonian mechanics), there is no such thing as a point source of gravity. In the usual case, the Force of gravity disappears at $r=0$, but since moving away from that position would still constitute a gain in PE (as you'd experience an opposing force), it can get more complicated to talk about the PE at or near $r=0$. Generally, beneath the surface of an object, the PE becomes dependent on the distribution of the mass of the object. The KE is again dependent on the total energy. But there is nothing very special at $r=0$. In fact, there are no particularly interesting points anywhere. At every point, the total energy is $E=PE+KE$, PE is a negative value that approaches 0 as $r\to\infty$, and KE is whatever is left over such that total energy is conserved.
-
About the special case where r is close to 0, I would add that usually M is also close to 0, else you can't reach it. (it may look satisfying in the formula for calculating v) – Ghislain Bugnicourt Sep 1 '14 at 16:30
@GhislainBugnicourt M isn't always close to zero, you could be inside the gravitating body (like a dyson sphere made entirely of neutrons or just boring a hole through a planet) – Jim Sep 1 '14 at 16:54
@JanHudec Serious clarification: are you rebutting my comment or saying I should mention M goes to 0 in point 3 (agreeing with Ghislain)? For the former, OK, I agree. For the latter, I think that would make the formula for v more misleading as it would indicate it takes less speed to escape from beneath the surface – Jim Sep 1 '14 at 19:37
@Jim: If you are below the surface, the material above you cancels some of the gravity and the forces perfectly balance in the centre, so effectively $M$ goes to $0$. – Jan Hudec Sep 1 '14 at 19:37
@Jim: Well, under the surface one would have to substitute proper formula for gravity inside a body. Trying to talk about what happens without that is somewhat misleading, yes. – Jan Hudec Sep 1 '14 at 19:42
Potential energy for a point mass (and also for a sphere) is not $PE = mgh$ (this is special case for a uniform field) but rather: $$PE = - \frac{GMm}{r}$$
where G is the gravitational constant, M and m are both masses and r is the distance between the masses. (in the case of a sphere, the distance is to the centre of the sphere)
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# Quick Answer: What Number Is 30% Of 80?
## What number is 20% of 35?
7What is 20 percent (calculated percentage %) of number 35.
## How do you find 30% of a number?
Once you have the decimal figure, multiply it by the number for which you seek to calculate the percentage; i.e., if you need to know 30 percent of 100, you convert 30 percent to a decimal (0.30) and multiply it by 100 (0.30 x 100, which equals 30).
## What number is 35 percent of 80?
28What is 35 percent (calculated percentage %) of number 80? Answer: 28.
## What is 30% of 50% as a percentage?
60Percentage Calculator: 30 is what percent of 50? = 60.
## What number is 95 percent of 60?
57Percentage Calculator: What is 95 percent of 60? = 57.
## What number is 30% of 100?
333.33333333333330 percent (calculated percentage %) of what number equals 100? Answer: 333.333333333333.
## What number is 40% of 80?
32What is 40 percent (calculated percentage %) of number 80? Answer: 32.
## What is 30 out of 100 as a percentage?
30%Convert fraction (ratio) 30 / 100 Answer: 30%
## What is passing mark out of 80?
The passing marks for theory paper out of 80 are 26, out of 70 marks students need to attain 23 marks and out of 60 marks, 19 marks are required to pass the examination. For practical examination, out of 40, 13 marks are required. Out of 30 marks, 9 marks are needed to pass the exam.
## What is the 33% of 70 marks?
What is 33 percent (calculated percentage %) of number 70? Answer: 23.1.
## How much is 30% off 200?
Percent Off Calculator 30% off 200 is 140.00 –
## What number is 28% of 80?
35%How much is 28 out of 80 written as a percent value? Convert fraction (ratio) 28 / 80 Answer: 35%
## What number is 80 percent of 60?
48Percentage Calculator: What is 80 percent of 60? = 48.
## What number is 20% of 30?
6What is 20 percent (calculated percentage %) of number 30? Answer: 6.
## What number is 25% of 80?
20Percentage Calculator: What is 25 percent of 80? = 20.
## What grade is a 60 out of 80?
75%Convert fraction (ratio) 60 / 80 Answer: 75%
## What percentage is 12 out of 80?
15%Convert fraction (ratio) 12 / 80 Answer: 15%
## What number is 80 percent of 30?
24What is 80 percent (calculated percentage %) of number 30? Answer: 24.
## What is the 33% of 80 marks?
Percentage Calculator: What is 33 percent of 80? = 26.4.
## What number is 90% of 80?
721 Answer. 72 is 90% of 80.
## What number is 15% of 80?
121 Answer. 12 is 15% of 80 .
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# show work step by step
Anonymous
### Question Description
1. f(x) = π₯ 3 β π₯ 2 β 8 x + 12
2. h(x) = 2π₯ 3 β 3π₯ 2 β 2 x + 3
3. g(x) = 36π₯ 4 β 13 π₯ 2 + 1
4. g(x) = π₯ 3 + 3π₯ 2 β 6 x β 8
### Unformatted Attachment Preview
NAME _____________________________________________ DATE ____________________________ PERIOD _____________ 2-4 Practice Zeros of Polynomial Functions List all possible rational zeros of each function. Then determine which, if any, are zeros. 1. f(x) = π₯ 3 β π₯ 2 β 8x + 12 2. h(x) = 2π₯ 3 β 3π₯ 2 β 2x + 3 3. g(x) = 36π₯ 4 β 13π₯ 2 + 1 4. g(x) = π₯ 3 + 3π₯ 2 β 6x β 8 Solve each equation. 5. 2π₯ 4 + 9π₯ 3 β 87π₯ 2 β 49x + 45 = 0 6. π₯ 3 β 5π₯ 2 β 17x = β21 7. Determine an interval in which all real zeros of f(x) = π₯ 4 + π₯ 3 β 7π₯ 2 β x + 6 must lie. Explain your reasoning using the upper and lower bound tests. Then find all real zeros. 8. Describe the possible real zeros of f(x) = π₯ 4 + 2π₯ 3 β 13π₯ 2 β 14x + 24. Write a polynomial function of least degree with real coefficients in standard form that has the given zeros. 9. β5, 1+ β11π 1β β11π 2 , 2 10. 1, β1, β3i, β β3i Use the given zero to find all real zeros of each function. Then write the linear factorization of p(x). 11. p(x) = π₯ 4 β 5π₯ 3 + 8π₯ 2 β 20x + 16; 2i 12. p(x) = π₯ 4 + 4π₯ 3 β 2π₯ 2 β 4x + 16; 1 + i 13. DRIVING An automobile moving at 12 meters per second on level ground begins to decelerate at a rate of 1.6 1 meters per second squared. The formula for the distance an object has traveled is d(t) = π£0 t + ππ‘ 2 , where π£0 is the 2 initial velocity and a is the acceleration. For what value(s) of t does d(t) = 40 meters? Chapter 2 23 Glencoe Precalculus ...
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## OLS: Ordinary Least Square Method
For those of you who love mathematics and would like to know from how the linear regression formula was derived, in this section of tutorial you will learn a powerful method called Ordinary Least Square (OLS). I assume that you know calculus to perform the OLS method. Knowing this method is important that you may learn to derive many regression formulas by yourselves.
is data of independent variable from observation
is the mean of
is data of dependent variable from observation
is the mean of
is the estimated of , that is represented by the regression model.
is the number of observation data
To perform ordinary least square method, you do the following steps:
1. Set a difference between dependent variable and its estimation:
2. Square the difference:
3. Take summation for all data
4. To get the parameters that make the sum of square difference become minimum, take partial derivative for each parameter and equate it with zero,
For example:
Find for model parameter for model estimation using Ordinary Least square!
The model only has two parameters, that is and .
We take partial derivative of the sum of square difference to the first parameter and equate it to zero . In taking the partial derivative, we assume , and are constant while is the only variable.
Equate it with zero we have
Actually, the two parameters, and , are the real constants and they can go out of the summation sign. Constant 2 is surely not equal to zero, thus we can cancel out to simplify.
We know that , thus we can simplify the last equation into
(1)
Now, we take partial derivative of the sum of square difference to the second parameter and equate it to zero . Similar to before, in taking partial derivative, we assume , and are constant, while is the only variable..
Equate it with zero we have
Actually, the two parameters, and , are the real constants and they can go out of the summation sign, Constant 2 is surely not equal to zero, thus we can cancel out to simplify.
We know that and , , thus we can further simplify the last equation into or,
(2)
Inputting equation (1) into equation (2), we have
Thus, the parameters of regression model are and
Notice that the slope is actually equivalent to the earlier formula of slope in this tutorial by simple algebra.
## Example
Find for model parameter for model estimation using Ordinary Least square!
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https://www.teachoo.com/5032/1301/Ex-7.5--10---Integrate-2x---3---(x2---1)-(2x---3)/category/Integration-by-partial-fraction---Type-3/
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Integration by partial fraction - Type 3
Chapter 7 Class 12 Integrals
Concept wise
### Transcript
Ex 7.5, 10 2 3 2 1 2 + 3 We can write the integrand as 2 3 2 1 2 + 3 = 2 3 1 + 1 2 + 3 = + 1 + 1 + 2 + 3 = 1 2 + 3 + + 1 2 + 3 + 1 + 3 1 + 1 2 + 3 By cancelling denominator 2 3 = 1 2 +3 + +1 2 +3 + 1 +1 By substituting x = 1, in (1) 2 3 = 1 1 2+3 + 1+1 2+3 + 1+1 1 1 5 = 2 1 + 0+ 0 5 = 2 = 5 2 Similarly Putting x = 1, in (1) 2 3 = 1 1 2+3 + 1+1 2+3 + 1 1 1+1 1 = A 0 + 2 5 + 0 1 = 10 = 1 10 Similarly Putting x = 3 2 , in (1) 2 3 2 3 = 3 2 1 2 3 2 +3 + 3 2 +1 2 3 2 +3 + 3 2 1 3 2 +1 6 = 0+ 0+ 1 2 5 2 6 = 5 4 = 24 5 Therefore 2 3 + 1 1 2 + 3 = 5 2 + 1 + 1 10 1 + 24 5 2 + 3 Hence we can write it as 2 3 + 1 1 2 + 3 = 5 2 + 1 + 1 10 1 + 24 5 2 + 3 = 5 2 log +1 1 10 log 1 24 5 1 2 log 2 +3 + = + + +
#### Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.
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# Important Questions for Class XII Board Exam – Physics – I
This article contains all the important theoretical questions for Physics – Class 12 CBSE Board Exam. The questions has been distributed chapter-wise. Set of these 10-15 questions per chapter will ensure more than 60% marks in Board Exam.
#### Electrostatics
1. Derive an expression for the electric field at a point on the axial position of an electric dipole.
2. Derive an expression for the electric field at a point on the equatorial position of an electric dipole.
3. Describe the Principle, construction and working of Van de Graff generator.
4. Derive an expression for the energy stored in a capacitor. Show that whenever two conductors share charges by bringing them into electrical contact, there is a loss of energy.
5. Derive an expression for the effective capacitance when capacitors are connected in (a) series and (b) parallel
6. Explain the principle of a capacitor and derive an expression for the capacitance of a parallel plate capacitor.
7. State Gauss theorem and apply it to find the electric field at a point due to (a) a line of charge (b) A plane sheet of charge (c) A Charged spherical conducting shell
##### Click Here for Important Questions for Class XII Board Exam – Physics – II
8. State Coulomb’s law and express it in vector form. Derive it using Gauss theorem.
9. Derive an expression for the torque on an electric dipole in a uniform electric field.
10. Derive an expression for the work done in rotating an electric dipole in a uniform electric field
11. Derive an expression for the energy stored (Potential Energy) in a dipole in a uniform electric field.
12. Derive an expression for the electrostatic potential energy of a system of point charges
13. Derive an expression for the capacitance of a parallel plate capacitor with (a) a dielectric slab (b) a metallic plate in between the plates of the capacitor.
14. Define electric potential at a point. Derive an expression for the electric potential at a point due to (a) a point charge (b) a system of point charges (c) a dipole.
15. Show that the work done in an electric field is independent of path.
16. What are dielectrics? Distinguish polar and nonpolar dielectrics. Define the term Polarization vector.
#### Current Electricity
1. Define drift velocity and derive an expression for it.
2. Derive the expression I=nAevd
3. Deduce Ohm’s law from elementary ideas and hence write an expression for resistance and resistivity.
4. Derive an expression for conductivity in terms of mobility
5. Explain the color coding of carbon resistors.
6. Derive an expression for the current in a circuit with external resistance R when (a) n identical cells of emf E and internal resistance r are connected in series (b) m identical cells are connected in parallel
7. State and explain Kirchhoff’s laws.
##### Click Here for Important Questions for Class XII Board Exam – Physics – II
1. State and explain the principle of Wheat Stone’s principle. Deduce it using Kirchhoff’s laws.
2. Describe how you will determine the resistance of a given wire using Meter Bridge.
3. Explain the principle of a potentiometer. Describe how will you determine (a) the ratio of emfs of two primary cells using potentiometer. (b) The internal resistance of a primary cell using potentiometer.
4. Explain the variation of resistance and resistivity with temperature and hence define temperature coefficient of resistance and resistivity.Click Here to see chapter-wise notes for Class 12 Board Exams of Physics.
#### Magnetic Effect of Current
1. State Biot- Savart law and apply it to find the magnetic field due to a circular loop carrying current at a point (a) at its centre (b) on the axis
2. State Ampere’s circuital law and apply it to find the magnetic field (a) inside a current carrying solenoid (b) inside a current carrying toroid
3. Apply Ampere’s circuital law to determine the magnetic field at a point due to a long straight current carrying conductor.
4. Derive an expression for the force on a current carrying conductor in a uniform magnetic field
5. Derive an expression for the force between long straight conductors carrying current and hence define 1 ampere.
6. Derive an expression for the torque on a current carrying loop in a uniform magnetic field.
7. Describe the principle construction and working of a Moving coil galvanometer.
8. Describe the conversion of a moving coil galvanometer into (a) Ammeter (b) Voltmeter
9. What is radial magnetic field? What is its importance in a moving coil galvanometer? How is a radial magnetic field realized in moving coil galvanometers?
10. Describe the principle construction and working of a cyclotron. Explain why an electron cannot be accelerated using a cyclotron.
11. Describe the motion of a charged particle in a magnetic field when it enters the field (a) perpendicular to the field lines (b) obliquely making and angle Θ with the field lines
12. Derive an expression for the magnetic dipole moment of a revolving electron and hence define Bohr magneton.
#### Electromagnetic Induction (EMI)
1. State and Explain Faraday’s laws of electromagnetic induction.
2. State Lenz’ Law and show that it is in accordance with the law of conservation of energy.
3. Use Lenz’ law to find the direction of induced emf in a coil when (a) a north pole is brought towards the coil (b) north pole taken away from the coil (c) A south pole is brought towards the coil and (d) a south pole is taken away from the coil, Draw illustrations in each case.
##### Click Here for Important Questions for Class XII Board Exam – Physics – II
1. What is motional emf. Deduce an expression for it. State Fleming’s right hand rule to find the direction of induced emf.
2. What are eddy currents? Describe the applications of eddy currents.
3. Explain the working of (a) Electromagnetic Brakes (b) Induction Furnace
4. Which physical quantity is called the INERTIA OF ELECTRICITY? Why is its called so?
5. Define self induction and self inductance. What is its unit? Write its dimensions.
6. Derive an expression for the self inductance of a long solenoid.
7. Explain the phenomenon of mutual induction and define mutual inductance. Write the unit and dimensions of mutual inductance.
8. What are the factors affecting mutual inductance of a pair of coils? Define coefficient of coupling.
9. Describe the various methods of producing induced emf. Derive an expression for the instantaneous emf induced in a coil rotated in a magnetic field.
10. What is displacement current? Explain its need.Click Here to see chapter-wise notes for Class 12 Board Exams of Biology
#### Alternative Current
1. Describe the principle construction and working of an AC generator. Draw neat labeled diagram
2. Define mean value of AC(over a half cycle) and derive an expression for it.
3. Define RMS value of AC and derive an expression for it.
4. Show that the average value of AC over a complete cycle is zero.
5. Show that the current and voltage are in phase in an ac circuit containing resistance only.
6. Deduce the phase relationship between current and voltage in an ac circuit containing inductor only.
7. Deduce the phase relationship between current and voltage in an ac circuit containing capacitor only.
8. Draw the phasor diagram showing voltage and current in LCR series circuit and derive an expression for the impedance
9. What do you mean by resonance in Series LCR circuit? Derive an expression for the frequency of resonance in LCR circuit.
10. Distinguish between resistance, reactance and impedance.
11. Define quality factor (Q factor) of resonance and derive an expression for it.
12. Describe the mechanism of electromagnetic oscillations in LC circuit and write expression for the frequency of oscillations produced.
13. Derive an expression for the average power in an ac circuit.
14. Define power factor. Deduce expression for it and explain wattless current?
15. Describe the principle construction theory and working of a transformer.
16. Describe the various losses in a transformer and explain how the losses can be minimized.
#### Electromagnetic Waves
1. Explain the inadequacy of Ampere’s circuital law
2. Describe Hertz experiment to demonstrate the production of electromagnetic waves
3. Write the properties of electromagnetic waves.
4. Write any five electromagnetic waves in the order of decreasing frequency and write any two properties and uses of each
5. Deduce an expression for velocity of em waves in vacuum
6. Establish the transverse nature of electromagnetic waves.
7. Compare the properties of electromagnetic waves and mechanical waves
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https://simple.wikipedia.org/wiki/Linear_differential_equation
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# Linear differential equation
In mathematics, a linear differential equation is a differential equation that is stated by a linear polynomial in the unknown function and its derivatives. It is an equation of the form:
${\displaystyle a_{0}(x)y+a_{1}(x)y'+a_{2}(x)y''\cdots +a_{n}(x)y^{(n)}=b(x)}$
where a0(x), ..., an(x) and b(x) are random differential functions that do not need to be linear. And y′, ..., y(n) are the derivatives of an unknown function y that follow after the variable x.
This equation is an ordinary differential equation (ODE). A linear differential equation may also be a linear partial differential equation (PDE), if the unknown function depends on several variables. And the derivatives that show in the equation are partial derivatives.
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https://metanumbers.com/6035
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## 6035
6,035 (six thousand thirty-five) is an odd four-digits composite number following 6034 and preceding 6036. In scientific notation, it is written as 6.035 × 103. The sum of its digits is 14. It has a total of 3 prime factors and 8 positive divisors. There are 4,480 positive integers (up to 6035) that are relatively prime to 6035.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 4
• Sum of Digits 14
• Digital Root 5
## Name
Short name 6 thousand 35 six thousand thirty-five
## Notation
Scientific notation 6.035 × 103 6.035 × 103
## Prime Factorization of 6035
Prime Factorization 5 × 17 × 71
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) 6035 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 6,035 is 5 × 17 × 71. Since it has a total of 3 prime factors, 6,035 is a composite number.
## Divisors of 6035
1, 5, 17, 71, 85, 355, 1207, 6035
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 7776 Sum of all the positive divisors of n s(n) 1741 Sum of the proper positive divisors of n A(n) 972 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 77.6853 Returns the nth root of the product of n divisors H(n) 6.20885 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 6,035 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 6,035) is 7,776, the average is 972.
## Other Arithmetic Functions (n = 6035)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 4480 Total number of positive integers not greater than n that are coprime to n λ(n) 560 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 792 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 4,480 positive integers (less than 6,035) that are coprime with 6,035. And there are approximately 792 prime numbers less than or equal to 6,035.
## Divisibility of 6035
m n mod m 2 3 4 5 6 7 8 9 1 2 3 0 5 1 3 5
The number 6,035 is divisible by 5.
## Classification of 6035
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (6035)
Base System Value
2 Binary 1011110010011
3 Ternary 22021112
4 Quaternary 1132103
5 Quinary 143120
6 Senary 43535
8 Octal 13623
10 Decimal 6035
12 Duodecimal 35ab
20 Vigesimal f1f
36 Base36 4nn
## Basic calculations (n = 6035)
### Multiplication
n×i
n×2 12070 18105 24140 30175
### Division
ni
n⁄2 3017.5 2011.67 1508.75 1207
### Exponentiation
ni
n2 36421225 219802092875 1326505630500625 8005461480071271875
### Nth Root
i√n
2√n 77.6853 18.2065 8.81392 5.70342
## 6035 as geometric shapes
### Circle
Diameter 12070 37919 1.14421e+08
### Sphere
Volume 9.20705e+11 4.57683e+08 37919
### Square
Length = n
Perimeter 24140 3.64212e+07 8534.78
### Cube
Length = n
Surface area 2.18527e+08 2.19802e+11 10452.9
### Equilateral Triangle
Length = n
Perimeter 18105 1.57709e+07 5226.46
### Triangular Pyramid
Length = n
Surface area 6.30834e+07 2.59039e+10 4927.56
## Cryptographic Hash Functions
md5 4639475d6782a08c1e964f9a4329a254 dd7db1ba26515303a98145d0ac80b5b2f4984035 4a081bff0bf93d06ee54ff45353e8b65bce23522ae6a45769f5968ff5956356e 9393e0370176c09cdac01c1323168566eb8e4f783dcc00cfeb2262f11f32acee27f6f8a9edec79fc2f183346401f6bc3444206c4f6fb67d4017b1cdf599f042b 959096bfa8c4826afb06e0bdfc1dd883a05bd990
| 1,442 | 4,041 |
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| 950,654,640 | 103,727 |
## Day 55 - Lesson 4.6
##### Learning Targets
• Use the multiplication rule for independent events to calculate probabilities.
• Calculate P(at least 1) using the complement rule and the multiplication rule for independent events.
• Determine if it is appropriate to use the multiplication rule for independent events in a given setting.
##### Answer Key:
The big idea here is about independence. Knowing whether or not there is a traffic jam on Tuesday does not change the probability of a traffic jam on Wednesday. In other words, a traffic jam on Wednesday does not depend on whether there was a traffic jam on Tuesday. On the other hand, knowing whether or not there is a snow day on Tuesday does change the probability of a snow day on Wednesday. In other words a snow day on Wednesday does depend on whether there was a snow day on Tuesday. If there is a snow day on Tuesday, it is much more likely that there might be a snow day on Wednesday (maybe a big snow storm dumps a ton of snow).
Students struggled with question #8. We pointed them back to question #7 to help them recognize they needed to use the complement rule here.
Most students will try to calculate the probability for the final question as (0.7)(0.6) = 0.42. We pointed out that the 0.6 probability for Friday would probably go up if we knew that Thursday is a snow day, so we can’t calculate this probability.
##### Teaching Tip:
In the Application and in the homework problems, students sometimes had trouble with P(none) in a given context. They wanted to find P(none) by taking the complement of P(all). These two events are not complements, as complementary events must comprise the whole sample space. What about P(1) or P(2) or P(3) etc?
## Next Day
Copyright © 2020 Stats Medic
Let's connect
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Dividing Money
## How to Divide Money
Jerry saved \$690 in 6 months. If he saved the same amount each month, how much did he save each month?
When dividing money, we follow the same steps as dividing decimals. We just add a dollar sign to our answer.
### Place Values in Money
Before we start dividing money, let's look at what the digits mean.
Digits to the left of the decimal are for dollars. Digits to the right are for cents.
### Dividing Money
Let's solve the problem about Jerry's savings.
Divide the total savings by the number of months.
\$690 ÷ 6 = ?
To divide the numbers, we use the long division method.
The steps are easy to do:
Divide
Multiply
Subtract
Check
Bring down
Let's get started. 😃
First, write the dividend inside the division bar.
Then we write the divisor to the left of the bar.
Tip: We don't have to write the dollar sign while we're computing. We can write it later.
Now, we start dividing.
We follow the steps of long division we learned earlier:
Make sure to write the dollar sign before the number. 💲
\$690 ÷ 6 = \$115
So, Jerry saved \$115 each month.
Great job.
### Another Example
Emma paid \$17.10 for a box of donuts. There were 6 donuts in it. How much does each donut cost?
This problem asks us to divide the total price by the number of donuts in the box.
\$17.10 ÷ 6 = ?
We'll use long division to solve this problem.
First, we arrange the numbers and get ready to divide.
Before dividing, write a decimal point in the quotient. Write this just above the decimal point in the dividend. ✏️
Now that we wrote the decimal point in the quotient, we can ignore it while we divide.
Next, follow the steps of long division to find your quotient, like this:
Our solution is now complete.
The quotient is 02.85.
We drop the leading zero, and just write \$2.85.
Remember to write the dollar sign before the number.
\$17.10 ÷ 6 = \$2.85
Now we know that each donut costs \$2.85.
Great work learning how to divide money. 👏
Now, start the practice.
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# 13000 Foot to Meter
Convert 13000 (thirteen thousand) Foots to Meters (ft to m) with our conversion calculator.
13000 Foots to Meters equals 3,962 m.
• Meter
• Kilometer
• Centimeter
• Millimeter
• Micrometer
• Nanometer
• Mile
• Yard
• Foot
• Inch
• Light Year
• Meter
• Kilometer
• Centimeter
• Millimeter
• Micrometer
• Nanometer
• Mile
• Yard
• Foot
• Inch
• Light Year
Convert 13000 Foots to Meters (ft to m) with our conversion calculator.
13000 Foots to Meters equals 3,962 m.
To convert 13000 feet to meters, you first need to know the conversion factor. One foot is equal to 0.3048 meters. Using this information, you can set up the conversion as follows:
$$[13000 \text{ feet} \times 0.3048 \frac{\text{meters}}{\text{foot}} = 3962.4 \text{ meters}]$$
Thus, 13000 feet is equal to 3962.4 meters.
The calculation is done based on the relationship between feet and meters. This relationship is defined internationally, with one foot officially being defined to be exactly 0.3048 meters. To find out how many meters are in a certain number of feet, you multiply the number of feet by 0.3048. This conversion is particularly useful in situations where measurements given in feet need to be related to metrics that are standard in countries using the metric system, including science and engineering contexts. The simplicity of this multiplication makes it straightforward to convert measurements from feet to meters, allowing for a clear understanding and comparison of lengths and distances across different measurement systems.
Items that are approximately 13000 feet in length include:
1. The height of many of the world's peaks:
• Several mountains and peaks around the world stand at or near this height, making them formidable challenges for climbers.
2. The length of some of the longest suspension bridges:
• Some suspension bridges, when measured from end to end including their approach spans, can reach lengths in this range.
3. Runways at large international airports:
• Some of the longest runways at major airports are approximately this length to accommodate the takeoff and landing of large, heavily-loaded aircraft.
4. Certain deep ocean trenches:
• Parts of the ocean floor, such as trenches or drop-offs, can plunge down to depths around 13000 feet, highlighting the vastness and mystery of the ocean.
5. The vertical drop of some of the largest waterfalls:
• Although no single waterfall has a vertical drop of 13000 feet, some of the world's largest waterfalls, when combining all of their cascades, come close to reaching similar elevation changes.
6. Large-scale engineering tunnels:
• Tunnels designed for roads, trains, or utilities might extend for several kilometers, with the longest reaching or exceeding this length.
7. The approximate cruising altitude of light to medium-sized aircraft:
• This is a typical cruising altitude for many passenger and cargo aircraft, striking a balance between efficiency and the avoidance of weather-related turbulence.
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# Can someone explain, in a simple way, why does wavelength affect diffraction?
I have already seen the following question: Why does wavelength affect diffraction?
But I still can't seem to understand the relationship. I understand that if the wavelength is the same size as the slit, the wave pattern will look like circles on the other side of the slit, which is maximum diffraction, because the slit will "function" as a point source.
But I don't understand why, if we reduce the wavelength, there's gonna be like more point sources in the same slit. What's the reason?
I hope I could make myself clear enough.
• Someone correct me if I'm wrong, but I think if the slit is too narrow, too much of the wave gets reflected; and if it's too wide, you get single slit interference: en.wikipedia.org/wiki/Diffraction#Single-slit_diffraction – Wood Nov 25 '16 at 22:40
• That's correct. But in this case of single slit interference, there's approximately 5 point sources in the space of the slit - but, technically, if you increase the wavelength, the number of point sources will reduce. The reason for that correlation is what I don't understand. – Roberto Valente Nov 25 '16 at 22:50
• "But in this case of single slit interference, there's approximately 5 point sources ..." In a particular visual representation someone drew five point sources. In reality this is a mathematical continuum. There are no discrete points, there is a whole wavefront. – dmckee Nov 26 '16 at 2:23
Forget about the "number of points". There are infinite number of points in the slit.
The explanation has to do with time. Each point of the slit corresponds to a different delay that light takes to reach a given target from that point.
For instance, if the target is sideways from the slit, some points of the slit will be closer (less delay) and some points will be further (more delay). The smaller the wavelength, the more oscillations that delay will contain. More oscillations mean more cancelling between them, because averaging over many oscillations adds up to zero. To summarize this, the smaller the wavelength, the more cancelling out of waves, thus the less light. This explains why a smaller wavelength tends to have less diffraction to the side.
The more you take a target to the side of the slit, the more delay you will add, thus less and less light.
Wavelength much larger than slit:
The slit is filled with infinitely many point sources, each one of which emits circular waves, as point sources always do. Those waves interfere constructively everywhere.
Wavelength much smaller than slit:
The slit is filled with infinitely many point sources, each one of which emits circular waves, as point sources always do. Those waves interfere constructively and destructively everywhere.
Any place receives waves from all the point sources, and those waves are not in phase, because of the different distances to different point sources. When some place seems to receive almost no waves, that's because the result of all the constructive and destructive interferences is almost zero there.
• thanks, I think I understand it better now. I have also found another link which I found helpful if anyone is interested here – Roberto Valente Nov 30 '16 at 1:02
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# A handbook of practical gauging
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Page iii - KEENE.— A Hand-Book of Practical Gauging: For the Use of Beginners, to which is added a Chapter on Distillation, describing the process in operation at the Custom- House for ascertaining the Strength of Wines. By JAMES B. KEENE, of HM Customs. 8vo.
Page 16 - Rule. — Multiply the area of the base by the perpendicular height, and the product will be the solid contents.
Page 13 - Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product will be the area.
Page 5 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a right angle ; and the straight line which stands on the other is called a perpendicular to it.
Page 17 - To find the solidity of a spheroid. Rule. — Multiply the square of the revolving axis by the fixed axis: and the product, multiplied by .5236, will give the solidity.
Page 12 - From half the sum of the three sides subtract each side ; multiply the half sum and the three remainders together, and the square root of the product will be the area required.
Page 18 - RULE.* Multiply the sum of the squares of the diameters of the two ends by the height of the frustum, and the product again by .3927, and it will give the solidity.
Page 5 - A Circle is a plane figure bounded by a curve line, called the Circumference, which is everywhere equidistant from a certain point within, called its Centre.
Page 16 - To find the solidity of a cone or pyramid. RULE . — Multiply the area of the base by ^ of its height.
Page 8 - A cone is a solid figure described by the revolution of a right angled triangle about one of the sides containing the right angle, which side remains fixed.
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Module 2: Computer Arithmetic
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1 Module 2: Computer Arithmetic 1 B O O K : C O M P U T E R O R G A N I Z A T I O N A N D D E S I G N, 3 E D, D A V I D L. P A T T E R S O N A N D J O H N L. H A N N E S S Y, M O R G A N K A U F M A N N P U B L I S H E R S
2 Complement Addition Subtraction Multiplication Division Arithmetic Operations 2
3 Introduction Numbers are represented by binary digits (bits): How are negative numbers represented? What is the largest number that can be represented in a computer world? What happens if an operation creates a number bigger than can be represented? What about fractions and real numbers? 3 A mystery: How does hardware really multiply or divide numbers?
4 Binary Numbers 4
5 Binary Numbers Binary Number System System Digits: 0 and 1 Bit (short for binary digit): A single binary digit LSB (least significant bit): The rightmost bit MSB (most significant bit): The leftmost bit 5 Upper Byte (or nybble): The right-hand byte (or nybble) of a pair Lower Byte (or nybble): The left-hand byte (or nybble) of a pair
6 Binary Equivalents Binary Numbers 1 Nybble (or nibble) = 4 bits 1 Byte = 2 nybbles = 8 bits 1 Kilobyte (KB) = 1024 bytes 1 Megabyte (MB) = 1024 kilobytes = 1,048,576 bytes 1 Gigabyte (GB) = 1024 megabytes = 1,073,741,824 bytes 6
7 Binary Numbers 7 Base 2
8 Binary Addition Rules of Binary Addition = = = = 0, and carry 1 to the next more significant bit Example: Carry
9 Binary Subtraction Rules of Binary Subtraction 0-0 = = 1, and borrow 1 from the next more significant bit 1-0 = = 0 Example: borrowed
10 Binary Multiplication Rules of Binary Multiplication 0 x 0 = 0 0 x 1 = 0 1 x 0 = 0 1 x 1 = 1, and no carry or borrow bits Example: 23 x 3 10 Another Method: Binary multiplication is the same as repeated binary addition
11 Binary division 11
12 2 s Complement Two's complement representation allows the use of binary arithmetic operations on signed integers, yielding the correct 2's complement results. Positive Numbers: Positive 2's complement numbers are represented as the simple binary. Negative Numbers: Negative 2's complement numbers are represented as the binary number that when added to a positive number of the same magnitude, will equals zero. 12
13 13 To represent positive and negative numbers look at the MSB (or the sign bit) MSB = 0 means positive MSB = 1 means negative
14 Step 1: Calculation of 2's Complement 14 invert the binary equivalent of the number by changing all of the ones to zeroes and all of the zeroes to ones (also called 1's complement) Step 2: Then add one. Example: = = Step1 : Step2 :
15 2's Complement Addition Two's complement addition follows the same rules as binary addition. Example: 5 + (-3) 5 = = Ignore (2)
16 2's Complement Addition Two's complement addition follows the same rules as binary addition. Example: 3 + (-5) 3 = = (-2)
17 2's Complement Subtraction 17 Two's complement subtraction is the binary addition of the minuhend to the 2's complement of the subtrahend (adding a negative number is the same as subtracting a positive one). Example : (-12) minuhend subtrahend
18 2's Complement Subtraction 18 Example : (-12) 7 = = (-5) CHECK! -5 = Step1 : Step2 : (5)
19 2's Complement Multiplication Two's complement multiplication follows the same rules as binary multiplication. Example : (-4) 4 = (-16) Ignore 4 = = x (-16) 19 CHECK! -16 = Step1 : Step2 : (16)
20 2's Complement Division Two's complement division is repeated 2's complement subtraction. 20 The 2's complement of the divisor is calculated, then added to the dividend. For the next subtraction cycle, the quotient replaces the dividend. This repeats until the quotient is too small for subtraction or is zero, then it becomes the remainder. The final answer is the total of subtraction cycles plus the dividend remainder Example : 6/ 3 = 2 divisor quotient
21 2's Complement Division Example: 6/3 Example: 7/3 6 + (-3) = (-3) = 0 6 / 3 = 2 Cycle 1 Cycle 2 The number of cycle (-3) = (-3) = 1 Cycle 1 Cycle 2 7 / 3 = 2 remainder 1
22 2's Complement Division 22 Example: 42/ (-6) = (-6) = (-6) = (-6) = 18 Cycle 1 Cycle 2 Cycle 3 Cycle (-6) = (-6) = (-6) = 0 42 / 6 = 7 Cycle 5 Cycle 6 Cycle 7
23 Sign Extension Extending a number representation to a larger number of bits. Example: 2 in 8 bit binary to 16 bit binary In signed numbers, it is important to extend the sign bit to preserve the number (+ve or ve) Example: -2 in 8 bit binary to 16 bit binary Sign bit Sign bit extended Sign bit
24 Detecting Overflow in Two Complement Numbers Overflow occurs when adding two positive numbers and the sum is negative, or vice versa A carry out occurred into the sign bit Overflow conditions for addition and subtraction 24
25 Overflow Rule for addition If 2 Two's Complement numbers are added, and they both have the same sign (both positive -7 or both = 1001 negative), -6 then = overflow 1010 occurs if and only if the result has the opposite sign. Adding two positive numbers must give a positive result Adding two negative numbers must give a negative result Overflow occurs Overflow never occurs when adding operands with different signs. Overflow occurs if 25 (+A) + (+B) = C ( A) + ( B) = +C (3) Example: Using 4-bit Two's Complement numbers ( 8 x +7) (-7) + (-6) = (-13) but Overflow (largest ve number is 8) The sign bit has changed to +ve
26 Overflow Rule for Subtraction If 2 Two's Complement numbers are subtracted, and their signs are different, then overflow occurs if and only if the result has the same sign as the subtrahend. subtrahend Overflow occurs if (+A) ( B) = C ( A) (+B) = +C result Example: Using 4-bit Two's Complement numbers ( 8 x +7) Subtract 6 from +7 (i.e. 7 (-6)) 26 result has the same sign as the subtrahend overflow happens
27 Overflow Rule for Subtraction If 2 Two's Complement numbers are subtracted, and their signs are different, then overflow occurs if and only if the result has the same sign as the subtrahend. Overflow occurs if (+A) ( B) = C ( A) (+B) = +C 7 = 0111 Example: Using 4-bit Two's Complement numbers ( 8 x +7) Subtract 6 from +7 (i.e. 7 (-6)) 27 Overflow occurs 6 = (-3) Result has same sign as subtrahend
29 Binary Multiplication 29
30 Two's Complement Multiplication There are a couple of ways of doing two's complement multiplication by hand. Remember that the result can require 2 times as many bits as the original operands. 30
31 Sign Extend Method" for Two's Complement Multiplication In 2's complement, sign extend both integers to twice as many bits. Then take the correct number of result bits from the least significant portion of the result A 4-bit example: 2 x (-3) 2= = = Sign extend to 8 bit x Correct answer underlined
32 " Partial Product Sign Extend Method " for Two's Complement Multiplication 32 Another way is to sign extend the partial products 1100 to the correct number of bits. x 0011 Sometimes we do have to make some adjustments Example 1: (-4) x 3-4 = = 0011 Sign extend to the correct number of bits (-12)
33 " Partial Product Sign Extend Method " for Two's Complement Multiplication Sometimes we do have to make some adjustments. If (+ve) x (+ve) then OK Normal stuff If (+ve) x (-ve) then get additive inverse of both And then Sign extend partial products Example: 3 x (-4) 33 If (-ve) x (+ve) then Sign extend partial products 3=0011 ;-4 = =1101; 4 =0100 If (-ve) x 1101 (-ve) then get x additive 0100 inverse of both (-12) Like the slide before (-4)x3
34 " Partial Product Sign Extend Method " for Two's Complement Multiplication Sometimes we do have to make some adjustments. 34 If (+ve) x (+ve) then OK -3= = If (+ve) x (-ve) then x 0011 get additive inverse of both 0011 And then 0011 Sign extend partial products (9) Example: 3 x (-4) If (-ve) x (+ve) then Sign extend partial products If (-ve) x (-ve) then get additive inverse of both Example: (-3)x(-3)
35 Signed Multiplication Another way to deal with signed numbers. 35 First convert the multiplier and multiplicand to positive numbers and then remember the original signs Leave the sign out of the calculation To negate the product only if the original signs disagree
36 1st Version of Multiplication Hardware Actual implementations are far more complex, and use algorithms that generate more than one bit of product each clock cycle. 32-bit multiplicand starts at right half of multiplicand register Algorithm Flows of 1st Version Multiplication Multiplier0 = 1 Start 1. Test Multiplier0 Multiplier0 = 0 Multiplicand 64 bits Shift left 1a. Add multiplicand to product and place the result in Product register 64-bit ALU Multiplier Shift right 32 bits 2. Shift the Multiplicand register left 1 bit Product 64 bits Write Control test 3. Shift the Multiplier register right 1 bit Product register is initialized at 0 Multiplicand register, product register, ALU are 64-bit wide; multiplier register is 32-bit wide 32nd repetition? No: < 32 repetitions Yes: 32 repetitions Done
37 Example of Multiplication 4 bits Example : 2 x 3 =? Multiplicand (MC) Multiplier (MP) Product (P) 2 x x 0011 Steps: 1a test multiplier (0 or 1) If 1 then P = P + MC If 0 then no operation 2 shift MC left 3 shift MP right All bits done? If still <max bit, repeat If = max bit, stop 37 Multiplier0 = 1 1a. Add multiplicand to product and place the result in Product register Start Test Multiplier0 2. Shift the Multiplicand register left 1 bit 3. Shift the Multiplier register right 1 bit 32nd repetition? Multiplier0 = 0 P = P + MC = = MC = MP = No: < 32 repetitions Max bit? NO repeat Yes: 32 repetitions Done
38 Iteration Step Multiplier (MP) Multiplicand (MC) Product (P) 0 Initial value a:1 P = P + MC 2: Shift MC left 3: Shift MP right 1a:1 P = P + MC 2: Shift MC left 3: Shift MP right 1a:0 no operation 2: Shift MC left 3: Shift MP right 1a:0 no operation 2: Shift MC left 3: Shift MP right Try with 5 x
39 Iteration Step Multiplier (MP) Multiplicand (MC) Product (P) 0 Initial value Example : 5 x 4 1a:0 no operation 2: Shift MC left : Shift MP right a:0 no operation 2: Shift MC left : Shift MP right a:1 P = P + MC : Shift MC left : Shift MP right a:0 no operation 2: Shift MC left : Shift MP right 0000 Try with 2 x (-3) 20 40
40 Iteration Step Multiplier (MP) Multiplicand (MC) Product (P) 0 Initial value a:1 P = P + MC Example : 2 x (-3) get additive inverse of both 2: Shift MC left : Shift MP right a:1 P = P + MC : Shift MC left : Shift MP right a:0 no operation 2: Shift MC left : Shift MP right a:0 no operation 2: Shift MC left : Shift MP right
41 Binary Division 42
42 1st Version of Division Hardware Divisor starts at left half of divisor register 32-bit divisor starts at left half of divisor register Divisor 64 bits Shift right Quotient register is initialized to be 0 43 Flows of 1st Version Division Start 1. Subtract the Divisor register from the Remainder register and place the result in the Remainder register Remainder > 0 Test Remainder Remainder < 0 64-bit ALU Remainder Write Control test Quotient Shift left 32 bits 2a. Shift the Quotient register to the left, setting the new rightmost bit to 1 2b. Restore the original value by adding the Divisor register to the Remainder register and place the sum in the Remainder register. Also shift the Quotient register to the left, setting the new least significant bit to 0 64 bits Remainder register is initialized with the dividend at right 3. Shift the Divisor register right 1 bit Divisor register, remainder register, ALU are 64-bit wide; quotient register is 32-bit wide 33rd repetition? No: < 33 repetitions Yes: 33 repetitions Done Algorithm
43 Example of Division 4 bits 44 Start 1. R = R - D Example : 7 / 2 =? Dividend (DD) Divisor (D) 7 / / 0010 Steps: 1 Remainder (R) = R D 2 test new R (>=0 or <0) 2a - If >=0 then R = no operation; Q = Shift left (add 1 at LSB) 2b - If <0 then R = D + R Q = Shift left (add 0 at LSB) 3 shift D right All bits done? If still <(max bit + 1), repeat If = (max bit+1), stop Quotient (Q) If R = R D = +ve 2a. Shift the Quotient register to the left, setting the new rightmost bit to 1 1. Subtract the Divisor register from the Remainder register and place the result in the Remainder register Remainder > 0 2a. R >= 0; R = no operation Q = Shift left (add 1 at LSB) 4. If not yet 33 repeat to Step 1 (new iteration) Test Remainder 3. D = Shift right 1 3. Shift the Divisor register right 1 bit 33rd repetition? If R = R D = -ve Remainder < 0 2b. R < 0; R = D + R Q = Shift left (add 0 at LSB) 2b. Restore the original value by adding the Divisor register to the Remainder register and place the sum in the Remainder register. Also shift the Quotient register to the left, setting the new least significant bit to 0 No: < 33 repetitions Yes: 33 repetitions Done
44 Iteration Step Quotient (Q) Divisor Divisor Remainder ( (D) starts at left half Example: 7/2 R) of divisor register 0 Initial value R = R - D R = R D = R + (-D) Negate b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R - D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R - D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right
45 Iteration Step Quotient (Q) Divisor (D) Remainder(R) R = R - D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R - D b. R >=0; R = no operation Q = Shift left (add 1 at LSB) D = Shift right R = R - D b. R >=0; R = no operation Q = Shift left (add 1 at LSB) D = Shift right Example: 7/2 = 3 remainder 1 47
46 Example: 6/3 Iteration Step Quotient (Q) Divisor (D) Remainder(R) 0 Initial value R = R - D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R - D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R - D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right
47 Iteration Step Quotient (Q) Divisor (D) Remainder (R) R = R - D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right R = R - D b. R >=0; R = no operation Q = Shift left (add 1 at LSB) D = Shift right R = R - D b. R < 0; R = D + R Q = Shift left (add 0 at LSB) D = Shift right Example: 6/3 = 2 49
48 Signed Division Signed divide: make both divisor and dividend positive and perform division negate the quotient if divisor and dividend were of opposite signs make the sign of the remainder match that of the dividend this ensures always dividend = (quotient divisor) + remainder quotient (x/y) = quotient ( x/y) e.g. 7 = = 3 2 1) 50
49 Module 2 Part 2 51 FLOATING POINT
50 Floating Point Representation Floating point aids in the representation of very big or very small fixed point numbers x Fixed point Floating point 976,000,000,000, x x 10-14
51 Floating Point Numbers 53 Significand/Fraction Exponent x Base/radix Significand and Exponent can be +ve or ve. Decimal numbers use the radix 10, binary use 2
52 Normalized and Unnormalized In generalized normalization (like in mathematics), a floating point number number is said to after be normalized the if the number after the radix radix point point is a non-zero is a value. Un-normalized floating number is when the number after the radix point is 0. Example: x x x x 10 5 unnormalized non-zero value. unnormalized normalized normalized
53 Normalization Process Normalization is the process of deleting the zeroes until a Move non-zero radix point value to is detected. the right (in this case 2 Example points) : 56 Move radix point to the left (in this x case points) x x x x x 10 6 A rule of thumb: moving the radix point to the right subtract exponent moving the radix point to the left add exponent
54 Floating Point Format for Binary Numbers In the beginning, the general form of floating point is : +/- 0.Mantissa x r +/- exponent In binary: +/- (sign) Exponent +/- (sign) Mantissa 1 word Mantissa = Significand The 2 sign bits are not good for design as it incurs extra costs
55 Biased Exponent A new value to represent the exponent without the sign bit is introduced 59 This will eliminate the sign for the exponent value that is the exponent will be positive. (indicative) +/- (sign) Biased Exponent Mantissa 1 word +/- (1 bit) E t (n bit) Mantissa, m
56 Biased Exponent 60 +/- (1 bit) E t (n bit) Mantissa, m Biased value, b = 2 n-1 Normalized exponent, e = E t - b Biased exponent, E t = e + b Where, E t = biased exponent n = bits of exponent format (i.e. the word format) This is used unless the IEEE standard is mentioned then is a different calculation
57 Conversion to Floating Point Number Change to binary (if given decimal number) Normalized the number Change the number to biased exponent Form the word (3 fields) 61
58 Example 3 62 Transform to floating point word using the following format (radix 2) Sign Biased exponent Significand 1 bit 4 bit 12 bit Step 1 : Change to binary This is the given word format x 2 = x 2 = x 2 = = = = x
59 Example 3 Step 2 : Normalized the number x Normalized exponent, e Step 3: Change the number to biased exponent 1 bit 4 bit 12 bit Biased value, b = 2 n-1 = = 8d = 1000b Biased exponent, E t = e + b = = x x
60 Example x Step 4 : Form the word (3 fields) 1 bit 4 bit 12 bit Padding Rule of thumb: -the biased exponent is always padded to the left - the significand (or mantissa) is always padded to the right
61 Floating-Point Representation Value in general form: (-1) S x F x 2 e In an 8-bit representation, we can represent: From 2 x to 2 x This is called single-precision = 1 word If anything goes above or under, then overflow and underflow happens respectively. One way to reduce this is to offer another format with a larger exponent use double precision (2 words) From 2 x to 2 x
62 IEEE 754 Floating-Point Standard 69 This 1 is made implicit to pack more bits into the significand
63 Normalized Scientific Notation In IEEE standard normalization (used in computers), a floating point number is said to be normalized if there is only a single non-zero before the radix point. Example: 70 there is only a single non-zero before the radix point normalized x B normalized x 2 011
64 Challenge of Negative Exponents Placing the exponent before the significand simplifies sorting sign of floating-point Exponent numbers using Significand integer comparison 1 bit instructions. 8 bit 23 bit However, using 2 s complement in the exponent field makes a negative exponent look like a big number. 72
65 Biased Notation 73 Bias In single precision is 127 In double precision 1023 Biased notation (-1)sign x (1 + Fraction) x 2 (exponent-bias)
66 74 To convert a decimal number to single (or double) precision floating point: Step 1: Normalize Step 2: Determine Sign Bit Step 3: Determine exponent IEEE 754 Conversion Step 4: Determine Significand
67 IEEE 754 Conversion : Example 1 Convert 10.4 d to single precision floating point. Step 1: Normalize x 2 = x 2 = x 2 = x 2 = x 2 = x 2 = For continuous results, take the 1 st pattern before it repeats itself 0.4 = = x x 2 3
68 IEEE 754 Conversion : Example 1 76 Step 2: Determine Sign Bit (S) Because (10.4) is positive, S = 0 3 is from 2 3 Step 3: Determine exponent Because its single precision bias = 127 Exponent = 3 + bias = = 130 d = b Step4: Determine Significand Drop the leading 1 of the significand x Then expand (padding) to 23 bits sign Exponent Significand
69 IEEE 754 Conversion : Example 2 Convert d to single precision floating point. Step 1: Normalize x 2 = x 2 = x 2 = = x x 2-1
70 IEEE 754 Conversion : Example 2 78 Step 2: Determine Sign Bit (S) Because (-0.75) is negative, S = 1 Step 3: Determine exponent Because its single precision bias = 127 Exponent = -1 + bias = = 126 d = b Step4: Determine Significand Drop the leading 1 of the significand -1.1 x Then expand (padding) to 23 bits sign Exponent Significand
71 IEEE 754 Conversion : Example 3 Convert d to double precision floating point. Step 1: Normalize x 2 = x 2 = x 2 = = x x 2-1
72 IEEE 754 Conversion : Example 3 80 Step 2: Determine Sign Bit (S) Because (-0.75) is negative, S = 1 Step 3: Determine exponent Because its double precision bias = 1023 Exponent = -1 + bias = = 1022 d = b Step4: Determine Significand Drop the leading 1 of the significand -1.1 x Then expand (padding) to 52 bits sign Exponent (11) Significand (52)
73 Converting Binary to Decimal Floating-Point What decimal number is represented by this single precision float? Extract the values: Sign = 1 81 Sign (1 bit) Exponent(8 bit) Significand(23 bit) Exponent = b = 129 d Significand Remember: Biased notation (-1)sign x (1 + Fraction) x 2 (exponent-bias) The Fraction = (0 x 2-1 ) + (1 x 2-2 ) + (0 x 2-3 ) = ¼ = ( ) The number = - (1.25 x 2 (exponent-bias) ) = - (1.25 x 2 2 ) = - (1.25 x 4) = -5.0
74 Module 2 Part 3 82 FLOATING-POINT OPERATIONS
76 Decimal Floating-Point Addition 84 Assume 4 decimal digits for significand and 2 decimal digits for exponent Step 1: Align the decimal point of the number that has the smaller exponent Step 2: add the significand Step 3: Normalize the sum check for overflow/underflow of the exponent after normalisation Step 4: Round the significand If the significand does not fit in the space reserved for it, it has to be rounded off Step 5: Normalize it (if need be)
77 Decimal Floating-Point Addition Example: d x d x 10-1 Step 1: Align the decimal point of the number that has the smaller exponent Make d x 10-1 to x = 1 x = 2 move 2 to left d x 10 1 Step 2: add the significand x d x x
78 Decimal Floating-Point Addition Example: d x d x 10-1 Step 3: Normalize the sum x x Step 4: Round the significand (to 4 decimal digits for significand) x x 10 2 Step 5: Normalize it (if need be) No need as its normalized
79 Decimal Floating-Point Addition Example: 0.5 d + ( d ) Adjusts the numbers Step 1: Align the decimal point of the number that has the smaller exponent b x b x b x b x 2-2 Make 1.11 b x 2-2 to x = -1 x = 1 move 1 to left b x 2-1
80 Decimal Floating-Point Addition Example: 0.5 d + ( d ) Step 2: add the significand Step 3: Normalize the sum Step 4: Round the significand (to 4 decimal digits for significand) Fits in the 4 decimal digits Step 5: Normalize it (if need be) No need as its normalized x x
81 Floating-Point Multiplication Flows 90
82 Floating-Point Multiplication 91 Assume 4 decimal digits for significand and 2 decimal digits for exponent Step 1: Add the exponent of the 2 numbers Step 2: Multiply the significands Step 3: Normalize the product check for overflow/underflow of the exponent after normalisation Step 4: Round the significand If the significand does not fit in the space reserved for it, it has to be rounded off Step 5: Normalize it (if need be) Step 6: Set the sign of the product
83 Floating-Point Multiplication 92 Example: (1.110 d x ) x (9.200 d x 10-5 ) Assume 4 decimal digits for significand and 2 decimal digits for exponent Step 1: Add the exponent of the 2 numbers 10 + (-5) = 5 If biased is considered 10 + (-5) = 132 Step 2: Multiply the significands x x 10 5
84 Floating-Point Multiplication Example: (1.110 d x ) x (9.200 d x 10-5 ) Step 3: Normalize the product x x 10 6 Step 4: Round the significand (4 decimal digits for significand) x 10 6 Still normalized Step 5: Normalize it (if need be) Step 6: Set the sign of the product x 10 6
85 Floating-Point Multiplication Example: (1.000 b x 2-1 ) x ( b x 2-2 ) 95 Assume 4 decimal digits for significand and 2 decimal digits for exponent Step 1: Add the exponent of the 2 numbers -1 + (-2) = -3 If biased is considered -1 + (-2) = 124 Step 2: Multiply the significands x x 10-3
86 Floating-Point Multiplication Example: (1.000 b x 2-1 ) x ( b x 2-2 Step 3: Normalize the product x 10-3 already normalized 96 Step 4: Round the significand (4 decimal digits for significand) x 10-3 Still normalized Step 5: Normalize it (if need be) Step 6: Set the sign of the product b x /32 d
87 Floating-Point ALU Sign Exponent Significand Sign Exponent Significand 97 Small ALU Compare exponents Exponent difference Control Shift right Shift smaller number right Big ALU Add 0 1 Increment or decrement 0 1 Shift left or right Normalize Rounding hardware Round Sign Exponent Significand
88 Accurate Arithmetic If a calculation exceeds the limits of the floating point scheme then CPU will flag this error. 98 If number is too tiny to be represented
89 Accurate Arithmetic : Truncation & Rounding 99 Some number have infinite decimal points (the irrational numbers) 1/3 d = Truncation is done to fit the decimal points into manageable units. Truncation is where decimal values beyond the truncation point are simply discarded and it can cause error in floating point calculations. Rounding :If you have a number such as then if you have to round this to 3 significant digits, the number becomes 3.46 A small error called the rounding error has occurred ***Note : the CPU will not flag any error when truncation and rounding occurs, as it is acting within its limits. programmers must assess if this will lead to significant errors
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# 2060
## 2,060 is an even composite number composed of three prime numbers multiplied together.
What does the number 2060 look like?
This visualization shows the relationship between its 3 prime factors (large circles) and 12 divisors.
2060 is an even composite number. It is composed of three distinct prime numbers multiplied together. It has a total of twelve divisors.
## Prime factorization of 2060:
### 22 × 5 × 103
(2 × 2 × 5 × 103)
See below for interesting mathematical facts about the number 2060 from the Numbermatics database.
### Names of 2060
• Cardinal: 2060 can be written as Two thousand and sixty.
### Scientific notation
• Scientific notation: 2.06 × 103
### Factors of 2060
• Number of distinct prime factors ω(n): 3
• Total number of prime factors Ω(n): 4
• Sum of prime factors: 110
### Divisors of 2060
• Number of divisors d(n): 12
• Complete list of divisors:
• Sum of all divisors σ(n): 4368
• Sum of proper divisors (its aliquot sum) s(n): 2308
• 2060 is an abundant number, because the sum of its proper divisors (2308) is greater than itself. Its abundance is 248
### Bases of 2060
• Binary: 1000000011002
• Base-36: 1L8
### Squares and roots of 2060
• 2060 squared (20602) is 4243600
• 2060 cubed (20603) is 8741816000
• The square root of 2060 is 45.3872228719
• The cube root of 2060 is 12.7239632707
### Scales and comparisons
How big is 2060?
• 2,060 seconds is equal to 34 minutes, 20 seconds.
• To count from 1 to 2,060 would take you about thirty-four minutes.
This is a very rough estimate, based on a speaking rate of half a second every third order of magnitude. If you speak quickly, you could probably say any randomly-chosen number between one and a thousand in around half a second. Very big numbers obviously take longer to say, so we add half a second for every extra x1000. (We do not count involuntary pauses, bathroom breaks or the necessity of sleep in our calculation!)
• A cube with a volume of 2060 cubic inches would be around 1.1 feet tall.
### Recreational maths with 2060
• 2060 backwards is 0602
• The number of decimal digits it has is: 4
• The sum of 2060's digits is 8
• More coming soon!
MLA style:
"Number 2060 - Facts about the integer". Numbermatics.com. 2022. Web. 5 July 2022.
APA style:
Numbermatics. (2022). Number 2060 - Facts about the integer. Retrieved 5 July 2022, from https://numbermatics.com/n/2060/
Chicago style:
Numbermatics. 2022. "Number 2060 - Facts about the integer". https://numbermatics.com/n/2060/
The information we have on file for 2060 includes mathematical data and numerical statistics calculated using standard algorithms and methods. We are adding more all the time. If there are any features you would like to see, please contact us. Information provided for educational use, intellectual curiosity and fun!
Keywords: Divisors of 2060, math, Factors of 2060, curriculum, school, college, exams, university, Prime factorization of 2060, STEM, science, technology, engineering, physics, economics, calculator, two thousand and sixty.
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University Calculus: Early Transcendentals (3rd Edition)
$\frac{dA}{dt}=\frac{1}{2}ab\cos\theta\frac{d\theta}{dt}$ $\frac{dA}{dt}=\frac{b}{2}(\sin\theta\frac{da}{dt}+a\cos\theta\frac{d\theta}{dt})$ $\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$
Given $A=\frac{1}{2}ab\sin\theta$ on taking a and b constant and differentiating the terms, we get: $\frac{dA}{dt}=\frac{1}{2}ab\cos\theta\frac{d\theta}{dt}$ taking b as a constant and differentiating A, we get: $\frac{dA}{dt}=\frac{b}{2}(\sin\theta\frac{da}{dt}+a\cos\theta\frac{d\theta}{dt})$ when no variables are constant and applying the triple product rule, we get: (uvw)'=u'vw+uv'w+uvw' $\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\frac{d\sin\theta}{dt})$ $\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$ Thus the final answer is: $\frac{dA}{dt}=\frac{1}{2}ab\cos\theta\frac{d\theta}{dt}$ $\frac{dA}{dt}=\frac{b}{2}(\sin\theta\frac{da}{dt}+a\cos\theta\frac{d\theta}{dt})$ $\frac{dA}{dt}=\frac{1}{2}(a\sin\theta\frac{db}{dt}+b\sin\theta\frac{da}{dt}+ab\cos\theta\frac{d\theta}{dt})$
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## College Physics (4th Edition)
The riders move through a distance of $16.8~m$
We can convert the angle to radians: $120^{\circ}\times \frac{\pi~rad}{180^{\circ}} = \frac{2\pi}{3}~rad$ We can find the distance through which the riders move: $d = \theta~r$ $d = (\frac{2\pi}{3}~rad)(8.0~m)$ $d = 16.8~m$ The riders move through a distance of $16.8~m$
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# 2003 AMC 10B Problems/Problem 19
The following problem is from both the 2003 AMC 12B #16 and 2003 AMC 10B #19, so both problems redirect to this page.
## Problem
Three semicircles of radius $1$ are constructed on diameter $\overline{AB}$ of a semicircle of radius $2$. The centers of the small semicircles divide $\overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?
$[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype("4 4"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label("A",A,W); label("B",B,E); label("1",(-1.5,-0.1),S); label("2",(0,-0.1),S); label("1",(1.5,-0.1),S);[/asy]$
$\textbf{(A) } \pi - \sqrt{3} \qquad\textbf{(B) } \pi - \sqrt{2} \qquad\textbf{(C) } \frac{\pi + \sqrt{2}}{2} \qquad\textbf{(D) } \frac{\pi +\sqrt{3}}{2} \qquad\textbf{(E) } \frac{7}{6}\pi - \frac{\sqrt{3}}{2}$
## Solution 1
$[asy] import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype("4 4"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label("A",A,W); label("B",B,E); label("1",(-1.5,-0.1),S); label("2",(0,-0.1),S); label("1",(1.5,-0.1),S); draw((1,0)--(0.5,0.866)); draw((0,0)--(0.5,0.866)); draw((-1,0)--(-0.5,0.866)); draw((0,0)--(-0.5,0.866));[/asy]$
By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into $\frac{5}{6}$ of a circle with radius $1$ and two equilateral triangles with side length $1$. This gives the area of the white region as $\frac{5}{6}\pi+\frac{2\cdot\sqrt3}{4}=\frac{5}{6}\pi+\frac{\sqrt3}{2}$. The area of the shaded region is the area of the white region subtracted from the area of the large semicircle. This is equivalent to $2\pi-\left(\frac{5}{6}\pi+\frac{\sqrt3}{2}\right)=\frac{7}{6}\pi-\frac{\sqrt3}{2}$.
Thus the answer is $\boxed{\textbf{(E)}\ \frac{7}{6}\pi-\frac{\sqrt3}{2}}$.
### Note
The reason why it is $\frac{5}{6}$ of a circle and why the triangles are equilateral are because, first, the radii are the same and they make up the equilateral triangles.
Secondly, the reason it is $\frac{5}{6}$ of a circle is because the middle sector has a degree of $180-2 \cdot 60 = 60$ and thus $\frac{60}{360}=\frac{1}{6}$ of a circle.
The other two have areas of $\frac{180-60}{360}=\frac{1}{3}$ of a triangle each.
Therefore, the total fraction of the circle(since they have the same radii) is $\frac{1}{6} + 2 \cdot \frac{1}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}.$
~mathboy282
Answer choices A and B are impossible since the area is obviously not a circle of radius 1 minus a triangle.
Answer choices C and D are also impossible since they are adding but we are subtracting.
That leaves us with only answer choice $\textbf{(E)}.$
~mathboy282
~savannahsolver
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## Algebra 2
### Course: Algebra 2>Unit 8
Lesson 1: Introduction to logarithms
# Relationship between exponentials & logarithms: graphs
Given a few points on the graph of an exponential function, Sal plots the corresponding points on the graph of the corresponding logarithmic function. Created by Sal Khan.
## Want to join the conversation?
• Hi,
This might be a silly question but how is y = log_b (X) the inverse of y=b^x? Isn't the inverse of y=b^x this log_b (y) = X?
Thanks
• I guess if you look at the function as y = f(x) = b^x. The inverse of the function would be x = f(y) = b^y. So, the logarithmic form of the inverse function is log_b(x) = y.
Ey that's just my take, might be wrong too lol.
• How did he know to use points 1,4, and 16 in the second chart? I thought that, given the question, he would have taken the x values from the existing points and plugged them into the 2nd function.
• For anyone who watches this video and has a similar question, think about it this way:
y = b^x can be rewritten as log_b(y) = x.
So for b = 4 (you can deduce that from the graph) and y = 1, 4, 16 (also deduced from the graph) you have
log_4(1) = 0
log_4(4) = 1
log_4(16) = 2
Which represent points (1, 0), (4, 1) and (16,2) the inverse of the points on the given graph!
• at around , why is sal taking the y values of the first table and using them as the x values of the second table?
• I had a little trouble on this too, but here's what I figured: The purpose of an exponential function is to find what value a number equals when it is raised to the power of x. However, the *purpose of a logarithm is to find the exponent (x value)* that gives you a certain amount when you raise a base by the unknown x value. Though it might sound complicated, if you try to define the purpose yourself, you'll end up with the same answer. For this reason, they are like the inverse functions of each other, just like multiplication and division.
In other words, the logarithm tries to lead you to the exponent needed to reach the value, while the exponential graph tries to lead you to the value given by the exponent's use.
Therefore, they are inverse operations as they undo each other. Furthermore, they reflect over the y=x line, and their coordinates are switched. If you check at , you'll see Sal agrees. If you check the background at this time, you'll also see how when the b= 4, everything fits together perfectly.
Hope this helps.
• Sorry about this but I have two questions. The first might be silly, but here goes: How does one solve for x when it is an exponent such as 2^x=1/64. I've been having to guess this whole time and try for answers on my calculator, because solving for x has not been working. I think I might be doing it wrong...
Second is, I didn't understand this video too well. I don't understand what the graph has
to do with any of this, even though I've been okay with everything pertaining to logarithms so far. Any answer is much appreciated. Thank you!
• When the variable is in the exponent, you need to use logarithms of whatever the base of the exponent is.
For 2^x = 1 / 64, the base is 2. Therefore, we'll be taking log base 2 of each side of the equation.
But before doing that, it's usually easiest to express both sides of the equation using the same base.
So, 2^x = 1 / 64 = 1 / 2^6 = 2^(-6)
log_2 ( 2^x ) = log_2 ( 2^-6 )
x = - 6
Hope this helps.
(I'll let someone else pick up on your question about the graph.)
• Hi, can you please explain how did you know that y=log_b(x) is inverse of y=b^x. a detailed answer would be appreciated. Thanks.
• The actual definition of a logarithm that proves this is several years ahead of the math you're having now, so it wouldn't mean much to you at this stage.
But, the short answer is that the logarithm was invented to be the inverse operation of an exponential function. In other words, there is a difficult mathematical process that was worked out over the course of about 80 years by Nicholas Mercator, Leonhard Euler and others. The term "log" is just a shorthand way of expressing this difficult bit of math.
So, a log is just a quick way of writing the inverse function of an exponential function. Thus, by definition, the log must be the inverse function of the exponential function.
So, the reason you are not given exactly what a log is or how to compute it by hand at this level of study is that it is very difficult math. Since we have calculators that can do this math for us, it is possible for people to use logs without having to master very advanced and difficult computations.
But, just for reference, here is one way to write the actual definition of a logarithm
log_b (x)= lim n→0 [x^(n) - 1] / [b^(n) - 1]
where b is the base of the logarithm.
• At , how do we know that b raised to the power 1 is equal to four. Is it a definite variable with its value
• It is a constant that we deduced from the graph that is given to us.
• I'm sorry but I'm confused by this video. I've understood everything so far but I got lost. How does this work?
• You have your input x in the Log function, and the output is graphed.
Like when x=2 on the equation and say it equals 5 when x=2
you would graph 2,5
(1 vote)
• To solve for other points, couldn't I figure out that b=4 from the pattern in y=b^x and go from there?
• From this set it is obvious that b=4, but the point of the exercise is to highlight the fact that the logarithmic function is an inverse of the exponential function, meaning that you still could have mapped the second set of points simply by reflecting over the line y = x, even if the values present did not provide a simple/obvious value for b.
• I've seen a notation such as 10 log (25/5) = 10 log (5). I'm confused about the "10" and the "log" placement. Is this the same as saying log_10 5? Another (very similar) notation is 20 log (0.1/2)=20 log (0.05). Again, why put the 10 and the 20 before the "log"? Thank You!
(1 vote)
• Whenever there's no base shown after "log", it means that the base is 10. A number before log simply means multiplication. Your example:
``20 log (0.1/2)``
is the equivalent of saying:
``20 * log_10 (0.1/2)``
| 1,643 | 6,388 |
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Courses
# Test: Numeric Entry- 4
## 15 Questions MCQ Test Section-wise Tests for GRE | Test: Numeric Entry- 4
Description
This mock test of Test: Numeric Entry- 4 for GRE helps you for every GRE entrance exam. This contains 15 Multiple Choice Questions for GRE Test: Numeric Entry- 4 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Numeric Entry- 4 quiz give you a good mix of easy questions and tough questions. GRE students definitely take this Test: Numeric Entry- 4 exercise for a better result in the exam. You can find other Test: Numeric Entry- 4 extra questions, long questions & short questions for GRE on EduRev as well by searching above.
*Answer can only contain numeric values
QUESTION: 1
### From a rectangular cardboard of length 21 cm and breadth 12 cm, a square whose diagonal is of length 3*sqrt(2) cm is cut out. Find the area of the remaining cardboard.
Solution:
Area of the cardboard = length*breath
= 21*12 = 252 sq.cm.
Diagonal of a square of side x cm is equal to x*sqrt(2) cm.
Hence,
side of square = 3 cm
Area of square = side*side = 3*3 = 9 sq.cm.
Area of the remaining cardboard
= 252-9
= 243 sq.cm.
*Answer can only contain numeric values
QUESTION: 2
### The height of a cone is trebled. The radius of the cone is doubled. By how many times will the volume of the cone increase?
Solution:
Let the original height be h and radius of the base be r.
Original volume will be = (1/3)*pi*r2*h
New height will be 3h New radius will be 2r New volume will be
= (1/3)*pi*(2r)2*(3h)
= (1/3)*pi*r2*h = (1/3)*pi*r^2*h*12
New volume = 12*original volume
Hence, the volume increases 12 times.
[pi=22/7, r2=r*r]
*Answer can only contain numeric values
QUESTION: 3
### A rectangular reservoir is 120 m long and 75 m high. The cross-sectional area of the pipe through which water flows is 0.04 sq.m. and the water level in the reservoir rises 2.4 meters in 18 hours. Find the speed of water flowing into the reservoir.
Solution:
Cross-sectional area of the pipe = 0.04 sq.m. Volume of water put into the reserviour in 18 hours
= 120*75*2.4
= 21600 cubic m
Volume of water put into the reserviour per hour = 21600/18 cubic m/hr
Required speed = Volume/area
= (21600/18)/0.04
= 30000 cubic m/hr
*Answer can only contain numeric values
QUESTION: 4
P(x-1,3) : P(x,4) = 1: 9, find x.
Solution:
P(x-1,3) : P(x,4)
= 1: 9 (x-1)!/(x-1-3)! : x!/(x-4)!
= 1:9 (x-1)!/(x-4)! : x(x-1)!/(x-4)!
= 1:9 1:x = 1:9
x = 9
*Answer can only contain numeric values
QUESTION: 5
If P(10,r) = 30240, then find r.
Solution:
P(10,r) = 30240 10!/(10-r)! = 30240
10!/(10-r)! = 3024*10 10!/(10-r)!
= 336*9*10 10!/(10-r)!
= 42*8*9*10 10!/(10-r)!
= 6*7*8*9*10 10!/(10-r)!
= 10!/5! 10!/(10-r)!
= 10!/(10-5)!
Hence,
r = 5
*Answer can only contain numeric values
QUESTION: 6
Find the sum of the first 20 terms of the AP 1, 4, 7, 10...
Solution:
The first term of the AP, a, is 1 and the common difference, d, is 3.
The sum of n terms is given by
Sn = (n/2)[2a+(n-1)d]
Sum of 20 terms
= (20/2)[2*1+(20-1)*3]
= 10(2+19*3)
= 10(59)
= 590
*Answer can only contain numeric values
QUESTION: 7
Tim buys two items at different rates for a total of Rs. 410 and sells one at a loss of 20% and the other at a gain of 25%. If both the items were sold at the same price, then find the cost price of the cheaper item.
Solution:
Let CP and SP be cost price and selling price respectively.
Let the CP of the first item be Rs.x and that of the second item be Rs.(410-x)
Selling price of the first item SP = CP(100-loss%)/100 = x(100-20)/100 = 8x/10
Selling price of the second item SP = CP(100+gain%)/100
= (410-x)(100+25)/100
= (410-x)(125/100)
Since both the items were sold at the same price,
we have 8x/10 = (410-x)(125/100) 80x
= 410*125 - 125x 80x + 125x
= 51250 205x=51250 x = 250
The second item would cost 410-250 = 160
The cheaper item costs Rs.160
*Answer can only contain numeric values
QUESTION: 8
Find the length of a parallel side of the trapezium if one of its parallel sides is 70 m long and its area is 2500 sq.m. The distance between the parallel sides is 40m.
Solution:
Let the unknown length be x m. Area of trapezium = (1/2)*sum of parallel sides*distance between parallel sides
2500 = (1/2)*(70+x)*40 70+x
= 2500*2/40 70+x
= 125 x = 125-70 = 55
The other side of the trapezium is 55 m long.
*Answer can only contain numeric values
QUESTION: 9
Find (b-a) when x4+ax2+bx+5 is exactly divisible be x2+3x+2. [x^4=x*x*x*x]
Solution:
x2+3x+2=(x+1)(x+2) (x+1) and (x+2) should be the factors of x4+ax2+bx+5
Putting x = -1 and x = -2,
we get
x4+ax2+bx+5 = (-1)4+a(-1)2+b(-1)+5=0
1+a-b+5=0 a-b = -6 ...(1)
x4+ax2+bx+5 = (-2)4+a(-2)2+b(-2) + 5=0 16 + 4a -2b+5=0 4a-2b = -21...(2)
Multiplying (1) by 4 and subtracting (2) from it,
we get
4a - 4b -4a +2b = -24 + 21 -2b = -3
b = 3/2 a = b-6 = 3/2-6
= (3-12)/2 = -9/2 b -a
= 3/2+9/2
= 12/2
= 6
*Answer can only contain numeric values
QUESTION: 10
x+1/x = 5. Find x3+1/x3. [x3=x*x*x]
Solution:
x+1/x = 5 (x+1/x)3
= 53 x^3+1/x3+3(x+1/x)
= 125 x3+1/x3
= 125 -3*5 =125-15
= 110 x3+1/x3
= 110
*Answer can only contain numeric values
QUESTION: 11
Pam lent Rs. 1800 to Ritu. She also lent Rs 2250 to Tammy for four years. They both returned the same interest charged at the same rate of interest of 3% per annum. For how many years did Ritu borrow the money?
Solution:
Let P, R and T be the principle, rate and time for the simple interest SI
SI = P*R*T/100
Since the SI is the same for both of them,
we have
1800*3*T/100
= 2250*3*4/100
T = 2250*3*4/(1800*3) = 5
The money was lent for 5 years
*Answer can only contain numeric values
QUESTION: 12
The length of the tangent drawn from an exterior point P to the circle is 63 cm. The point P is at a distance of 65 cm from the centre of the circle. Find the radius of the circle.
Solution:
Let the centre of the circle be the point O and let the point at which the tangent drawn from point P meets the circle be T.
PTO will be a right triangle right angled at T.
PT = 63 cm and PO = 65 cm.
Applying Pythagoras theorem to the triangle PTO and squaring both the sides,
we get PO2 = PT2+TO2 652=632+TO2 4225 = 3969 + TO2 TO2
= 4225-3969 = 256
TO = 16 cm
The radius of the circle is 16 cm.
[PO2=PO*PO]
*Answer can only contain numeric values
QUESTION: 13
A hollow cylinder of height 3m is melted to form a solid cylinder of the same height. The external and internal radii of the hollow cylinder are 29 cm and 20 cm respectively. Find the radius of the solid cylinder.
Solution:
Volume of cylinder = π*r2*h,
where r is the radius and h is the height. Internal radius = 20 cm and external radius = 29 cm
Volume of metal used
= π*(29)2*h - pi(20)2*h
= π*h[841-400]
= π*h*441
Volume of solid cylinder
= π*h*441
= π*h*R2,
where R is the radius of the solid cylinder
R2 = 441 R = sqrt(441) = 21
The radius of the solid cylinder is 21 cm.
[π=22/7, r2=r*r]
*Answer can only contain numeric values
QUESTION: 14
A shopkeeper sells each book for Rs 1134 after giving a discount of 19% on the marked price. Had he sold the books at the printed price, he would have earned a profit of 40%. Find the cost price of each book.
Solution:
Let MP be the marked price, SP be the selling price and CP be the cost price.
MP = 100*SP/(100-discount%)
= 100*1134/(100-19) = 1400
If the books were sold at the primted price the selling price would be Rs1400 CP = 100*SP/(100+profit%)
= 100*1400/(100+40)
= 100*1400/140 =1000
The cost price of each book is Rs.1000
*Answer can only contain numeric values
QUESTION: 15
A metallic cylinder was melted and a sphere was formed from the metal. The diameter of the cylinder is 6 cm and its length is 4 cm. Find the radius of the sphere.
Solution:
Volume of the cylinder =π*r2*h,
where r is the radius and h is the height of the cylinder.
Radius of the cylinder = 6/2 = 3cm
Volume = π*32*4 = pi*36
Volume of sphere = 4/3*π*r3,
where r is the radius π*36 = 4/3*π*r3 36*3/4 = r3
r3 = 27
r = 3cm
The radius of the sphere is 3 cm.
[π=22/7, r2=r*r]
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# Just Research...........
• Jun 3rd 2008, 10:43 PM
Apostolos
Just Research...........
I am doing a research relatively with anti-divisors and the “opposite” of them numbers .
Actually I am trying to count (to find a formula which counts) the number of pairs*of type (“anti divisor odd composite number less than half of an integer a”---“odd composite number greater than the half of an integer a”) such as the addition of them has as result the integer a.
For example :24=9+15
9<24/2 is odd comp. and does not divide 24
and 24>15>24/2 is and odd comp.
So there is one pair * of integers of the this specific type for the number 24...
I have done some work on that but I didn’t yet found one way to count exactly the number of ways which an integer can be written on this way.
the only thing I know about that is a lower bound for this number but its not really useful.I t would be more useful to know the upper bound value
Does anyone knows anything about that?(Thinking)
Thanks!!
• Jun 4th 2008, 10:40 PM
CaptainBlack
Quote:
Originally Posted by Apostolos
I am doing a research relatively with anti-divisors and the “opposite” of them numbers .
Actually I am trying to count (to find a formula which counts) the number of pairs*of type (“anti divisor odd composite number less than half of an integer a”---“odd composite number greater than the half of an integer a”) such as the addition of them has as result the integer a.
For example :24=9+15
9<24/2 is odd comp. and does not divide 24
and 24>15>24/2 is and odd comp.
So there is one pair * of integers of the this specific type for the number 24...
I have done some work on that but I didn’t yet found one way to count exactly the number of ways which an integer can be written on this way.
the only thing I know about that is a lower bound for this number but its not really useful.I t would be more useful to know the upper bound value
Does anyone knows anything about that?(Thinking)
Thanks!!
Let $O_{1/2}(n)$ be the be one less than the number of odds less than $n/2$ (so we are not counting $1$ as it is a unit and we are going to be interested in composits and primes only). and let $p_c(x)$ be a prime counting function that gives the number of primes less than $x$.
Also let $AD(n)$ denote the number of pairs $(a,b)$ with $a$ and $b$ composite, $a and $n=a+b$. Then:
$AD(n) \le O_{1/2}(n)- p_c(n/2)$
If I have done this right, the right hand side of this inequality should be the number of odd composites less than $n/2$.
RonL
• Jun 5th 2008, 04:29 AM
NonCommAlg
Quote:
Originally Posted by Apostolos
I am doing a research relatively with anti-divisors and the “opposite” of them numbers .
Actually I am trying to count (to find a formula which counts) the number of pairs*of type (“anti divisor odd composite number less than half of an integer a”---“odd composite number greater than the half of an integer a”) such as the addition of them has as result the integer a.
For example :24=9+15
9<24/2 is odd comp. and does not divide 24
and 24>15>24/2 is and odd comp.
So there is one pair * of integers of the this specific type for the number 24...
I have done some work on that but I didn’t yet found one way to count exactly the number of ways which an integer can be written on this way.
the only thing I know about that is a lower bound for this number but its not really useful.I t would be more useful to know the upper bound value
Does anyone knows anything about that?(Thinking)
Thanks!!
the problem is very difficult! an upper bound that can be proved quite easily is this:
$f(n) \leq \min \{[n/4] + \omega_1(n) + 1 - \tau_1(n) -\pi (n/2 - 1), [n/4]+ \pi (n/2) - \pi (n) \},$
where $f$ is the fuction you're looking for, $[ \ \ ]$ is the floor function, $\omega_1(n)$ is the number
of (distinct) odd prime factors of n, $\tau_1(n)$ is the number of odd divisors of n, and finally
$\pi(x)$ is the prime counting function.
| 1,053 | 3,913 |
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The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A231135 Number of (n+1)X(4+1) black-square subarrays of 0..2 arrays with no element equal to a strict majority of its diagonal and antidiagonal neighbors, with values 0..2 introduced in row major order 1
16, 182, 2260, 27171, 336004, 4066129, 50257244, 608468617, 7520563372, 91054483047, 1125418461348, 13625913937795, 168414092245220, 2039060342079409, 25202456511185596, 305136757252909097 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Column 4 of A231137 LINKS R. H. Hardin, Table of n, a(n) for n = 1..210 FORMULA Empirical: a(n) = 175*a(n-2) -4017*a(n-4) +34311*a(n-6) -146236*a(n-8) +322472*a(n-10) -291040*a(n-12) +116032*a(n-14) -24064*a(n-16) +1024*a(n-18) EXAMPLE Some solutions for n=4 ..x..0..x..1..x....x..0..x..1..x....x..0..x..1..x....x..0..x..0..x ..0..x..2..x..0....1..x..0..x..2....2..x..0..x..1....0..x..1..x..2 ..x..1..x..0..x....x..2..x..2..x....x..2..x..0..x....x..2..x..2..x ..1..x..1..x..1....0..x..1..x..0....1..x..1..x..2....1..x..1..x..1 ..x..2..x..0..x....x..0..x..2..x....x..2..x..1..x....x..2..x..0..x CROSSREFS Sequence in context: A199018 A204608 A279282 * A230992 A016298 A288720 Adjacent sequences: A231132 A231133 A231134 * A231136 A231137 A231138 KEYWORD nonn AUTHOR R. H. Hardin, Nov 04 2013 STATUS approved
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Last modified May 29 07:13 EDT 2023. Contains 363029 sequences. (Running on oeis4.)
| 676 | 1,758 |
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# [_\f't\ '( V\ \V\OVI
Math 1010 Intermediate Algebra Group Project
1P
Background Information:
Linear Programming is a technique used for optimization of a real-world situation. Examples of
optimization include maximizing the number of items that can be manufactured or minimizing
the cost of production. The equation that represents the quantity to be optimized is called the
objective function, since the objective of the process is to optimize the value. In this project the
objective is to maximize the number of people who will be reached by an advertising campaign.
The objective is subject to limitations or constraints that are represented by inequalities.
Limitations on the number of items that can be produced, the number of hours that workers are
available, and the amount of land a farmer has for crops are examples of constraints that can
realistic goal. In this project one ofthe constraints will be based on an advertising budget.
Graphing the system of inequalities based on the constraints provides a visual representation of
the possible solutions to the problem. If the graph is a closed region, it can be shown that the
values that optimize the objective function will occur at one of the "corners" of the region.
The Problem:
In this project your group will solve the following situation:
radio and on TV. The business plans to purchase at least 60 total ads and they want to have
each. The advertising budget is \$4320. It is estimated that each radio ad will be heard by 2000
listeners and each TV ad will be seen by 1500 people. How many of each type of ad should be
purchased to maximize the number of people who will be reached by the advertisements?
Modeling the Problem:
Let X be the number of radio ads that are purchased and Y be the number of TV ads.
1. Write down a lim;ar inequality for the total number of desired ads.
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linear inequality that expresses this fact.
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4. There are two more constraints that must be met. These relate to the fact that there
cannot be s negative numbers of advertisements. Write the two inequalities that model
these constraints:
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5. Next, write down the function for the number of people that will be exposed to the
P = :2J)D0 x. -t- ! ?o o y
You now have four linear inequalities and an objective function. These together describe the
situation. This combined set of inequalities and objective function make up what is known
mathematically as a linear programming problem. Write all of the inequalities and the
objective function together below. This is typically written as a list of constraints, with the
objective function last.
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6. To solve this problem, you will need to graph the intersection of all five inequalities
on one common XY plane. Do this on the grid below. Have the bottom left be the
origin, with the horizontal axis representing X and the vertical axis representing Y. Label
the axes with what they represent and label your lines as you graph them.
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7. The shaded region in the above graph is called the feasible region. Any (x, y) point in the region
corresponds to a possible number of radio and TV ads that will meet all the requirements of the
problem. However, the values that will maximize the number of people exposed to the ads will
occur at one of the vertices or corners of the region. Your region should have three corners. Find
the coordinates of these corners by solving the appropriate system of linear equations. Be sure to
show your work and label the (x, y) coordinates of the corners in your graph.
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of people who will be exposed to the ad.
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applying math. Be specific.
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A024202 a(n) = [ (3rd elementary symmetric function of S(n))/(first elementary symmetric function of S(n)) ], where S(n) = {first n+2 odd positive integers}. 1
1, 11, 38, 96, 205, 385, 662, 1068, 1635, 2401, 3410, 4706, 6339, 8365, 10840, 13826, 17391, 21603, 26536, 32270, 38885, 46467, 55108, 64900, 75941, 88335, 102186, 117604, 134705, 153605, 174426, 197296, 222343, 249701, 279510, 311910, 347047, 385073 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,2 LINKS Robert Israel, Table of n, a(n) for n = 1..10000 FORMULA Empirical g.f.: x*(x^4-5*x^3-7*x-1) / ((x-1)^5*(x^2+x+1)). - Colin Barker, Aug 15 2014 From Robert Israel, Dec 30 2016: (Start) a(n) = floor(A024197(n)/(n+2)^2) = floor(n*(n+1)*(n^2+3*n+1)/6). a(n) = (n^4+4*n^3+4*n^2+n-4)/6 if n == 1 (mod 3). Otherwise a(n) = n*(n+1)*(n^2+3*n+1)/6. The empirical g.f. can be obtained from this. (End) MAPLE f:= proc(n) if n mod 3 = 1 then (n^4+4*n^3+4*n^2+n-4)/6 else n*(n+1)*(n^2+3*n+1)/6 fi end proc: map(f, [\$1..100]); # Robert Israel, Dec 30 2016 MATHEMATICA Table[Floor[n*(n + 1)*(n^2 + 3*n + 1)/6], {n, 1, 50}] (* G. C. Greubel, Dec 30 2016 *) PROG (PARI) for(n=1, 25, print1(floor(n*(n+1)*(n^2+3*n+1)/6), ", ")) \\ G. C. Greubel, Dec 30 2016 CROSSREFS Cf. A024197. Sequence in context: A010002 A143109 A007585 * A213775 A133258 A288745 Adjacent sequences: A024199 A024200 A024201 * A024203 A024204 A024205 KEYWORD nonn AUTHOR STATUS approved
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Question
Sun February 24, 2013 By:
the period of oscillation of a simple pendulum is T=2??L/g,L is about 10 cm and is known to 1 mm accuracy.The period of oscillation is about .5s.The time for 100 oscillation is measured with a wrist watch of 1 s resolution.what is the accuracy in the determination of g?
Sun February 24, 2013
The only factors concerned with T are of L and g
thus we can write
g = K L/T2
where K is a constant
taking logatithm on both sides
log g = log K + log L +2log T
∆g/g = ∆L/L + 2* ∆T/T
In terms of percentage,
(∆g/g)*100 = (∆L/L)*100 + 2* (∆T/T)*100
Here the resolution or accuracy are that values that could occur as errors .
Time period T= Time of one oscillation= t/n=Total time/ no of oscillation-
Percentage error in L =
(∆L/L)*100 = 100 *(0.1/10) =1%
Percentage error in T =
(∆T/T)*100 = 100 *(1/50) =2%
Percentage error in g =
(∆g/g)*100 = 1% +2 *2%= 5%
Related Questions
Fri January 18, 2013
What is Absolute Error Plz Explain with Examples.
Tue October 09, 2012
Home Work Help
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# How to Find the Inverse of a Function
••• kimberrywood/iStock/GettyImages
Print
To find an inverse function in math, you must first have a function. It can be almost any set of operations for the independent variable x that yields a value for the dependent variable y. In general, to determine the inverse of a function of x, substitute y for x and x for y in the function, then solve for x.
#### TL;DR (Too Long; Didn't Read)
In general, to find the inverse of a function of x, substitute y for x and x for y in the function, then solve for x.
## Inverse Function Defined
The mathematical definition of a function is a relation (x, y) for which only one value of y exists for any value of x. For example, when the value of x is 3, the relation is a function if y has only one value, such as 10. The inverse of a function takes the y values of the original function as its own x values, and produces y values that are the original function’s x values. For example, if the original function returned the y values 1, 3 and 10 when its x variable had the values 0, 1 and 2, the inverse function would return y values 0, 1 and 2 when its x variable had the values 1, 3 and 10. Essentially, an inverse function swaps the x and y values of the original. In mathematical language, if the original function is f(x) and the inverse is g(x), then
g(f(x)) = x
## Algebra Approach for Inverse Function
To find the inverse of a function involving the two variables, x and y, replace the x terms with y and the y terms with x, and solve for x. As an example, take the linear equation, y = 7x − 15.
y = 7x - 15 \quad \text{(Original function)} \\ \,\\ x = 7y - 15 \quad \text{(Replace y with x and x with y)}\\ \,\\ x + 15 = 7y - 15 + 15 \quad \text{(Add 15 to both sides.)} \\ \,\\ x + 15 = 7y \quad \text{(Simplify)} \\ \,\\ \frac{x + 15}{7} = \frac{7y}{7} \quad\text{(Divide both sides by 7.)} \\ \,\\ \frac{x + 15}{7} = y \quad\text{(Simplify)}
The function, (x + 15) / 7 = y is the inverse of the original.
## Inverse Trigonometric Functions
To find the inverse of a trigonometric function, it pays to know about all the trig functions and their inverses. For example, if you want to find the inverse of y = sin(x), you need to know that the inverse of the sine function is the arcsine function; no simple algebra will get you there without arcsin(x). The other trig functions, cosine, tangent, cosecant, secant and cotangent, have the inverse functions arccosine, arctangent, arccosecant, arcsecant and arccotangent, respectively. For example, the inverse of y = cos(x) is y = arccos(x).
## Graph of Function and Inverse
The graph of a function and its inverse is interesting. When you plot the two curves, then draw a line corresponding to the function, y = x, you’ll notice that the line appears as a “mirror.” Any curve or line below y = x is “reflected” symmetrically above it. This is true for any function, whether polynomial, trigonometric, exponential or linear. Using this principle, you can graphically illustrate the inverse of a function by graphing the original function, drawing the line at y = x, then drawing the curves or lines needed to create a “mirror image” that has y = x as an axis of symmetry.
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## Algebra: A Combined Approach (4th Edition)
${-7,7}$
Step 1: $3|y|-4=17$ Step 2: Adding $4$ to both sides, $3|y|-4+4=17+4$. Step 3: $3|y|=21$ Step 4: Dividing both sides by $3$, $\frac{3|y|}{3}=\frac{21}{3}$. Step 5: $|y|=7$ Step 6: Using the absolute value property, y=7 or y=-7. Step 7: Therefore, the solution set is ${-7,7}$.
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Chapter3
Chapter3 - Chapter 3 Examples of Functions Obvious is the...
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Chapter 3 Examples of Functions Obvious is the most dangerous word in mathematics. E. T. Bell 3.1 M¨obius Transformations The first class of functions that we will discuss in some detail are built from linear polynomials. Definition 3.1. A linear fractional transformation is a function of the form f ( z ) = az + b cz + d , where a, b, c, d C . If ad bc = 0 then f is called a M¨obius 1 transformation . Exercise 11 of the previous chapter states that any polynomial (in z ) is an entire function. From this fact we can conclude that a linear fractional transformation f ( z ) = az + b cz + d is holomorphic in C \ d c (unless c = 0, in which case f is entire). One property of M¨ obius transformations, which is quite special for complex functions, is the following. Lemma 3.2. M¨obius transformations are bijections. In fact, if f ( z ) = az + b cz + d then the inverse function of f is given by f 1 ( z ) = dz b cz + a . Remark. Notice that the inverse of a M¨ obius transformation is another M¨ obius transformation. Proof. Note that f : C \ { d c } C \ { a c } . Suppose f ( z 1 ) = f ( z 2 ), that is, az 1 + b cz 1 + d = az 2 + b cz 2 + d . As the denominators are nonzero, this is equivalent to ( az 1 + b )( cz 2 + d ) = ( az 2 + b )( cz 1 + d ) , 1 Named after August Ferdinand M¨ obius (1790–1868). For more information about M¨ obius, see http://www-groups.dcs.st-and.ac.uk/ history/Biographies/Mobius.html . 23
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CHAPTER 3. EXAMPLES OF FUNCTIONS 24 which can be rearranged to ( ad bc )( z 1 z 2 ) = 0 . Since ad bc = 0 this implies that z 1 = z 2 , which means that f is one-to-one. The formula for f 1 : C \ { a c } C \ { d c } can be checked easily. Just like f , f 1 is one-to-one, which implies that f is onto. Aside from being prime examples of one-to-one functions, M¨ obius transformations possess fas- cinating geometric properties. En route to an example of such, we introduce some terminology. Special cases of M¨ obius transformations are translations f ( z ) = z + b , dilations f ( z ) = az , and in- versions f ( z ) = 1 z . The next result says that if we understand those three special transformations, we understand them all. Proposition 3.3. Suppose f ( z ) = az + b cz + d is a linear fractional transformation. If c = 0 then f ( z ) = a d z + b d , if c = 0 then f ( z ) = bc ad c 2 1 z + d c + a c . In particular, every linear fractional transformation is a composition of translations, dilations, and inversions. Proof. Simplify. With the last result at hand, we can tackle the promised theorem about the following geometric property of M¨ obius transformations. Theorem 3.4. M¨obius transformations map circles and lines into circles and lines. Proof. Translations and dilations certainly map circles and lines into circles and lines, so by the last proposition, we only have to prove the theorem for the inversion f ( z ) = 1 z . Before going on we find a standard form for the equation of a straight line. Starting with ax + by = c (where z = x + iy ), let α = a + bi . Then α z = ax + by + i ( ay bx ) so α z + α z = α z + α z = 2 Re( α z ) = 2 ax + 2 by . Hence our standard equation for a line becomes α z + α z = 2 c, or Re( α z ) = c. (3.1) Circle case : Given a circle centered at z 0 with radius r
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# Math
posted by .
For a period of time as an alligator grows, its mass is proportional to the cube of its length. When the alligator's length changes by 15.8%, its mass increases by 17.3kg. Find its mass at the end of this process.
• Math -
mass=k*length^3
mass+17.3=k(length*1.158)^3
k*length^3=k(length^3 ( 1.55)-17.3
.55K*length^3=17.3
mass*.55=17.3
massfinalthen=17.3/.55 + 17.3
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### Course: Basic geometry and measurement>Unit 3
Lesson 2: Angle introduction
# Angle basics review
Review what an angle is and how an angle is measured.
### What is an angle?
An angle is two rays that share a $\text{vertex}$.
Intersecting lines or line segments also form angles.
## Practice: Identifying an angle
Problem 1
Which of the following figures is an angle?
## Measuring angles
Angles are measured in degrees.
The wider an angle is open, the greater its measure. If you compare these two angles, the first one is open wider.
So, the measure of this angle
is greater than the measure of this angle.
## Practice set: Which angle is greater?
Problem 2A
Which angle measure is greater?
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• What is a Vertex?
• the vertex of an angle is the point where the rays or line segments meet.
• So, do angles relate to shapes like squares and circles?
• they do
• why is B used more than A in these questions?
• I'm not sure.
thank you for making me notice that
• At how do I practice to get better at noticing angles, can khan academy help me with that
• A place where two rays meet is called an angle.
• why are angels measured in degress
• the babylonians (an ancient civilization, like Greeks) invented the 360 degree value which is the maximum an angle can be (which is a complete circle and after that, it kinda repeats on itself and the value of the angle gets even bigger). It's based on their base 60 counting (/number) system. There also exist other ways like radian ( angles are measured in pi) and tau (basically, "2*radians" tau = 2pi.)
• how do you measure an angle
• ABC angle would measure how far A to C is
• what is a plane figure
• 1000
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# Linear Algebra.Linear Transformation.Help
• Mar 5th 2009, 09:50 AM
ypatia
Linear Algebra.Linear Transformation.Help
Suppose T is the linear transformation on R^3 that takes each point (x,y,z) to (x+y+z, x+y,z). Describe what T^-1 does to the point (x,y,z). Thanks in advance.
• Mar 5th 2009, 11:25 AM
Inverse Transformation
Hello ypatia
Quote:
Originally Posted by ypatia
Suppose T is the linear transformation on R^3 that takes each point (x,y,z) to (x+y+z, x+y,z). Describe what T^-1 does to the point (x,y,z). Thanks in advance.
Do you know how to find the inverse of a 3x3 matrix? There are quite a few web-sites that will explain the method, if you don't. Just Google 'Inverse of a 3x3 matrix'.
In the question you have here, the transformation can be written in matrix form as
$\begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}\be gin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}x+y+z\\x+y\\z\end{pmatrix}$
So, to find the inverse transformation, you'll need the inverse of the matrix $\begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}$, which is $\begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}$
To find its effect on $(x, y, z)$, just do the matrix multiplication:
$\begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix } = \begin{pmatrix}z\\y-z\\x-y\end{pmatrix}$
So there's your answer: $T^{-1}:(x,y,z) \rightarrow (z,y-z,x-y)$
• Mar 5th 2009, 11:43 AM
ypatia
Quote:
Hello ypatiaDo you know how to find the inverse of a 3x3 matrix? There are quite a few web-sites that will explain the method, if you don't. Just Google 'Inverse of a 3x3 matrix'.
In the question you have here, the transformation can be written in matrix form as
$\begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}\be gin{pmatrix}x\\y\\z\end{pmatrix} = \begin{pmatrix}x+y+z\\x+y\\z\end{pmatrix}$
So, to find the inverse transformation, you'll need the inverse of the matrix $\begin{pmatrix}1&1&1\\1&1&0\\1&0&0\end{pmatrix}$, which is $\begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}$
To find its effect on $(x, y, z)$, just do the matrix multiplication:
$\begin{pmatrix}0&0&1\\0&1&-1\\1&-1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix } = \begin{pmatrix}z\\y-z\\x-y\end{pmatrix}$
So there's your answer: $T^{-1}:(x,y,z) \rightarrow (z,y-z,x-y)$
Is this right?? "T is not injective and, as such, it does not have an inverse. For an example of the lack of injectivity, T(1,0,1) -->(2,1,1) and T(2,-1,1) -->(2,1,1)." Because if this is right contradicts to the above solution of T^-1. Thanks in advance
• Mar 5th 2009, 12:20 PM
Plato
The matrix is $
\left( {\begin{array}{*{20}c}
1 & 1 & 1 \\
1 & 1 & 0 \\
0 & 0 & 1 \\
\end{array} } \right)
$
which is singular.
• Mar 5th 2009, 01:14 PM
Transformation matrix
Quote:
Originally Posted by Plato
The matrix is $
\left( {\begin{array}{*{20}c}
1 & 1 & 1 \\
1 & 1 & 0 \\
0 & 0 & 1 \\
\end{array} } \right)
$
which is singular.
Sorry, yes. I mapped $(x,y,z)$ onto $(x+y+z, x+y, x)$, of course.
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https://www.enotes.com/homework-help/find-altitiude-ad-cf-traingle-abc-whose-vertices-3-384177
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# find the altitiude of AD BE and CF in traingle ABC whose vertices are A(-3,4) B(3,6) C(2,9)..also show that AD BE and CF are concurrent?u may change the verices value..jus need to kno the method
sciencesolve | Teacher | (Level 3) Educator Emeritus
Posted on
You need to evaluate the area of triangle `Delta ABC` using the following formula, such that:
`A = (1/2)*|[(-3,4,1),(3,6,1),(2,9,1)]|`
`A = (1/2)*|(-18 + 27 + 8 - 12 + 27 - 12)| => A = 10`
You also may evaluate the area of triangle using the following formula, such that:
`A = (b*h)/2`
b represents the length of one base of triangle
h represents the height of triangle
`A = (BC*AD)/2 = (AC*BE)/2 = (AB*CF)/2`
You need to evaluate the lengths BC,AC,AB, using distance formula such that:
`AB = sqrt((x_B - x_A)^2 + (y_B - y_A)^2)`
`AB =sqrt((3+3)^2 + (6-4)^2) => AB =sqrt40 = 2sqrt10`
`BC = sqrt((x_C - x_B)^2 + (y_C - y_B)^2)`
`BC = sqrt((2-3)^2 + (9-6)^2) =>BC = sqrt10`
`AC = sqrt((x_C - x_A)^2 + (y_C - y_A)^2)`
`AC = sqrt((2+3)^2 + (9-4)^2) => AC = sqrt50 = 5sqrt2`
You need to evaluate AD such that:
You need to evaluate BE such that:
`A = (AC*BE)/2 => 20 = 5sqrt2*BE => BE = 2sqrt2`
You need to evaluate CF such that:
`A = (AB*CF)/2 => 20 = 2sqrt10*CF => CF = 10/sqrt10 = sqrt10`
Hence, evaluating the lengths of the heights of triangle `Delta ABC` , yields `AD = 2sqrt10, BE = 2sqrt2, CF = sqrt10.`
user4803277 | eNotes Newbie
Posted on
plz post fig also..andi how the area of traingle ABC is equal to that of A=bc.ad /2 ..xplain plz.or if any other method...
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https://www.physicsforums.com/threads/calculate-max-acceleration-such-that-a-does-not-slide.870579/
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# Calculate max acceleration such that A does not slide
1. Homework Statement
Friction ≤ μR
## The Attempt at a Solution
:[/B]
Ok so I need help for (ii) and this is what I did:
I thought that in order for A not to slide, the net force on B ≤ Friction between the boxes
450a ≤ 0.2(200g)
a≤0.89
however, when I tried this,
200a≤0.2(200g)
a≤2
What's the correct physics behind this?
BvU
Homework Helper
Friction ≤ μR might be correct. You don't say what R is ?
What ##\mu## did you find in part (i) (just checking ) ?
450 a ≤ 0.2(200g) would that be 200 g like in 200 times the acceleration from gravity ? Again, you don't say what 200 g stands for...
Friction ≤ μR might be correct. You don't say what R is ?
What ##\mu## did you find in part (i) (just checking ) ?
450 a ≤ 0.2(200g) would that be 200 g like in 200 times the acceleration from gravity ? Again, you don't say what 200 g stands for...
Hi. R is the normal reaction force, sorry for not stating it
I got μ=0.7 for part (i)
200g is the normal reaction force of A due to its weight , a is the acceleration of the the whole body
BvU
Homework Helper
So if you draw a free body diagram of A, you get a maximum friction force of mA g Newton. The force to accelerate block A should therefore not be bigger than that. And that force is mA times the acceleration. The mass of B does not come into the equations, only the maximum force it can exert on A.
So if you draw a free body diagram of A, you get a maximum friction force of mA g Newton. The force to accelerate block A should therefore not be bigger than that. And that force is mA times the acceleration. The mass of B does not come into the equations, only the maximum force it can exert on A.
Ah I see. So the force that actually accelerates A is the net force, P-(friction between B and floor), right?
BvU
Homework Helper
So the force that actually accelerates A is friction force between A and B. that is the only horizontal force that works on A.
the net force, P-(friction between B and floor),
No. If you want to see why not, draw a free body diagram of box B !
So the force that actually accelerates A is friction force between A and B. that is the only horizontal force that works on A.
No. If you want to see why not, draw a free body diagram of box B !
But how does frictional force accelerates A? I thought there must be some forward force?
But how does frictional force accelerates A? I thought there must be some forward force?
The static friction between the boxes will accelerate box ##A##.
There is a forward force acting on ##A##. Imagine this, instead of saying that box ##B## is moving forward. You could reverse this and say that box ##A## is moving backwards. It is just saying that box ##A## is moving backward relative to box ##B##.
And What is the friction direction if an object is trying to move backward? The direction is Forward. The static friction wants box ##A## to keep up with box ##B##.
Another way to imagine this (which I like the most), You could put two fingers on your left hand. Now try to move your left hand in any direction. You will feel a tendency or a force acting on your fingers in the same direction.
Now you may ask why is it static friction not kinetic? Well, you want it to move with the same acceleration, don't you? so as long as the acceleration of box ##A## equals the acceleration of box. We can say that at any moment box ##A## is still relative to box ##B##. You could visualize this two by watching two cars racing each other. The drivers see each other as if they were standing still but they see their surrounding move.
Last edited:
BvU
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Spring into money math with this great worksheet that challenges your child to think of the cost of things and work with subtracting and dividing decimals.
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Batter Up! Compare Batting Averages of Major League Baseball Players
Help your students understand the math behind America's favorite pastime, baseball! With this worksheet, students will compare batting average statistics to practice subtracting decimals to the thousandths.
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Presidential Problems at The Pizza Place
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Presidential Problems at The Pizza Place
Your students will practice adding and subtracting decimal numbers to help total the cost of customers' meals at Anita Pepperoni's Pizzeria.
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# Resources tagged with: Generalising
Filter by: Content type:
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### There are 77 results
Broad Topics > Thinking Mathematically > Generalising
### Hypotenuse Lattice Points
##### Age 14 to 16 Challenge Level:
The triangle OMN has vertices on the axes with whole number co-ordinates. How many points with whole number coordinates are there on the hypotenuse MN?
### Beelines
##### Age 14 to 16 Challenge Level:
Is there a relationship between the coordinates of the endpoints of a line and the number of grid squares it crosses?
### A Tilted Square
##### Age 14 to 16 Challenge Level:
The opposite vertices of a square have coordinates (a,b) and (c,d). What are the coordinates of the other vertices?
### Polycircles
##### Age 14 to 16 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
### Pareq Calc
##### Age 14 to 16 Challenge Level:
Triangle ABC is an equilateral triangle with three parallel lines going through the vertices. Calculate the length of the sides of the triangle if the perpendicular distances between the parallel. . . .
### Pick's Theorem
##### Age 14 to 16 Challenge Level:
Polygons drawn on square dotty paper have dots on their perimeter (p) and often internal (i) ones as well. Find a relationship between p, i and the area of the polygons.
### Attractive Tablecloths
##### Age 14 to 16 Challenge Level:
Charlie likes tablecloths that use as many colours as possible, but insists that his tablecloths have some symmetry. Can you work out how many colours he needs for different tablecloth designs?
### AMGM
##### Age 14 to 16 Challenge Level:
Can you use the diagram to prove the AM-GM inequality?
### Converging Means
##### Age 14 to 16 Challenge Level:
Take any two positive numbers. Calculate the arithmetic and geometric means. Repeat the calculations to generate a sequence of arithmetic means and geometric means. Make a note of what happens to the. . . .
### Winning Lines
##### Age 7 to 16
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Lower Bound
##### Age 14 to 16 Challenge Level:
What would you get if you continued this sequence of fraction sums? 1/2 + 2/1 = 2/3 + 3/2 = 3/4 + 4/3 =
### Games Related to Nim
##### Age 5 to 16
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Generally Geometric
##### Age 16 to 18 Challenge Level:
Generalise the sum of a GP by using derivatives to make the coefficients into powers of the natural numbers.
### Pair Products
##### Age 14 to 16 Challenge Level:
Choose four consecutive whole numbers. Multiply the first and last numbers together. Multiply the middle pair together. What do you notice?
### In a Spin
##### Age 14 to 16 Challenge Level:
What is the volume of the solid formed by rotating this right angled triangle about the hypotenuse?
### One, Three, Five, Seven
##### Age 11 to 16 Challenge Level:
A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses.
### Painted Cube
##### Age 14 to 16 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Semi-square
##### Age 14 to 16 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Nim-like Games
##### Age 7 to 16 Challenge Level:
A collection of games on the NIM theme
### Jam
##### Age 14 to 16 Challenge Level:
A game for 2 players
### Nim
##### Age 14 to 16 Challenge Level:
Start with any number of counters in any number of piles. 2 players take it in turns to remove any number of counters from a single pile. The loser is the player who takes the last counter.
### Sums of Pairs
##### Age 11 to 16 Challenge Level:
Jo has three numbers which she adds together in pairs. When she does this she has three different totals: 11, 17 and 22 What are the three numbers Jo had to start with?”
### More Twisting and Turning
##### Age 11 to 16 Challenge Level:
It would be nice to have a strategy for disentangling any tangled ropes...
### Arithmagons
##### Age 11 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges?
### Multiplication Square
##### Age 14 to 16 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### For Richer for Poorer
##### Age 14 to 16 Challenge Level:
Charlie has moved between countries and the average income of both has increased. How can this be so?
### Harmonic Triangle
##### Age 14 to 16 Challenge Level:
Can you see how to build a harmonic triangle? Can you work out the next two rows?
### Searching for Mean(ing)
##### Age 11 to 16 Challenge Level:
If you have a large supply of 3kg and 8kg weights, how many of each would you need for the average (mean) of the weights to be 6kg?
### Cuboid Challenge
##### Age 11 to 16 Challenge Level:
What's the largest volume of box you can make from a square of paper?
### Sliding Puzzle
##### Age 11 to 16 Challenge Level:
The aim of the game is to slide the green square from the top right hand corner to the bottom left hand corner in the least number of moves.
### Mystic Rose
##### Age 14 to 16 Challenge Level:
Use the animation to help you work out how many lines are needed to draw mystic roses of different sizes.
### Partly Painted Cube
##### Age 14 to 16 Challenge Level:
Jo made a cube from some smaller cubes, painted some of the faces of the large cube, and then took it apart again. 45 small cubes had no paint on them at all. How many small cubes did Jo use?
### Multiplication Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Generating Triples
##### Age 14 to 16 Challenge Level:
Sets of integers like 3, 4, 5 are called Pythagorean Triples, because they could be the lengths of the sides of a right-angled triangle. Can you find any more?
##### Age 11 to 16
Dave Hewitt suggests that there might be more to mathematics than looking at numerical results, finding patterns and generalising.
### Odd Differences
##### Age 14 to 16 Challenge Level:
The diagram illustrates the formula: 1 + 3 + 5 + ... + (2n - 1) = n² Use the diagram to show that any odd number is the difference of two squares.
### Steel Cables
##### Age 14 to 16 Challenge Level:
Some students have been working out the number of strands needed for different sizes of cable. Can you make sense of their solutions?
### Fractional Calculus III
##### Age 16 to 18
Fractional calculus is a generalisation of ordinary calculus where you can differentiate n times when n is not a whole number.
### Janine's Conjecture
##### Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Jam
##### Age 14 to 16 Challenge Level:
To avoid losing think of another very well known game where the patterns of play are similar.
##### Age 14 to 16 Challenge Level:
A counter is placed in the bottom right hand corner of a grid. You toss a coin and move the star according to the following rules: ... What is the probability that you end up in the top left-hand. . . .
### Gnomon Dimensions
##### Age 14 to 16 Challenge Level:
These gnomons appear to have more than a passing connection with the Fibonacci sequence. This problem ask you to investigate some of these connections.
### Building Gnomons
##### Age 14 to 16 Challenge Level:
Build gnomons that are related to the Fibonacci sequence and try to explain why this is possible.
### Pinned Squares
##### Age 14 to 16 Challenge Level:
What is the total number of squares that can be made on a 5 by 5 geoboard?
### Pentanim
##### Age 7 to 16 Challenge Level:
A game for 2 players with similarities to NIM. Place one counter on each spot on the games board. Players take it is turns to remove 1 or 2 adjacent counters. The winner picks up the last counter.
### Rational Roots
##### Age 16 to 18 Challenge Level:
Given that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.
### Magic Squares II
##### Age 14 to 18
An article which gives an account of some properties of magic squares.
### Chocolate 2010
##### Age 14 to 16 Challenge Level:
First of all, pick the number of times a week that you would like to eat chocolate. Multiply this number by 2...
### Equilateral Areas
##### Age 14 to 16 Challenge Level:
ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.
### Of All the Areas
##### Age 14 to 16 Challenge Level:
Can you find a general rule for finding the areas of equilateral triangles drawn on an isometric grid?
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https://byjus.com/question-answer/if-tangent-to-dfrac-x-2-a-2-dfrac-y-2-b-2-1-meets/
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Question
# If tangent to x2a2+y2b2=1 meets the axes at A and B, then the locus of mid point of AB isa2x2+b2y2=2a2x2+b2y2=4a2x2+b2y2=1none of these
Solution
## The correct option is B a2x2+b2y2=4Let the equation of tangent to the ellipse x2a2+y2b2=1 be xacosθ+ybsinθ=1 So, the points A=(acosθ,0), B=(0,bsinθ) Let the mid point of AB be (h,k) ⇒a2cosθ=h, b2sinθ=k ⇒2cosθ=ah,2sinθ=bk ∴ The required locus is a2x2+b2y2=4
Suggest corrections
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https://educationexpert.net/mathematics/2371149.html
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25 February, 15:05
# what is the least number that must be added to 2000 so that the sum is divisible exactly 10,12,16and 18? do it step by step
+1
1. 25 February, 16:07
0
The integers divisible by any set of positive integers are the multiples of their LCM
let us first write the factored form of each
10 = 2*5
12 = 2*2*3
16 = 2*2*2*2
18 = 2 x3*3
Now we will find lcm of these numbers
LCM = 2*2*2*2*3*3*5 = 720
The multiples of 720 are divisible by 10,12,16 and 18.
2000/720 = 2.777777 ...
The least integer greater than that is 3, so 3*720 = 2160 is the least integer greater than 2000 that is divisible by 10,12,16 and 18.
so if we need to find what must be added to 2000 so that the sum is divisible by 10,12,16 and 18, we must subtract 2000 from 2160
2160-2000=160
so we must add 160 to 2000 so that the sum is divisible exactly 10,12,16and 18
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http://blog.covertbay.com/2008/07/four-members-of-band.html
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No longer updated!
## Wednesday, July 16, 2008
### Four members of a band ...
Four members of a band are walking to a night concert. They decide to take a shortcut, but must cross a bridge. Luckily they have one flashlight. Because of the varying size of their instruments, it takes each member a different amount of time to cross the bridge - it takes the first person one minute, the second person two minutes, the third person five minutes and the fourth person ten minutes. They must cross the bridge in pairs, travelling at the slower speed so if the one minute person went with the ten minute person, it would take a total of ten minutes. Since there is only one flashlight, one person must come back across the bridge, then another pair can cross. They only have 17 minutes to cross the bridge and still get to the concert on time. What order should they cross to get everyone across and get to the concert?
Answer: First, the one minute person and the two minute person must cross the bridge, for a total of two minutes. Then the one minute person should come back with flashlight - total of three minutes. The five minute person and the ten minute person cross together next, making the total thirteen minutes. Now the two minute person goes back and (total now fifteen minutes) and gets the one minute person and they cross together bringing the total to seventeen minutes.
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https://convertoctopus.com/4176-months-to-hours
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## Conversion formula
The conversion factor from months to hours is 730.485, which means that 1 month is equal to 730.485 hours:
1 mo = 730.485 hr
To convert 4176 months into hours we have to multiply 4176 by the conversion factor in order to get the time amount from months to hours. We can also form a simple proportion to calculate the result:
1 mo → 730.485 hr
4176 mo → T(hr)
Solve the above proportion to obtain the time T in hours:
T(hr) = 4176 mo × 730.485 hr
T(hr) = 3050505.36 hr
The final result is:
4176 mo → 3050505.36 hr
We conclude that 4176 months is equivalent to 3050505.36 hours:
4176 months = 3050505.36 hours
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 hour is equal to 3.2781453627736E-7 × 4176 months.
Another way is saying that 4176 months is equal to 1 ÷ 3.2781453627736E-7 hours.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that four thousand one hundred seventy-six months is approximately three million fifty thousand five hundred five point three six hours:
4176 mo ≅ 3050505.36 hr
An alternative is also that one hour is approximately zero times four thousand one hundred seventy-six months.
## Conversion table
### months to hours chart
For quick reference purposes, below is the conversion table you can use to convert from months to hours
months (mo) hours (hr)
4177 months 3051235.845 hours
4178 months 3051966.33 hours
4179 months 3052696.815 hours
4180 months 3053427.3 hours
4181 months 3054157.785 hours
4182 months 3054888.27 hours
4183 months 3055618.755 hours
4184 months 3056349.24 hours
4185 months 3057079.725 hours
4186 months 3057810.21 hours
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https://www.smore.com/0mm0s
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# Math
### By: Annabelle Harris period 5-6
Step 1: find the common denominator
Step 2: change the fraction by multiplying the numerator by the same number you multiplied the denominator by unless the denominators are the same number
Step 3: add the numerators together
Step 4: Simplify if the answer is not in simplest form
Step 5: You Did It!!!!!!!!!!!!!!!!!!
## Example:
1/4 +2/4 = 3/4
2/3+2/5=1 1/15
5x3=15
2x5=10
2x3=6
6/15+10/15= 16/15
Simplify
16/15= 1 1/15
## Subtracting Fractions
Step 1: find the common denominator
Step 2: change the fraction by multiplying the numerator by the same number you multiplied the denominator by unless the denominators are the same number
Step 3: subtract the numerators together
Step 4: Simplify if the answer is not in simplest form
Step 5: You Did It!!!!!!!!!!!!!!!!!!
2/3-1/3=1/3
3/4-2/5=1 3/20
4x5=20
3x5=15
2x4=8
15/20-8/20=23/20
Simplify
23/20=1 3/20
## Multiplying Fractions
Step1: multiply across
Step2: Simplify
3/4x2/5=3/10
3x2=6
4x5=20
6/20
Simplify
1,6,2
1,20,2
6/2=3
20/2=10
3/10
## Dividing Fractions
Step1: KEEP the first number the same
Step2: Change the division problem to multiplication
Step3: FLIP the second number
Step4: multiply
Step5: simplify
## Example:
2/3/3/4=8/9
2/3/4/3=
2x4=8
3x3=9
8/9
There is nothing to simplify
Multiplyin' Fractions
Step 1: Line the decimals up in vertical columns
Step 2: Add or subtract the decimals like a normal addition or subtraction problems
Step 3: your decimal doesn't move carry it down
1.1 +
1.1
-----
2.2
3.3 -
2.1
-----
1.2
## Multiplying Decimals
Step 1: Line up your decimals in vertical columns
Step 2: Multiply like a normal multiplication problem
Step 3: Count the digits behind the decimal
Step 4: place your decimal to the left how ever many digits you counted
5.1 x
1.5
-----
.55
## Dividing Decimals
Step 1: Move the decimal that is outside the house to make it a whole number
Step 2: Move the decimal in the house the same number of digits as you did to the first decimal
Step 3: Divide like a normal division problem
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https://www.edplace.com/worksheet_info/maths/keystage2/year5/topic/877/1873/how-many-factors-1
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# How Many Factors? (1)
In this worksheet, students state how many factors a number has by looking at its "factor spider"!
Key stage: KS 2
Curriculum topic: Number: Multiplication and Division
Curriculum subtopic: Identify Multiples and Factors
Difficulty level:
### QUESTION 1 of 10
A number such as 6 has 4 factors, shown in pairs on the legs of this factor spider.
1 x 6 and 2 x 3 both equal 6.
The number 12 has 6 factors, shown on the legs of this factor spider.
Look at the factor spider for the number 9, which has 3 factors.
Note that the factor 3 should only be counted once.
Want to understand this further and learn how this links to other topics in maths?
Why not watch this video?
How many factors does the following number have?
5
How many factors does the following number have?
6
How many factors does the following number have?
8
How many factors does the following number have?
3
How many factors does the following number have?
1
How many factors does the following number have?
4
How many factors does the following number have?
10
How many factors does the following number have?
14
How many factors does the following number have?
11
How many factors does the following number have?
12
• Question 1
How many factors does the following number have?
5
2
EDDIE SAYS
Prime 5: 1, 5
• Question 2
How many factors does the following number have?
6
4
EDDIE SAYS
6: 1, 2, 3, 6
• Question 3
How many factors does the following number have?
8
4
EDDIE SAYS
8: 1, 2, 4, 8
• Question 4
How many factors does the following number have?
3
2
EDDIE SAYS
Prime 3: 1, 3
• Question 5
How many factors does the following number have?
1
1
EDDIE SAYS
Square number
• Question 6
How many factors does the following number have?
4
3
EDDIE SAYS
Square number 4: 1, 2, 4
• Question 7
How many factors does the following number have?
10
4
EDDIE SAYS
10: 1, 2, 5, 10
• Question 8
How many factors does the following number have?
14
4
EDDIE SAYS
14: 1, 2, 7, 14
• Question 9
How many factors does the following number have?
11
2
EDDIE SAYS
Prime number 11: 1, 11
• Question 10
How many factors does the following number have?
12
6
EDDIE SAYS
12: 1, 2, 3, 4, 6, 12
---- OR ----
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# Search by Topic
#### Resources tagged with Making and proving conjectures similar to Square Mean:
Filter by: Content type:
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### There are 30 results
Broad Topics > Using, Applying and Reasoning about Mathematics > Making and proving conjectures
### To Prove or Not to Prove
##### Age 14 to 18
A serious but easily readable discussion of proof in mathematics with some amusing stories and some interesting examples.
### DOTS Division
##### Age 14 to 16 Challenge Level:
Take any pair of two digit numbers x=ab and y=cd where, without loss of generality, ab > cd . Form two 4 digit numbers r=abcd and s=cdab and calculate: {r^2 - s^2} /{x^2 - y^2}.
### Multiplication Square
##### Age 14 to 16 Challenge Level:
Pick a square within a multiplication square and add the numbers on each diagonal. What do you notice?
### Janine's Conjecture
##### Age 14 to 16 Challenge Level:
Janine noticed, while studying some cube numbers, that if you take three consecutive whole numbers and multiply them together and then add the middle number of the three, you get the middle number. . . .
### Loopy
##### Age 14 to 16 Challenge Level:
Investigate sequences given by $a_n = \frac{1+a_{n-1}}{a_{n-2}}$ for different choices of the first two terms. Make a conjecture about the behaviour of these sequences. Can you prove your conjecture?
### How Old Am I?
##### Age 14 to 16 Challenge Level:
In 15 years' time my age will be the square of my age 15 years ago. Can you work out my age, and when I had other special birthdays?
### What's Possible?
##### Age 14 to 16 Challenge Level:
Many numbers can be expressed as the difference of two perfect squares. What do you notice about the numbers you CANNOT make?
### Always a Multiple?
##### Age 11 to 14 Challenge Level:
Think of a two digit number, reverse the digits, and add the numbers together. Something special happens...
##### Age 14 to 16 Challenge Level:
Explore the relationship between quadratic functions and their graphs.
### Triangles Within Squares
##### Age 14 to 16 Challenge Level:
Can you find a rule which relates triangular numbers to square numbers?
### A Little Light Thinking
##### Age 14 to 16 Challenge Level:
Here is a machine with four coloured lights. Can you make two lights switch on at once? Three lights? All four lights?
### Multiplication Arithmagons
##### Age 14 to 16 Challenge Level:
Can you find the values at the vertices when you know the values on the edges of these multiplication arithmagons?
### Triangles Within Pentagons
##### Age 14 to 16 Challenge Level:
Show that all pentagonal numbers are one third of a triangular number.
### Triangles Within Triangles
##### Age 14 to 16 Challenge Level:
Can you find a rule which connects consecutive triangular numbers?
### Problem Solving, Using and Applying and Functional Mathematics
##### Age 5 to 18 Challenge Level:
Problem solving is at the heart of the NRICH site. All the problems give learners opportunities to learn, develop or use mathematical concepts and skills. Read here for more information.
### Polycircles
##### Age 14 to 16 Challenge Level:
Show that for any triangle it is always possible to construct 3 touching circles with centres at the vertices. Is it possible to construct touching circles centred at the vertices of any polygon?
### Close to Triangular
##### Age 14 to 16 Challenge Level:
Drawing a triangle is not always as easy as you might think!
##### Age 14 to 16 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a non-convex quadrilateral.What do you notice about the quadrilateral PQRS and its area?
##### Age 14 to 16 Challenge Level:
The points P, Q, R and S are the midpoints of the edges of a convex quadrilateral. What do you notice about the quadrilateral PQRS as the convex quadrilateral changes?
### Alison's Mapping
##### Age 14 to 16 Challenge Level:
Alison has created two mappings. Can you figure out what they do? What questions do they prompt you to ask?
### Charlie's Mapping
##### Age 11 to 14 Challenge Level:
Charlie has created a mapping. Can you figure out what it does? What questions does it prompt you to ask?
### Helen's Conjecture
##### Age 11 to 14 Challenge Level:
Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?
### Happy Numbers
##### Age 11 to 14 Challenge Level:
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
### Dice, Routes and Pathways
##### Age 5 to 14
This article for teachers discusses examples of problems in which there is no obvious method but in which children can be encouraged to think deeply about the context and extend their ability to. . . .
### Few and Far Between?
##### Age 14 to 18 Challenge Level:
Can you find some Pythagorean Triples where the two smaller numbers differ by 1?
### Consecutive Negative Numbers
##### Age 11 to 14 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Curvy Areas
##### Age 14 to 16 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Exploring Simple Mappings
##### Age 11 to 14 Challenge Level:
Explore the relationship between simple linear functions and their graphs.
### Epidemic Modelling
##### Age 14 to 18 Challenge Level:
Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths.
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# math
posted by .
An arithmetic series has third term 11. The ninth term is 5 times the second term. Find the common difference.
• math -
Let the first term be called a and the common difference be called b.
a + 2b = 11
a + 8b = 5*(a + b) = 5a + 5b
The last equation can be rewritten
4a - 3b = 0
4a + 8b = 44
11b = 44
b = 4
a = 11 - 2b = 3
The series is 3,7,11,15,19,23,27,31,35...
Note that 35 (the ninth term) is 5 times 7
## Similar Questions
1. ### Arithmetic Sequence
Given the third term of an arithmetic sequence less than the fourth term by three. The seventh term is two times the fifth term. Find the common difference and the first term.
2. ### Math
The fifth term of an arithmetic progression is three times the second term,and the third term is 10.a)What is the first term,b)the common difference and c)the 15th term?
3. ### math
the fourth term of an arithmetic progression is equal to 3 times the first term and the seventh term exceeds twice the third term by 1 find the first term and the common differrence
4. ### Algebra
True or False 1. – 5, – 5, – 5, – 5, – 5, … is an arithmetic sequence. 2. In an arithmetic sequence, it is possible that the 13th term is equal to its 53rd term. 3. In an arithmetic sequence, the common difference is computed …
5. ### math
The sum of the first 10 term of an arithmetic sequence is 145 and the sum of the fourth and ninth term is five times the third term. Determine the first term and constant difference.
6. ### math
if 4th term of an arithmetic progression is 3 times the 1st term nd 7th term exceeds twice the third term by 1. find the first term and common difference
7. ### Arithmetic progession
The third term of an A.P is 10times more than the second term. Find the sum of the eight and fifteen term of the A.P if the seventh term is seven times the first term
8. ### Math
The sum of third and ninth terms of arithmetic series is 20 and the difference between the twelfth and fourth term is 32.determine the value of the first term and the constant difference.
9. ### arithmetic
in an arithmetic progression the 13th term is 27 and the 7th term is three times the second term find;the common difference,the first term and the sum of the first ten terms
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The fourth term of an arithmetic sequence is less than the fifth term by 3. The seventh term is three times the fifth term. Find the common difference and the first term
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| 681,242,916 | 9,705 |
# Search by Topic
#### Resources tagged with Investigations similar to Shapes in the Alphabet:
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Stage:
Challenge level:
### Making Squares
##### Stage: 2
Investigate all the different squares you can make on this 5 by 5 grid by making your starting side go from the bottom left hand point. Can you find out the areas of all these squares?
### Tiles on a Patio
##### Stage: 2 Challenge Level:
How many ways can you find of tiling the square patio, using square tiles of different sizes?
### My New Patio
##### Stage: 2 Challenge Level:
What is the smallest number of tiles needed to tile this patio? Can you investigate patios of different sizes?
### Pebbles
##### Stage: 2 and 3 Challenge Level:
Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time?
### Stairs
##### Stage: 1 and 2 Challenge Level:
This challenge is to design different step arrangements, which must go along a distance of 6 on the steps and must end up at 6 high.
### Polo Square
##### Stage: 2 Challenge Level:
Arrange eight of the numbers between 1 and 9 in the Polo Square below so that each side adds to the same total.
### Cubes Here and There
##### Stage: 2 Challenge Level:
How many shapes can you build from three red and two green cubes? Can you use what you've found out to predict the number for four red and two green?
### 28 and It's Upward and Onward
##### Stage: 2 Challenge Level:
Can you find ways of joining cubes together so that 28 faces are visible?
### More Pebbles
##### Stage: 2 and 3 Challenge Level:
Have a go at this 3D extension to the Pebbles problem.
### Triangle Shapes
##### Stage: 1 and 2 Challenge Level:
This practical problem challenges you to create shapes and patterns with two different types of triangle. You could even try overlapping them.
### Fit These Shapes
##### Stage: 1 and 2 Challenge Level:
What is the largest number of circles we can fit into the frame without them overlapping? How do you know? What will happen if you try the other shapes?
### Halving
##### Stage: 1 Challenge Level:
These pictures show squares split into halves. Can you find other ways?
### Triangle Relations
##### Stage: 2 Challenge Level:
What do these two triangles have in common? How are they related?
### Bean Bags for Bernard's Bag
##### Stage: 2 Challenge Level:
How could you put eight beanbags in the hoops so that there are four in the blue hoop, five in the red and six in the yellow? Can you find all the ways of doing this?
### Room Doubling
##### Stage: 2 Challenge Level:
Investigate the different ways you could split up these rooms so that you have double the number.
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Sort the Street
##### Stage: 1 Challenge Level:
Sort the houses in my street into different groups. Can you do it in any other ways?
##### Stage: 2 Challenge Level:
I like to walk along the cracks of the paving stones, but not the outside edge of the path itself. How many different routes can you find for me to take?
### It Figures
##### Stage: 2 Challenge Level:
Suppose we allow ourselves to use three numbers less than 10 and multiply them together. How many different products can you find? How do you know you've got them all?
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In this investigation, you must try to make houses using cubes. If the base must not spill over 4 squares and you have 7 cubes which stand for 7 rooms, what different designs can you come up with?
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Cut differently-sized square corners from a square piece of paper to make boxes without lids. Do they all have the same volume?
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Arrange your fences to make the largest rectangular space you can. Try with four fences, then five, then six etc.
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An activity making various patterns with 2 x 1 rectangular tiles.
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If we had 16 light bars which digital numbers could we make? How will you know you've found them all?
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When newspaper pages get separated at home we have to try to sort them out and get things in the correct order. How many ways can we arrange these pages so that the numbering may be different?
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Ana and Ross looked in a trunk in the attic. They found old cloaks and gowns, hats and masks. How many possible costumes could they make?
### Crossing the Town Square
##### Stage: 2 and 3 Challenge Level:
This tricky challenge asks you to find ways of going across rectangles, going through exactly ten squares.
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In how many ways can you stack these rods, following the rules?
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You cannot choose a selection of ice cream flavours that includes totally what someone has already chosen. Have a go and find all the different ways in which seven children can have ice cream.
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Suppose there is a train with 24 carriages which are going to be put together to make up some new trains. Can you find all the ways that this can be done?
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Investigate how this pattern of squares continues. You could measure lengths, areas and angles.
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If you have three circular objects, you could arrange them so that they are separate, touching, overlapping or inside each other. Can you investigate all the different possibilities?
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## A certain computer company produces two different monitors, P and Q. In 2010, what was the net profit from the sale of
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### A certain computer company produces two different monitors, P and Q. In 2010, what was the net profit from the sale of
by BTGmoderatorLU » Wed Jan 12, 2022 5:56 pm
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## Global Stats
Source: Official Guide
A certain computer company produces two different monitors, P and Q. In 2010, what was the net profit from the sale of the two monitors?
1) Of the company's expenses in 2010, rent and utilities totaled $500,000 2) In 2010, the company sold 50,000 units of monitor P at$300 per unit and 30,000 units of monitor Q at $650 per unit The OA is E ### GMAT/MBA Expert GMAT Instructor Posts: 16084 Joined: 08 Dec 2008 Location: Vancouver, BC Thanked: 5254 times Followed by:1267 members GMAT Score:770 ### Re: A certain computer company produces two different monitors, P and Q. In 2010, what was the net profit from the sale by [email protected] » Thu Jan 13, 2022 9:47 am ## Timer 00:00 ## Your Answer A B C D E ## Global Stats BTGmoderatorLU wrote: Wed Jan 12, 2022 5:56 pm Source: Official Guide A certain computer company produces two different monitors, P and Q. In 2010, what was the net profit from the sale of the two monitors? 1) Of the company's expenses in 2010, rent and utilities totaled$500,000
2) In 2010, the company sold 50,000 units of monitor P at $300 per unit and 30,000 units of monitor Q at$650 per unit
The OA is E
Target question: In 2010, what was the net profit from the sale of the two monitors?
Key concept: profit = revenue - expenses
Statement 1: Of the company's expenses in 2010, rent and utilities totaled $500,000. This just tells us that a PORTION of the company's expenses =$500,000
In addition to not knowing the TOTAL expenses, we also don't know the TOTAL revenue
Statement 1 is NOT SUFFICIENT
Statement 2: In 2010, the company sold 50,000 units of monitor P at $300 per unit and 30,000 units of monitor Q at$650 per unit.
From this information, we COULD determine the company's TOTAL revenue (although we wouldn't waste time doing so on test day)
However, even if we know the company's total revenue, we still have no information about the expenses.
Statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
To answer the target question, we need the companies TOTAL expenses and its TOTAL revenue.
Statement 1 tells us that a PORTION of the company's expenses is \$500,000
Statement 2 tells the company's TOTAL revenue
Since we still don't know the company's TOTAL revenue, we can't answer the target question with certainty
The combined statements are NOT SUFFICIENT
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# Moment of Inertia
Science > Rotational Motion > You are Here
Rigid Body:
• A rigid body is one whose geometric shape and size remains unchanged under the action of any external force.
#### Axis of rotation:
• When a rigid body performs the rotational motion, the particles of the body moves in circles. The centres of these circles lie on a straight line called the axis of rotation, which is fixed and perpendicular to the plane of circles.
• The particles on the axis of rotation are stationary.
#### Moment of Inertia:
• The moment of inertia of a rigid body about a given axis is, defined as the sum of the products of the mass of each and every particle of the body and the square of its distance from the given axis.
• S.I. unit of moment of inertia is kg m² and C.G.S. system it is g cm². Dimensions of the moment of inertia are [M1L2T0]
#### Explanation:
• If the system of masses consists of a number of point masses m1, m2, m3, m4, ……..msituated at a distance of r1, r2, r3, r4, ……..rfrom the axis of rotation, then by definition of moment of Inertia we have
• Instead of assuming body to be composed of discrete masses, it can be considered to be composed of continuous matter (mass); then the process of summation should be replaced by integration with proper limits.
#### Physical Significance of Moment of Inertia:
• Newton’s first law of motion is also called a law of inertia indicates that a body is unable to change by itself its state of rest or state of uniform motion along a straight line. This property of inertness is known as inertia. It is an inherent property of matter. It is due to the inertia, a body opposes any change in its state of rest or of uniform motion in a straight line.
• In the translational motion, the mass of a body is a measure of its inertia. Greater the mass, larger is the inertia, greater is the force required to produce a given linear acceleration in it.
• In rotational motion, the moment of inertia of a body is a measure of its inertia. Greater the moment of inertia, larger is the torque required to produce a given angular acceleration in it.
• Thus moment of inertia in the rotational motion is analogous to the mass in translational motion because it plays the same role in rotational motion as the mass plays in translational motion. This is clear from the following table
Sr.No. Translational motion Rotational motion 1 Linear Momentum p = m v Angular Momentum L = Iω 2 Force F = m a Torque τ = Iα 3 Kinetic Energy K.E. = ½mv² Kinetic Energy K.E. = ¼mv²
Radius of Gyration:
• The radius of gyration of a rigid body about a given axis is defined as the distance from the axis at which the whole mass of the body must be supposed to be concentrated so that this imaginary point mass has the same moment of Inertia as the actual body, about the given axis.
• The radius of gyration is denoted by letter ‘K’. As it is a distance, its S.I. unit is m and C.G..S. unit is cm.
• Its dimensions are [M0L1T0]
• The expression for moment of inertia in terms of the radius of gyration is I = MK².
Where I Moment of inertia of a body
M = Mass of the body
K = Radiation of gyration of the body
#### Physical Significance of Radius of Gyration:
• Moment of inertia depends on the mass of the body, the distribution of mass about the axis of rotation and the position of the axis of rotation.
• These factors can be separated by expressing the moment of inertia as a product of the mass and the square of a particular distance from the axis of rotation. This particular distance is called as radius of gyration (K) of the body.
• By definition of radius of gyration
• From the above explanation we can conclude that, the radius of gyration is the measure of distribution of the mass of the body about the given axis
#### Example – 1:
• Four point masses 1kg, 2 kg, 3 kg and 4 kg are located at the corners A, B, C and D of a square ABCD of side 1m. Find the moment of inertia and radius of gyration in each of the following cases when axis of rotation is
• passing through A and perpendicular to the plane of ABCD
• passing through O, the centre of the square and perpendicular to plane of ABCD
• along the side AB
• along diagonal AC
• Solution:
• Given: m1 = 1 kg, m2 = 2 kg, m3 = 3 kg, m4 = 4 kg, AB = BC = CD = AD = 1 m
• Axis passing through A and perpendicular to the plane of ABCD
Sr.No. Mass Distance from A 1 m1 = 1 kg r1 = 0 2 m2 = 2 kg r2= 1 m 3 m3 = 3 kg r3= √2 m 4 m4 = 4kg r4 = 1m
IA = ∑ miri² = m1r1² + m2r2² + m3r3² + m4r4²
∴ IA = (1)(0)² + (2)(1)² + (3)(√2)² + (4)(1)²
∴ IA = 0 + 2 + 6 + 4 = 10 kg m²
M = m1 + m2 + m3 + m4 = 1 + 2 +3 + 4 = 10 kg
By definition of radius of gyration
IA = MKA²
∴ 10 = 10 KA²
∴ KA² = 1
KA = 1 m
Ans: Moment of inertia of system about axis passing through A and perpendicular to the plane of ABCD is 10 kg m² and corresponding radius of gyration is 1 m.
• Axis passing through O, the centre of the square and perpendicular to plane of ABCD
Sr.No. Mass Distance from O 1 m1 = 1 kg r1 = √2 /2 m 2 m2 = 2 kg r2= √2 /2 m 3 m3 = 3 kg r3= √2 /2 m 4 m4 = 4kg r4 = √2 /2 m
IO = ∑ miri² = m1r1² + m2r2² + m3r3² + m4r4²
∴ IO = (1)(√2 /2)² + (2)(√2 /2)² + (3)(√2 /2)² + (4)(√2 /2)²
∴ IO = (1)(0.5)² + 2(0.5)² + 3(0.5)² + 4(0.5)² = 10(0.5)² = 10 × 0.25 = 2.5 kg m²
M = m1 + m2 + m3 + m4 = 1 + 2 +3 + 4 = 10 kg
By definition of radius of gyration
IA = MKO²
∴ 2.5 = 10 KO²
∴ KO² = 0.25
∴ KO = 0.5 m
Ans: Moment of inertia of system about axis passing through O and perpendicular to the plane of ABCD is 2.5 kg m² and corresponding radius of gyration is 0.5 m.
• Axis passing through AB:
Sr.No. Mass Distance from AB 1 m1 = 1 kg r1 = 0 m 2 m2 = 2 kg r2= 0 m 3 m3 = 3 kg r3= 1 m 4 m4 = 4kg r4 = 1 m
IAB = ∑ miri² = m1r1² + m2r2² + m3r3² + m4r4²
∴ IAB = (1)(0)² + (2)(0)² + (3)(1)² + (4)(1)²
∴ IAB = 0 + 0 + 3 + 4 = 7 kg m²
M = m1 + m2 + m3 + m4 = 1 + 2 +3 + 4 = 10 kg
By definition of radius of gyration
IAB = MKAB²
∴ 7 = 10 KAB²
∴ KAB² = 0.7
KAB = 0.837 m
Ans: Moment of inertia of system about side AB is 75 kg m² and corresponding radius of gyration is 0.837m.
• Axis passing through Diagonal AC:
Sr.No. Mass Distance from AC 1 m1 = 1 kg r1 = 0 m 2 m2 = 2 kg r2= √2 /2 m 3 m3 = 3 kg r3= 0 m 4 m4 = 4kg r4 = √2 /2 m
IAC = ∑ miri² = m1r1² + m2r2² + m3r3² + m4r4²
∴ IAC = (1)(0)² + (2)(√2 /2)² + (3)(0)² + (4)(√2 /2)²
∴ IAC = 0 + (2)(0.5)² + 0 + 4(0.5)² = 7 kg m²
M = m1 + m2 + m3 + m4 = 1 + 2 +3 + 4 = 10 kg
By definition of radius of gyration
IAC = MKAC²
∴ 7 = 10 KAC²
∴ KAC² = 0.7
KAC = 0.837 m
Ans: Moment of inertia of system about diagonal AC is 75 kg m² and corresponding radius of gyration is 0.837m.
#### Example – 2:
• Three point masses 1 kg, 2 kg and 3 kg are located at the vertices A, B and C of an equilateral triangle ABC of side 1m. Find the moment of inertia and radius of gyration in each of the following cases when axis of rotation is
• passing through A and perpendicular to the plane of triangle ABC.
• passing through centroid G of the triangle and perpendicular to the plane of triangle ABC
• along the side AB
• Solution:
• Given: m1 = 1 kg, m2 = 2 kg, m3 = 3 kg, , AB = BC = AC = 1 m
• Axis passing through A and perpendicular to the plane of triangle ABC
Sr.No. Mass Distance from A 1 m1 = 1 kg r1 = 0 m 2 m2 = 2 kg r2= 1 m 3 m3 = 3 kg r3= 1m
IA = ∑ miri² = m1r1² + m2r2² + m3r3²
∴ IA = (1)(0)² + (2)(1)² + (3)(1)²
∴ IA = 0 + 2 +3 = 5 kg m²
M = m1 + m2 + m3 = 1 + 2 +3 = 6 kg
By definition of radius of gyration
IA = MKA²
∴ 5 = 6 KA²
∴ KA² = 5/6 = 0.8333
KA = 0.913 m
Ans: Moment of inertia of system about A is 5 kg m² and
corresponding radius of gyration is 0.913 m.
• Passing through centroid G of the triangle and perpendicular to the plane of triangle ABC
For equilateral triangle GA = GB = GC = side/√ = 1 / √ = 0.577 m
Sr.No. Mass Distance from A 1 m1 = 1 kg r1 = 0.577 m 2 m2 = 2 kg r2= 0.577 m 3 m3 = 3 kg r3= 0.577 m
IG = ∑ miri² = m1r1² + m2r2² + m3r3²
∴ IG = (1)(0.577)² + (2)(0.577)² + (3)(0.577)²
∴ IG = 1/3 + 2/3 +3/3 = 6/3 = 2 kg m²
M = m1 + m2 + m3 = 1 + 2 +3 = 6 kg
By definition of radius of gyration
IG = MKG²
∴ 2 = 6 KG²
∴ KG² = 2/6 = 0.3333
KG = 0.577 m
Ans: Moment of inertia of system about G is 2 kg m² and corresponding radius of gyration is 0.577 m.
• Along the side AB
Sr.No. Mass Distance from A 1 m1 = 1 kg r1 = 0 m 2 m2 = 2 kg r2= 0 m 3 m3 = 3 kg r3= √3 /2=0.866 m
IAB = ∑ miri² = m1r1² + m2r2² + m3r3²
∴ IAB = (1)(0)² + (2)(0)² + (3)(0.866)²
∴ IAB = 0 + 0 + 1.25 = 1.25 kg m²
M = m1 + m2 + m3 = 1 + 2 +3 = 6 kg
By definition of radius of gyration
IAB = MKAB²
∴ 1.25 = 6 KAB²
∴ KAB² = 1.25/6 = 0.2083
KAB = 0.456 m
Ans: Moment of inertia of system about G is 1.25 kg m² and corresponding radius of gyration is 0.456 m.
#### Example – 3:
• A light thin uniform rod of length 2 m has two bodies stuck to its ends. The mass of each body is 100 g. Find the moment of inertia of the system about a transverse axis passing through i) the centre of mass of the rod ii) one end of the rod.
• Case – I: Axis through the centre of mass of the rod
Mass (m) Distance from axis of rotation (r) m1 r1 = l/2 = 2/2 = 1m m2 r2 = l/2 = 2/2 = 1m
• Case – II: Axis through the centre of mass of the rod
Mass (m) Distance from axis of rotation (r) m1 r1 = l = 2 m m2 r2 = 0 m
Ans: The M.I. of the system about the axis passing through the centre is 0.2 kg m2
and about the axis passing through the end is 0.4 kg m2.
#### Kinetic Energy of a Rotating Body:
• Consider a rigid body rotating about axis passing through point O and perpendicular to the plane of paper in anticlockwise sense as shown. Consider infinitesimal element at P of mass dm in the plane of the paper. Let distance of point P from the axis of rotation be r.
The moment of inertia of the rigid body is given by
• As the body is a rigid body, the element at P will start moving in a circular motion. Let v be its tangential or linear velocity.
Then by concepts of circular motion
v = r ω ……….. .. (2)
Due to this velocity, the element at P will possess kinetic energy which is given by
From equation (2) and (3)
• Similarly, we can find kinetic energy of each and every infinitesimal element in the body. Total kinetic energy of body can be found by integrating both sides of above equation.
This is an expression for the kinetic energy of a rotating body about a given axis.
#### Relation between kinetic energy of a rotating body and the frequency of its rotation
Substituting ω = 2πn in equation (5) we get
The quantities in the bracket are constant.
∴ E α n²
Thus, the kinetic energy of rotating body about a given axis is directly proportional to the square of the frequency of the body.
#### Relation between kinetic energy of rotating a body and its angular momentum
• The kinetic energy of rotating body about a given axis is given by
#### Torque Acting on Rotating Body:
• Consider a rigid body rotating about an axis passing through point O and perpendicular to the plane of the paper in an anticlockwise sense as shown. Let us consider infinitesimal element at P of mass dm in the plane of the paper. Let distance of this element from the axis of rotation be r.
The moment of Inertia of the rigid body is given by
• Let τ be the external torque acting on the body. Under the influence of this torque let the body rotates with angular acceleration α. Let dF be the magnitude of external force acting on the element, in the plane of page and at right angles to position vector of point P, then by Newton’s Second Law of motion,
dF = dm . a ……….. (2)
where ‘a’ is the magnitude of linear acceleration (tangential acceleration) of the element.
By concept of circular motion, we know that
a = r α …………. (3)
From Equations (2) and (3)
dF = dm . r. α ……… (4)
Let dτ be the torque acting on the element, then magnitude of the torque is given by
dτ = r , dF ……………… (5)
From equations (4) and (5)
dτ = r , dm . r. α
dτ = r ² dm α
• Similarly, we can find torque on each and every infinitesimal element in the body. The total torque acting on the body can be found by integrating both sides of above equation
• As body is rigid body angular acceleration (α) of all the elements is the same and it is taken out of integration sign
From equations (1) and (6)
τ = Iα
This is an expression for torque acting on rotating body.
#### Example – 3:
• A disc of mass 50 Kg and radius 0.4 m is capable of rotation about an axis passing through its centre and at right angles to its plane. If a constant torque of 12 Nm acts on it, find the angular acceleration caused in the disc.
• Solution:
• Given: Mass of disc = M = 50 kg, radius of disc = R = 0.4 m, Torque = τ = 12 Nm
• To Find: Angular acceleration = α = ?
The M.I. of a disc about an axis passing through its centre and at right angles to its plane is given by
I = ½MR² = ½ × 50 × 0.4² = 25 × 0.16 = 4 kg m²
We have torque, τ = I α
∴ α = τ / I = 12/4 = 3 rad/s²
Ans: Angular acceleration = 3 rad/s²
#### Example – 4:
• A constant couple of 25 Nm acts on a flywheel of mass 100 kg and radius of gyration 25 cm. What is the resulting angular acceleration?
• Solution:
• Given: Mass of flywheel (disc) = M = 100 kg, radius of gyration = K = 25 cm = 0.25 m, Torque = τ = 25 Nm
• To Find: Angular acceleration = α = ?
The M.I. of a body in terms of radius of gyration is given by
I = MK² = 100 × 0.25² = 100 × 0.0625 = 6.25 kg m²
We have torque, τ = I α
∴ α = τ / I = 25/6.25 = 4 rad/s²
Ans: Angular acceleration = 4 rad/s²
#### Example – 5:
• A constant torque of magnitude 5000 Nm acting on a body increases its angular velocity from 4 rad/s to 20 rad/s in 8 s. Calculate the M.I. of the body about the axis of rotation.
• Solution:
• Given: Initial angular speed = ω1 = 4 rad/s, Final angular speed = ω2 = 20 rad/s, Time taken = t = 8 s, Torque = τ = 5000 Nm
• To Find: Moment of Inertia = I = ?
α = (ω– ω1)/t = (20 – 4)/8 = 2 rad/s²
We have torque, τ = I α
∴ I = τ/ α = 5000/2 = 2500 kg m²
Ans: Moment of inertia of body is 2500 kg m²
#### Example – 6:
• The speed of rotation of the body increases from 60 rpm too 90 rpm in 1 minute. Calculate the torque acting on the body, if its M.I. is 500 kgm².
• Solution:
• Given: Initial angular speed = N1 = 60 r.pm., Final angular speed = N2 = 90 r.pm., Time taken = t = 1 min = 60 s, Moment of Inertia = I = kgm²
• To Find: Torque = τ = ?
We have torque, τ = I α
∴ τ = 500 × 0.0524 = 26.18 Nm
Ans: Torque acting is 26.18 Nm
#### Example – 7:
• A solid sphere of radius 25 cm and mass 25 kg rotates about a fixed axis passing through its centre. If its angular velocity changes from 2π rad/s to 12π rad/s in 5 s, calculate the torque applied.
• Solution:
• Given: Radius of sphere = R = 25 cm = 0.25 m, Mass of sphere = 25 kg, Initial angular speed = ω1 = 2π rad/s, Final angular speed = ω2 = 12π rad/s, Time taken = t = 5 s,
• To Find: Torque = τ = ?
M.I. of sphere about a fixed axis passing through its centre is given by
α = (ω– ω1)/t = (12π – 2π)/5 = 2π = 2 × 3.142 = 6.284 rad/s²
We have torque, τ = I α
∴ τ = 0.625 × 6.284 = 3.928 Nm
Ans: Torque required is 3.928 Nm
#### Example – 8:
• The angular velocity of a disc rotating in its plane changes from 2 rad/s to 10 rad/s in one minute when a constant torque of 2 Nm applied. What is the M. I. of the disc?
• Solution:
• Given: Initial angular speed = ω1 = 2 rad/s, Final angular speed = ω2 = 10 rad/s, Time taken = t = 1 min = 60 s, Torque = τ = 2 Nm
• To Find: Moment of inertia of the disc = I = ?
α = (ω– ω1)/t = (10 – 2)/60 = 8/60 = 0.1333 rad/s²
We have torque, τ = I α
∴ I = τ / α = 2/0.1333 = 15 kg m²
Ans: Moment of inertia of the disc is 15 kg m²
#### Example – 9:
• A flywheel in the form of a disc is initially at rest. When a constant torque acts upon it for one minute, it attains a speed of 300 r.p.m. Calculate the torque if the mass and radius of the disc are 10 kg and 0.4 m respectively.
• Solution:
• Given: Radius of disc = R = 0.4 m Mass of disc = 10 kg, Initial angular speed = N1 = 0 rad/s, Final angular speed = N2 = 300 r.p.m., Time taken = t = 1 min = 60 s,
• To Find: Torque = τ = ?
M.I. of disc about an axis passing through its centre and perpendicular to plane is given by
I = ½MR² = ½ × 10×0.4² = 0.8 kg m²
ω= 2πN2/ 60 = (2× π × 300)/60 = 10π rad/s
α = (ω– ω1)/t = (10π – 0)/60 = π/6 = 0.5237 rad/s²
We have torque, τ = I α
∴ τ = 0.8 × 0.5237 = 0.419 Nm
Ans: Torque required is 0.419 Nm
#### Statement :
• The moment of inertia of a rigid body about any axis is equal to the sum of its moment of inertia about a parallel axis through its centre of mass and the product of the mass of the body and the square of the distance between the two axes.
#### Explanation :
• Consider a rigid body of mass M rotating about an axis passing through point O and perpendicular to the plane of the paper. Let IO be the moment of inertia of the body about the axis passing through O and perpendicular to the plane of the paper. Let IG be the moment of inertia of the body about the axis passing through the centre of mass of the body (G) and parallel to given axis passing through O.
• Let ‘h’ be the distance between the two axes i.e. (OG) = h. Then by principle of parallel axes’
IO = IG + Mh²
This is the mathematical statement of the principle of parallel axes.
#### Proof :
• Consider the infinitesimal element of the body of mass ‘dm’ at P, in the plane of the paper. Join OP, OG and GP. Draw a perpendicular from P on extended line OG, such that PD ⊥ OG.
By definition Of moment of inertia
Applying Pythagoros theorem to Δ OPD
OP² = OD² + PD²
∴ OP²2 = (OG + GD)² + PD²
∴ OP² = OG²+2OG..GD + GD² + PD² …(3)
In DPGD, by Pythagoros theorem,
PG² = GD² + PD² …….. (4)
From-equations (3) and (4)
OP² = OG² + 2 OG..GD + PG²
Multiplying both sides of above equation by dm
dm.OP² = dm.OG² + 2 dm.OG.GD + dm.PG²
integrating both sides
IO = IG + Mh² (Proved)
#### Statement:
• Moment of inertia of a rigid plane lamina about an axis perpendicular to its plane is equal to the sum of its moment of inertia about any two mutually perpendicular axes in its plane and meeting in the point where the perpendicular axis cuts the lamina.
#### Explanation:
• Consider a thin rigid plate or layer or lamina (in this case thickness of the body is less compared to its surface area). Let OX and OY be two mutually perpendicular axes in the plane of a rigid lamina, intersecting at O. Let OZ be, other axes which is passing through point O and is perpendicular to the plane of the lamina. Let lx , ly and Iz be the moments of Inertia of the lamina about OX, OY and OZ axes respectively.
• Then by principle of perpendicular axes,
lz = lx + ly
This is the mathematical statement of the principle of perpendicular axes.
#### Proof:
• Consider infinitesimal element in the plane of lamina having mass dm situated at point P. Let the coordinates of point P be (x, y). Join OP and draw PM and PN perpendicular on OX and OY, respectively.
Then OM = NP = x and MP = ON = y, Let OP = r
Moment of inertia of the lamina about Z- axis is given by
Similarly, M.I. of the lamina about X-axis is given by
Similarly Moment of Inertia of the lamina about Y-axis la given by
Applying Pythagoros theorem to triangle OPM
OP² = OM² + PM²
∴ z² = x² + y²
Multiplying both sides of equation by dm and integrating
From equation (1), (2), (3) and (4)
lz = lx + ly (Proved)
Science > Rotational Motion > You are Here
### 2 Comments
1. Sanket Talwekar
How did u find distance in every case?
• Hemant More
Each side of the square is 1 m. Hence by Pythagoras theorem diagonal is √2 m. and all the distances should be measured from the point through which the axis of rotation passes.
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# What is negative 7 over 8 plus 3 over 4 equal?
Updated: 12/18/2022
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it equals negative 1/8
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8 and 3/4
### Does negative eight plus negative three equal?
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### What does positive 5 plus negative 3 plus negative 5 equal?
5 + (-3) + (-5) = -3
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### What does a Minus plus a plus equal?
Negative plus a negative is a negative. Ex. -1+-1=-2 Negative times a negative is a positive. -2*-2=4 But a negative [minus] plus a positive [plus] can be minus, zero or plus: For example, -3 + 2 = -1 -3 + 3 = 0 -3 + 4 = +1
negative 30
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# Local Systems: The connection perspective
Welcome to the next installation of my series on local systems. In this post I’ll be talking about connections. This post should require less sophistication than the last few — no schemes, no functors — I’ll almost be coming at the subject afresh. There will be another post later, explaining how you might get to connections if you started out thinking about the infinitesimal site.
To start out with, let’s talk about derivatives; ordinary, single variable calculus derivatives. We have a function $f$ of a variable $x$. Then the derivative of $f$ is the function $f'(x) := \lim_{h \to 0} (f(x+h) - f(x))/h$. There are two directions in which we might want to generalize this idea. The first is to work with functions on a manifold, on a space which has no inherent coordinate system. This is the subject of your standard Calculus on Manifolds course, and I am going to assume that my readers are at least vaguely familiar with it. The second is to work, not with functions, but with sections of vector bundles. That’s our subject in this post.
So, let’s think about a vector bundle $V$ on the line $\mathbb{R}$, and let $\sigma$ be a section of $V$. If we want to define $\sigma'$, we need to subtract $\sigma(x+h)$ and $\sigma(x)$, two vectors which live in different fibers. To think of it another way, we need to distinguish between $f(x)$, the point in the fiber over $x$, and $f(x)$, the constant function which assigns the same value at every point. Suppose that, for any $v \in V_x$, we had a local section $c_v$ of $V$ with $c_v(x)=v$; we think of $c_v$ as a constant function. Then we could define $\sigma'(x) = \lim_{h \to 0} \left( \sigma(x+h) - c_{\sigma(x)}(x+h) \right)/h$.
A local system gives us the constant functions $c_v$. (Indeed, in definitions A.2 and B.6, we took a local system to be the constant functions, along with the data of certain maps between them.) Today, we will take the fundamental object to be the operation of derivation, and see how to build everything else from it.
We start with a special case, and then build up to the whole. Let $V$ be a vector bundle on $\mathbb{R}$. We’ll write $x$ for the coordinate on $\mathbb{R}$. Of course, $V$ can be trivialized, but for the moment we don’t want to trivialize it. A connection on $V$ is a map $\nabla$ from sections of $V$ to sections of $V$, such that
(1) For any two sections $\sigma$ and $\tau$ of $V$, we have $\nabla(\sigma + \tau) = \nabla(\sigma) + \nabla(\tau)$.
(2) For any section $\sigma$ of $V$, and any scalar-valued function $f$ on $X$, we have $\nabla(f \sigma) = (\partial f/\partial x) \sigma + f \nabla(\sigma)$.
If we do trivialize $V$, then the operation of taking the derivative with respect to $x$ obeys these axioms. The point is that these are axioms which hold without talking about any choice of a trivialization. If you’re an algebraist, the following might help you: a derivation is a map from an algebra (to something); a connection is a map from a module.
A section $\sigma$ of $V$ is locally constant if $\nabla(\sigma)=0$. So this is how to go from connections to the more sheafy perspectives which are focused on locally constant sections.
I said early on that a local system is a vector bundle with isomorphisms between different fibers. How, in the setting of connections, do we build an isomorphism between one fiber of $V$ and another? Let $a$ and $b$ be two points of $\mathbb{R}$ and let $v_0$ be a point in the fiber $V_a$. Solve the differential equation $\nabla(\sigma)=0$, with initial condition $\sigma(x)=v_0$. The isomorphism between $V_a$ and $V_b$ will send $v_0$ to $\sigma(v_0)$. So, this is how to go from a connection to the path groupoid approach, when our as space is $\mathbb{R}$. Later, when we work on more complicated spaces, we’ll have to keep track of the path along which we solve the differential equation, but everything else will look the same.
Let’s see this differential equation in coordinates. Choosing an arbitrary trivialization of $V$ in order to write things down, we have $\nabla(u) = \partial u/\partial x + A(x) \cdot u$ where $A(x)$ is an $n \times n$ matrix, varying with $x$. (Exercise!) So we have to solve the differential equation $u' = - A u$. If $n=1$, this has the solution $u(t) = e^{-\int_a^{t} A(x) dx} \cdot v_0$; for larger $n$, one usually cannot give a closed form solution.
I have now presented all the main ideas, in the case where our space is the real line. We will now move to the case of an arbitrary manifold $X$. This will introduce two difficulties: we won’t have a natural coordinate system on $X$, and there is a genuinely new phenomenon that happens when $X$ has dimension larger than one — the possibility of curvature.
Let’s start by addressing the lack of coordinates; this will just be a matter of careful notation. Let $D$ be a vector field on $X$; we will think of this as a derivation on the scalar-valued functions on $X$. Then we want a way, $\nabla_D$, of differentiating with respect to $D$. This should obey
(1′) $\nabla_D(a\sigma + b\tau) = a\nabla_D(\sigma) + b\nabla_D(\tau)$, where $a$ and $b$ are real constants and
(2′) $\nabla_D(f \sigma) = D(f) \sigma + f \nabla_D(\sigma)$.
We also need a condition on how this depends on $D$:
(0) $\nabla_{fD+gE}(\sigma) = f \nabla_D(\sigma) + g \nabla_E(\sigma)$, where $f$ and $g$ are scalar valued functions.
You now have reached the definition of a connection. I always found it hard to remember the difference between (1′) and (0).
Why are the coefficients $a$ and $b$ just constants, while $f$ and $g$ get to be functions? Unfortunately, wordpress won’t let me do a hidden section inside a hidden section, so click on this asterisk to read the explanation*.
We won’t be interested in all connections, we will only be interested in the integrable ones. (Also known as flat, or zero-curvature.) This will be easiest to explain in coordinates. Let $x_1$, $x_2$, …, $x_d$ be coordinates on $X$ near $x$. Then we expect $\nabla_{\partial/\partial x_i} \nabla_{\partial/\partial x_j} \sigma$ to equal $\nabla_{\partial/\partial x_j} \nabla_{\partial/\partial x_i} \sigma$. After all, if we were doing honest differentiation of functions, we would have $\partial^2 f / \partial x_i \partial x_j = \partial^2 f / \partial x_j \partial x_i$.
A connection is said to be integrable if we have $\nabla_{\partial/\partial x_j} \nabla_{\partial/\partial x_i} \sigma$ for some (equivalently any) set of coordinates. An equivalent condition is that $\nabla_D \nabla_E \sigma - \nabla_E \nabla_D \sigma = \nabla_{[D,E]} \sigma$ for any vector fields $D$ and $E$.
It turns out that, if $\nabla$ is an integrable connection, then the differential equation $\nabla(\sigma)=0$ is uniquely solvable for any initial condition $\sigma(x) = v_0$. These solutions give a local trivialization of our vector bundle, just like in the one dimensional case we discussed above.
If $\nabla$ is not integrable, then we can still solve $\latex \nabla (\sigma)=0$ along any path. But making even topologically trivial changes to the path may change the result. In other words, you get holonomy, not just monodromy.
Next up: some other ways to think about integrability, such as $D$-modules, curvature and the deRham complex. Then, the relationship between connections and the infinitesimal perspective on local systems.
* Suppose we have a vector field $D$, and a section $\sigma$, and we want to compute the value of $\nabla_D(\sigma)$ at a point $x \in X$. Then it is enough to know $D$ at $x$. However it is not enough to know $\sigma$ at $x$; we have to know the first order variation of $\sigma$.
This is part of a more general distinction that everyone should learn at some point. Suppose that $V$ and $W$ are two vector bundles on $X$, and $\mathcal{V}$ and $\mathcal{W}$ are the sheaves of sections of $X$. Suppose we have a map $\phi:\mathcal{V} \to \mathcal{W}$. If $\phi(au+bv) = a \phi(u) + b \phi(v)$ for $a$ and $b$ constants, then $\phi$ is called a “map of sheaves” and the stalk of $\phi(v)$ will depend on the stalk of $v$. But if $\phi(fu+gv) = f \phi(u) + g \phi(v)$, then $\phi$ is called a “map of $\mathcal{O}_X$-modules”, and the fiber of $\phi(v)$ will depend only on the fiber of $v$. In other words, $\phi$ will be induced by a map of vector bundles $V \to W$. So $\nabla$ has the first kind of linearity in $\sigma$ and the second (more local) kind in $D$.
As another application of the above ideas, if $\nabla$ and $\nabla'$ are two connection on the same vector bundle $V$, then $\nabla - \nabla'$ is a map of $\mathcal{O}_X$-modules from $V$ to itself. Hey, I just solved on of your exercises for you! (New exercise: which one!)
## 7 thoughts on “Local Systems: The connection perspective”
1. john mangual says:
Oh cool, so you can construct a bundle which is $S^1 \times \mathbb{C}$ but for which the constant functions are $\sigma_z(\theta) = e^{i k \theta} z$ for some $k \in \mathbb{R}$. In fact, any connection of them form $A : S^1 \to SU(1)$ has to be homotopic to one of these, no? If we require there be no holonomy, then $k \in \mathbb{Z}$. Are these classified by some kind of K-theory?
2. John,
I think you meant to say that one can construct a trivial bundle on S^1 with fiber C, and can give a connection whose flat local sections have the exponential form (but the parameter k can be complex). SU(1) is a trivial group.
Since S^1 is one-dimensional, any connection on this bundle is flat, and the resulting monodromy representation is the same as one of the exponential form. The Riemann-Hilbert correspondence gives an equivalence between coherent sheaves with flat (aka integrable) connection and locally constant sheaves of vector spaces, so these connections are classified by representations of the fundamental group into automorphisms of the fiber. In the one-dimensional case, these are parametrized by homomorphisms from Z into $\mathbb{C}^\times$, i.e., elements of $\mathbb{C}^\times$.
3. john mangual says:
Yeah, I meant $U(1)$. Since the base space, $S^1$ is 1-dimensional, we can’t have two independent vector fields so any connection is flat. The monodromy representation of $\pi_1(S^1) = \mathbb{Z}$ is determined by the image of the generator and can be any member $\mathbb{C}$.
Actually, we can’t have 0 because $\int_0^{x_1} A(x) dx$ would have to be $-\infty$.
My point was that by choosing $A(x) = A$, the resulting constant sections are $u(x) = e^{ Ax } u_0$, we can have constant sections that wind around the origin in $\mathbb{C}$ by choosing $A$ to be purely imaginary. Then we can look at the set of connections whose monodromy representation is trivial. These are classified by the number of times they wind around. Could I even say there is a map from the space of connections to the fundamental group of the orthogonal group of the fiber?
I would love to see a post on the Riemann-Hilbert correspondence.
4. john mangual says:
“formula does not parse” is just the integral get when solving for the constant sections. it’s the exponential of something finite so it can’t be zero.
I fixed it: you were missing curly braces on the exponent –NS
5. Justin says:
Did the “next up: some other ways to think about integrability, such as -modules, curvature and the deRham complex” ever happen?
I’d like to know.
6. David Speyer says:
Nope, this is how far I got before life intervened. Sorry!
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# Why isn't the derivative of $e^x$ equal to $xe^{(x-1)}$?
When we take a derivative of a function where the power rule applies, e.g. $x^3$, we multiply the function by the exponent and subtract the current exponent by one, receiving $3x^2$. Using this method, why is it that the derivative for $e^x$ equal to itself as opposed to $xe^{x-1}$? I understand how the actual derivative is derived (through natural logs), but why can’t we use the original method to differentiate? Furthermore, why does the power rule even work? Thank you all in advance for all of your help.
#### Solutions Collecting From Web of "Why isn't the derivative of $e^x$ equal to $xe^{(x-1)}$?"
You are confusing things. If I define $f : \mathbb{R} \to \mathbb{R}$ by $f(x)=x^n$ this is very different from defining $g: \mathbb{R} \to \mathbb{R}$ by $g(x)= a^x$, note that in the first one the exponent is not varying and on the other function the variable appears on the exponent.
For the first function the derivative is just $f'(x) = nx^{n-1}$, for the second one things are different. First it turns out that first you need to define what it means to raise something to a real number (notice that the usual definition doesn’t work, what would mean multiplying a number by itself $\pi$ times?), in that case for reasons that I won’t explain here we define this function as:
$$a^x = e^{x\ln a}$$
In that case, if we know how to differentiate $e^x$ (and usually when we construct this, we already know), we’ll have the following:
$$(\ln \circ g)(x)=x \ln a$$
Now the chain rule gives:
$$\ln'(g(x))g'(x)=\ln a$$
However $\ln'(x) = 1/x$ because of the construction of $\ln$ and $g(x)=a^x$ so tha we have:
$$\frac{1}{a^x}g'(x)=\ln a \Longrightarrow g'(x) = a^x \ln a$$
Notice that there was a crucial appeal to the definition $a^x = e^{x \ln a}$. To know why we define things this way look at Spivak’s Calculus, there’s an entire chapter devoted to all the constructions about logs and exponentials.
Here’s an analogy that may help. The derivative of $\frac{x}{3}$ is $\frac{1}{3}$, while the derivative of $\frac{3}{x}$ is $-\frac{3}{x^2}$. The quotient rule gives different results when you swap the numerator and denominator; the same thing happens for powers.
$$e^x=1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!} \dots$$
Now I give you freedom to differentiate the RHS w.r.t $x$. You will still get the same RHS.
The different is that in $x^n$ (or more generally, $x^\alpha$) it’s the base which varies and the exponent stays fixed, while in $e^x$ (or more generally $\alpha^x$) it’s the exponent which varies while the base stays fixed.
Let’s look at a generalization of this. Assume that both can vary, i.e. we look at $$f(x) = b(x)^{p(x)} \text{.}$$
for some functions $b$ (for base) and $p$ (for power, since e for exponent is taken). Let’s also assume that $b(x) > 0$, to avoid having to deal with negative basis. By writing this as $$f(x) = h(b(x),p(x)) \quad\text{where}\quad h(u,v) = u^v = e^{v\log u}$$
the structure of this becomes more explicit. Now, if we do know that $\frac{de^x}{dx} = e^x$, by extensive use of the chain rule we can find \begin{aligned} h_u(u,v) &= \frac{\partial h}{\partial u} = \frac{v}{u}e^{v\log u}\\ h_v(u,v) &= \frac{\partial h}{\partial v} = e^{v\log u}\log u\\ f'(x) &= h_u(b(x),p(x))b'(x) + h_v(b(x),p(x))p'(x) \\ &= b(x)^{p(x)} \left(b'(x)\frac{p(x)}{b(x)} + p'(x)\log b(x) \right) \\ &= b(x)^{p(x)-1} \left(b'(x)p(x) + p'(x)b(x)\log b(x) \right) \end{aligned}
Now let’s apply this to $x^n$, i.e. we set $b(x)=x$ and $p(x)=n$, and get $$f'(x) = x^{n-1}\left(n + 0\cdot x\log x\right) = nx^{n-1} \text{.}$$
So we got the rule for powers of $x$ from the rule that $\left(e^x\right)’ = e^x$, plus the chain rule. We can also set $b(x)=e$ and $p(x)=x$ and get back that $$f'(x) = e^{x-1}\left(0\cdot x + 1\cdot e\log e\right) = e^{x-1}\left(e\right) = e^x \text{,}$$
but of course we expected that since we used that rule to derive our formula.
Recall the definition of the derivative:
$$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}.$$
Now let’s first apply this to $f(x)=x^a$, with $a$ real:
$$(x^a)’ = \lim_{h\rightarrow 0}\frac{(x+h)^a – x^a}{h} = \lim_{h\rightarrow 0}x^a\frac{(1+h/x)^a – 1}{h}.$$
Using the Taylor expansion
$$(1+y)^a = 1 + ay + a(a-1)\frac{y^2}{2} + \ldots\qquad \text{for y<1},$$
we get
$$(x^a)’ = \lim_{h\rightarrow 0}x^a\frac{1 + ah/x + \mathcal{O}(h^2/x^2) – 1}{h} = x^{a-1}\left[a + \lim_{h\rightarrow 0}\mathcal{O}(h/x)\right] = ax^{a-1}.$$
Now, let’s do the same for $f(x)=e^x$:
$$(e^x)’ = \lim_{h\rightarrow 0}\frac{e^{x+h}-e^x}{h} = \lim_{h\rightarrow 0}e^x\frac{e^{h}-1}{h}.$$
Since
$$e^h = 1 + h + \frac{h^2}{2} + \ldots,$$
we get
$$(e^x)’ = \lim_{h\rightarrow 0}e^x\frac{1 + h + \mathcal{O}(h^2) -1}{h} = e^x\left[1 + \lim_{h\rightarrow 0}\mathcal{O}(h)\right] = e^x.$$
In applying the power rule, chain rule and other rules for derivatives, it critically matters how the independent variable $x$, with respect to which the derivative is being taken, appears in the formula.
The power rule does not simply assert that we look at the syntax of the formula, find powers, and then bring them down as coefficients and decrement them by one!
The power rule says that we do this for terms where the derivation variable $x$ is being raised to something. If we are differentiating with respect to $x$, and there is a term of the form $x^a$, then the power rule applies. It does not apply to some $a^b$ where $x$ does not even occur (basically a constant term with respect to $x$) nor to some $a^x$ where $x$ occurs, but not in the correct position for the rule to apply.
The power rule is that $x^a$ differentiates into $ax^{a-1}$.
More generally, in combination with the chain rule, $f(x)^a$ goes to $af(x)^{a-1}f'(x)$. In the special case of $f(x) = x$, $f'(x)$ is just 1.
One way to approach an understanding of why the power rule works is to go back to first principles of differentiation and work with the following limit:
$$\lim_{h \to 0}\frac{(x+h)^a – x^a}{h}$$
You can see where the above is headed right away, because the $(x + h)^a$ term, when expanded, will produce an $x^a$ term, which will cancel with the $-x^a$ term, and so the overall polynomial reduces in degree by one.
After the $x^a$ term $(x + h)^a$ produces a ax^{a-1}h term. The other terms don’t matter because they produce higher powers of $h$, thus:
$$\lim_{h \to 0}\frac{(x^a + ax^{a-1}h + Khx^{a-2}h^2 + \ldots + h^a) – x^a}{h}$$
We don’t care what K is; it’s some constant that depends on $a$ (which row of Pascal’s triangle we are in to pick the coefficients for the expansion of the binomial). That term and all the others denoted by $\ldots$ will disappear. First, the $x^a – x^a$ cancels:
$$\lim_{h \to 0}\frac{ax^{a-1}h + Khx^{a-2}h^2 + \ldots + h^a}{h}$$
Then we do the division to distribute the denominator $h$ into the numerator:
$$\lim_{h \to 0}ax^{a-1} + Khx^{a-2}h + \ldots + h^{a-1}$$
Now, since this is a limit as $h$ approaches zero, all the terms that have $h$, or a power of $h$ disappear, leaving us with the limit just being $ax^{a-1}$, which is the power rule:
$$\lim_{h \to 0}ax^{a-1} + Khx^{a-2}h + \ldots + h^{a-1} = ax^{a-1}$$
So we did not even have to compute the full expansion of $(x + h)^a$, because we know that the third and subsquent terms all contain $h$ and therefore disappear in the limit.
Another piece of intuition about derivatives is to look at difference series. For instance take the succession of squares: 1, 4, 9, 16, 25, 36, 49, 64 and compute the deltas between them. They are: 3, 5, 7, 9, 11, 13, 15. Hey look, $n^2$ became $2n + 1$. The power is reduced from quadratic to linear, and the coefficient doubled.
The rule $\frac d{dx}x^k=kx^{k-1}$ for $k\geq1$ integer is a straightforward application of the basic case $\frac d{dx}x=1$ and the rule
$\frac d{dx}(fg)=f\frac d{dx}g+g\frac d{dx}f$ reiterated $k-1$ times.
The problem with $e^x$ is that it doesn’t admit an expression as product of simpler functions.
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# How do you solve x^2 + 8 = 28?
Jul 20, 2015
color(green)(x=+-2sqrt5
#### Explanation:
${x}^{2} + 8 = 28$
${x}^{2} = 28 - 8$
${x}^{2} = 20$
$x = \pm \sqrt{20}$
color(green)(x=+-2sqrt5
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+0
# rectangle
0
39
1
If the base is doubled, then which of the following statements about its area will be true?
Mar 7, 2021
#1
+31208
+1
Original area 2 x 5 = 10 cm2
Now double the base from 5 to 10
New area = 2 x 10 = 20 cm2 = 2 times the original area
Mar 7, 2021
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# Advanced Higher Physics Unit 1 - PowerPoint PPT Presentation
Advanced Higher Physics Unit 1. Rotational Dynamics. Using moments. The spanner exert a moment or turning effect on the nut. Turning point. distance from force to turning point. force. If the moment is big enough, it will unscrew the nut.
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### Advanced Higher Physics Unit 1
Rotational Dynamics
The spanner exert a moment or turning effect on the nut.
Turning point
distance from force to turning point
force
If the moment is big enough, it will unscrew the nut.
If not, there are two ways of increasing the moment.
• Increase the distance from the Force to the pivot-apply a force
• at the end of the spanner or use a longer spanner.
Turning point
distance from force to turning point
force
If the same force is applied over a greater distance, a larger moment
is produced.
2. Increasing the Force applied-push/pull harder or get someone
stronger to do it!
Turning point
distance from force to turning point
force
If a greater force is applied over the same distance,
a larger moment is produced.
The moment of a force is given by:
F is the force applied, measured in Newton (N)
d is the distance from turning point, measured in metres (m)
The moment of a force is therefore measured in
Newton metres (Nm).
A force is applied to the rim of a disc which can rotate around its centre axis. In this case the moment of a force is called the Torque.
F
r
In data booklet
F is the force applied, measured in Newton (N)
r is the radius of the circle, measured in metres (m)
T is the Torque associated with force F,
measured in Newton metres (Nm).
Inertia can be defined as resistance to change in motion.
In linear motion, MASS is a measure of an object’s inertia
(since a large mass needs a large force to produce an acceleration).
In angular motion, we use MOMENT OF INERTIA.
• The moment of inertia of an object is its resistance to change
• in angular motion.
• The moment of inertia depends on:
• The mass of an object
• How the mass is distributed about the axis of rotation.
Consider a mass m at a distance r from the axis of rotation. The moment of inertia can be calculated using:
r
In data booklet
m
m is the mass of the object, measured in kg
r is the distance from the axis of rotation, measured in m
I is the moment of inertia measured in kgm²
All the mass can be considered to be at the same distance from the axis.
Masses on a very light rod
r
Wheel with heavy rim and very light spokes
r
In these cases I=mr² where m is the total mass.
With:
ltotal length of the rod
m total mass of the rod
The moment of inertia for a rodrotating about end is 4 times bigger
than the moment of inertia for a rod rotating about centre as it is
harder to do so.
This is because there are now more particles at a greater distance
from the axis of rotation.
In data booklet
r
Where m is the total mass of the disc
r is the radius of the circle
Where m is the total mass of the sphere
r is the radius of the sphere
(all the moment of inertia formulas can be found
in the data booklet in Additional Relationships)
Newton 2nd Law
An unbalanced Torque will produce an angular acceleration.
In data booklet
With I, the moment of inertia in kgm²
α, the angular acceleration in radsˉ²
T, the Torque in Nm
The angular momentum is defined as the moment of the linear momentum.
r
m
For this particle of mass m:
The linear momentum p = mv
v
w
The angular momentum = the moment of p = mvr = mr²w, since v=rw.
In data booklet
With L angular momentum measured in kgm²sˉ¹.
A rigid body is an object in which all the individual parts have the
same angular velocity w.
The angular momentum of this body is the total of the angular
momenta of its particles:
w is constant as all particles must be rotating at the same rate.
In data booklet
In the absence of external Torque, the total angular momentum
before impact equal the total angular momentum after impact.
Not in data booklet
Before:
After:
Total angular momentum before = total angular momentum after
Iwheelw0 = w (Iwheel + Imud)
The moment of inertia of the wheel and the mud after impact is larger
than the moment of inertia of the wheel before impact.
Therefore the angular velocity of the wheel is smaller after impact.
Example: pupil spinning on a chair
After:
Before:
Pupil draws arms in
Pupil pushes arms out
Total angular momentum before = total angular momentum after
Ioutw1 = Iinw2
Iin is smaller than Iout because the particles are closer to the axis
of rotation.
Therefore w2 is larger than w1.
Example: mass dropped on a turntable
Before:
After:
Axis of rotation
Total angular momentum before = total angular momentum after
Idiscw1 = w2(Idisc+Imass)
The moment of inertia of the disc and the mass after impact is larger
than the moment of inertia of the disc before impact.
Therefore the angular velocity of the disc is smaller after impact.
In data booklet
With I moment of inertia, measured in kgm²
w angular velocity, measured in radsˉ¹
Erot rotational kinetic energy, measured in J
Not in data booklet
With T, Torque measured in Nm
θ, angular displacement in rad
Ew, work done in J
In the absence of frictional torque:
Not in data booklet
Ep = mgh
Erot = ½Iw²
h
w
Ek = ½mv²
v
Potential energy at top = total of linear and angular kinetic energy
mgh = ½mv² + ½Iw²
The equations of angular motion are similar to those of linear motion.
linear motion
angular motion
10 kg
6m
• Find:
• Total I
• Unbalanced T
• Driving T
• Driving F
• Final w
• Deceleration when driving T removed
• Time taken to move to rest.
Frictional torque = 1000 Nm
time = 5s
r
• Find:
• Driving Torque
• Moment of inertia of solid disc
• Angular acceleration
• Angular displacement
• Angular velocity
F
Solid disc mass = 20 kg
Axle radius = 1cm
Force applied to axle = 4000 N
Frictional Torque = 30 Nm
Cord length = 50 cm
• A turntable of mass 5kg and radius 25 cm is rotating at 10 radsˉ¹.
• A metal ring of mass 2 kg and radius 10 cm is dropped over the centre of the turntable.
• Find the new angular velocity of the system.
• Using rotational energy determine whether this is an elastic or an inelastic situation.
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# Bari's question at Yahoo! Answers regarding minimizing the surface area of a silo
#### MarkFL
Staff member
Here is the question:
Can you solve this Calculus problem?
I need help.
Solve using Related Rates or Optimization.
Bob is building a silo to store 1000 ft^3 of hay. Find the minimum amount of material and the dimensions of the silo to achieve it. Since his silo is a cylinder topped with a hemisphere, the formulas are as follows:
V=pi x r^2 x h + (2/3)pi x r^3
SA= 2 x pi x r x h + 2 x pi x r ^2
I have posted a link there to this thread so the OP can view my work.
#### MarkFL
Staff member
Hello Bari,
The volume $V$ of the silo is given by:
$$\displaystyle V=\pi hr^2+\frac{2}{3}\pi r^3$$
The surface area $S$ is given by:
$$\displaystyle S=2\pi hr+2\pi r^2$$
Solving the first equation for $h$, we obtain:
$$\displaystyle h=\frac{3V-2\pi r^3}{3\pi r^2}$$
Substituting for $h$ into the second equation, we obtain the surface area as a function of one variable, $r$:
$$\displaystyle S(r)=2Vr^{-1}+\frac{2\pi}{3}r^2$$
Differentiating with respect to $r$ and equating the result to zero, we find:
$$\displaystyle S'(r)=-2Vr^{-2}+\frac{4\pi}{3}r=0$$
Solving for $r$, we obtain the critical value:
$$\displaystyle r=\left(\frac{3V}{2\pi} \right)^{\frac{1}{3}}$$
Using the second derivative test to determine the nature of the extremum at this critical value, we find:
$$\displaystyle S''(r)=4Vr^{-3}+\frac{4\pi}{3}>0\,\forall\,0<r$$
Hence, the extremum is a minimum. And so the dimensions if the silo which minimizes the surface area are:
$$\displaystyle r=\left(\frac{3V}{2\pi} \right)^{\frac{1}{3}}$$
$$\displaystyle h=\frac{3V-2\pi\left(\frac{3V}{2\pi} \right)}{3\pi\left(\frac{3V}{2\pi} \right)^{\frac{2}{3}}}=0$$
Another method we could use is optimization with constraint, via Lagrange multipliers.
We have the objective function:
$$\displaystyle S(h,r)=2\pi hr+2\pi r^2$$
Subject to the constraint:
$$\displaystyle g(h,r)=\pi hr^2+\frac{2}{3}\pi r^3-V=0$$
which yields the system:
$$\displaystyle 2\pi r=\lambda\left(\pi r^2 \right)\implies \lambda=\frac{2}{r}$$
$$\displaystyle 2\pi h+4\pi r=\lambda\left(2\pi hr+2\pi r^2 \right)\implies \lambda=\frac{h+2r}{r(h+r)}$$
This implies:
$$\displaystyle \frac{2}{r}=\frac{h+2r}{r(h+r)}$$
$$\displaystyle 2hr+2r^2=hr+2r^2$$
$$\displaystyle hr=0$$
We know $r\ne0$, hence we must have $h=0$, and so the constraint becomes:
$$\displaystyle \frac{2}{3}\pi r^3-V=0\implies r=\left(\frac{3V}{2\pi} \right)^{\frac{1}{3}}$$
Note: The fact the $h=0$ follows from the fact that a spherical shape encloses more volume per surface area than a cylinder, and so the silo winds up simply being a dome.
Now, using the given data:
$$\displaystyle V=1000\text{ ft}^3$$
the dimensions we seek are:
$$\displaystyle r=\left(\frac{3\cdot1000\text{ ft}^3}{2\pi} \right)^{\frac{1}{3}}=\sqrt[3]{\frac{1500}{\pi}}\text{ ft}\approx7.816\text{ ft}$$
$$\displaystyle h=0\text{ ft}$$
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# Finding closed or numeric value for $\int_0^{\infty } \frac{x \csc (a x)}{x^2+b^2} \, dx$
According to Gradshteyn 3.747-3 $$\int_0^{\infty } \frac{x \csc (a x)}{x^2+b^2} \, dx=\frac{\pi}{2\sinh(ab)}$$ $$b>0$$
so I'm trying to get some numeric results from Mathematica 12.
a = 6; b = 12;
NIntegrate[(x/((x^2 + b^2) Sin[a x])), {x, 0, Infinity}, MaxRecursion -> 12]
However this does not match the values given by
Pi/(2*Sinh[a*b])
So it is not clear where I'm going wrong.
• How did you evaluate the NIntegrate for undefined parameters a,b ? Please show some code! Aug 3, 2020 at 15:44
• @Ulrich Neumann I just did a = 6; b = 12; but I also just put them in manually too to be sure. Aug 3, 2020 at 15:46
• There are singularities in the integral when a*x = Pi *k for some integer k, and when the intergral goes through the point x = b Aug 3, 2020 at 15:46
• @flinty is there any code to deal with these? Aug 3, 2020 at 15:49
• ^ 3-747.3 seems to be discussed here on page 9. researchgate.net/publication/… Aug 3, 2020 at 17:04
I want to focus on the numerics here. There is a way to do this integral taking infinite number of singularities into account.
One can split the integral into the domains containing only one singularity at $$a x_n=\pi n$$, i.e. $$x\in[x_n-\frac{\pi}{2a},x_n+\frac{\pi}{2a}]$$ and an integral in the interval $$x\in[0,\frac{\pi}{2a}]$$. On the last step we use NSum with some options. The options are very important. Consider first my failed attempt:
Clear[f]
f[n_?NumericQ,a_,b_]:=NIntegrate[(x/((x^2+b^2) Sin[a x])),{x,(π n)/a-π/(2a),(π n)/a,(π n)/a+π/(2a)},WorkingPrecision->100,Method->PrincipalValue]
f0[a_,b_]:=NIntegrate[(x/((x^2+b^2) Sin[a x])),{x,0,π/(2a)},WorkingPrecision->100]
f0[6,12]+NSum[f[n,6,12],{n,1,∞},Method>"AlternatingSigns",WorkingPrecision->100]
Out[1]= 6.39989549924364176901258523623081516506764870738550937643852237103123602088582134002479513849115*10^-6
The numerical value seems to be highly accurate and well converged, but is different from the analytic one
e[a_,b_]:=N[Pi/(2*Sinh[a*b]),100]
e[6,12]
Out[2]= 1.690235331526788818439805170791473807196429480676031631266783609275102725281157127348346908376558298*10^-31
The discrepancy can be corrected by changing the default NSumTerms value. Let us introduce the relative error as a function of this parameter
relErr[a_,b_,k_]:=Abs[f0[a,b]+NSum[f[n,a,b],{n,1,∞},
Method->"AlternatingSigns",
WorkingPrecision->100,
NSumTerms->k]
-e[a,b]]/Abs[e[a,b]]//N
and try a few values:
relErr[6,12,21]
Out[3]= 3.78639*10^25
relErr[6,12,22]
Out[4]= 2.38766*10^-68
relErr[6,12,25]
Out[5]= 2.3903*10^-68
Thus, it is important to explicitly consider the first 22 terms!
• It should be noticed that f0[6, 12] + NSum[f[n, 6, 12], {n, 1, \[Infinity]}, WorkingPrecision -> 100] performs 0.*10^-7. Aug 3, 2020 at 20:42
• You don't take into account the singularity at the origin in your f0[a_,b_]. Aug 3, 2020 at 20:45
• @user64494 Thank you for checking other options. 1. According to the documentation one needs to "use a method for alternating series to get a very precise sum approximation". 2. This is a removable singularity (the limit $x\rightarrow 0$) is finite. Aug 3, 2020 at 20:51
• -1. In you answer you use e[a_,b_]:=N[Pi/(2*Sinh[a*b]),100] . Imagine you don't know it. Such sort answer does not make a good impression. Aug 4, 2020 at 4:04
• I'd like to add that the NSumTerms->k option means you consider a truncated sum and that sum is a sum for the principal value of the integral over $(0,\frac {\pi k} a)$. This is not the required result. Aug 4, 2020 at 4:30
Using the substitution u==a x the identity is transformed to Integrate[u/((u^2 + ab^2) Sin[u]), {u, 0, Infinity}] == Pi/(2 Sinh[ab])
The integral is singular at u==k Pi, k=1,2,... and the integration range is splitted into subintervals (similar to @yarchik answer) containing only one singularity (thanks to @flinty and @ ChipHurst comments):
int[ab_?NumericQ, n_?NumericQ] :=NIntegrate[u/((u^2 + ab^2) Sin[u]) , {u, 0, Pi/2}] +
Sum[ NIntegrate[u/((u^2 + ab^2) Sin[u]) , {u, k Pi - Pi/2, k Pi + Pi/2},Method -> PrincipalValue, Exclusions -> {u == k Pi}], {k, 1, n} ]
This finite sum matchs quite well:
Plot[{int[ab, 10], Pi/(2 Sinh[ab])}, {ab, 0, 1},PlotStyle -> {{Thickness[0.01],Lighter[Blue]}, Red},
PlotLabel ->"{int[ab,10],\!$$\*FractionBox[\(Pi$$, $$2\\\ Sinh[ab]$$]\)}",AxesLabel -> {"ab ", None}]
• This is a modification of the @yarchik's approach. Aug 4, 2020 at 7:48
• +1. int[ab, Infinity] works well to me. I think you give a correct numerical answer. Aug 4, 2020 at 8:20
• @user64494 Thanks, the "modification" shows a straightforward approach. Ticky idea int[ab, Infinity] Aug 4, 2020 at 8:22
• I changed my mind. In fact, it does not work: int[72, Infinity] returns the input. I was missed by the plot of Pi/(2 Sinh[ab]. Aug 4, 2020 at 8:25
• @yarchik Your solution is great! I tried to avoid such high values WorkingPrecision->100 in my answer. You can't see convergence in my comment, but a trend of decaying values. Aug 5, 2020 at 11:38
Following the @flinty's comment, we obtain
Residue[x/(x^2 + b^2)/Sin[a*x], {x, π/a*n}, Assumptions -> n ∈ Integers]
(*((-1)^-n n π)/(a^2 b^2 + n^2 π^2)*)
Sum[%, {n, -∞, ∞}]
(*0*)
and
2*π*I*Residue[x/(x^2 + b^2)/Sin[a*x], {x, I*b}]
(*π Csch[a b]*)
Now, making use of the Jordan's lemma, we conclude that $$PV\int_{-\infty}^\infty\frac x {(x^2+b^2)\sin(ax)}\,dx= \pi \text{csch}(a b).$$ It remains to apply the parity of the integrand.
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Proving properties of electric fields using Gauss’s Theorem | Quantum Science Philippines
## Proving properties of electric fields using Gauss’s Theorem
Use Gauss’s theorem and $\oint \vec{E} \cdot d\vec{l} = 0$ to prove the following:
(a) Any excess charge placed on a conductor must lie entirely on its surface. (A conductor by definition contains charges capable of moving freely under the action of applied electric fields.)
Solution:
Suppose that the field were initially nonzero. Since this is a conductor, any charges in the interior would move in response to the field. After a time, this process stops since the moving charges produce currents which dissipate energy. The final configuration would then have charges arranged so that the interior is zero. Recall that in equilibrium, the electric field inside a conductor is zero. Since $\vec{E} = 0$ everywhere inside the conductor, then from Gauss’s Law, $\oint \vec{E} \cdot \hat{n} da =4\pi \int_V \rho(\vec{x}) d^3 x = o$, the charge density $\rho (\vec{x}) = o$ everywhere in the interior. Therefore, every point inside a conductor has zero charge, and any excess charge can only reside on the surface of the conductor.
(b) A closed, hollow conductor shields its interior from fields due to charges outside, but does not shield its exterior from the fields due to charges placed inside it.
Solution:
Part 1. Consider the charge exterior to the conductor which produces an electric field, as shown in the figure.
The electric field in the conductor is zero, with induced charge densities on the exterior and interior surfaces of the conductor.
i) Imagine moving a charge on the interior surface from point A to point B along path 2 which goes throughout the conductor itself. Since $\vec{E} =0$ in the conductor, $\int_2 \vec{E} \cdot d\vec{l} = 0$ along this path.
ii) Move the same charge from A to B along path 1, in the interior cavity of the conductor. Since the electrostatic field is conservative, $\vec{E} \cdot d\vec{l} = 0$ along its path.
This must be true also for any path in the interior. So generally, $\vec{E} =0$ in the interior. Therefore the conductor shields its interior from field due to charge placed outside.
Part 2. Consider a positive charge $Q$ placed inside a hollow conductor as shown in the figure.
The charge induces a charge density in the interior surface of the conductor in such a way that the electric field in the interior of the conductor is zero. Assuming that the conductor is charge neutral, this means that there is an induced charge density on the exterior surface of total charge Q. If we apply Gauss’s Law to the Gaussian surface G surrounding the conductor, the total charge enclosed is still $Q$. Therefore, there is an electric field outside the conductor.
(c) The electric field at the surface of a conductor is normal to the surface and has a magnitude $\sigma / \epsilon_0$, where $\sigma$ is the charge density per unit area on the surface.
Solution:
Note that in equilibrium, the field at exterior surface must be normal to the surface, so that the tangential component is zero. The magnitude of the field is derived using Gauss’s Law with a Gaussian pillbox which cuts through the surface. The electric field is zero on the conducting side of the pillbox. So,
$\oint \vec{E} \cdot \hat{n} da =EA$, with $A$ the area on the surface.
$E A = (\frac{q}{\epsilon_0} )$
Rearranging, we get
$E =(\frac{q}{A} ) (\frac{1}{\epsilon_0} )$.
Define $\sigma$ as the charge per unit area $q/A$,
$E = (\frac{\sigma}{\epsilon_0} )$.
References:
Classical Electrodynamics, John David Jackson, 3rd Edition, Chapter 1.
Introduction to Classical Electrodynamics, David Griffiths, Chapter 2.
University Physics, Young and Freedman, 11th Edition, Chapter 24.
Gauss’s Law, wikipedia.com.
—————————–
Adelle is currently pursuing her MS Physics degree at the Mindanao State University- Iligan Institute of Technology in Iligan City.
### 4 Responses to “Proving properties of electric fields using Gauss’s Theorem”
1. philippine information Says:
philippine information…
[…]Proving properties of electric fields using Gauss’s Theorem | Quantum Science Philippines[…]…
2. Bob Says:
a combination of mathematics and physics
I need a remedial school
3. Paul Abbott Says:
Your figure for part 2 is incorrect. The positive charge Q on the outside surface is uniformly distributed, irrespective of the position of the charge Q inside the hollow conductor.
4. David Says:
Your “proof” for part I of (b) holds verbatim even in the case that there is charge *interior* to the conductor, and therefore can not be correct.
I am guessing that the mistake is due to singularities in crossing over the boundary of the interior surface of the conductor
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# Table of 270
## IntroductionThe signs cross (×), asterisk (*) and dot (.) represent multiplication. We are most likely to use the cross when writing in our notebooks. In computer languages and algebra, the asterisk and dot are used. The multiplication table of 270 is included in the Table of 270, and it depicts the addition of the same number when multiplied by another number.The result of multiplying any number by one is the number itself. Under multiplication, one is known as the identity element.When any number is multiplied by zero, the result is always zero.In the case of the multiplication of integers, we need to see the sign of the integers. Multiplication of two positive integers:This Story also ContainsIntroductionTable Of ContentsWhat Is 270 TableMultiplication FormulaMethod Of SolvingSolved QuestionsTableTwo positive integers upon multiplication give a positive productMultiply two negative integers:Two negative integers upon multiplication give a positive product
• What is 270 table
• Multiplication formula
• Method of solving
• Solved questions
• Table
• Frequently asked questions
## What Is 270 Table
When the 3 digit number 270 is multiplied continuously with consecutive numbers in their increasing order. This results in a comprehensive table of 270.
## Multiplication Formula
[Multiplicand × Multiplier = Product] is the multiplication formula, where:
The first number is the multiplicand (factor).
The multiplier is the second number (factor).
Product: The result of multiplying the multiplicand by the multiplier.
The symbol for multiplication: "×” (which connects the entire expression)
## Method Of Solving
One-digit numbers can be multiplied easily using multiplication tables, but for larger numbers, we divide the numbers into columns based on their respective place values, such as ones, tens, hundreds, thousands, and so on. Multiplication problems are classified into two types:
1. Multiplication without regrouping
2. Multiplication with regrouping
Multiplication without regrouping
Multiplying two numbers without regrouping involves smaller numbers that do not require a carry-over to the next higher place value. It is the basic level that can help a learner understand the fundamentals of multiplication before progressing to higher-level problems such as regrouping.
Multiplication with regrouping
Multiplying more than two numbers with regrouping results in a 2-digit product. In this type of multiplication, we must carry the result to the next higher place value.
## Solved Questions
Q.1: If a person purchases two oranges per day, how many oranges will the person purchase in 270 days?
Ans: Given,
the Number of oranges purchased in a day = 2.
A person will purchase it in 270 days.
=
270 \times 2=540
Q2.What is the value of 270 times 11?
Ans: From the table of 270, the value of 270 times 11 is
270 \times 11=2970
Q3.Determine how many times we should multiply 270 to get 2700.
Ans: Using the table of 270, we know that
270 \times 10=2700
As a result, we must multiply 270 by ten to get 2700.
Q4.What is the value of 270 times 15 plus 15 minus 12?
Ans: From the table of 270, 270 times 15 plus 15 minus 12
\begin{aligned}
& 270 \times(15+15-12) \\
& =270 \times 18
\end{aligned}
## Table
Using multiplication
270 \times 1=270 270 \times 2=540 270 \times 3=810 270 \times 4=1080 270 \times 5=1350 270 \times 6=1620 270 \times 7=1890 270 \times 8=2160 270 \times 9=2430 270 \times 10=2700
Using repeated addition 270 270 + 270 = 540 270 + 270 + 270 = 810 270 + 270 + 270 + 270 = 1080 270 + 270 + 270 + 270 + 270 = 1350 270 + 270 + 270 + 270 + 270 + 270 = 1620 270 + 270 + 270 + 270 + 270 + 270 + 270 = 1890 270 + 270 + 270 + 270 + 270 + 270 + 270 + 270 = 2160 270 + 270 + 270 + 270 + 270 + 270 + 270 + 270 + 270 = 2430 270 + 270 + 270 + 270 + 270 + 270 + 270 + 270 + 270 + 270 = 2700
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# volts, amps watts formula
Thus, to find amps substitute … Formula: DC amps to watts Watt = Ampere x Volt AC single phase amps to watts Watt = Ampere x Volt x Power factor AC three phase amps to watts (With line to line voltage) Watt = √3 x Ampere x Line to line RMS voltage x Power factor AC three phase amps to watts (With line to neutral voltage) Watt = 3 x Ampere x Line to neutral RMS voltage x Power factor The formula for Volts is Watts divided by Amps. If you need a volts to watts calculator, our electrical conversion calculator has you covered! To use the chart, cover the W in the chart with a finger and use the remaining visible chart calculation of V multiplied by A. The formula is (A)* (V) = (W). T he formula for Watts is Volts times Amps. You know that P = 100 W, and V = 6 V. So, you can rearrange the equation to solve for I and substitute in the numbers. What are Watts? This comes from the equation P = I * V. Where P is the power in Watts, I is the current in Amps and V is the voltage in Volts. Power(W) = 12 V × 2 A. What's the difference between a volt, amp, and watt? That's it! flash_on Energy conversion straighten Length conversion functions Number conversion flash_on Power conversion ac_unit Temperature conversion fitness_center Weight conversion flash_on Voltage conversion settings_input_antenna Frequency conversion color_lens Color conversion image Image conversion flash_on Electrical calculation For example, if you have a current of 2 A and a voltage of 5 V, the power is 2A * 5V = 10W. The real power P in watts (W) is equal to the apparent power S in volt-amps (VA), times the power factor PF: P (W) = S (VA) × PF. The SI unit of voltage is a volt, the unit of amperage is an ampere (usually shortened to amp), and the unit of power is a watt. Determine the watts in a power source. It can sometimes be difficult to locate voltage, amperage, and wattage ratings on a user manual or spec sheet. The convention of using watts, amps, and volts. A = W / (3 × PF × V) How to convert amps to watts So amps are equal to watts divided by 3 times power factor times volts. The number of watts is equal to amps multiplied by volts. Advertisement. To use the chart, cover the W in the chart with a finger and use the remaining visible chart calculation of V multiplied by A. You will need to know the amps and the volts in the power source. The needed factor will be calculated for you when you click on the Calculate button for that table. For example, let’s convert 12 volts to watts, for a DC circuit with 2 amps of current. VA to watts calculation formula. Voltage (V) Current (I) Resistance (R) Power (P) Each of these quantities are measured using different units: Voltage is measured in volts (V) Current is measured in amps (A) Resistance is measured in ohms (Ω) Power is measured in watts (W) Electrical power, or the wattage of an electrical system, is always equal to the voltage multiplied by the current. Watts measure the power that is actually generated within the electrical system. The voltage V in volts (V) is equal to the power P in watts (W) divided by the current I in amps (A): The voltage V in volts (V) is equal to the square root of the power P in watts … Power is measured in watts and voltage is measured in volts. To determine the wattage, use a simple multiplication formula. Why is your power bill in kilowatt-hours and your battery bank in milliamp-hours? So watts are equal to volt-amps times the power factor. The formula for Watts is Volts times Amps. Simply fill in two of the blank boxes below and click “Calculate” to convert amps to volts or watts. watts = volt-amps × PF. Using the equation I = P/V, we can calculate how much current in amps would be required to get 100 watts out of this 6-volt bulb. or. In the four tables below, you may enter two of the four factors in Ohm's Law. What is the real power in watts when the apparent power is 3000 VA and the power factor is 0.8? amps = watts / (3 × PF × volts) or. This comes from the equation P = I * V. Where P is the power in Watts, I is the current in Amps and V is the voltage in Volts. Using our sample panel data, 12 Volts multiplied by 5 Amps equals 60 Watts. The ampere (or amps) is the amount of electricity used. Power(W) = 24 watts. When talking about the fundamental Ohm's law, we consider a few physical quantities: resistance R, voltage V, and amperage I.Electric current can also be a source of power P so that it can release or transport some energy. Watts = Amps x Volts Examples: 10 Amps x 120 Volts = 1200 Watts; 5 Amps x 240 Volts = 1200 Watts; Amps = Watts / Volts Examples: 4160 Watts / 208 Volts = 20 Amps; 3600 Watts / 240 Volts = 15 Amps; Volts = Watts / Amps Examples: 2400 Watts / 20 Amps = 120 Volts; 2400 Watts / 10 Amps = 240 Volts The conversion of Watts to Volts at fixed amperage is governed by the equation Volts = Watts/Amps For example 100 watts/10 amps = 10 volts Converting Volts to Watts The conversion of Volts to Watts at fixed amperage is governed by the equation Watts = Amps x Volts For example 1.5 amps * … Solution: P = 3000VA × 0.8 = 2400W . Firstly, rewrite the initial watts to amps formula into the amps to watts equation: P = 3 * V * PH * I, As long as all of the values have desired units, simply input them into the formula: P = 3 * 100 V * 0.9 * 15 A = 4050 W, The outcome can be expressed as: "15 amps to watts, for the VLN equal to 100 volts is 4050 W." How to … The power output of the light bulb is 100 watts. For example, if you have a current of 2 A and a voltage of 5 V, the power is 2A * 5V = 10W. A single-phase vacuum has an AC voltage of 127 volts (LN), 4.3 Amps and a power factor of 0.92, how many watts does the vacuum have? Voltage measures the force or pressure of the electricity. Example. The formula is (A)* (V) = (W). W = VA × PF. In other words, watt=amp X volt. This is the formula to convert voltage to wattage: Power(W) = Voltage(V) × Current(A) Thus, to solve for wattage, simply multiply the voltage by the current in amps. Using our sample panel data, 12 Volts multiplied by 5 Amps equals 60 Watts. T he formula for Volts is Watts divided by Amps. They are Power (P) or (W), measured in Watts, Voltage (V) or (E), measured in Volts, Current or Amperage (I), measured in Amps , and Resistance (R) measured in Ohms. Derived from the formula V x A = W, a watt is the rate of power flow that results from amps flowing through a volt’s electromotive force.
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Task 1-1 linear regression notes
Keywords: network IPython Python
1, Explanation of linear regression model
1. What is a linear model
Representation data can be described by linetype relationship. In the case of formulation, y=wx+b. The process of modeling is the process of finding w, b.
However, due to the deviation of real data, unbiased estimation cannot be obtained, and only a model with the smallest deviation can be found.
2. Solution of linear model
Basic concepts:
Dataset = training set + test set = sample + label
The sample is represented by characteristics (the characteristics related to house price include area, old and new, etc.)
The model hopes to establish the relationship between these characteristics and house price, so as to minimize the error between the predicted value and the real value, so that the predicted house price (predicted value) and the actual house price (real value) are closest
loss function
In model training, we need to measure the error between the predicted value and the real value. Generally, we choose a non negative number as the error, and the smaller the value is, the smaller the error is. A common choice is the square function. Its expression in evaluating the sample error with index iii is
l(i)(w,b)=12(y^(i)−y(i))2, l^{(i)}(\mathbf{w}, b) = \frac{1}{2} \left(\hat{y}^{(i)} - y^{(i)}\right)^2, l(i)(w,b)=21(y^(i)−y(i))2,
L(w,b)=1n∑i=1nl(i)(w,b)=1n∑i=1n12(w⊤x(i)+b−y(i))2. L(\mathbf{w}, b) =\frac{1}{n}\sum_{i=1}^n l^{(i)}(\mathbf{w}, b) =\frac{1}{n} \sum_{i=1}^n \frac{1}{2}\left(\mathbf{w}^\top \mathbf{x}^{(i)} + b - y^{(i)}\right)^2. L(w,b)=n1i=1∑nl(i)(w,b)=n1i=1∑n21(w⊤x(i)+b−y(i))2.
Optimization function
When the loss function can not be solved, it needs to be solved iteratively to approach the real value. In the optimization algorithm of numerical solution, mini batch stochastic gradient descent is widely used in deep learning. Its algorithm is very simple: first select the initial value of a group of model parameters, such as random selection; then iterate the parameters many times, so that each iteration may reduce the value of the loss function. In each iteration, a small batch (B \ match {B} b) consisting of a fixed number of training data samples is randomly and uniformly sampled, and then the derivative (gradient) of model parameters related to the average loss of data samples in small batch is calculated. Finally, the product of this result and a preset positive number is used as the reduction of model parameters in this iteration.
(w,b)←(w,b)−η∣B∣∑i∈B∂(w,b)l(i)(w,b) (\mathbf{w},b) \leftarrow (\mathbf{w},b) - \frac{\eta}{|\mathcal{B}|} \sum_{i \in \mathcal{B}} \partial_{(\mathbf{w},b)} l^{(i)}(\mathbf{w},b) (w,b)←(w,b)−∣B∣ηi∈B∑∂(w,b)l(i)(w,b)
Learning rate: η \ eta η represents the step size that can be learned in each optimization
Batch size: B \ batch {B} B is the batch size in small batch calculation
To summarize, there are two steps to optimize the function:
• (i) Initialization of model parameters, generally using random initialization;
• (ii) we iterate on the data several times, and update each parameter by moving the parameter in the negative gradient direction.
2, Explanation of linear regression code
1. Vector computing
Fast vector calculation
One way to add vectors is to add them scalar by element. c[i] = a[i] + b[i]
Another way to add vectors is to add these two vectors directly. d = a + b
2. Realization of linear regression model from zero
# import packages and modules
%matplotlib inline
import torch
from IPython import display
from matplotlib import pyplot as plt
import numpy as np
import random
print(torch.__version__)
Generate data
Suppose the model parameters w and b generate data, in order to make the data conform to the real situation, add fluctuations
Generate a data set of 1000 samples. The following is the linear relationship used to generate data:
price=warea⋅area+wage⋅age+b \mathrm{price} = w_{\mathrm{area}} \cdot \mathrm{area} + w_{\mathrm{age}} \cdot \mathrm{age} + b price=warea⋅area+wage⋅age+b
# set input feature number
num_inputs = 2
# set example number
num_examples = 1000
# set true weight and bias in order to generate corresponded label
true_w = [2, -3.4]
true_b = 4.2
#Generate 1000x2 random numbers as eigenvalues
features = torch.randn(num_examples, num_inputs,
dtype=torch.float32)
#According to the values of w and b, the corresponding labels of features are generated
labels = true_w[0] * features[:, 0] + true_w[1] * features[:, 1] + true_b
#Increase interference
labels += torch.tensor(np.random.normal(0, 0.01, size=labels.size()),
dtype=torch.float32)
Using images to show data
#Relationship between data feature [1] and label
plt.scatter(features[:, 1].numpy(), labels.numpy(), 1);
Divide data sets into small batches
#
def data_iter(batch_size, features, labels):
num_examples = len(features)
indices = list(range(num_examples))
random.shuffle(indices) # random read 10 samples
for i in range(0, num_examples, batch_size):
j = torch.LongTensor(indices[i: min(i + batch_size, num_examples)]) # the last time may be not enough for a whole batch
yield features.index_select(0, j), labels.index_select(0, j)
batch_size = 10
for X, y in data_iter(batch_size, features, labels):
print(X, '\n', y)
break
• Procedure execution sequence:
for gets the iteratable object from the function data ITER and calls the next method.
Run data ITER to yield and return. Yield here is equivalent to return
After the value of for is given to X and y, proceed to the next iteration and continue the for loop in data item
• Relevant knowledge:
generator : the function with yield is a generator (generator is a special kind of iterator), not a function. This generator has a function that is next function. Next is equivalent to the number generated in "next step". This time, the next start is executed in the place where next stops. So when next is called, the generator stops in the next step Stop at the beginning, and then return the number to be generated after meeting yield. This step is over.
The essence of the for item in iteratable loop is to obtain the iterator of iteratable object iteratable through the iter() function, and then call the next() method to get the next value and assign it to item. When encountering the exception of StopIteration, the loop ends.
reference:
Iterator() function and next() function for in... The essence of circulation
Understanding the for loop in Python : there are steps for the for loop to call the next method
Initialize model parameters
w = torch.tensor(np.random.normal(0, 0.01, (num_inputs, 1)), dtype=torch.float32)
b = torch.zeros(1, dtype=torch.float32)
All the sensors have the. Requirements "grad property, which can be set. After that, you can call backward() to find the derivation.
Definition model
Define the training model for training parameters:
price=warea⋅area+wage⋅age+b \mathrm{price} = w_{\mathrm{area}} \cdot \mathrm{area} + w_{\mathrm{age}} \cdot \mathrm{age} + b price=warea⋅area+wage⋅age+b
def linreg(X, w, b):
Define loss function
We use the mean square error loss function:
l(i)(w,b)=12(y^(i)−y(i))2, l^{(i)}(\mathbf{w}, b) = \frac{1}{2} \left(\hat{y}^{(i)} - y^{(i)}\right)^2, l(i)(w,b)=21(y^(i)−y(i))2,
def squared_loss(y_hat, y):
return (y_hat - y.view(y_hat.size())) ** 2 / 2
Define optimization function
Here, the optimization function uses random gradient descent of small batch:
(w,b)←(w,b)−η∣B∣∑i∈B∂(w,b)l(i)(w,b) (\mathbf{w},b) \leftarrow (\mathbf{w},b) - \frac{\eta}{|\mathcal{B}|} \sum_{i \in \mathcal{B}} \partial_{(\mathbf{w},b)} l^{(i)}(\mathbf{w},b) (w,b)←(w,b)−∣B∣ηi∈B∑∂(w,b)l(i)(w,b)
def sgd(params, lr, batch_size):
for param in params:
param.data -= lr * param.grad / batch_size # ues .data to operate param without gradient track
train
When the data set, model, loss function and optimization function are defined, they can prepare for model training.
# super parameters init
lr = 0.03#Learning rate
num_epochs = 5#Training cycle
net = linreg
loss = squared_loss
# training
#Carry out 5 rounds of training, each round of training is solved in batches, and the accuracy of 5 rounds of results is averaged
for epoch in range(num_epochs): # training repeats num_epochs times
# in each epoch, all the samples in dataset will be used once
# X is the feature and y is the label of a batch sample
for X, y in data_iter(batch_size, features, labels):
l = loss(net(X, w, b), y).sum()
# calculate the gradient of batch sample loss
# Using small batch random gradient descent to item model parameters model solution
sgd([w, b], lr, batch_size)
# reset parameter gradient
#Model deviation
train_l = loss(net(features, w, b), labels)
print('epoch %d, loss %f' % (epoch + 1, train_l.mean().item()))
Automatic derivation Here l.backward() gets the derivative of loss for w and b, and x.grad gets the derivative
3. Simple implementation of pytorch
import torch
from torch import nn
import numpy as np
torch.manual_seed(1)
print(torch.__version__)
torch.set_default_tensor_type('torch.FloatTensor')
Generate data set
Generating datasets here is exactly the same as in a zero based implementation.
num_inputs = 2
num_examples = 1000
true_w = [2, -3.4]
true_b = 4.2
features = torch.tensor(np.random.normal(0, 1, (num_examples, num_inputs)), dtype=torch.float)
labels = true_w[0] * features[:, 0] + true_w[1] * features[:, 1] + true_b
labels += torch.tensor(np.random.normal(0, 0.01, size=labels.size()), dtype=torch.float)
import torch.utils.data as Data
batch_size = 10
# combine featues and labels of dataset
dataset = Data.TensorDataset(features, labels)
# put dataset into DataLoader
dataset=dataset, # torch TensorDataset format
batch_size=batch_size, # mini batch size
shuffle=True, # whether shuffle the data or not
)
for X, y in data_iter:
print(X, '\n', y)
break
Definition model
Define the class, network structure and propagation mode of line network
class LinearNet(nn.Module):
def __init__(self, n_feature):
super(LinearNet, self).__init__() # call father function to init
self.linear = nn.Linear(n_feature, 1) # function prototype: torch.nn.Linear(in_features, out_features, bias=True)
def forward(self, x):
y = self.linear(x)
return y
net = LinearNet(num_inputs)
print(net)
LinearNet(
(linear): Linear(in_features=2, out_features=1, bias=True)
)
# ways to init a multilayer network
# method one
net = nn.Sequential(
nn.Linear(num_inputs, 1)
# other layers can be added here
)
# method two
net = nn.Sequential()
# method three: put the neural network layer into the dictionary and pass it in
from collections import OrderedDict
net = nn.Sequential(OrderedDict([
('linear', nn.Linear(num_inputs, 1))
# ......
]))
print(net)#All neural networks
print(net[0])#First floor
Initialize model parameters
# 2. Normal distribution - N(mean, std)
# torch.nn.init.normal_(tensor, mean=0, std=1)
# 3. Constant - fixed value val
# torch.nn.init.constant_(tensor, val)
nn.init.constant_(w, 0.3)
from torch.nn import init
init.normal_(net[0].weight, mean=0.0, std=0.01)
init.constant_(net[0].bias, val=0.0) # or you can use net[0].bias.data.fill_(0) to modify it directly
Define loss function
for param in net.parameters():
print(param)
loss = nn.MSELoss() # nn built-in squared loss function
# function prototype: torch.nn.MSELoss(size_average=None, reduce=None, reduction='mean')
Define optimization function
import torch.optim as optim
optimizer = optim.SGD(net.parameters(), lr=0.03) # Built in random gradient descent function
print(optimizer) # function prototype: torch.optim.SGD(params, lr=, momentum=0, dampening=0, weight_decay=0, nesterov=False)
train
The two cycles, period and batch, y.view and reshape in numpy are the same, making the data arranged horizontally
num_epochs = 3
for epoch in range(1, num_epochs + 1):
for X, y in data_iter:
output = net(X)
l = loss(output, y.view(-1, 1))
`
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Triangle Equations Formulas Calculator
Mathematics - Geometry
Equilateral Triangle
Inputs:
length of side (a)
|
Conversions:
a = 0 = 0
Solution:
Inscribed Circle Radius (r) = HAS NOT BEEN CALCULATED
Change Equation
Select an equation to solve for a different unknown
Scalene Triangle:
No sides have equal length
No angles are equal
Scalene Triangle Equations
These equations apply to any type of triangle. Reduced
equations for equilateral, right and isosceles are below.
Perimeter Semiperimeter Area Area Base Height Angle Bisector of side a Angle Bisector of side b Angle Bisector of side c Median of side a Median of side b Median of side c Altitude of side a Altitude of side b Altitude of side c Circumscribed Circle Radius Inscribed Circle Radius
Law of Cosines
length of side a angle of A
Equilateral Triangle:
All three sides have equal length
All three angles are equal to 60 degrees
Equilateral Triangle Equations
Perimeter Semiperimeter Area Altitude Median Angle Bisector Circumscribed Circle Radius Inscribed Circle Radius
Right Triangle:
One angle is equal to 90 degrees
Right Triangle Equations
Pythagorean Theorem Perimeter Semiperimeter Area Altitude of a Altitude of b Altitude of c Angle Bisector of a Angle Bisector of b Angle Bisector of c Median of a Median of b Median of c Inscribed Circle Radius Circumscribed Circle Radius
Isosceles Triangle:
Two sides have equal length
Two angles are equal
Isosceles Triangle Equations
Perimeter Semiperimeter Area Altitudes of sides a and c Altitude of side b Median of sides a and c Median of side b Angle Bisector of sides a and c Angle Bisector of side b Circumscribed Circle Radius Inscribed Circle Radius
Where
P = Perimeter s = Semiperimeter a = Length of side a b = Length of side b c = Length of side c h = Altitude m = Median A = Angle A B = Angle B C = Angle C t = Angle bisector R = Circumscribed Circle Radius r = Inscribed Circle Radius
Reference - Books: 1) Max A. Sobel and Norbert Lerner. 1991. Precalculus Mathematics. Prentice Hall. 4th ed.
by Jimmy Raymond
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Lecture 10
Lecture 10 - Today Physics 231 General University Physics...
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1 Physics 231 General University Physics Spring 2007 Lecture 10 Today ± Normal Force ± Application of Newton’s II Law ² Free Body Diagram ² Examples ² More than one object Normal Force ± The normal force (table on monitor) is the reaction of the force the monitor exerts on the table ± Normal means perpendicular, in this case ± How does the force on the TV from the table `know’ to be exactly equal to the weight of TV? ± Notation N N nf ≡≡ G G G Normal Force ± Normal force from compression ± Demo: compress spring/rubber ball/golf ball ± Demo: Bridge (Weight on Meter Stick) ± Discussion: Sitting on Sofa/Block on Table Direction of Normal Force ± Discussion: Bow and Arrow, Meter Stick ± Discussion: Push Book against Wall ± Normal is perpendicular Application of Newton’s II Law ± Assumption ² Objects can be modeled as point particles ± Include only the external forces action on the object ² Do not include reaction forces 123 ... = net i FF F F F Σ = +++ G GGGG / net aF m = G G
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2 Free-Body Diagram ± Identify all forces acting on the object (not by the object) ± Draw a coordinate system ± Draw the object as a dot at the origin ± Draw vectors for each identified force ± Draw and label the net force Free-Body Diagram: Example ± Skier Pulled Uphill Free-Body Diagram: Example ± An ice block propelled across frozen lake Free-Body Diagram: Example ± An elevator elevates upward Objects in Equilibrium ± If the acceleration of an object that can be modeled as a particle is zero, the object is said to be in equilibrium ± Mathematically, the net force acting on the object is zero 0 0 and 0 xy F FF = == ∑∑ Problem ± In an electricity experiment, a 1g plastic ball is
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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 Angles. 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
### Angles Starters:
Air Traffic Control: Work out which aircraft are in danger of colliding from their positions and direction of travel. An exercise in understanding bearings.
Angle Estimates: Estimate the sizes of each of the angles then add your estimates together.
Pie Chart: An exercise in estimating what the sectors of a pie chart represent.
Sectors: Work out which sectors fit together to make complete circles. Knowledge of the sum of the angles at a point will help find more than one correct solution to this puzzle.
Square Angles: Find a trapezium, a triangle and a quadrilateral where all of the angles are square numbers.
### Angles Advanced Starters:
Angle Thinking: Find the range of possible angles, x, for which tan x > cos x > sin x
Geometry Snack: Find the value of the marked angle in this diagram from the book Geometry Snacks
Hands Together: The hands of a clock are together at midnight. At what time are they next together?
Triangle or Quadrilateral: Can a quadrilateral have a straight angle?
Index of Advanced Starters
#### Snooker Angles
An online game for one or two players requiring an ability to estimate angles.
The short web address is:
Transum.org/go/?to=snookerangles
### Curriculum for Angles:
#### Year 5
Pupils should be taught to know angles are measured in degrees: estimate and compare acute, obtuse and reflex angles more...
Pupils should be taught to draw given angles, and measure them in degrees (°) more...
Pupils should be taught to identify:
- angles at a point and one whole turn (total 360°)
- angles at a point on a straight line and a turn (total 180°)
- other multiples of 90° more...
Pupils should be taught to use the properties of rectangles to deduce related facts and find missing lengths and angles more...
#### Year 6
Pupils should be taught to draw 2-D shapes using given dimensions and angles more...
Pupils should be taught to compare and classify geometric shapes based on their properties and sizes and find unknown angles in any triangles, quadrilaterals, and regular polygons more...
Pupils should be taught to recognise angles where they meet at a point, are on a straight line, or are vertically opposite, and find missing angles. more...
#### Years 7 to 9
Pupils should be taught to draw and measure line segments and angles in geometric figures, including interpreting scale drawings more...
Pupils should be taught to describe, sketch and draw using conventional terms and notations: points, lines, parallel lines, perpendicular lines, right angles, regular polygons, and other polygons that are reflectively and rotationally symmetric more...
Pupils should be taught to derive and illustrate properties of triangles, quadrilaterals, circles, and other plane figures [for example, equal lengths and angles] using appropriate language and technologies more...
Pupils should be taught to apply the properties of angles at a point, angles at a point on a straight line, vertically opposite angles more...
Pupils should be taught to understand and use the relationship between parallel lines and alternate and corresponding angles more...
Pupils should be taught to derive and use the sum of angles in a triangle and use it to deduce the angle sum in any polygon, and to derive properties of regular polygons more...
Pupils should be taught to apply angle facts, triangle congruence, similarity and properties of quadrilaterals to derive results about angles and sides, including Pythagoras’ Theorem, and use known results to obtain simple proofs more...
#### Years 10 and 11
Pupils should be taught to {apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results} more...
### Feedback:
Comment recorded on the 19 June 'Starter of the Day' page by Nikki Jordan, Braunton School, Devon:
"Excellent. Thank you very much for a fabulous set of starters. I use the 'weekenders' if the daily ones are not quite what I want. Brilliant and much appreciated."
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"I think this is a brilliant website as all the students enjoy doing the puzzles and it is a brilliant way to start a lesson."
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"I used this with my bottom set in year 9. To engage them I used their name and favorite football team (or pop group) instead of the school name. For homework, I asked each student to find a definition for the key words they had been given (once they had fun trying to guess the answer) and they presented their findings to the rest of the class the following day. They felt really special because the key words came from their own personal information."
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"Find the starters wonderful; students enjoy them and often want to use the idea generated by the starter in other parts of the lesson. Keep up the good work"
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"We all love your starters. It is so good to have such a collection. We use them for all age groups and abilities. Have particularly enjoyed KIM's game, as we have not used that for Mathematics before. Keep up the good work and thank you very much
Best wishes from Inger Kisby"
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Could we have some on angles too please?"
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I rate this site as a 5!"
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### Notes:
Pupils should understand that angles represent an amount of turning and be able to estimate the size of angle. When constructing models and drawing pupils should be able to measure and draw angles to the nearest degree and use appropriate language associated with angles.
Pupils should know the angle sums of polygons and that of angles at a point and on a straight line. They will learn about angles made in circles by chords, radii and tangents and recognise the relationships between them.
Pupils will work with angles using trigonometry, transformations and bearings. In exams pupils are often instructed that while non-exact answers should be given to three significant figures, angle answers should be given to one decimal place.
### Angles Teacher Resources:
Angle Theorem Kim's Game: A memory game to be projected to help the whole class revise the circle angle theorems.
Angle Theorems: Diagrams of the angle theorems with interactive examples.
Circle Theorems: Diagrams of the circle theorems to be projected onto a white board as an effective visual aid.
Geometry Toolbox: Create your own dynamic geometrical diagrams using this truly amazing tool from GeoGebra.
Kite Maths: Can you make a kite shape from a single A4 size sheet of paper using only three folds?
Plane Bearings: A visual aid designed to help pupils estimate three figure bearings.
### Angles Activities:
Angle Chase: Find all of the angles on the geometrical diagrams.
Angle Parallels: Understand and use the relationship between parallel lines and alternate and corresponding angles.
Angle Points: Apply the properties of angles at a point, angles on a straight line and vertically opposite angles.
Angles in a Triangle: A self marking exercise involving calculating the unknown angle in a triangle.
Circle Theorem Pairs: A pairs game based around ten theorems about the angles made with chords, radii and tangents of circles.
Circle Theorems Exercise: Show that you understand and can apply the circle theorems with this self marking exercise.
Estimating Angles: Estimate the size of the given acute angles in degrees.
Geometry Toolbox: Create your own dynamic geometrical diagrams using this truly amazing tool from GeoGebra.
Measuring Angles: Measure the size of the given angles to within two degrees of their actual value.
Online Logo: An online version of the Logo programming language with 30 mathematical challenges.
Pie Charts: Develop the skills to construct and interpret pie charts in this self-marking set of exercises.
Polygon Angles: A mixture of problems related to calculating the interior and exterior angles of polygons.
Proof of Circle Theorems: Arrange the stages of the proofs for the standard circle theorems in the correct order.
Scale Drawings: Measure line segments and angles in geometric figures, including interpreting scale drawings.
Snooker Angles: An online game for one or two players requiring an ability to estimate angles.
Finally there is Topic Test, a set of 10 randomly chosen, multiple choice questions suggested by people from around the world.
Alternatively, for the more advanced student, there is an ever-growing collection of Exam-Style Questions with worked solutions on the topic of Angles.
### Angles Investigations:
Tessellations: Drag the shapes onto the canvas to create tessellating patterns and investigate the laws of tessellations.
### Angles Videos:
Angle Properties Song: A song about the angles made with parallel lines.
Ruler and Compass Construction: A demonstration of standard ruler and compass constructions
The 50 Cent Riddle: A 50 cent coin has 12 equal sides. If you place two coins next to each other on a table, what is the angle formed between the two coins?
### Angles Worksheets/Printables:
Angle Chase Worksheets: A set of printable Angle Chase sheets on which pupils fill in the missing angles.
Mearsuring Lines and Angles: Practice using a ruler and protractor on this worksheet with answers provided.
Polybragging Cards: Printable cards for the Polybragging game. Use the properties of polygons to win.
### Angles External Links:
Links to other websites containing resources for Angles 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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#### Angle Chase
Find all of the angles on the geometrical diagrams.
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Tuesday, January 21, 2014
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# Interval arithmetic¶
The mpi type holds an interval defined by a pair of mpf values. Arithmetic on intervals uses conservative rounding so that, if an interval is interpreted as a numerical uncertainty interval for a fixed number, any sequence of interval operations will produce an interval that contains what would be the result of applying the same sequence of operations to the exact number.
## Basic operations¶
You can create an mpi from a number (treated as a zero-width interval) or a pair of numbers. Strings are treated as exact decimal numbers (note that a Python float like 0.1 generally does not represent the same number as its literal; use '0.1' instead):
>>> from mpmath import *
>>> mp.dps = 15
>>> mpi(3)
mpi(mpf('3.0'), mpf('3.0'))
>>> print mpi(3)
[3.0, 3.0]
>>> print mpi(2, 3)
[2.0, 3.0]
>>> print mpi(0.1) # probably not what you want
[0.10000000000000000555, 0.10000000000000000555]
>>> print mpi('0.1') # good, gives a containing interval
[0.099999999999999991673, 0.10000000000000000555]
The fact that '0.1' results in an interval of nonzero width proves that 1/10 cannot be represented using binary floating-point numbers at this precision level (in fact, it cannot be represented exactly at any precision).
Some basic examples of interval arithmetic operations are:
>>> print mpi(0,1) + 1
[1.0, 2.0]
>>> print mpi(0,1) + mpi(4,6)
[4.0, 7.0]
>>> print 2 * mpi(2, 3)
[4.0, 6.0]
>>> print mpi(-1, 1) * mpi(10, 20)
[-20.0, 20.0]
Intervals have the properties .a, .b (endpoints), .mid, and .delta (width):
>>> x = mpi(2, 5)
>>> x.a
mpf('2.0')
>>> x.b
mpf('5.0')
>>> x.mid
mpf('3.5')
>>> x.delta
mpf('3.0')
Intervals may be infinite or half-infinite:
>>> print 1 / mpi(2, inf)
[0.0, 0.5]
The in operator tests whether a number or interval is contained in another interval:
>>> mpi(0, 2) in mpi(0, 10)
True
>>> 3 in mpi(-inf, 0)
False
Division is generally not an exact operation in floating-point arithmetic. Using interval arithmetic, we can track both the error from the division and the error that propagates if we follow up with the inverse operation:
>>> print 1 / mpi(3)
[0.33333333333333331483, 0.33333333333333337034]
>>> print 1 / (1 / mpi(3))
[2.9999999999999995559, 3.0000000000000004441]
The same goes for computing square roots:
>>> print sqrt(mpi(2)) ** 2
[1.9999999999999995559, 2.0000000000000004441]
By design, interval arithmetic propagates errors, no matter how tiny, that would get rounded off in normal floating-point arithmetic:
>>> print mpi(1) + mpi('1e-10000')
[1.0, 1.000000000000000222]
Interval arithmetic uses the same precision as the mpf class; if mp.dps = 50 is set, all interval operations will be carried out with 50-digit precision. Of course, interval arithmetic is guaranteed to give correct bounds at any precision (in the absence of bugs!), but a higher precision makes the intervals narrower and hence more accurate:
>>> mp.dps = 5
>>> print mpi(pi)
[3.141590118, 3.141593933]
>>> mp.dps = 30
>>> print mpi(pi)
[3.14159265358979...793333, 3.14159265358979...797277]
## Interval functions¶
It should be noted that the support for interval arithmetic in mpmath is still somewhat primitive, but the standard arithmetic operators +, -, *, /, ** and sqrt() should work correctly. It is also possible to call the functions exp(), log(), sin(), cos(), tan() with interval arguments. Here are some examples:
>>> mp.dps = 15
>>> print mpi(0.5, 1.5) ** mpi(0.5, 1.5)
[0.35355339059327373086, 1.837117307087383633]
>>> print exp(mpi(0))
[1.0, 1.0]
>>> print exp(mpi(-inf, inf))
[0.0, +inf]
>>>
>>> print exp(mpi(-inf, 0))
[0.0, 1.0]
>>> print exp(mpi(0, inf))
[1.0, +inf]
>>> print exp(mpi(0, 1))
[1.0, 2.7182818284590455349]
>>>
>>> print log(mpi(1))
[0.0, 0.0]
>>> print log(mpi(0,1))
[-inf, 0.0]
>>> print log(mpi(0,inf))
[-inf, +inf]
>>> print log(mpi(2))
[0.69314718055994528623, 0.69314718055994539725]
>>>
>>> print sin(mpi(100, inf))
[-1.0, 1.0]
>>> print cos(mpi('-0.1', '0.1'))
[0.99500416527802570954, 1.0]
Interval support will be added to more functions in the future.
## Establishing inequalities¶
Interval arithmetic can be used to establish inequalities such as . The left-hand and right-hand sides in this inequality agree to over 30 digits, so low-precision arithmetic may give the wrong result:
>>> mp.dps = 25
>>> exp(pi*sqrt(163)) < (640320**3 + 744)
False
The answer should be True, but the rounding errors are larger than the difference between the numbers. To get the right answer, we can use interval arithmetic to check the sign of the difference between the two sides of the inequality. Interval arithmetic does not tell us the answer right away if we keep mp.dps = 25, but it is honest enough to admit it:
>>> print exp(pi * sqrt(mpi(163))) - (640320**3 + 744)
...
[-0.0000003576..., 0.00000029429...]
There is both a negative and a positive endpoint, so we cannot tell for certain whether the true difference is on one side or the other of zero. The solution is to increase the precision until the answer is strictly one-signed:
>>> mp.dps = 35
>>> print exp(pi * sqrt(mpi(163))) - (640320**3 + 744)
...
[-7.499493...e-13, -7.499237...e-13]
## Linear algebra¶
Intervals may be used in matrices. See the matrix documentation page for details.
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# An electric toy car with a mass of 2 kg is powered by a motor with a voltage of 4 V and a current supply of 2 A. How long will it take for the toy car to accelerate from rest to 1 m/s?
Oct 5, 2017
This is a problem for the use of the principle of conservation of energy. The car's kinetic energy when its speed is up to $1 \frac{m}{s}$ is
$K E = \frac{1}{2} \cdot 2 k g \cdot {\left(1 \frac{m}{s}\right)}^{2} = 1 k g \cdot {m}^{2} / {s}^{2} = 1 J$
So, how long does it take that motor to deliver 1 J of energy? The power being consumed by the motor is given by
$P = E \cdot I = 4 V \cdot 2 A = 8 W$
$P o w e r \cdot t i m e = \text{energy}$, therefore $P o w e r = \text{energy"/"time}$.
The unit Watt is equivalent to the combination of units $\frac{\text{Joule}}{\sec}$. Therefore the motor is consuming power from the voltage supply at the rate of $8 \frac{J}{s}$.
So
$8 \frac{J}{s} \cdot t = 1 J$
$t = \frac{1 \cancel{J}}{8 \frac{\cancel{J}}{s}} = 0.125 s$
Note that we are ignoring the fact that the motor will warm up which wastes some of the power. Ignoring such losses simplify the problem. So it will get more complicated in the future - but, don't panic.
I hope this helps,
Steve
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How to Graph Relationships
+ 3
Posted Oct 24 2009 09:48 PM
Whenever a relationship between two variables is discovered and defined, we can use one variable to guess another. Drawing a regression line allows you to picture the relationship and make predictions.
So, you've just been named assistant regional manager of ice cream sales for 10,000 square feet of prime beachfront retail space along the shores of Sunflower Lake in northeast Kansas. Congratulations! You have a lot of responsibility and many strategic decisions to make about how to maximize profit. One dilemma that you will confront is whether to even open. Being open costs money and uses resources, and if you will sell few ice cream cones that day, it probably won't be worth it to even unlock the service window of your brightly painted plywood shack.
If only there were some way to magically know how good business will be on any given day. As an amateur statistician, you assume there must be a scientific way to guess how many cones will sell without having to actually open for business and test the market for the day. You're in luck. There is a way to make estimates of the value or score on some variable (such as ice cream sales) by using other information.
The key is that the other information must come from a variable that is related to the variable of interest. By drawing a line that shows the relationship among your variables for the days you know, you can look at the line as it extends into the future (or the past) for the days you do not know and guess what will happen. Such a graphic tool is called a regression line.
Drawing a Picture of the Future
Observant folks often discover correlations between variables. The usefulness of knowing that a relationship exists goes beyond descriptive statistics, however.
Imagine that you have data on the activities around Sunflower Lake. Among other things, you have collected information about the amount of ice cream sales under the former assistant regional manager of ice cream sales (in number of ice cream cones sold) and the high temperature for each day (in degrees Fahrenheit). The correlation coefficient that represents the relationship between heat and craving for ice cream should be positive and fairly large. That is, as the heat increases, sales probably increase.
Intuitively, it makes sense that with some experience, you could look at the thermometer and get a sense of how busy the ice cream stand is going to be that day. Once you know that there is a positive or negative relationship between two variables, it makes sense that knowing the score on one will give you a general idea of what the score is on the other.
Once you find a relationship between two variables like this, it is reasonable to assume that the relationship between your two variables is linear. In other words, if you produce a graph with all the possible values of one variable as the X-axis (the horizontal line along the bottom) and all the possible values of the other variable as the Y-axis (the vertical line along the side) and then plot each pair of scores, the resulting dots form an essentially straight line.
Connecting the Dots
Figure 1. shows a way to graph the relationship between the temperature and ice cream sales at the beach.
Graph A places dots to represent both values on the two variables, based on historic information you have collected. For instance, the lowest dot means that at 70 degrees, 50 ice cream cones were sold. At 90 degrees, 60 cones were sold. There is a clear pattern here, and the relationship looks like a straight line. For every 10-degree jump in temperature, sales go up 5 cones. For every 1-degree change in temperature, there is a 1/2-cone increase in sales. Graph B draws a line based on this rule. The line goes through every dot.
In Figure 2, analyze Graph B to get a sense of the power of a regression equation. The line includes territory that is not sampled by the data. For instance, we do not have data for 100-degree days. With the regression equation, though, we can estimate what sales might be. If we place a dot on the line at the 100-degree mark, it appears to match up with the 65-cones mark. Using this regression equation, we could estimate that on 100-degree days, 65 cones would be sold. We could do the same for cooler days. Our graph suggests that on a 60-degree day, 45 cones would be sold.
Playing "What If?"
The relationship between heat and cone sales can be expressed mathematically. Our data for graphs A and B in Figure 2, “A linear relationship between sales and temperature” look like this:
High temperature Ice cream cones sold 70 50 80 55 90 60
So, let's see how we could build an equation that describes the relationship using numbers. Regression lines are statistical tools, after all. Notice that if we start with 70 degrees, we get 50 cones. If we enter 70 into our formula, we want 50 to be the output. We also want 80 to get us 55, and 90 to get us 60.
I played around with different possibilities using these values in an attempt to figure out what must be done to the input number to get the correct output number. I noticed that the "ice cream cones sold" value was always smaller than the temperature variable, so I wanted an equation that would shrink the temperature. Linear equations require a constant (some value to use in every equation) in order to produce a straight line, so I needed to have a constant in my equation as well. Rather than use trial and error, you could also enter this data into a statistics program, such as SPSS, or a spreadsheet, such as Excel, to produce the correct components. I found that this formula works well:
Cones Sold = 15 + (Temperature x.50)
Tip
Algebraically, if you begin with a constant and then add some standard amount that is altered only through basic mathematical functions, such as multiplication, you will define a straight line that can be graphed.
"What if?" is a fun game to play with regression lines. Enter a value in one end and a guess comes out the other end; you can get an answer even for unrealistic scenarios. Throw some crazy value onto the line, such as 200 degrees, and you can still get an estimate for cone sales: 115!
The regression equation for this relationship would describe a line that could be drawn to show this relationship visually. With real data, the relationship is seldom as clear as it is in our example. (The correlation for our small fictional data set is a perfect 1.0.)
Tip
In statistics, regression formulas make use of the correlation coefficient, the means, and the standard deviations of both groups of variable scores, regardless of the strength of the relationship in the data set.
Why It Works
The accuracy of these sorts of regression estimates depends on a couple important factors. First, the relationship between variables must be fairly large. Small relationships produce dots all over the place in patterns that aren't straight at all, and a regression line drawn through such a mess misses a lot of dots and is not accurate. Unfortunately, in the social sciences, we don't find very many really strong relationships, so regression predictions tend to produce a certain number of errors. In statistics, errors come with the territory.
Second, the relationship must be at least sort of linear. As in our ice cream cone example, if the nature of the relationship changes somewhere along the regression line, the regression line will miss some of the data. Fortunately, most relationships in the natural world are linear or at least close to it.
Where It Doesn't Work
The actual relationship might not be exactly linear, but if it is essentially so, then regression analysis works pretty well. For example, with our ice cream example, maybe there is a certain increase in sales for every degree jump in the temperature. If that increase is the same regardless of where we are on the scale, we'll see a linear relationship. It is possible, though, that sales jump once a certain temperature is reached. Perhaps once it is over 90 degrees at the beach, people really flock to get relief.
Graphs C and D in Figure 3 show what happens if the true relationship isn't exactly linear.
Following the requirements of linear regression, the regression equation always produces a straight line and, in this case, two of the dots fall right on it, but one does not. This line does a decent job of explaining the data by picturing the relationship, but because the relationship is not linear, the regression equation makes some errors.
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When an equation is having two variables with every variables in first degree such that variables are not multiplied with each other then it is called as a linear equations in two variables.
This chapter covers the following concepts-
-Pair of linear equations in two variables
-Graphical methods of solution of a pair of linear equations
-Algebraic methods of solving a pair of linear equations- substitution method, elimination method, cross multiplication method
-equations reducible to a pair of linear equations in two variables.
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### Lecture 19: Linear Equation In Two Variable Part- 19
00:34:00
Published 18-May-2017 Bilingual
Linear Equation in Two Variables | CBSE Class 10 Maths (Class X) This lesson has been made very interactive by Sameer Kohli Sir. Watching this video you will feel as if you are in his class. He has shared tips to learn all the methods.
The topics and sub-topics in Chapter 3 Pair of Linear Equations in Two Variables
1 Introduction,
2 Pair of Linear Equations in Two Variables,
3 Graphical Method of Solution of a pair of Linear Equations,
4 Algebraic Methods of Solving a Pair of Linear Equations,
-Substitution Method
-Elimination Method
-Cross-Multiplication Method
5 Equations Reducible to a Pair of Linear Equations in Two Variables and
6 Summary
#### Sameer Kohli
Mr. Kohli, founder of Jupiter Education Planet, is a visionary in the field of education and holds an experience of more than 15 years as a Mathematics and Science teacher. Owing to his student-friendly teaching style, he has been able to build an unflinching reputation for himself. He is not only an excellent teacher who has delivered exceptional results throughout his career, but also has a deep insight into child and parent psychology. Over the years, he has acquired expertise in the areas of motivation and counselling of parents and children to help them take better decisions. With a passion to change lives for the better, Mr. Kohli knows the right approach to help children in optimum decision making at both personal and professional level.
##### Qualification
B.Sc. (Hon.) (D.U.)
10 years
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# Four dice are thrown simultaneously
Four dice are thrown simultaneously. The probability that $$4$$ and $$3$$ appear on two of the dice given that $$5$$ and $$6$$ appear on the other two dice is:
a) $$1/6$$
b) $$1/36$$
c) $$12/51$$
d) None of these
Since the events are independent, I feel the probability is $$1/6 \times 1/6 = 1/36$$
• Conditional probability, namely, $P(x) = P($each dice has $3, 4, 5$ and $6$ respectively $)/P(5$ and $6$ appear on $2$ dices$)$, where x is our event we want to find probability for. – Makina Nov 16 '18 at 8:26
• Given the current list of choices, the correct answer is d). – kludg Nov 16 '18 at 9:24
• Die is singular; dice is plural; dices is the third person singular form of the verb to dice, meaning to cut into small cubes. – N. F. Taussig Nov 16 '18 at 10:39
Even if there are only two dice, the probability of observing $$4$$ and $$3$$ is not $$1/6 \times 1/6$$, but rather, $$1/18$$, since the sample space is the set of ordered pairs $$(a,b)$$ where each $$a, b \in \{1, 2, 3, 4, 5, 6\}$$. Thus there are two desired outcomes $$(4,3)$$, $$(3,4)$$ out of $$6^2 = 36$$ possible outcomes.
When there are four dice, two of which you are told are $$5$$ and $$6$$, you must reason carefully and precisely. Given the set of all outcomes of four dice rolls, select those that show at least one $$5$$ and at least one $$6$$. Of these, how many show $$3$$ and $$4$$ on the other two dice?
Doing mathematics is not about "feelings." It is about showing and justifying your reasoning. Describing your calculation without providing a sound basis for why you are doing what you are doing, is not math.
• But isn't it given that two events - 3 and 4 have surely happened? Then why to worry about that? – Archer Nov 16 '18 at 8:10
• I now think that the answer should be 1/18 – Archer Nov 16 '18 at 8:10
• Why don't you actually try the calculation in the correct way, instead of just claiming that the answer "should" be something that isn't even one of the answer choices? – heropup Nov 16 '18 at 8:24
The tricky part of the solution is to find the number of outcomes such that 2 dice land $$5$$ and $$6$$; it can be done using inclusions/exclusions.
Let $$\Omega$$ be the set of all outcomes, of size $$6^4$$. Let $$S_5$$ be the set of outcomes that contain no $$5$$'s, of size $$5^4$$. Let $$S_6$$ be the set of outcomes that contain no $$6$$'s, of size $$5^4$$. Let $$S_{5,6}$$ be the set of outcomes that contain no $$5$$'s no $$6$$'s, of size $$4^4$$.
Using inclusion/exclusion principle the number of outcomes that contain at least one $$5$$ and at least one $$6$$ is $$6^4-5^4-5^4+4^4=302$$ The rest is simple, and the answer is $$12/151$$
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# Need Little Help
• Mar 30th 2010, 06:57 AM
kap
Need Little Help
Hello Everybody,
I need help in solving this question. The equation is:
$\displaystyle 5x^2 + 4x + 1 \equiv 0$
The question is to:
determine all the solutions in Z(17) and in Z(20).
[In Z(17): 9 and 14, in Z(20): none ]
NOTE::
The Z(17) is suppose to be the whole numbers symbol with subscript 17 but i don't know how to do it in Latex. The same applies to Z(20).
• Mar 30th 2010, 07:30 AM
novice
Quote:
Originally Posted by kap
Hello Everybody,
I need help in solving this question. The equation is:
$\displaystyle 5x^2 + 4x + 1 \equiv 0$
The question is to:
determine all the solutions in Z(17) and in Z(20).
[In Z(17): 9 and 14, in Z(20): none ]
NOTE::
The Z(17) is suppose to be the whole numbers symbol with subscript 17 but i don't know how to do it in Latex. The same applies to Z(20).
$\displaystyle 5x^2 + 4x + 1 \not \equiv 0$
• Mar 30th 2010, 08:31 AM
kap
• Mar 30th 2010, 09:48 AM
novice
Quote:
Originally Posted by kap
You have no real solution but complex.
• Mar 30th 2010, 11:11 AM
Soroban
Hello, kap!
I never studied modulo arithmetic formally,
. . but I just "invented" a method . . .
Quote:
Solve: .$\displaystyle 5x^2 + 4x + 1 \:\equiv\: 0 \text{ (mod 17)}$
We have: .$\displaystyle 5x^2 + 4x + 1 \:\equiv\:0\text{ (mod 17)}$
Multiply by 7: .$\displaystyle 35x^2 + 28x + 7 \:\equiv\:0\text{ (mod 17)}$
. . . . . . . . . . . $\displaystyle x^2 + 11x + 24 \:\equiv\:0\text{ (mod 17)}$
. . . . . . . . . . . $\displaystyle (x + 8)(x+3) \:\equiv\:0\text{ (mod 17)}$
Therefore: .$\displaystyle \begin{Bmatrix}\;x &\equiv& -8 & \equiv & 9 & \text{(mod 17)}\;\\ x &\equiv& -3 & \equiv & 14 & \text{(mod 17)}\end{Bmatrix}$
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## Plot a histogram of sample hypnotized, Basic Statistics
Assignment Help:
Computer Project for Student Exam Scores
The data in the excel spread sheet titled Data for Hypothesis Testing in the column titled Scores from Student Population represents more than a thousand math exam scores (the entire population of test takers.) The column titled Sample Forced to 3 hrs Nightly Study Halls represents a group of students that were forced to attend math study halls for one semester and then they were retested. The column titled Sample Hypnotized to Perform Better represents students that were hypnotized and conditioned to use their spare time to study math and then retested.
Prepare a report in word, with title page, which includes graphs and calculations when appropriate to address all of the questions that are outlined below.
1. Evaluate the Scores fromStudent Population,Forced to 3 hrs Nightly Study Halls, and Sample Hypnotized to Perform Better for OUTLIERS. The exam was on a scale of 0-100% with a 20 point bonus.
a. Arrange the data in order and find Q1 and Q3
b. Find the interquartile range: IQR=Q3-Q1
c. Eliminate any data that is smaller than Q1-1.5(IQR)
or larger than Q3+1.5(IQR). The data remaining after screening will be used for the remaining analysis.
How many outliers were in the Scores from Student Population data set?
How many outliers were in the Forced to 3 hrs Nightly Study Halls data set?
How many outliers were in the Sample Hypnotized to Perform Better data set?
2. Determine if all three Columns of Data are normal.
a. Report the Pearson's index PI of skewness for Scores from Population and explain the acceptable range.
b. Report the Pearson's index PI of skewness for Sample Forced to 3 hrs Nightly Study Hallsand explain the acceptable range.
c. Report the Pearson's index PI of skewness for Sample Hypnotized to Performand explain the acceptable range.
d. Plot a Histogram of Scores from Population and refer to its normality.
e. Plot a Histogram of Sample Forced to 3 hrs Nightly Study Halls refer to its normality.
f. Plot a Histogram of Sample Hypnotized to Perform refer to its normality.
3. Confidence intervals z scores.
a. Calculate a 95% confidence interval from the population mean. Using the Sample Forced to 3 hrs Nightly Study Halls data. Does the mean of the Sample Forced to 3 hrs Nightly Study Halls data fall within the confidence interval of the mean of the Scores from the Population.
b. Calculate a 95% confidence interval from the population mean. Using the Sample Hypnotized to Perform Better data. Does the mean of the Sample Hypnotized to Perform Better data data fall within the confidence interval of the mean of the Scores from the Population.
4. Confidence intervals t scores.
a. Calculate a 95% confidence interval from the Sample Forced to 3 hrs Nightly Study Halls mean. Does the mean of the population data fall within the confidence interval of the mean of the Sample Forced to 3 hrs Nightly Study Halls.
b. Calculate a 95% confidence interval from the Sample Hypnotized to Perform Better data. Does the mean of the Population data fall within the confidence interval of the mean of the Sample Hypnotized to Perform Better.
5. Hypothesis Testing.
a. Write a null hypothesis and alternate hypothesis that would go along with the data (Sample Forced to 3 hrs Nightly Study Halls) that was collected.
b. Write a null hypothesis and alternate hypothesis that would go along with the data (Sample Hypnotized to Perform Better) that was collected.
c. Using a level of significance of α=.05 for these Two-Tailed hypothesis, make conclusions about which hypotheses to accept and which to reject using z scores.
6. Hypothesis Testing.
a. Write a null hypothesis and alternate hypothesis that would go along with the data (Sample Forced to 3 hrs Nightly Study Halls) that was collected.
b. Write a null hypothesis and alternate hypothesis that would go along with the data (Sample Hypnotized to Perform Better) that was collected.
c. Using a level of significance of α=.05 for these Two-Tailed hypothesis, make conclusions about which hypotheses to accept and which to reject using t scores.
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Problems with logaritm
• Oct 21st 2008, 09:23 PM
blubla
Problems with logaritm
Hello everyone! I have a few problem regarding Log.
First of all, it is a homework about Logarithm. I provide an image of my homework here:
http://img443.imageshack.us/img443/8...thmsbj4.th.jpg
(The ones squared with red are the ones I have problem with solving)
The main problem I have with Logs are the exponents. I know how to convert from logarithm form to exponent form and vice versa. But when solving for a variable exponent, it becomes difficult.
This is what I have done:
http://img75.imageshack.us/img75/9156/log1kf8.th.jpg
In problem 57.
I can solve a.
But b, c, and d are different.
In b. when solving on exponential form, how can I make a base with a variable exponent equal to a fraction (or decimal)?
In c. when solving on exponential form, how can I make a base with a variable exponent equal to a root?
The same goes with d.
In problem 58.
I could solve a. But it seems that b, c, and d, are solved differently. All of them contain negative exponents. How can I make a base with an exponent equal to another positive number if the exponent is negative?
Thanks in advance for the help!
• Oct 21st 2008, 10:17 PM
earboth
Quote:
Originally Posted by blubla
Hello everyone! I have a few problem regarding Log.
First of all, it is a homework about Logarithm. I provide an image of my homework here:
...
(The ones squared with red are the ones I have problem with solving)
The main problem I have with Logs are the exponents. I know how to convert from logarithm form to exponent form and vice versa. But when solving for a variable exponent, it becomes difficult.
This is what I have done:
...
to b):
$\displaystyle \log_2\left(\dfrac1{16}\right)=\log_2\left(\dfrac1 {2^4}\right)=\log_2\left(2^{-4}\right)=-4$
The other two problems have to be done similarly.
to c):
$\displaystyle \log_2(\sqrt{8})=\log_2(\sqrt{2^3})=\log_2((2^3)^{ \frac12})=\log_2(2^{\frac32})=\dfrac32$
The other two problems have to be done similarly.
to d):
$\displaystyle \log_{\frac12}(32)=\log_{\frac12}(2^5)=\log_{\frac 12}(((2^{-1})^{-1})^5)=\log_{\frac12}\left(\left(\dfrac12\right)^{-5}\right) = -5$
The other two problems have to be done similarly.
• Oct 21st 2008, 10:40 PM
earboth
Quote:
Originally Posted by blubla
Hello everyone! I have a few problem regarding Log.
...
In problem 57.
I can solve a.
But b, c, and d are different.
In b. when solving on exponential form, how can I make a base with a variable exponent equal to a fraction (or decimal)?
In c. when solving on exponential form, how can I make a base with a variable exponent equal to a root?
The same goes with d.
In problem 58.
I could solve a. But it seems that b, c, and d, are solved differently. All of them contain negative exponents. How can I make a base with an exponent equal to another positive number if the exponent is negative?
Thanks in advance for the help!
If you have the equation
$\displaystyle x^a = c$ ....... which you want to solve for x you only have to take both sides to the power $\displaystyle \dfrac1a$
That means: $\displaystyle x^a = c~\implies~\left(x^a\right)^{\frac1a} = c^{\frac1a}~\implies~x^1=c^{\frac1a}$
to 58 b):
$\displaystyle x^{-3} = 2~\implies~\left(x^{-3}\right)^{-\frac13} = 2^{-\frac13}~\implies~x=\dfrac1{\sqrt[3]{2}}$
to 58 c):
$\displaystyle x^{-8} = \dfrac1{16} ~\implies x^{-8} = \dfrac1{2^4} \implies x^{-8} = 2^{-4} \implies$ $\displaystyle \left(x^{-8}\right)^{-\frac18} = \left(2^{-4}\right)^{-\frac18} \implies x=2^{\frac12} \implies x=\sqrt{2}$
to 58 d):
Keep in mind that 27 = 3³. Therefore the solution is $\displaystyle x = \dfrac13$
• Oct 23rd 2008, 12:05 AM
blubla
That helped me a lot. Thanks!(Nod)
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215328 (number)
215,328 (two hundred fifteen thousand three hundred twenty-eight) is an even six-digits composite number following 215327 and preceding 215329. In scientific notation, it is written as 2.15328 × 105. The sum of its digits is 21. It has a total of 7 prime factors and 24 positive divisors. There are 71,744 positive integers (up to 215328) that are relatively prime to 215328.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 6
• Sum of Digits 21
• Digital Root 3
Name
Short name 215 thousand 328 two hundred fifteen thousand three hundred twenty-eight
Notation
Scientific notation 2.15328 × 105 215.328 × 103
Prime Factorization of 215328
Prime Factorization 25 × 3 × 2243
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 7 Total number of prime factors rad(n) 13458 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 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 215,328 is 25 × 3 × 2243. Since it has a total of 7 prime factors, 215,328 is a composite number.
Divisors of 215328
24 divisors
Even divisors 20 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 24 Total number of the positive divisors of n σ(n) 565488 Sum of all the positive divisors of n s(n) 350160 Sum of the proper positive divisors of n A(n) 23562 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 464.034 Returns the nth root of the product of n divisors H(n) 9.13878 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 215,328 can be divided by 24 positive divisors (out of which 20 are even, and 4 are odd). The sum of these divisors (counting 215,328) is 565,488, the average is 23,562.
Other Arithmetic Functions (n = 215328)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 71744 Total number of positive integers not greater than n that are coprime to n λ(n) 17936 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 19183 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 71,744 positive integers (less than 215,328) that are coprime with 215,328. And there are approximately 19,183 prime numbers less than or equal to 215,328.
Divisibility of 215328
m n mod m 2 3 4 5 6 7 8 9 0 0 0 3 0 1 0 3
The number 215,328 is divisible by 2, 3, 4, 6 and 8.
Classification of 215328
• Arithmetic
• Refactorable
• Abundant
Expressible via specific sums
• Polite
• Non-hypotenuse
Base conversion (215328)
Base System Value
2 Binary 110100100100100000
3 Ternary 101221101010
4 Quaternary 310210200
5 Quinary 23342303
6 Senary 4340520
8 Octal 644440
10 Decimal 215328
12 Duodecimal a4740
20 Vigesimal 16i68
36 Base36 4m5c
Basic calculations (n = 215328)
Multiplication
n×y
n×2 430656 645984 861312 1076640
Division
n÷y
n÷2 107664 71776 53832 43065.6
Exponentiation
ny
n2 46366147584 9983929826967552 2149819641781269037056 462916363825477099211194368
Nth Root
y√n
2√n 464.034 59.9377 21.5415 11.6579
215328 as geometric shapes
Circle
Diameter 430656 1.35295e+06 1.45664e+11
Sphere
Volume 4.18206e+16 5.82654e+11 1.35295e+06
Square
Length = n
Perimeter 861312 4.63661e+10 304520
Cube
Length = n
Surface area 2.78197e+11 9.98393e+15 372959
Equilateral Triangle
Length = n
Perimeter 645984 2.00771e+10 186480
Triangular Pyramid
Length = n
Surface area 8.03085e+10 1.17662e+15 175815
Cryptographic Hash Functions
md5 c846a06e6b88b6e3fb269be93c2c5680 1920abf52b18445be3d44a16df56f60e9c1c18bf 5febc2d6f034a64ceb317ae19b40e7839839df095ab52f8b067e6bc1a51df100 550f35a3e3daf38fce403692763e7a6ff7442fd1a4a25eeb58edb0d6c65a5e5aa9fdb955419dd49826faf5d1eac45e8c4baf2fc194de9d3662df8fcb2a792799 59df1de958802433a8ad33a49bf040cda8d4697c
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0
# The largest prime number that can be written as two of prime numbersr and as the difference of two prime numbers?
Updated: 11/4/2022
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Q: The largest prime number that can be written as two of prime numbersr and as the difference of two prime numbers?
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Related questions
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### What is the difference between rational and irrational in math?
Rational numbers are numbers that can be written as a fraction. Irrational numbers cannot be expressed as a fraction.
### Is the difference of rational numbers a rational number?
Rational numbers are numbers that can be written as a fraction. Irrational numbers cannot be expressed as a fraction. All natural numbers are rational.
98
### Difference between real and rational number?
Rational numbers are numbers that can be written as a fraction. Real numbers are any number, including irrationals.
### How does rational number different from fractional number?
There is no difference. Rational numbers are numbers that can be written as a fraction. Irrational numbers cannot be expressed as a fraction.
### Is the difference of any 2 rational numbers a rational number?
Yes it is. Rational numbers are numbers that can be written as a fraction. Irrational numbers cannot be expressed as a fraction.
36
### What is the difference from rational numbers and a number?
A rational number can be written as (one whole number) divided by (another whole number). An irrational number can't.
### Could you find the sum of all seven digit numbers that can be written using only ones and zeros?
If we use numbers that can be written with only ones and zeros, then we're working with binary numbers. The largest binary number that can be written with seven bits is "127". When you remember to include 'zero', that makes a total of 128 different numbers. The sum of the (smallest + largest) is (0 + 127) = 127 The sum of the (second smallest + second largest) = (1 + 126) = 127 The sum of the (third smallest + third largest) = (2 + 125) = 127 Each sum built in this way is 127, and there are (half of 128) = 64 of them. So the sum of all 128 numbers is (64 times 127) = 8,128
5 = 2+3 = 7-2
### Using each digit one time and only one time What are the three largest numbers that can be written using the digits 1 4 7 0 5 and 9?
The three largest numbers (in descending order) would be... 975410, 975401 & 975104
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# How prove concave sequence inequality $\left(\sum_{k=1}^{n}a_{k}\right)^2\ge\frac{3n-c}{4}\sum_{k=1}^{n}a^2_{k}$
let concave sequence $\{a_{n}\}$,such $a)_{n}\ge 0$,and such $$\dfrac{a_{i-1}+a_{i+1}}{2}\le a_{i},i=1,2,\cdots,n-1$$ where $a_{0}=0$.
show that $\exists c>0$ such that for every postive integer number $n$ have $$\left(\sum_{k=1}^{n}a_{k}\right)^2\ge\dfrac{3n-c}{4}\sum_{k=1}^{n}a^2_{k}$$
It is said the $\dfrac{3n}{4}$ is best constant .But I can't prove it
In 1988 IMO problem have this: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=365107&sid=c799d4bad33b8626717f999951038390#p365107
and I found this Iran Team selection 2010 have simarler problem:see
and 2010 china TST test have http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1457433&sid=c799d4bad33b8626717f999951038390#p1457433
and I find this sequence a paper:http://link.springer.com/article/10.1007%2FBF01085887#page-1
but for my problem I can't it,Thank you
I show below that for a fixed $n$, we have $\bigg(\sum_{i=1}^na_i\bigg)^2 \geq d_{n-2} \sum_{i=1}^na_i^2$ where $d_n=\frac{(1+2+3+\ldots+n)^2}{1^2+2^2+\ldots+n^2}=\frac{3n(n+1)}{2n+1}$. The constant $d_{n-2}$ is easily seen to be optimal (take $a_i=i-1$ for $i<n$ and $a_n=0$). So you are left with the problem of finding the best constant $c$ such that $\frac{3(n-2)(n-1)}{2(2n-3)} \geq \frac{3n-c}{4}$ for any $n\geq 2$. It is easy to compute that the best $c$ is $c=6$.
Let us put (for $1\leq k\leq n$)
$$b_k=\left\lbrace \begin{array}{lcl} a_1 & {\rm if} & k=1 \\ 2a_k-(a_{k-1}+a_{k+1}) & {\rm if} & 1 < k < n \\ a_n & {\rm if} & k=n \\ \end{array} \right.\tag{1}$$
Expressing everything in terms of the new variables $b_k$, the initial inequality is equivalent to the fact that a certain polynomial in the $b_k$ with nonnegative coefficients is nonnegative :
A little computation shows that, for $1\leq i\leq n$, $$a_i=\left\lbrace\begin{array}{lcl}b_1 & {\rm if} & i=1 \\ \frac{1}{n-1}\bigg((n-i)b_1+\sum_{j=2}^{i-1}(n-i)(j-1)b_j+\sum_{j=i}^{n-1}(n-j)(i-1)b_j +(i-1)b_n\bigg) & {\rm if} & 1 < i < n \\ b_n & {\rm if} & i=n \\ \end{array} \right.\tag{2}$$ whence $$\sum_{i=1}^na_i=\frac{1}{2}\Bigg(nb_1+\Bigg(\sum_{j=2}^{n-1}(j-1)(n-j)b_j\Bigg)+nb_n\Bigg)\tag{3}$$ so that $\bigg(\sum_{i=1}^na_i\bigg)^2=\sum_{1\leq i \leq j \leq n}C_1(i,j)b_ib_j$, with $$C_1(i,j)=\frac{1}{4}\times\left\lbrace \begin{array}{lcl} n^2 & {\rm if} & i=j=1 \ \text{or} \ i=j=n \\ (j-1)^2(n-j)^2 & {\rm if} & i=j\not\in\lbrace1,n \rbrace \\ 2n(j-1)(n-j)(2(j-1)n-(j^2-1)) & {\rm if} & i\neq j,i=1, j< n \\ 2n^2 & {\rm if} & i\neq j,i=1, j= n \\ 2(i-1)(n-i)(j-1)(n-j) & {\rm if} & i\neq j,i>1, j< n \\ 2(i-1)(n-i)n & {\rm if} & i\neq j,i>1, j=n \\ \end{array} \right.\tag{4}$$ and also $\sum_{i=1}^na_i^2=\sum_{1\leq i \leq j \leq n}C_2(i,j)b_ib_j$, with $$C_2(i,j)=\frac{1}{6(n-1)}\times\left\lbrace \begin{array}{lcl} n(2n-1) & {\rm if} & i=j=1 \ \text{or} \ i=j=n \\ (j-1)(n-j)(2(j-1)n-(2j^2-2j-1)) & {\rm if} & i=j\not\in\lbrace1,n \rbrace \\ (j-1)(n-j)(2(j-1)n-(j^2-1)) & {\rm if} & i\neq j,i=1, j< n \\ 2n(n-2) & {\rm if} & i\neq j,i=1, j= n \\ (i-1)(n-j)(2(j-1)n-(j^2+i^2-2i-1)) & {\rm if} & i\neq j,i>1, j< n \\ 2(i-1)(n+i-2)(n-i) & {\rm if} & i\neq j,i>1, j=n \\ \end{array}\right.\tag{5}$$ Checking each one of the six cases in turn, we see that $C_1(i,j) \geq d_{n-2}C_2(i,j)$ for any $i \leq j$, which finishes the proof.
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https://socratic.org/questions/what-is-the-cartesian-form-of-1-25pi-8
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# What is the Cartesian form of ( -1, (25pi)/8 ) ?
Apr 8, 2016
$\left(0.924 , 0.383\right)$
#### Explanation:
$x = r \cos \theta , y = r \sin \theta$
$R = - 1 , \theta = \frac{25 \pi}{8}$
$x = - 1 \cos \left(\frac{25 \pi}{8}\right) , y = - 1 \sin \left(\frac{25 \pi}{8}\right)$
$x \approx 0.924 , y \approx 0.383$
$\left(0.924 , 0.383\right)$
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https://www.physicsforums.com/threads/consider-a-pendulum-of-length-d-2-calculate-the-electric-field-inside-capacitor.518910/
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# Consider a pendulum of length d/2 calculate the electric field inside capacitor
1. Aug 3, 2011
### blueyellow
1. The problem statement, all variables and given/known data
consider a parallel-plate capacitor with square plates of side L and distance d (<<L) between them, charged with charges +Q and -Q. The plates of the capacitor are horizontal, with the lowest lying on the x-y plane, and the orientation is such that their sides are parallel to the x and y axis, respectively
consider a simple pendulum of length d/2 and mass m, hanging vertically from the centre of the top plate, that can oscillate in the x-z plane. Calculate the electric field inside the capacitor, and the electrical force acting on the mass if it has charge q (the field produced by the charge q itself can be neglected, and the polarity of the capacitor is such that the force will act in the negative z direction).
2. Relevant equations
E=-grad V
=[q/(4pi*epsilon0)][1/(r^2)]*(r-hat)
3. The attempt at a solution
should i just use the above equation, and sort of ignore the r-hat at the end? would the answer to what the electric field is be zero?
then would the electrical force be F=Bqv?
but what would i put in for B and v?
2. Aug 3, 2011
### Mike Pemulis
Your attempt at a solution is almost completely wrong. Your second equation,
E =[q/(4pi*epsilon0)][1/(r^2)]*(r-hat)
is the equation for the electric field created by a single point charge. If you distribute many electric charges on the surface of an object, such as a capacitor, the electric field is different; not surprisingly, the formula for the electric field depends on the shape of the object. It is possible to use your formula, and some calculus, to calculate the electric field created by charges distributed on any shape. Your book should include an explanation of how to do this. Alternatively, the formula for an electric field inside a parallel-plate capacitor is used very often, so you can probably find the formula in your book or online.
Now, this gives you the electric field, but you still need to find the force on the charged particle. The one you quote, F = Bqv, is not the one you need. That is actually the force from a magnetic field on a moving charged particle. There are many equations that start with "F;" you need to find the one that applies to this problem, which means you have to read the text around the equation, not just the equation itself. Again, the formula you need can be found in your book or Wikipedia.
3. Aug 14, 2011
### blueyellow
I found another equation:
F(subscript 1 2)=[1/(4pi*epsilon)]*[[q(subscript1)*q(subscript2)]/[(r-subscript1 2)^2]]*(r-hat-subscript1 2)
but I'm not sure how this can be used since there is only one charge mentioned in the question and the equation involves two charges
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook
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# OEIS A084639
last edited by 1 year, 8 months ago
re: my comment made to this sequence, here is my attempt to xplain it.
okay, see attached file. sorry to make anything not clear enough, I seem to think at a speed different than I can write !)
I began with small bitstrings of 3, 4 and 5 bits as you can see, and varying # of bit transitions ("1x in file means 1 transition, 2x two transitions, etc)
essentially look at the orange shading, I looked for the "adjustment" required to make things work out for what the maximal decimal value of the bit string could be as a f{# set bits, # bit transitions}
So, for xample, treating a N=4 bitstring, can form a matrix of # set bits and # bit transitions
SO, we can have
1110 (3 set bits, 1 transition)
1101 (3 set bits, 2 transition)
1100 (2 set bits, 1 transition)
1010 (2 set bits, 3 transition)
1001 (2 set bits, 2 transition)
etc
I concluded the max value = N-1 left aligned bits + (2^k -2) - adjustment/correction as per my first email [ it's only an "adjustment" from the "incorrect" partial form, it's just part of the formula ...]
where k= (N - #set bits). By left aligned bits, see cells C20-C22. For 2 set bits in a 4 bit string, the upper bound (max) is 1100 obviously or 12. For that case, it has 1 transition, so k=4-2, so 2^k-2=2, as you'll see in those cells C20-C22. This yields A+B=12+2=14, which is the absolute max one can see for N=4 (given the initial restriction that not all bits are set) ie 1110=14, so there is no adjustment. However, for 2 bit transitions, the upper bound is 1101=13, 1 less than 14. For 3 transitions, the upper bound is 1010=10, 4 less than 14. Etc
this generated a sequence 0,1,4,9,20,41,84,169... as the correction ("C"), actually a subtraction.
I then searched OEIS and got the hit to A084639. I then immed tried the 30 bit one and it matched, so formula seems correct (the A+B+C) though it should have read A+B-C in my earlier email I think
SO, the final bound is (N-1 left aligned bits ) + 2^(N -#set bits-1) - A084639(ptr) with ptr=#bit transitions-1
- - -
I think my comments saying how to use WA to generate entries was editted out but one merely follows the pattern as:
for a(32), the input is (note it's 33 bits in each part)
111111111111111111111111111111110 base 2 - 101010101010101010101010101010101 base 2
(one just adds or subtracts a 1 from the front part, and mimics the alternating pattern for the back part )
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1. ## simplify logarithm
hi
i have the following
ln t = B1 + B2.T + B3.X + B4.X^2 + B5.X^3
where B1,B2,B3,B4,B5 are constants and X = ln y
i need this in a form t = C1 (y^C2) (e^(-C3/T))
where C1,C2,C3 are constants.
ln and e are natural log terms.
can this be done?
thanks
jas
p.s. i know e^(B3.X) = e^(B3. ln y) =e^(ln y^B3) = y^B3
but i'm not sure how to get the other terms B4.X^2 = B4 (ln y) (ln y)
and B5.X^3 = B5 (ln y) (ln y) (ln y)
in a similar form of y^??? (or if its possible!!)
2. Originally Posted by jas
hi
i have the following
ln t = B1 + B2.T + B3.X + B4.X^2 + B5.X^3
where B1,B2,B3,B4,B5 are constants and X = ln y
i need this in a form t = C1 (y^C2) (e^(-C3/T))
where C1,C2,C3 are constants.
ln and e are natural log terms.
can this be done?
thanks
jas
p.s. i know e^(B3.X) = e^(B3. ln y) =e^(ln y^B3) = y^B3
but i'm not sure how to get the other terms B4.X^2 = B4 (ln y) (ln y)
and B5.X^3 = B5 (ln y) (ln y) (ln y)
in a similar form of y^??? (or if its possible!!)
Taking the exponential of both sides,
$\displaystyle t= e^{B1+ B2T+ B3X+ B4X^2+ B5X^3}= e^{B1}(e^{T})^B2(e^x)^{B3}(e^X)^{2B4}(e^X)^{3B5}$
Since x= ln y, $\displaystyle e^X= e^{ln y}= y$ so that is the same as $\displaystyle t= e^{B1}e^{B2T}y^{B3}y^{2B4}y^{3B5}$$\displaystyle = e^{B1}e^{B2T}y^{B3+ 2B4+ 3B5}$.
So taking $\displaystyle C1= e^{B1}$, $\displaystyle C2= B3+ 2B4+ 3B5$, [tex], and $\displaystyle C3= -B3$ you will have $\displaystyle t= C1(y^{C2})e^{-C3T}$. I do not believe you can choose the constants so that it is $\displaystyle e^{-C/T}$ rather than $\displaystyle e^{-CT}$.
3. by
ln t = B1 + B2.T + B3.X + B4.X^2 + B5.X^3
i mean
ln t = B1 + B2.T + B3.(ln y) + B4.(ln y)(ln y) + B3.(ln y)(ln y)(ln y)
so (ln y)(ln y) is not the same as (ln y^2) = 2 ln y ?
so i don't know if the simplification you have is true?
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https://www.jiskha.com/questions/1630211/a-500-n-tightrope-walker-stands-at-the-center-of-the-rope-such-that-each-half-of-the-rope
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Math
A 500-N tightrope walker stands at the center of the rope such that each half of the rope makes an angle of 10.0◦ with the horizontal. What is the tension in the rope?
1. 👍
2. 👎
3. 👁
1. T*sin(180-10) + T*sin10 = 500.
0.174T + .174T = 500
0.347T = 500
T = 1440 N.
1. 👍
2. 👎
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# Simple harmonic motion centered at the origin
I'm doing a problem on the simple harmonic motion but I didn't understand it really well.
The problem says: "A point describes a simple harmonic motion centered at the origin, the period is T=0.628s. For t = 0 point's position is x=0.15m etc..."
What it means with "harmonic motion is centered at the origin"? I thought that x=0 when t=0 but this is not right apparently.
• -1. Of course it is not right! The question states that $t=0$ when $x=0.15$. It cannot also be at $x=0$ when $t=0$. It cannot be at two places at the same time. – sammy gerbil Nov 14 '17 at 12:21
The solution is symmetric about $x=0$. Thus, starting from $$x(t)= x_0\cos(\omega t+ \phi)$$ you can derive $\omega$ and the phase $\phi$ from the data of your problem.
A solution not symmetric about the origin would be of the form $$x(t)= A+ x_0\cos(\omega t+ \phi)\, .$$ The harmonic motion would then be "centered" about $A$.
• @Peto this kind of language occurs because one could choose to have $x=0$ at the point where the sho is attached to a wall (if a spring-mass system). Then the equilibrium position would be at some position $\ne 0$. – ZeroTheHero Nov 13 '17 at 16:45
It means that the SHM, in other words the body's oscillations such that its $a \propto x$, are oscillations that are about the origin. It does not necessarily imply that at $t = 0$, $x = 0$. For example, think of a pendulum which starts with from either the left or the right side to the bob's mean position. At $t = 0$, it is starting from a position, $x \neq 0$.
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# What is 10th of an inch?
## What is 10th of an inch?
0.0001 inches
Tenths. In machining, where the thou is often treated as a basic unit, 0.0001 inches (2.54 micrometres) can be referred to as “one tenth”, meaning “one tenth of a thou” or “one ten thousandth”. (The metric comparison is discussed below.)
How do you calculate tenths in inches?
Converting Tenths of a Foot to Inches To calculate how many inches in the measurement you have, multiply the decimal by 12. For example, if the measurement is 100.2 feet, multiply 0.2 by 12 to get 2.4 inches.
### What is 1/2 inch as a decimal?
0.500
The decimal equivalents of eights, sixteenths, thirty-seconds and sixty-fourths of an inch.
Inches
fractional decimal
1/2 0.500
5/8 0.625
3/4 0.750
Read along the ruler, starting at the “zero” mark, until you reach the far edge of whatever you’re measuring. Note where that edge falls between the centimeter marks (the large, numbered marks on the tenths or metric side of any ruler).
## How much is 4 tenths in inches?
Inches to Decimals of a Foot Calculator
Inch Decimal of a Foot
2 inches 0.167
3 inches 0.250
4 inches 0.333
5 inches 0.417
What is 0.1 on a ruler?
Decimal Rulers utilizing inches normally have marks or graduations of 0.1″ (1/10″) and 0.01″ (1/100)”, some rulers may also have 0.020″ (1/50″) graduations. In the drawing below, we show a ruler with graduations or marks of 0.1″ (1/10″) and 0.01″ (1/100″).
### How do you read a tenth on a tape measure?
How do you read a tenth of an inch on a ruler?
## What is one tenths of an inch in inches?
One tenth of an inch = .1 inches. What is 42 inches to the nearest tenth? 42 inches to the nearest tenths is 42.0 inches. What is 5 tenths of an inch?
How do tenths of a foot round to a half inch?
Generally, tenths of a foot are only accurate to about a half-inch. For example, any number between 0.05 feet to 0.14 feet both round out to 0.1 feet. Did you find this page helpful? How to Convert Decimals Into Feet, Inches and Fractions…
### What is the tenth of a foot in centimeters?
Explanation: 112 of a foot = 1 inch, 1″ or 2.54 cm. 110 of an inch (2.54 mm) was fairly commonly used. The tenth of a foot is an American measurement.
How do you convert inches to fractions?
To convert 2.4 inches into the fractions used on a measuring tape, multiply the decimal by 16. For example 0.4 times 16 is 6.4, or approximately 6/16ths of an inch. Next, simply the fraction by dividing it by the common denominator.
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# How do you find abs( -3 i )?
Oct 27, 2016
$\left\mid - 3 i \right\mid = 3$
#### Explanation:
The absolute value of a number is its distance from the origin,
So if we take our complex number with cartesian coordinates,
$\left(0 , - 3\right)$ or $0 - 3 i$
to work out its distance we use Pythagoras's theorem.
${c}^{2} = {a}^{2} + {b}^{2}$
with c=abs(("a , b)")
so to find out c,
$c = \sqrt{{a}^{2} + {b}^{2}}$
$c = \sqrt{{\left(0\right)}^{2} + {\left(- 3\right)}^{2}}$
$c = \sqrt{9}$
$c = 3$
$\left\mid - 3 i \right\mid = 3$
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