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## Calculus 8th Edition
Published by Cengage
# Chapter 1 - Functions and Limits - 1.1 Four Ways to Represent a Function - 1.1 Exercises - Page 22: 59
#### Answer
$A=\frac{\sqrt{3}}{4}x^2$, $x>0$
#### Work Step by Step
We have an equilateral triangle with sides $x$ and we need to find the area. We know that: $A=1/2*base*height=\frac{1}{2}xh$ We need to eliminate $h$. We construct a right triangle in the middle of the equilateral triangle with sides $h$, $x$, and $1/2x$. We use the Pythagorean Theorem with these three sides: $(\frac{1}{2}x)^2+h^2=x^2$ $h^2=x^2-(\frac{1}{2}x)^2$ $h=\pm\sqrt{x^2-(\frac{1}{2}x)^2}$ $h=+\sqrt{\frac{3}{4}x^2}=\frac{\sqrt{3}}{2}x$ (We eliminate the negative because lengths must be positive.) We plug in $h$ in the area formula: $A=\frac{1}{2}x\frac{\sqrt{3}}{2}x=\frac{\sqrt{3}}{4}x^2$ The domain is $x>0$ because lengths must be positive.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# Financial Mathematics
1. Nov 9, 2005
### playboy
How do you find the annual effective rate of interest?
The question reads: You lend a freind $15 000 to be amortized by semiannual payments for 8 years, with interest at j2 = 9%. You deposit each payment in an account paying J12 = 7%. What annual effective rate of interest have you earned over the entire 8-year period? Ans = 8.17% Hmmm... i have absolutly no idea how to get the annuale effective rate of interest. My TA showed, (in another question) that its something like (1 + i)^n = 1 + r and solve for r? Please help somebody Thanks 2. Nov 10, 2005 ### hotvette Conceptually, it works like this. There is initial outlay of$15,000. The payments that come in annually are immediately invested. At the end of 8 years there is a total value of all investments. The 'effective' interest rate is the equivalent rate at which the initial outlay would compound at to achieve the same final result after 8 years. It might help to draw out a time line and treat each pmt and ensuing investment as a separate problem. Find out how much each is worth after the 8 years is up, sum the totals together, and then it's a straightforward back solution for a std compound interest problem.
By the way, you have 2 identical posts. If this was intentional, pls avoid that in the future.
P.S. One of the most useful classes (in terms of constantly using the material learned) I took in graduate school was called "Engineering Economy".
Last edited: Nov 10, 2005
3. Nov 10, 2005
### playboy
No, that was not intentional, i didn't know i did that :S... I will avoid that in the future!
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Module 8: ac amplifiers based on jfets (multisim)
MULTISIM
Introduction
This experiment explores the uses of JFET as AC signal amplifiers. It also describes techniques for measuring the input and output impedance of circuits.
Remember that your lab report will need to include your measurements, calculations, screenshots, etc. as indicated at the end of this outline.
Procedure
1. Common Source Amplifier
1.1 Build the common source amplifier shown in Figure 8.1
Figure 8. 1: Common source amplifier
1.2 Using the oscilloscope, measure the voltage gain of the amplifier defined as Av = Vout/Vin.
2. Measurement of Input impedance
In this section we will learn a very useful technique to measure the input impedance of any circuit. This technique is based on placing a known resistor in series with the input of the circuit and measuring the voltage drop across the new resistor. This technique can also be used in live circuits and not just
simulations.
Figure 8. 2: Circuit to measure input impedance
2.1 Build the circuit shown in Figure 8.2 You will notice that this is the same circuit used in Figure 8.2 with the extra R4 resistor added to the input.
2.2 Measure with the oscilloscope the voltage at node V1 and the voltage at node V4.
2.3 Calculate input impedance as follows:
3. Measurement of output impedance
The following technique can be used to measure the output impedance of a circuit. In practical circuits, the best value of the load resistor must be selected by trial and error.
3.1 Measure Vout as shown in the circuit from Figure 8.1. We will name this voltage Vout
3.2 Connect a load resistor of 1 kΩ at the output of the same circuit. Measure the voltage across the load. We will call this voltage Vload
3.3 Calculate output impedance as:
(in the case of using a different value for the load resistor, change the 10 kΩ value in the equation to the appropriate value of resistor used).
Laboratory Report
Create a laboratory report using Word or another word processing software that contains at least these elements: – Introduction: what is the purpose of this laboratory experiment? – Results for each section : Measured and calculated values, calculations, etc. following the outline. Include screenshots for the circuits and waveforms as necessary — You can press Alt + Print_Screen inside Multisim or if using Windows 7, you can use the “Snipping tool”. Either way, you can paste these figures into your Word processor.
– Conclusion : What area(s) you had difficulties with in the lab; what did you lean in this experiment; how it applies to your coursework and any other comments.
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# 3.5: Frequency and Alternators
In the last chapter, we learned the term cycle means from the point in a waveform to where the waveform starts to repeat itself. When we discuss the term frequency, we are referring to how many cycles can occur in one second. Frequency is measured in hertz (shout out to Heinrich Hertz) or CPS (cycles per second). Two factors affect the frequency in an alternator: rotation speed and the number of poles.
Figure 52. Sine wave cycle
## Rotation speed
As the armature rotates through the field, it starts to create a waveform (as we saw in the last chapter). One full mechanical rotation of the armature creates one full sine wave on a two-pole alternator. If the two-pole alternator spins three complete revolutions in one second, it will create three full sine waves in that one second. We would say that the frequency is at three cycles per second or three hertz (as the cool kids say).
A machine’s rotational speed is measured in rotations per minute or RPM. However, we are not concerned with minutes, but rather, with seconds when dealing with frequency. Therefore, RPM must be converted to rotations per second (RPS). As there are 60 seconds in a minute, all we have to do is to divide the RPM by 60 to convert it to RPS.
For example, if the armature is spinning at a rate of 1800 RPM on a two-pole alternator, we can say that it is spinning at 30 rotations per second. If this alternator has two poles, then in one second it will generate 30 cycles of voltage. It then could be said to have a frequency of 30 cycles per second or 30 Hertz. The frequency of an alternator is directly proportional to the rotational speed of the alternator.
## Number of Poles
If we add poles to the alternator, we can change the frequency. In a two-pole alternator, Side A of the armature (Figure 53) passes from north to south, and then south to north, to create one complete sine wave. I f we add two more poles, as in Figure 54, then Side A of the armature will move past two north poles and two south poles in one full mechanical revolution.
Figure 53. Two pole alternator
Two full sine waves are created in one complete mechanical revolution. If a two-pole alternator creates one cycle of voltage in one second (or one hertz of frequency), a four pole alternator will create two cycles of voltage in one second (or two hertz).
The frequency of an alternator is directly proportional to the number of poles in the alternator.
Figure 54. Four pole alternator
## Formula time!
Knowing that rotation speed is directly proportional to frequency and that the number of poles is directly proportional to frequency, we can use a formula. The formula looks like this:
$f= \dfrac{P}{2} \times \dfrac{N}{60} \tag{Frequency formula}$
where…
• $$f$$ = frequency in hertz
• $$P$$ = number of poles
• $$N$$ = rotational speed in RPM
We divide the number of poles by two because there will always be a set of two poles. You can’t have a north pole without a south. We divide the RPM by 60 because we are concerned with rotations per second, not rotations per minute. The formula in Figure 56 can be combined to look like this:
$f = \dfrac{PN}{120} \tag{Combined frequency formula}$
Video! This video will walk you through how frequency is related to the RPM and the number of poles of an alternator.
A YouTube element has been excluded from this version of the text. You can view it online here: https://pressbooks.bccampus.ca/trigf...ricians/?p=278
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# Inequalities math homework problem
1. Oct 9, 2007
### xCanx
A rectangular solid is to be constructed with a special kind of wire along all
the edges. The length of the base is to be twice the width of the base. The
height of the rectangular solid is such that the total amount of wire used (for
the whole figure) is 40 cm. Find the range of possible values for the width of
the base so that the volume of the figure will lie between 2 cm3 and 4 cm3.
Can someone show me how to start off?
2. Oct 9, 2007
### symbolipoint
w=width, L=length, h=height;
w+L+h=40, L=2w
wLh>2 and wLh<4
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## 17Calculus Integrals - Volume of Revolution Using The Washer-Disc Method
##### 17Calculus
This page covers single volume integrals when an area is rotated about a vertical or horizontal line. The area is defined by equations in the form $$y=f(x)$$ or $$x=f(y)$$ and we use the washer (disc) method. For other ways to calculate volume, see the links in the related topics panel.
If you want a full video lecture on this topic, we recommend this video and this instructor.
### Prof Leonard - Volume of Solids By Disks and Washers Method [2hr-47mins-48secs]
video by Prof Leonard
Alternate Names Disc Method Disk Method Ring Method
We choose to use the term washer-disc method to refer to this technique. We think it covers the two most commonly used and most descriptive names. This, of course, is a personal preference for this site and you need to check with your instructor to see what they require.
What Is A Washer?
If you are not familiar with a washer (other than to wash clothes), this wiki page has pictures and explains what a washer is. In short, it is a disc with a circular hole in it whose center is the same as the full disc.
Overview
When calculating the volume of rotation, there are 3 factors that determine how to set up the integral.
1. method (washer-disc or cylinder-shell)
2. axis of rotation
3. function (graph and form of the equations)
On this page, we discuss the washer-disc method where the axis of rotation will always be either an axis or a straight line that is parallel to one of the axes. However, before we discuss the rotation of an area, we need to know how to describe an area in the plane. This is a critical step to setting up your integral correctly. If you didn't completely understand this from the main volume integrals page, you can go over it again here.
### Describing A Region In The xy-Plane
To describe an area in the xy-plane, the first step is to plot the boundaries and determine the actual region that needs to be described. There are several graphing utilities listed on the tools page. Our preference is to use the free program winplot (used to plot these graphs; we used gimp to add labels and other graphics). However, graphing by hand is usually the best and quickest way.
We use the graph to the right to facilitate this discussion. A common way to describe this area is the area bounded by $$f(x)$$ (red line), $$g(x)$$ (blue line) and $$x=a$$ (black line).
[Remember that an equation like $$x=a$$ can be interpreted two ways, either the point x whose value is a or the vertical line. You should be able to tell what is meant by the context.]
Okay, so we plotted the boundaries and shaded the area to be described. Now, we need to choose a direction to start, either vertically or horizontally. We will show both ways, starting with vertically, since it is more natural and what you are probably used to seeing. Also, this area is easier to describe vertically than horizontally (you will see why as you read on).
Vertically
Our first step is to draw a vertical arrow on the graph somewhere within the shaded area, like we have done here. Some books draw an example rectangle with the top on the upper graph and the bottom on the lower graph. That is the same idea as we have done with the arrow.
Now we need to think of this arrow as starting at the left boundary and sweeping across to the right boundary of the area. This sweeping action is important since it will sweep out the area. As we think about this sweeping, we need to think about where the arrow enters and leaves the shaded area. Let's look our example graph to demonstrate. Think about the arrow sweeping left to right. Notice that it always enters the area by crossing $$g(x)$$, no matter where we draw it. Similarly, the arrow always exits the area by crossing $$f(x)$$, no matter where we draw it. Do you see that?
But wait, how far to the right does it go? We are not given that information. What we need to do is find the x-value where the functions $$f(x)$$ and $$g(x)$$ intersect. You should be able to do that. We will call that point $$(b,f(b))$$. Also, we will call the left boundary $$x=a$$. So now we have everything we need to describe this area. We give the final results below.
Vertical Arrow
$$g(x) \leq y \leq f(x)$$
arrow leaves through $$f(x)$$ and enters through $$g(x)$$
$$a \leq x \leq b$$
arrow sweeps from left ($$x=a$$) to right ($$x=b$$)
Horizontally
We can also describe this area horizontally (or using a horizontal arrow). We will assume that we can write the equations of $$f(x)$$ and $$g(x)$$ in terms of $$y$$. ( This is not always possible, in which case we cannot describe the area in this way. ) For the sake of this discussion, we will call the corresponding equations $$f(x) \to F(y)$$ and $$g(x) \to G(y)$$.
Let's look at the graph. Notice we have drawn a horizontal arrow. Just like we did with the vertical arrow, we need to determine where the arrow enters and leaves the shaded area. In this case, the arrow sweeps from the bottom up. As it sweeps, we can see that it always crosses the vertical line $$x=a$$. However, there is something strange going on at the point $$(b,f(b))$$. Notice that when the arrow is below $$f(b)$$, the arrow exits through $$g(x)$$ but when the arrow is above $$f(b)$$, the arrow exits through $$f(x)$$. This is a problem. To overcome this, we need to break the area into two parts at $$f(b)$$.
Lower Section - - This section is described by the arrow leaving through $$g(x)$$. So the arrow sweeps from $$g(a)$$ to $$g(b)$$.
Upper Section - - This section is described by the arrow leaving through $$f(x)$$. The arrow sweeps from $$f(b)$$ to $$f(a)$$.
The total area is the combination of these two areas. The results are summarized below.
Horizontal Arrow
lower section
$$a \leq x \leq G(y)$$
arrow leaves through $$G(y)$$ and enters through $$x=a$$
$$g(a) \leq y \leq g(b)$$
arrow sweeps from bottom ($$y=g(a)$$) to top ($$y=g(b)$$)
upper section
$$a \leq x \leq F(y)$$
arrow leaves through $$F(y)$$ and enters through $$x=a$$
$$f(b) \leq y \leq f(a)$$
arrow sweeps from bottom ($$y=f(b)$$) to top ($$y=f(a)$$)
Type 1 and Type 2 Regions
Some instructors may describe regions in the plane as either Type 1 or Type 2 (you may see II instead of 2). As you know from the above discussion, some regions are better described vertically or horizontally. Type 1 regions are regions that are better described vertically, while Type 2 regions are better described horizontally. The example above was a Type 1 region.
Here is a quick video clip going into more detail on Type 1 and Type 2 regions.
### Krista King Math - type I and type 2 regions [1min-39secs]
video by Krista King Math
washer-disc method
x-axis rotation
y-axis rotation
Now we will discuss each of these plots separately and explain each part of the plots.
Getting Started
Here are some key things that you need to do and know to get started.
1. Draw a rough plot of the area that is being rotated. This is usually best done by hand since you will need to label it.
2. Decide what method you will use, washer-disc or cylinder-shell.
3. On the rough plot from point 1, label the axis of rotation and draw a representative rectangle somewhere in the area.
4. Label R and r.
Once those steps are done, you are ready to set up your integral.
The volume integral using the washer-disc method is based on the volume of a washer or disc. Let's think a bit about the volume of a washer-disc. If we start with a full disc (no hole in the middle), the volume is the surface area times the thickness. Since the disc is a circle, the area of a circle is $$\pi R^2$$ where $$R$$ is the radius of the circle. The volume is $$\pi R^2 t$$ where $$t$$ is the thickness. We choose to use a capital R here as the radius of the disc.
Now, with a washer, we take the disc we just discussed and put a circular hole in it with it's center the same as the full disc. (Think of a CD or DVD disc.) Now the volume is reduced by what we have taken out of the center. This empty space has volume $$\pi r^2 t$$, where $$r$$ is the radius of the small hole. The thickness, $$t$$, is the same as the full disc.
So now we have what we need to put together an equation for the washer-disc with a hole in the middle. We take the volume of the full disc and subtract the volume of the hole to get $$V = \pi R^2 t - \pi r^2 t = \pi t(R^2-r^2)$$. [Notice the special case when there is no hole in the middle, can be thought of as $$r=0$$ giving the volume of the disk as just $$V=\pi R^2 t$$.]
summary of the washer-disc method
the representative rectangle is perpendicular to axis of revolution
$$R$$ is the distance from the axis of rotation to the far end of the representative rectangle
$$r$$ is the distance from the axis of rotation to the closest end of the representative rectangle
x-axis rotation equation
$$\displaystyle{ V = \pi \int_{a}^{b}{R^2-r^2~dx} }$$
y-axis rotation equation
$$\displaystyle{ V = \pi \int_{c}^{d}{R^2-r^2~dy} }$$
Note - Notice that $$R$$ and $$r$$ are distances, so they are always positive (although since we square them, the sign doesn't make any difference in the equations).
washer-disc method with x-axis rotation
If you feel like you need further explanation of this, here is a video that tries to explain this method by drawing in three dimensions. In this video, notice that the axis of rotation runs along one side of the figure and, consequently, $$r=0$$.
### Khan Academy - Disc Method [10min-4secs]
Okay, now let's work some problems using the washer-disc method revolving an area about the x-axis.
x-axis rotation practice
area: $$y=1/x, x=1, x=3$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, x=1, x=3$$ revolved about the x-axis. Give your answer in exact terms.
$$2\pi/3$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, x=1, x=3$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3487 video solution
$$2\pi/3$$
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area: $$y=x^2, y=4x-x^2$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=4x-x^2$$ revolved about the x-axis. Give your answer in exact terms.
$$V = 32\pi/3$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=4x-x^2$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3494 video solution
$$V = 32\pi/3$$
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area: $$y=\sqrt{x-1}, y=0, x=5$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x-1}, y=0, x=5$$ revolved about the x-axis. Give your answer in exact terms.
$$V=8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x-1}, y=0, x=5$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Krista King Math - 324 video solution
video by Krista King Math
$$V=8\pi$$
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area: $$y=x^2, x=y^2$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about the x-axis. Give your answer in exact terms.
$$\displaystyle{V=\frac{3\pi}{10}}$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about the x-axis. Give your answer in exact terms.
Solution
This problem is solved by two different instructors.
### Krista King Math - 877 video solution
video by Krista King Math
$$\displaystyle{V=\frac{3\pi}{10}}$$
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area: $$y=x^3, y=x, x \geq 0$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x \geq 0$$ revolved about the x-axis. Give your answer in exact terms.
$$V=4\pi/21$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x \geq 0$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### PatrickJMT - 1174 video solution
video by PatrickJMT
$$V=4\pi/21$$
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area: $$y=\sqrt{x}, y \geq 0, x=4$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y \geq 0, x=4$$ revolved about the x-axis. Give your answer in exact terms.
$$8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y \geq 0, x=4$$ revolved about the x-axis. Give your answer in exact terms.
Solution
This problem is solved by two different instructors.
In the second video, he doesn't finish the integration, so here are the details.
$$\displaystyle{ \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi [(4^2)/2 - (0^2)/2] = 8\pi}$$
### PatrickJMT - 1359 video solution
video by PatrickJMT
$$8\pi$$
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area: $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ revolved about the x-axis. Give your answer in exact terms.
$$V = 9\pi^2$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=2-\sin(x)$$, $$x=0$$, $$x=2\pi$$, $$y=0$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Dr Chris Tisdell - 2003 video solution
video by Dr Chris Tisdell
$$V = 9\pi^2$$
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area: $$y^2=x-2, x=5$$ in the first quadrant axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2=x-2, x=5$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
$$4.5\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2=x-2, x=5$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2272 video solution
video by Michel vanBiezen
$$4.5\pi$$
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area: $$y=x^2, y=x$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the x-axis. Give your answer in exact terms.
$$2\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2273 video solution
video by Michel vanBiezen
$$2\pi/15$$
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area: $$y=\sqrt{9-x^2}$$ in the first quadrant axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{9-x^2}$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
$$V=18\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{9-x^2}$$ in the first quadrant revolved about the x-axis. Give your answer in exact terms.
Solution
### MIP4U - 2276 video solution
video by MIP4U
$$V=18\pi$$
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area: $$y=\sqrt{x}, y=x^2$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=x^2$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Khan Academy - 1184 video solution
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area: $$y=\sqrt{x}, y\geq0, x=1$$ axis of rotation: x-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y\geq0, x=1$$ revolved about the x-axis. Give your answer in exact terms.
Solution
### Khan Academy - 1182 video solution
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Derive the equation for the volume of a sphere of radius r using the washer-disc method.
Problem Statement
Derive the equation for the volume of a sphere of radius r using the washer-disc method.
Solution
### Khan Academy - 1183 video solution
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Looking back at the plots, you will notice that the axis of revolution is always a coordinate axis, either the x-axis or the y-axis. A twist you will see is when the axis of revolution is another line. On this site, we will discuss only axes that are parallel to one of the coordinate axes.
In this case, the equations that will change are the ones that describe the distance from the axes of rotation. We suggest that you set up a sum from the parallel coordinate axis to the axis of rotation and then solve for whatever variable you need. This concept, especially, requires you to think over in your mind several times and look at examples.
parallel to x-axis rotation practice
area: $$y=x^2, x=0, y=4$$ axis of rotation: $$y=4$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the $$y=4$$. Give your answer in exact terms.
$$V = 256\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the $$y=4$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3490 video solution
$$V = 256\pi/15$$
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area: $$y=2-x^2, y=1$$ axis of rotation: $$y=1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2-x^2, y=1$$ revolved about the $$y=1$$. Give your answer in exact terms.
$$V = 16\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2-x^2, y=1$$ revolved about the $$y=1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3491 video solution
$$V = 16\pi/15$$
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area: $$y=x, y=x^2$$ axis of rotation: $$y=2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x, y=x^2$$ revolved about $$y=2$$. Give your answer in exact terms.
$$V = 8\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x, y=x^2$$ revolved about $$y=2$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3497 video solution
$$V = 8\pi/15$$
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area: $$y=1/x, y=0, x=1, x=3$$ axis of rotation: $$y=-1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, y=0, x=1, x=3$$ revolved about $$y=-1$$. Give your answer in exact terms.
$$V = 2\pi/3 + 2\pi\ln(3) = 2\pi(\ln(3)+1/3)$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1/x, y=0, x=1, x=3$$ revolved about $$y=-1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3498 video solution
$$V = 2\pi/3 + 2\pi\ln(3) = 2\pi(\ln(3)+1/3)$$
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area: $$f(x)=x, g(x)=x^2-3x$$ axis of rotation: $$y=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=x, g(x)=x^2-3x$$ revolved about $$y=5$$. Give your answer in exact terms.
Hint
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=x, g(x)=x^2-3x$$ revolved about $$y=5$$. Give your answer in exact terms.
$$\displaystyle{ V = \frac{1472\pi}{15} }$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$f(x)=x, g(x)=x^2-3x$$ revolved about $$y=5$$. Give your answer in exact terms.
Hint
Solution
1. Draw the plot and the example rectangle and label r and R. See the hint or the animation. We drew the example rectangle perpendicular to the axis of revolution since we were told to use the washer-disc method. The distance r is the distance from the axis of revolution to closest end of the rectangle. The distance R is from the axis of revolution to the farthest end of the example rectangle.
2. Choose The Integral - The volume integral we need is $$V = \pi\int_a^b{R^2-r^2~dx}$$. We chose this integral because we are told to use the washer-disc method, so we need an integral with r and R. We integrate with respect to x since the example rectangle is vertical and consequently it moves horizontally, sweeping in the x-direction.
3. Determine r and R - From the plot, let's start on the axis of revolution and move down to the x-axis of revolution. This distance is 5 units. Moving back up to the end of the rectangle that lands on $$y=x$$, we move y units. What we are left with is r, so $$r=5-y$$. However, we need to replace y with expression for y in terms of x. Since $$y=x$$ is the line that we are working with, we have $$r=5-x$$.
Determining the expression for R is a bit trickier. Starting on the axis of revolution, we move down to the x-axis which is 5 units. However, when $$x < 3$$ we need to go a little further in the same direction to get the full distance R. Let's put that aside for a minute and think about the part of the graph for $$x > 3$$. In this case, we need to go back up y units, so $$R=5-y$$. This looks the same as r but in this case, we are landing on the plot $$y=x^2-3x$$. So $$R=5-(x^2-3x) = -x^2+3x+5$$. Let's plug in a few values and compare the values to graph to see if they match.
$$x=3$$ $$R=-3^2+3(3)+5=5$$ 𞀄 $$x=4$$ $$R=-4^2+3(4)+5=1$$ 𞀄
So far, so good. Let's plug in a few values $$x < 3$$ and see what we get.
$$x=0$$ $$R=-0^2+3(0)+5=5$$ 𞀄 $$x=2$$ $$R=-2^2+3(2)+5=7$$ 𞀄
So it looks like we have the correct equation for R. We can do the same with r to check our equation. This does not guarantee that we have the right equations but it may give an indication if they are incorrect. I usually check both endpoints and at least one other point, two other points is even better.
4. Set up and evaluate the integral - If we look at the animation above, we can tell that the rectangle sweeps across the area from $$x=0$$ to $$x=4$$. So our integral is $$\displaystyle{ V = \pi \int_0^4{ (-x^2+3x+5)^2 - } }$$ $$\displaystyle{ (5-x)^2 ~ dx }$$. Let's evaluate it.
$$\displaystyle{ V = \pi\int_0^4{ (-x^2+3x+5)^2 - (5-x)^2 ~ dx } }$$ $$\displaystyle{ V = \pi\int_0^4{ (x^4-3x^3-5x^2-3x^3+ } }$$ $$9x^2+15x-5x^2+15x+25) -$$ $$(25-10x+x^2) ~dx$$ $$\displaystyle{ V = \pi\int_0^4{ x^4-6x^3-2x^2+40x ~dx } }$$ $$\displaystyle{ V = \pi\left[ \frac{x^5}{5} - \frac{6x^4}{4} -\frac{2x^3}{3}+\frac{40x^2}{2} \right]_0^4 }$$ $$\displaystyle{ V = \pi\left[ \frac{4^5}{5} - \frac{6(4)^4}{4} -\frac{2(4)^3}{3}+\frac{40(4)^2}{2} \right] - 0 }$$ $$\displaystyle{ V = \pi\left[ \frac{1024}{5} - \frac{1536}{4} -\frac{128}{3}+\frac{640}{2} \right] }$$ $$\displaystyle{ V = \pi \frac{1472}{15} }$$
$$\displaystyle{ V = \frac{1472\pi}{15} }$$
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area: $$y=x^2, x=y^2$$ axis of rotation: $$y=1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$y=1$$. Give your answer in exact terms.
$$\displaystyle{V=11\pi/30}$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$y=1$$. Give your answer in exact terms.
Solution
### Krista King Math - 1172 video solution
video by Krista King Math
$$\displaystyle{V=11\pi/30}$$
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area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=-2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=-2$$. Give your answer in exact terms.
$$V=25\pi/21$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=-2$$. Give your answer in exact terms.
Solution
In the video, he set up the integral but did not evaluate it. His integral was $$\displaystyle{ \int_{0}^{1}{\pi[(2+x)^2 - \pi(2+x^3)^2]dx} }$$. This evaluates to $$\displaystyle{ \pi \left[ x^2 + \frac{x^3}{3} - x^4 - \frac{x^7}{7} \right]_{0}^{1} }$$
### PatrickJMT - 1175 video solution
video by PatrickJMT
$$V=25\pi/21$$
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area: $$y=x^3, y=x, x\geq0$$ axis of rotation: $$y=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=5$$. Give your answer in exact terms.
$$V=97\pi/42$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=x, x\geq0$$ revolved about $$y=5$$. Give your answer in exact terms.
Solution
### PatrickJMT - 1176 video solution
video by PatrickJMT
$$V=97\pi/42$$
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area: $$y=x^2, y=x$$ axis of rotation: $$y=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about $$y=5$$. Give your answer in exact terms.
$$23\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about $$y=5$$. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2275 video solution
video by Michel vanBiezen
$$23\pi/15$$
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Now, let's work some problems with the y-axis as the axis of rotation. Here is the plot that contains all the information you need to work these problems.
washer-disc method with y-axis rotation
y-axis rotation practice
area: $$y=x^2, x=0, y=4$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the y-axis. Give your answer in exact terms.
$$V = 8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=0, y=4$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3488 video solution
$$V = 8\pi$$
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area: $$y=x^{2/3}, x=0, y=1$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^{2/3}, x=0, y=1$$ revolved about the y-axis. Give your answer in exact terms.
$$V = \pi/4$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^{2/3}, x=0, y=1$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3489 video solution
$$V = \pi/4$$
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area: $$y^2 = x, x=2y$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2 = x, x=2y$$ revolved about the y-axis. Give your answer in exact terms.
$$V = 64\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y^2 = x, x=2y$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3496 video solution
$$V = 64\pi/15$$
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area: $$y=\sqrt{x}, y=0, x=3$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the y-axis. Give your answer in exact terms.
$$V = 36\pi\sqrt{3}/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3495 video solution
$$V = 36\pi\sqrt{3}/5$$
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area: $$x=2\sqrt{y}, x=0, y=9$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=2\sqrt{y}, x=0, y=9$$ revolved about the y-axis. Give your answer in exact terms.
$$V=162\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=2\sqrt{y}, x=0, y=9$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Krista King Math - 343 video solution
video by Krista King Math
$$V=162\pi$$
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area: $$y=1-x^2, y=0$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1-x^2, y=0$$ revolved about the y-axis. Give your answer in exact terms.
$$\displaystyle{V=\frac{\pi}{2}}$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=1-x^2, y=0$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Krista King Math - 878 video solution
video by Krista King Math
$$\displaystyle{V=\frac{\pi}{2}}$$
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area: $$y=x^2, y=5, x=0$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=5, x=0$$ revolved about the y-axis. Give your answer in exact terms.
$$25\pi/2$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=5, x=0$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2271 video solution
video by Michel vanBiezen
$$25\pi/2$$
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area: $$y=x^2, y=x$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the y-axis. Give your answer in exact terms.
$$\pi/6$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=x$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2274 video solution
video by Michel vanBiezen
$$\pi/6$$
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area: $$y=\sqrt{x}$$ on $$[0,4]$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Unless otherwise instructed, use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}$$ on $$[0,4]$$ revolved about the y-axis. Give your answer in exact terms.
$$V=32\pi/5$$
Problem Statement
Unless otherwise instructed, use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}$$ on $$[0,4]$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### MIP4U - 2277 video solution
video by MIP4U
$$V=32\pi/5$$
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area: $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ revolved about the y-axis. Give your answer in exact terms.
$$V=5\pi/24$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1, y \geq 1/2, x \geq 0$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2280 video solution
video by Michel vanBiezen
$$V=5\pi/24$$
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area: $$y=-3x+6$$, x-axis, y-axis axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=-3x+6$$, the x-axis and the y-axis revolved about the y-axis. Give your answer in exact terms.
$$V=8\pi$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=-3x+6$$, the x-axis and the y-axis revolved about the y-axis. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2281 video solution
video by Michel vanBiezen
$$V=8\pi$$
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area: $$y=x^2, y=0, x=1$$ axis of rotation: y-axis method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, y=0, x=1$$ revolved about the y-axis. Give your answer in exact terms.
Solution
### Khan Academy - 1185 video solution
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parallel to y-axis rotation practice
area: $$y=\sqrt{x}, y=0, x=3$$ axis of rotation: $$x=3$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the $$x=3$$. Give your answer in exact terms.
$$V = 24\pi\sqrt{3}/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about the $$x=3$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3492 video solution
$$V = 24\pi\sqrt{3}/5$$
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area: $$x=y^2, x=1$$ axis of rotation: $$x=1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=y^2, x=1$$ revolved about $$x=1$$. Give your answer in exact terms.
$$V = 16\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x=y^2, x=1$$ revolved about $$x=1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3493 video solution
$$V = 16\pi/15$$
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area: $$y=\sqrt{x}, y=0, x=3$$ axis of rotation: $$x=6$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about $$x=6$$. Give your answer in exact terms.
$$V = 84\pi\sqrt{3}/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=\sqrt{x}, y=0, x=3$$ revolved about $$x=6$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3499 video solution
$$V = 84\pi\sqrt{3}/5$$
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area: $$y=x^2, x=y^2$$ axis of rotation: $$x=-1$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$x=-1$$. Give your answer in exact terms.
$$V = 29\pi\/30$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^2, x=y^2$$ revolved about $$x=-1$$. Give your answer in exact terms.
Solution
### The Organic Chemistry Tutor - 3500 video solution
$$V = 29\pi\/30$$
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area: $$y=x^3, y=0, x=1$$ axis of rotation: $$x=2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=0, x=1$$ revolved about $$x=2$$. Give your answer in exact terms.
$$V = 3\pi/5$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=x^3, y=0, x=1$$ revolved about $$x=2$$. Give your answer in exact terms.
Solution
### Krista King Math - 1173 video solution
video by Krista King Math
$$V = 3\pi/5$$
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area: $$y=2\sqrt{x}, y=0, x=4$$ axis of rotation: $$x=5$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2\sqrt{x}, y=0, x=4$$ revolved about $$x=5$$. Give your answer in exact terms.
$$V=832\pi/15$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$y=2\sqrt{x}, y=0, x=4$$ revolved about $$x=5$$. Give your answer in exact terms.
Solution
### MIP4U - 1916 video solution
video by MIP4U
$$V=832\pi/15$$
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area: $$x^2+y^2=1$$, x-axis, y-axis axis of rotation: $$x=2$$ method: washer-disc
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1$$, the x-axis and the y-axis, revolved about $$x=2$$. Give your answer in exact terms.
$$V=(\pi/12)[3\pi-4]$$
Problem Statement
Use the washer-disc method to calculate the volume of rotation of the area bounded by $$x^2+y^2=1$$, the x-axis and the y-axis, revolved about $$x=2$$. Give your answer in exact terms.
Solution
### Michel vanBiezen - 2279 video solution
video by Michel vanBiezen
$$V=(\pi/12)[3\pi-4]$$
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Resistors and Ohm's Law Video Lessons
Concept
# Problem: a) Explain the distinction between an ohmic and non-ohmic material, in terms of how the current and resistance behave as the voltage difference across the material is changed.b) Now, imagine a single-loop circuit with a battery, two wires, and a 10 Ohm resistor. The wires are also ohmic, but with a resistance much smaller than the 10 Ohm resistor. Despite the disparity in resistance, the current in the wire is the same as that through the resistor since they are in series. Using Ohm's Law, explain how this uniformity in current relates to (or arises from) the individual potential differences across the wire and resistor.
###### FREE Expert Solution
a)
Ohmic materials are materials that obey Ohm's law: show linear relationship between the current and the voltage, and their resistances do not change with the variation in voltage.
83% (26 ratings)
###### Problem Details
a) Explain the distinction between an ohmic and non-ohmic material, in terms of how the current and resistance behave as the voltage difference across the material is changed.
b) Now, imagine a single-loop circuit with a battery, two wires, and a 10 Ohm resistor. The wires are also ohmic, but with a resistance much smaller than the 10 Ohm resistor. Despite the disparity in resistance, the current in the wire is the same as that through the resistor since they are in series. Using Ohm's Law, explain how this uniformity in current relates to (or arises from) the individual potential differences across the wire and resistor.
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# Quick Algorithm: Get Ideal Size (Square like) For a Board Game Having an Arbitrary (but Even) Number of Fields
Say you are developing a game like Chess, Go, Checkers, Tic-Tac-Toe or Memory. In each of those games the game board is a rectangle looking playfield of different size (rows x columns). Tic-Tac-Toe is 3×3, Checkers is 8×8, while Go can be 19×19 or 13×13 and similar.
In a game with an arbitrary number of game fields you might want to have the board look as closely to square as possible (rectangle where height and width are the same). Think of Memory. Let’s say we have 24 cards, that is 12 pairs. If you want to place them in a rectangle looking grid, most similar to square, you would go for 4 x 6 (or 6 x 4) board size (as it would look more square like than 3 x 8 and 2 x 12 or 1 x 24 would be too wide).
Therefore, the question: having an arbitrary number of game field pairs, what is the ideal, most square looking, grid size?
And the answer is an algorithm (some math knowledge required) like this one:
TGridSize = record
Rows, Columns : integer;
end;
function CalcGridSize(const numberOfPairs : Cardinal) : TGridSize;
//look for ideal rectangle dimensions (square is ideal)
//to present fields, number of fields = 2 * numberOfPairs
var
fieldCount : integer;
i : integer;
begin
fieldCount := 2 * numberOfPairs;
result.Rows := 1;
result.Columns := fieldCount;
if Sqrt(fieldCount) = Trunc(Sqrt(fieldCount)) then
begin
result.Rows := Trunc(Sqrt(fieldCount));
result.Columns := Trunc(Sqrt(fieldCount));
Exit;
end;
for i := Trunc(Sqrt(fieldCount)) downto 2 do
if (fieldCount mod i) = 0 then
begin
result.Rows := i;
result.Columns := fieldCount div i;
Exit;
end;
end; (*CalcGridSize*)
And here are some results:
var
gridSize : TGridSize;
i : integer;
begin
for i := 1 to 20 do
begin
gridSize := CalcGridSize(i);
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# 6011 (number)
6011 is an odd four-digits prime number following 6010 and preceding 6012. In scientific notation, it is written as 6.011 × 103. The sum of its digits is 8. It has a total of one prime factor and 2 positive divisors. There are 6,010 positive integers (up to 6011) that are relatively prime to 6011.
## Basic properties
• Is Prime? yes
• Number parity odd
• Number length 4
• Sum of Digits 8
• Digital Root 8
## Name
Name six thousand eleven
## Notation
Scientific notation 6.011 × 103 6.011 × 103
## Prime Factorization of 6011
Prime Factorization 6011
Prime number
Distinct Factors Total Factors Radical ω 1 Total number of distinct prime factors Ω 1 Total number of prime factors rad 6011 Product of the distinct prime numbers λ -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ -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 8.70135 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 6011 is 6011. Since it has only one prime factor, 6011 is a prime number.
## Divisors of 6011
2 divisors
Even divisors 0 2 1 1
Total Divisors Sum of Divisors Aliquot Sum τ 2 Total number of the positive divisors of n σ 6012 Sum of all the positive divisors of n s 1 Sum of the proper positive divisors of n A 3006 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G 77.5306 Returns the nth root of the product of n divisors H 1.99967 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 6011 can be divided by 2 positive divisors (out of which none is even, and 2 are odd). The sum of these divisors (counting 6011) is 6012, the average is 3006.
## Other Arithmetic Functions (n = 6011)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ 6010 Total number of positive integers not greater than n that are coprime to n λ 6010 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π ≈ 789 Total number of primes less than or equal to n r2 0 The number of ways n can be represented as the sum of 2 squares
There are 6,010 positive integers (less than 6011) that are coprime with 6011. And there are approximately 789 prime numbers less than or equal to 6011.
## Divisibility of 6011
m n mod m
2 1
3 2
4 3
5 1
6 5
7 5
8 3
9 8
6011 is not divisible by any number less than or equal to 9.
## Classification of 6011
• Arithmetic
• Prime
• Deficient
### Expressible via specific sums
• Polite
• Non hypotenuse
• Prime Power
• Square Free
## Base conversion 6011
Base System Value
2 Binary 1011101111011
3 Ternary 22020122
4 Quaternary 1131323
5 Quinary 143021
6 Senary 43455
8 Octal 13573
10 Decimal 6011
12 Duodecimal 358b
16 Hexadecimal 177b
20 Vigesimal f0b
36 Base36 4mz
## Basic calculations (n = 6011)
### Multiplication
n×y
n×2 12022 18033 24044 30055
### Division
n÷y
n÷2 3005.5 2003.67 1502.75 1202.2
### Exponentiation
ny
n2 36132121 217190179331 1305530167958641 7847541839599391051
### Nth Root
y√n
2√n 77.5306 18.1823 8.80515 5.69888
## 6011 as geometric shapes
### Circle
Radius = n
Diameter 12022 37768.2 1.13512e+08
### Sphere
Radius = n
Volume 9.09764e+11 4.5405e+08 37768.2
### Square
Length = n
Perimeter 24044 3.61321e+07 8500.84
### Cube
Length = n
Surface area 2.16793e+08 2.1719e+11 10411.4
### Equilateral Triangle
Length = n
Perimeter 18033 1.56457e+07 5205.68
### Triangular Pyramid
Length = n
Surface area 6.25827e+07 2.55961e+10 4907.96
## Cryptographic Hash Functions
md5 e3b80d30a727c738f3cff0941f6bc55a 83a08a7fdc4059c7ac2e647e546add90679b9ac4 c37fd5582393747ef03b83ad095a5650d2f5335acc65eaa7db54c4b2a21d1092 3dc2b7303e1342405a887f123ed4e5615ff7f0a3a805f9e157d23fb8fedee7bd4eb3302d39c67b8b66304f3309bb8e0720dcac9649bfa295eb8cc8e7a4451918 ac22fadd6e29870db1b5de10b3d7f60f5317d167
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Skip to content
# Plot Multiple Data Sets on the Same Chart in Excel
• Last Updated : 29 Jun, 2021
Sometimes while dealing with hierarchical data we need to combine two or more various chart types into a single chart for better visualization and analysis. This type of chart having multiple data sets is known as “Combination charts”.
In this article, we are going to see how to make combination charts from a set of two different charts in Excel using the example shown below.
Example: Consider a famous coaching institute that deals with both free content in their YouTube channel and also have their own paid online courses. There are broadly two categories of students in this institute :
1. The students who enrolled in the coaching but are learning from YouTube free video content.
2. The students who enrolled as well as bought paid online courses.
So, the institute asked their Sales Department to make a statistical chart about how many paid courses from a pool of courses which the institute deals with were sold from the year 2014 to the last year 2020 and also show the percentage of students who have enrolled in these paid courses.
### Table :
Here, the first data is “Number of Paid courses sold” and the second one is “Percentage of Students enrolled”. Now our aim is to plot these two data in the same chart with different y-axis.
### Implementation :
Follow the below steps to implement the same:
Step 1: Insert the data in the cells. After insertion, select the rows and columns by dragging the cursor.
Step 2: Now click on Insert Tab from the top of the Excel window and then select Insert Line or Area Chart. From the pop-down menu select the first “2-D Line”.
From the above chart we can observe that the second data line is almost invisible because of scaling. The present y-axis line is having much higher values and the percentage line will be having values lesser than 1 i.e. in decimal values. Hence, we need a secondary axis in order to plot the two lines in the same chart. In Excel, it is also known as clustering of two charts.
The steps to add a secondary axis are as follows :
1. Open the Chart Type dialog box
`Select the Chart -> Design -> Change Chart Type`
Another way is :
`Select the Chart -> Right Click on it -> Change Chart Type`
2. The Chart Type dialog box opens. Now go to the “Combo” option and check the “Secondary Axis” box for the “Percentage of Students Enrolled” column. This will add the secondary axis in the original chart and will separate the two charts. This will result in better visualization for analysis purposes.
The combination chart with two data sets is now ready. The secondary axis is for the “Percentage of Students Enrolled” column in the data set as discussed above.
Now various formatting can be carried out in this secondary axis using the Format Axis window on the right corner of Excel.
`Select the secondary Axis -> Right Click -> Format Axis -> Format Axis Dialog Box`
Changing the Bounds of Secondary Axis
You can further format the above chart by making it more interactive by changing the “Chart Styles”, adding suitable “Axis Titles”, “Chart Title”, “Data Labels”, changing the “Chart Type” etc. It can be done using the “+” button in the top right corner of the Excel chart.
Finally, after all the modification, the chart with multiple data sets looks like :
We can infer from the above chart that in the year 2019, the percentage of students who enrolled in the online paid courses are relatively less but in 2020 more students have enrolled in paid courses than free content on YouTube.
My Personal Notes arrow_drop_up
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### Home > CCA2 > Chapter 4 > Lesson 4.2.3 > Problem4-95
4-95.
Solve the equations below.
1. $\sqrt { x + 15 } = 5 + \sqrt { x }$
Square both sides.
$\left(\sqrt{x+15}\right)^2=\left(5+\sqrt{x}\right)^2$
$x+15=\left(5+\sqrt{x}\right)\left(5+\sqrt{x}\right)$
$x+15=25+10\sqrt{x}+x$
Isolate the square root of $x$ on one side of the equation.
$-10=10\sqrt{x}$
Divide both sides by $10$.
$-1=\sqrt{x}$
What can you take the square root of and get $−1$?
No real solutions.
1. $(y−6)^2+10=3y$
Expand the $(y−6)^2$.
Rearrange the equation so that it equals zero.
Solve by factoring and using the Zero Product Property, or use the Quadratic Formula.
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Purchase Solution
# Functions: K-T Condition
Not what you're looking for?
Consider the following program:
Maximize f(x,y)=x^2+4xy+y^2
subject to g(x,y)=x^2+y^2-1=0
##### Solution Summary
A function is maximized using Kuhn-Tucker condition. The maximized function results are determined.
##### Solution Preview
Solution. Let us denote the gradient vector of the function f(x,y) by Df(x,y). We rewrite the original program as follows.
Minimize F(x,y)=-f(x,y)=-x^2-4xy-y^2
. g(x,y)=x^2+y^2-1=0.
Since DF(x,y)=(-2x-4y,-4x-2y)', Dg(x,y)=(2x,2y)', by K-T ...
Solution provided by:
###### Education
• BSc , Wuhan Univ. China
• MA, Shandong Univ.
###### Recent Feedback
• "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
• "excellent work"
• "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
• "Thank you"
• "Thank you very much for your valuable time and assistance!"
##### Multiplying Complex Numbers
This is a short quiz to check your understanding of multiplication of complex numbers in rectangular form.
##### Geometry - Real Life Application Problems
Understanding of how geometry applies to in real-world contexts
##### Exponential Expressions
In this quiz, you will have a chance to practice basic terminology of exponential expressions and how to evaluate them.
##### Graphs and Functions
This quiz helps you easily identify a function and test your understanding of ranges, domains , function inverses and transformations.
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### A: Number and Operations
#### A.1: Students will understand numerical concepts and mathematical operations
A.1.1: Understand numbers, ways of representing numbers, relationships among numbers, and number systems.
A.1.1.1: Exhibit an understanding of the place-value structure of the base-ten number system by:
A.1.1.1.a: reading, modeling, writing, and interpreting whole numbers up to 10,000
A.1.1.1.b: comparing and ordering numbers up to 1,000
A.1.1.1.c: recognizing the position of a given number in the base-ten number system and its relationship to benchmark numbers such as 10, 50, 100, 500
A.1.1.2: Use whole numbers by using a variety of contexts and models (e.g., exploring the size of 1,000 by skip-counting to 1,000 using hundred charts or strips 10 or 100 centimeters long).
A.1.1.4: Identify the relationship among commonly encountered factors and multiples (e.g., factor pairs of 12 are 1 x 12, 2 x 6, 3 x 4; multiples of 12 are 12, 24, 36).
A.1.1.5: Use visual models and other strategies to recognize and generate equivalents of commonly used fractions and mixed numbers (e.g., halves, thirds, fourths, sixths, eighths, and tenths).
A.1.1.6: Demonstrate an understanding of fractions as parts of unit wholes, parts of a collection or set, and as locations on a number line.
A.1.1.7: Use common fractions for measuring and money (e.g., using fractions and decimals as representations of the same concept, such as half of a dollar = 50 cents)
A.1.2: Understand the meaning of operations and how they relate to one another.
A.1.2.1: Use a variety of models to show an understanding of multiplication and division of whole numbers (e.g., charts, arrays, diagrams, and physical models [i.e., modeling multiplication with a variety of pictures, diagrams, and concrete tools to help students learn what the factors and products represent in various contexts]).
A.1.2.2: Find the sum or difference of two whole numbers between 0 and 10,000.
A.1.2.3: Solve simple multiplication and division problems (e.g., 135 รท.(ٱ= 5
A.1.2.4: Identify how the number of groups and the number of items in each group equals a product.
A.1.2.5: Demonstrate the effects of multiplying and dividing on whole numbers (e.g., to find the total number of legs on 12 cats, 4 represents the number of each [cat] unit, so 12 x 4 = 48 [leg] units).
A.1.3: Compute fluently and make reasonable estimates.
A.1.3.1: Choose computational methods based on understanding the base-ten number system, properties of multiplication and division, and number relationships.
A.1.3.2: Use strategies (e.g., 6 x 8 is double 3 x 8) to become fluent with the multiplication pairs up to 10 x 10.
A.1.3.4: Demonstrate reasonable estimation strategies for measurement, computation, and problem solving.
### B: Algebra
#### B.1: Students will understand algebraic concepts and applications.
B.1.1: Understand patterns, relations, and functions.
B.1.1.5: Recognize and use the commutative property of multiplication (e.g., if 5 x 7 = 35, then what is 7 x 5?).
B.1.1.6: Create, describe, and extend numeric and geometric patterns including multiplication patterns.
B.1.1.7: Represent simple functional relationships:
B.1.1.7.a: solve simple problems involving a functional relationship between two quantities (e.g., find the total cost of multiple items given the cost per unit)
B.1.2: Represent and analyze mathematical situations and structures using algebraic symbols.
B.1.2.2: Recognize and use the commutative and associative properties of addition and multiplication (e.g., "If 5 x 7 = 35, then what is 7 x 5? And if 5 x 7 x 3 = 105, then what is 7 x 3 x 5?").
B.1.2.3: Explore the ways that commutative, distributive, identity, and zero properties are useful in computing with numbers.
B.1.3: Use mathematical models to represent and understand quantitative relationships.
B.1.3.1: Model problem situations with objects and use representations such as pictures, graphs, tables, and equations to draw conclusions.
### C: Geometry
#### C.1: Students will understand geometric concepts and applications.
C.1.1: Analyze characteristics and properties of two- and three-dimensional geometric shapes and develop mathematical arguments about geometric relationships.
C.1.1.1: Describe and compare the attributes of plane and solid geometric figures to show relationships and solve problems:
C.1.1.1.a: identify, describe, and classify polygons (e.g., pentagons, hexagons, and octagons)
C.1.1.1.b: identify lines of symmetry in two-dimensional shapes
C.1.1.1.c: explore attributes of quadrilaterals (e.g., parallel and perpendicular sides for the parallelogram, right angles for the rectangle, equal sides and right angles for the square)
C.1.1.1.d: identify right angles
C.1.2: Specify locations and describe spatial relationships using coordinate geometry and other representational systems.
C.1.2.2: Use ordered pairs to graph, locate specific points, create paths, and measure distances within a coordinate grid system.
C.1.3: Apply transformations and use symmetry to analyze mathematical situations.
C.1.3.1: Predict and describe the results of sliding, flipping, and turning two-dimensional shapes.
C.1.3.2: Identify and describe the line of symmetry in two- and three-dimensional shapes.
### D: Measurement
#### D.1: Students will understand measurement systems and applications.
D.1.1: Understand measurable attributes of objects and the units, systems, and process of measurement.
D.1.1.3: Identify time to the nearest minute (elapsed time) and relate time to everyday events.
D.1.2: Apply appropriate techniques, tools, and formulas to determine measurements.
D.1.2.2: Estimate measurements.
### E: Data Analysis and Probability
#### E.1: Students will understand how to formulate questions, analyze data, and determine probabilities.
E.1.1: Formulate questions that can be addressed with data and collect, organize, and display relevant data to answer them.
E.1.1.1: Collect and organize data using observations, measurements, surveys, or experiments.
E.1.1.2: Represent data using tables and graphs (e.g., line plots, bar graphs, and line graphs).
E.1.1.3: Conduct simple experiments by determining the number of possible outcomes and make simple predictions:
E.1.1.3.a: identify whether events are certain, likely, unlikely, or impossible
E.1.1.3.b: record the outcomes for a simple event and keep track of repetitions
E.1.1.3.d: use the results to predict future events
E.1.2: Select and use appropriate statistical methods to analyze data.
E.1.2.1: Apply and explain the uses of sampling techniques (e.g., observations, polls, tally marks) for gathering data.
E.1.4: Understand and apply basic concepts of probability.
E.1.4.1: Discuss the degree of likelihood of events and use terminology such as "certain," "likely," "unlikely".
Correlation last revised: 1/20/2017
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information.
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Annette Sabin
## Answered question
2022-01-10
Which operation could we perform in order to find the number of milliseconds in a year?
### Answer & Explanation
Hector Roberts
Beginner2022-01-11Added 31 answers
The amount of milliseconds in a year must be determined.
First of all, milli means thousandth and thus one second contains 1000 milliseconds.
Then, there are 60 seconds in one minute and thus the number of milliseconds in a minute is then obtain by multiplying the number of milliseconds in a second by the number of seconds in a minute.
Then, there are 60 minutes in one hour and thus the number of milliseconds in a minute by the number of milliseconds in a minute by the number of minutes in a hour.
Then, there are 24 hours in a day and thus the number of milliseconds in a day is then obtain by multiplying the number of milliseconds in an hour by the number of hours in a day.
Finally, we assume that there are 365 days in a year. The number of milliseconds in a year is then by multiplying the number of milliseconds in a day by the number of days in a year.
Then mark the correct action: $1000\cdot 60\cdot 60\cdot 24\cdot 365$
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The Fano tetrahedron
The "Fano tetrahedron" is built from Fano planes. We start with a description of the Fano plane, which you can skip if you're already familiar with it.
The Fano plane
Diagram 1. The Fano plane
A Fano plane is is a finite projective plane of seven points and seven lines, with three points on every line and three lines through every point. It is a balanced block design, with seven blocks (the lines) and seven elements (the points). It can be specified as the balanced block design 2-(7,7,3,3,1); and as the last value of this balanced block design is 1, it is a Steiner system, and as such its descriptor can be written S(2,3,7). Any two points lie on one and only one line, and any two lines pass through one and only one point.
A Fano plane is shown to the right. Three of the lines are shown in dark gray, the in red, and one in green – this has no mathematical significance, but it may make the diagram easier to understand. Its vertices are labelled with numbers, which should be regarded, not as integers, but as bit-strings (001, 010, 011 etc.). Wherever two points on a line are labelled p and q, the third point on that line is labelled p xor y.
Building the Fano tetrahedron
Diagram 2a. A tetrahedral net built from four Fano planes Diagram 2b. The tetrahedral net with six more lines added.
Just as a tetrahedron is built from four triangles, a Fano tetrahedron is built from four Fano planes. First we build the "net" of the tetrahedron, as shown in diagram 2a. A tetrahedron can be built from this net by folding it along the lines labelled 3, 5 and 6, and uniting the pairs and triplet of vertices with the same label.
But what we have now is not a balanced block design or Steiner system, for two reasons. We need to add one more point, labelled 15, in the centre of the tetrahedron, with seven lines through it, joining each vertex (1,2,4,8) to its antipodal face (14,13,11,7) and each edge (3,5,6) to its antipodal edge (12,10,9). Also we need to add six more lines, shown in diagram 2b in blue and purple.
The diagrams do not show the central vertex with its label "15". You'll have to imagine it, and the lines that pass through it.
As with the Fano plane in diagram 1, each pair of points lies on one line. It is therefore a Steiner syetem, S(2,3,15). The labels on the points are again such that the xor of any two points is the label on the third point on their line.
Catalogue of lines
Below is a table cataloging the 35 lines (blocks) of the Fano tetrahedron, and listing the elements (Vertices, Edges, Faces, and the Centre) of the tetrahedron that lie on them. The others are easy to understand, but the blue and magenta lines may seem irregular. In fact they join a pair of faces and the edge antipodal to their mutual edge. The blue edges and the magenta ones are equivalent, but look different when drawn on the net in the diagram.
Colour of lines
in diagram
Elements
on line
Number of
such lines
Description
GreyV E V6edges
RedV F E12medians
GreenE E E4in-circles
BlueF E F6lie on great circles
of the tetrahedron
Magenta
(not shown)V C F4pass through the centre,
joining antipodes
E C E3
Total35
Another view
Diagram 3. No longer a net, but the tetrahedron drawn with one vertex (its apex, if you like) at infinity in all directions.
Diagram 3 shows another view of the Fano tetrahedron. Each edge is now shown only once. The apex vertex, labeled "8", is at infinity.
The magenta lines, which lie along great circles, ought to be straight, but would then coincide with red lines. So instead, they have been drawn with slight S-shaped curves.
Further simplexes
Just as we can use the Fano plane to build the Fano tetrahedron, we can use the Fano tetrahedron to build the Fano 5-cell, with 5 vertices, 31 numbered points, and 155 lines, the Steiner system S(2,3,31); and so on.
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# Simplify 264.96*(0.577-0.0924/b)*(1+0.2592/b)
264.96⋅(0.577-0.0924b)⋅(1+0.2592b)
Apply the distributive property.
(264.96⋅0.577+264.96(-0.0924b))⋅(1+0.2592b)
Multiply 264.96 by 0.577.
(152.88192+264.96(-0.0924b))⋅(1+0.2592b)
Multiply 264.96(-0.0924b).
Multiply -1 by 264.96.
(152.88192-264.960.0924b)⋅(1+0.2592b)
Combine -264.96 and 0.0924b.
(152.88192+-264.96⋅0.0924b)⋅(1+0.2592b)
Multiply -264.96 by 0.0924.
(152.88192+-24.482304b)⋅(1+0.2592b)
(152.88192+-24.482304b)⋅(1+0.2592b)
Move the negative in front of the fraction.
(152.88192-24.482304b)⋅(1+0.2592b)
Expand (152.88192-24.482304b)(1+0.2592b) using the FOIL Method.
Apply the distributive property.
152.88192(1+0.2592b)-24.482304b(1+0.2592b)
Apply the distributive property.
152.88192⋅1+152.881920.2592b-24.482304b(1+0.2592b)
Apply the distributive property.
152.88192⋅1+152.881920.2592b-24.482304b⋅1-24.482304b⋅0.2592b
152.88192⋅1+152.881920.2592b-24.482304b⋅1-24.482304b⋅0.2592b
Simplify and combine like terms.
Simplify each term.
Multiply 152.88192 by 1.
152.88192+152.881920.2592b-24.482304b⋅1-24.482304b⋅0.2592b
Multiply 152.881920.2592b.
Combine 152.88192 and 0.2592b.
152.88192+152.88192⋅0.2592b-24.482304b⋅1-24.482304b⋅0.2592b
Multiply 152.88192 by 0.2592.
152.88192+39.62699366b-24.482304b⋅1-24.482304b⋅0.2592b
152.88192+39.62699366b-24.482304b⋅1-24.482304b⋅0.2592b
Multiply -1 by 1.
152.88192+39.62699366b-24.482304b-24.482304b⋅0.2592b
Multiply -24.482304b⋅0.2592b.
Multiply 0.2592b and 24.482304b.
152.88192+39.62699366b-24.482304b-0.2592⋅24.482304b⋅b
Multiply 0.2592 by 24.482304.
152.88192+39.62699366b-24.482304b-6.34581319b⋅b
Raise b to the power of 1.
152.88192+39.62699366b-24.482304b-6.34581319b1b
Raise b to the power of 1.
152.88192+39.62699366b-24.482304b-6.34581319b1b1
Use the power rule aman=am+n to combine exponents.
152.88192+39.62699366b-24.482304b-6.34581319b1+1
152.88192+39.62699366b-24.482304b-6.34581319b2
152.88192+39.62699366b-24.482304b-6.34581319b2
152.88192+39.62699366b-24.482304b-6.34581319b2
Combine the numerators over the common denominator.
152.88192+39.62699366-24.482304b-6.34581319b2
152.88192+39.62699366-24.482304b-6.34581319b2
Subtract 24.482304 from 39.62699366.
152.88192+15.14468966b-6.34581319b2
Simplify 264.96*(0.577-0.0924/b)*(1+0.2592/b)
## Our Professionals
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#### We are MathExperts
Solve all your Math Problems: https://elanyachtselection.com/
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# How do you write H(x) = 2/3|x-5| as a piecewise function?
Aug 25, 2017
$H \left(x\right) = - \frac{2}{3} \left(x - 5\right) , x < 5$
$H \left(x\right) = \frac{2}{3} \left(x - 5\right) , x \ge 5$
#### Explanation:
If x is less than 5, the number within the absolute value will be negative. Thus, if we wish to rid ourselves of the absolute value symbol, which as one recalls changes any negative expression within it positive, we must find a way to make $\frac{2}{3} \left(x - 5\right)$ positive when $x < 5$.
This is most easily accomplished by multiplying the expression by -1 when $x < 5$ ; in this case, since it will only be applied when $\frac{2}{3} \left(x - 5\right) < 0$, it will turn the expression positive. The breakpoint is a at x=5, because at x=5 the expression is equal to 0, and as x increases the expression becomes more positive.
Thus, we are left with
$H \left(x\right) = - \frac{2}{3} \left(x - 5\right) , x < 5$
$H \left(x\right) = \frac{2}{3} \left(x - 5\right) , x \ge 5$
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# what is a probability????????????????????????????????????????
Gowri sankar
292 Points
8 years ago
HAI GURAVAIAH,
THE PROBABILITY IS DIFINED BY THE QUALITY OF BEING PROBABLE IS CALLED THE PROBABILITY.........................................................................
317 Points
8 years ago
Hello Guravaiah
Probability is the chance that something will happen - how likely it is that some event will happen. Sometimes you can measure a probability with a number like "10% chance of rain", or you can use words such as impossible, unlikely, possible, even chance, likely and certain. Example: "It is unlikely to rain tomorrow".
Prabhakar ch
577 Points
8 years ago
Dear Guravaiah
Probability is the measure of the likelihood that an event will occur. Probability is quantified as a number between 0 and 1 (where 0 indicates impossibility and 1 indicates certainty).
mohan
216 Points
8 years ago
dear sivagiri......
probility is a quantified as a number between 0 and 1.the higher the probolity of an event,the more certain we are that the event will occur.....
N JYOTHEESWAR
342 Points
8 years ago
Probability is the chance that something will happen how likely it is that some event will happen. Sometimes you can measure a probability or you can use words such as impossible, unlikely, possible, even chance, likely and certain.
SAI SARDAR
1700 Points
8 years ago
Guravaiah, Probability is the measure of likliness.And the number of ways for reaching some thing.all the best.
sudarshan
174 Points
8 years ago
Probability is the chance that something will happen - how likely it is that some event will happen. Sometimes you can measure a probability with a number like "10% chance of rain", or you can use words such as impossible, unlikely, possible, even chance, likely and certain. Example: "It is unlikely to rain tomorrow".
raj
383 Points
8 years ago
hello guravaProbability is the chance that something will happen - how likely it is that some event will happen. Sometimes you can measure a probability with a number like "10% chance of rain", or you can use words such as impossible, unlikely, possible, even chance, likely and certain. Example: "It is unlikely to rain tomorrow".
T Dileep
100 Points
8 years ago
the extent to which an event is likely to occur ,measured by the ratio of the favourable cases to the whole number of casespossible.
T.kumar
281 Points
8 years ago
Probability is the measure of the likelihood that an event will occur. Probability is quantified as a number between 0 and 1 (where 0 indicates impossibility and 1 indicates certainty). The higher the probability of an event, the more certain we are that the event will occur.
Dheeru chowdary
100 Points
8 years ago
dear guravaih,
probabilty is the nmeasure of likehood i.e., how probable it will happens.................k
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# Introduction and key concepts
Page 2 / 3
Amplitude
The amplitude is the maximum displacement of a particle from its equilibrium position.
## Investigation : amplitude
Fill in the table below by measuring the distance between the equilibrium and each peak and troughs in the wave above. Use your ruler to measure the distances.
Peak/Trough Measurement (cm) a b c d e f
2. Are the distances between the equilibrium position and each peak equal?
3. Are the distances between the equilibrium position and each trough equal?
4. Is the distance between the equilibrium position and peak equal to the distance between equilibrium and trough?
As we have seen in the activity on amplitude, the distance between the peak and the equilibrium position is equal to the distance between the trough and the equilibrium position. This distance is known as the amplitude of the wave, and is the characteristic height of wave, above or below the equilibrium position. Normally the symbol $A$ is used to represent the amplitude of a wave. The SI unit of amplitude is the metre (m).
If the peak of a wave measures $2\phantom{\rule{2pt}{0ex}}m$ above the still water mark in the harbour, what is the amplitude of the wave?
1. The definition of the amplitude is the height of a peak above the equilibrium position. The still water mark is the height of the water at equilibrium and the peak is $2\phantom{\rule{2pt}{0ex}}m$ above this, so the amplitude is $2\phantom{\rule{2pt}{0ex}}m$ .
## Investigation : wavelength
Fill in the table below by measuring the distance between peaks and troughs in the wave above.
Distance(cm) a b c d
2. Are the distances between peaks equal?
3. Are the distances between troughs equal?
4. Is the distance between peaks equal to the distance between troughs?
As we have seen in the activity on wavelength, the distance between two adjacent peaks is the same no matter which two adjacent peaks you choose. There is a fixed distance between the peaks. Similarly, we have seen that there is a fixed distance between the troughs, no matter which two troughs you look at. More importantly, the distance between two adjacent peaks is the same as the distance between two adjacent troughs. This distance is called the wavelength of the wave.
The symbol for the wavelength is $\lambda$ (the Greek letter lambda ) and wavelength is measured in metres ( $m$ ).
The total distance between $4$ consecutive peaks of a transverse wave is $6\phantom{\rule{2pt}{0ex}}m$ . What is the wavelength of the wave?
1. From the sketch we see that 4 consecutive peaks is equivalent to 3 wavelengths.
2. Therefore, the wavelength of the wave is:
$\begin{array}{ccc}\hfill 3\lambda & =& 6\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\hfill \\ \hfill \lambda & =& \frac{6\phantom{\rule{0.166667em}{0ex}}\mathrm{m}}{3}\hfill \\ & =& 2\phantom{\rule{0.166667em}{0ex}}\mathrm{m}\hfill \end{array}$
## Investigation : points in phase
Fill in the table by measuring the distance between the indicated points.
Points Distance (cm) A to F B to G C to H D to I E to J
What do you find?
In the activity the distance between the indicated points was the same. These points are then said to be in phase . Two points in phase are separate by an integer (0,1,2,3,...) number of complete wave cycles. They do not have to be peaks or troughs, but they must be separated by a complete number of wavelengths.
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
Got questions? Join the online conversation and get instant answers!
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# 3.7: Practice SD Formula and Interpretation
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You may or may not understand the importance of calculating and understanding the variation of your data. In some data sets, the data values are concentrated closely near the mean; in other data sets, the data values are more widely spread out from the mean. The most common measure of variation, or spread, is the standard deviation. The standard deviation is a number that measures how far data values are from their mean.
##### The Standard Deviation
• provides a numerical measure of the overall amount of variation in a data set, and
• can be used to determine whether a particular data value is close to or far from the mean.
There are a couple common kinds of questions that standard deviations can answer, in addition being foundational for later statistical analyses. First, a standard deviation helps understand the shape of a distribution. Second, a standard deviation can show if a score is extreme.
### Describing the Shape of a Distribution
The standard deviation provides a measure of the overall variation in a data set.
The standard deviation is always positive or zero. The standard deviation is small when the data are all concentrated close to the mean, exhibiting little variation or spread. Distributions with small standard deviations have a tall and narrow line graph. The standard deviation is larger when the data values are more spread out from the mean, exhibiting more variation. Distributions with large standard deviations may have a wide and flat lin graph, or they may be skewed (with the outlier(s) making the standard deviation bigger).
Suppose that we are studying the amount of time customers wait in line at the checkout at supermarket A and supermarket B. the average wait time at both supermarkets is five minutes. At supermarket A, the standard deviation for the wait time is two minutes; at supermarket B the standard deviation for the wait time is four minutes.
Because supermarket B has a higher standard deviation, we know that there is more variation in the wait times at supermarket B. Overall, wait times at supermarket B are more spread out from the average; wait times at supermarket A are more concentrated near the average.
### Identifying Extreme Scores
The standard deviation can be used to determine whether a data value is close to or far from the mean.
Suppose that Rosa and Binh both shop at supermarket A. Rosa waits at the checkout counter for seven minutes and Binh waits for one minute. At supermarket A, the mean waiting time is five minutes and the standard deviation is two minutes. The standard deviation can be used to determine whether a data value is close to or far from the mean.
Rosa waits for seven minutes:
• Seven is two minutes longer than the average of five; two minutes is equal to one standard deviation.
• Rosa's wait time of seven minutes is two minutes longer than the average of five minutes.
• Rosa's wait time of seven minutes is one standard deviation above the average of five minutes.
Binh waits for one minute.
• One is four minutes less than the average of five; four minutes is equal to two standard deviations.
• Binh's wait time of one minute is four minutes less than the average of five minutes.
• Binh's wait time of one minute is two standard deviations below the average of five minutes.
• A data value that is two standard deviations from the average is just on the borderline for what many statisticians would consider to be far from the average. Considering data to be far from the mean if it is more than two standard deviations away is more of an approximate "rule of thumb" than a rigid rule. In general, the shape of the distribution of the data affects how much of the data is further away than two standard deviations. (You will learn more about this in later chapters.)
The number line may help you understand standard deviation. If we were to put five and seven on a number line, seven is to the right of five. We say, then, that seven is one standard deviation to the right of five because $$5 + (1)(2) = 7$$.
If one were also part of the data set, then one is two standard deviations to the left of five because $$5 + (-2)(2) = 1$$.
• In general, a value = mean + (#ofSTDEV)*(standard deviation)
• where #ofSTDEVs = the number of standard deviations
• #ofSTDEV does not need to be an integer
• One is two standard deviations less than the mean of five because: $$1 = 5 + (-2)(2)$$. (The numbers in parentheses that touch should be multiplied)
The equation value = mean + (#ofSTDEVs)*(standard deviation) can be expressed for a sample and for a population.
• sample: $$x = \bar{x} + \text{(#ofSTDEV) \times (s)}$$
• Population: $$x = \mu + \text{(#ofSTDEV) \times (s)}$$
The lower case letter s represents the sample standard deviation and the Greek letter $$\sigma$$ (sigma, lower case) represents the population standard deviation.
The symbol $$\bar{x}$$ is the sample mean and the Greek symbol $$\mu$$ is the population mean.
### Calculating the Standard Deviation
If $$x$$ is a number, then the difference "$$x$$ – mean" is called its deviation. In a data set, there are as many deviations as there are items in the data set. The deviations are used to calculate the standard deviation. If the numbers belong to a population, in symbols a deviation is $$x - \mu$$. For sample data, in symbols a deviation is $$x - \bar{x}$$.
The procedure to calculate the standard deviation depends on whether the numbers are the entire population or are data from a sample. The calculations are similar, but not identical. Therefore the symbol used to represent the standard deviation depends on whether it is calculated from a population or a sample. The lower case letter s represents the sample standard deviation and the Greek letter $$\sigma$$ (sigma, lower case) represents the population standard deviation. If the sample has the same characteristics as the population, then s should be a good estimate of $$\sigma$$.
To calculate the standard deviation, we need to calculate the variance first. The variance is the average of the squares of the deviations (the $$x - \bar{x}$$ values for a sample, or the $$x - \mu$$ values for a population). The symbol $$\sigma^{2}$$ represents the population variance; the population standard deviation $$\sigma$$ is the square root of the population variance. The symbol $$s^{2}$$ represents the sample variance; the sample standard deviation s is the square root of the sample variance. You can think of the standard deviation as a special average of the deviations.
If the numbers come from a census of the entire population and not a sample, when we calculate the average of the squared deviations to find the variance, we divide by $$N$$, the number of items in the population. If the data are from a sample rather than a population, when we calculate the average of the squared deviations, we divide by n – 1, one less than the number of items in the sample.
##### Formulas for the Sample Standard Deviation
$s = \sqrt{\dfrac{\sum(X-\bar{X})^{2}}{n-1}} \nonumber$
For the sample standard deviation, the denominator is $$n - 1$$, that is the sample size MINUS 1.
## Practice!
##### Example $$\PageIndex{1}$$
In a fifth grade class at a private school, the teacher was interested in the average age and the sample standard deviation of the ages of her students. The following data are the ages for a sample of n = 20 fifth grade students. The ages are rounded to the nearest half year in Table $$\PageIndex{1}$$, but first let's talk about the context.
1. Who was the sample? Who could this sample represent (population)?
The sample is the 20 fifth graders from a private school. The population could be all fifth graders from private schools?
1. What was measured?
Age, in years, was measured. This is the DV, the outcome variable.
Table $$\PageIndex{1}$$- Ages of a sample of 20 fifth graders
9
9.5
9.5
10
10
10
10
10.5
10.5
10.5
10.5
11
11
11
11
11
11
11.5
11.5
11.5
1. What is the mean?
$\bar{x} = \dfrac{(9+9.5+9.5+10+10+10+10+10.5+10.5+10.5+10.5+11+11+11+11+11+11+11.5+11.5+11.5)}{20} = 10.525 = 10.53 \nonumber$
The average age is 10.53 years, rounded to two places.
1. What is the standard deviation?
The variance may be calculated by using a table. Then the standard deviation is calculated by taking the square root of the variance. We will explain the parts of the table after calculating s.
Table $$\PageIndex{1}$$- Ages of One Fifth Grade Class
Data Deviations Deviations2
x (X – $$\bar{X}$$) (X– $$\bar{X})^2$$
9 $$9 – 10.525 = –1.525$$ $$(–1.525)^2 = (-1.525 \times -1.525) = 2.325625$$
9.5 $$9.5 – 10.525 = –1.025$$ $$(–1.025)^2 = (–1.025 \times –1.025) = 1.050625$$
9.5 $$9.5 – 10.525 = –1.0.25$$ $$(–1.025)^2 = 1.050625$$
10 $$10 – 10.525 = –0.525$$ $$(–0.525)^2 = (–0.525 \times –0.525)= 0.275625$$
10 $$10 – 10.525 = –0.525$$ $$(–0.525)^2 = 0.275625$$
10 $$10 – 10.525 = –0.525$$ $$(–0.525)^2 = 0.275625$$
10 $$10 – 10.525 = –0.525$$ $$(–0.525)^2 = 0.275625$$
10.5 $$10.5 – 10.525 = –0.025$$ $$(–0.025)^2 = (–0.025 \times –0.025)= 0.000625$$
10.5 $$10.5 – 10.525 = –0.025$$ $$(–0.025)^2 = 0.000625$$
10.5 $$10.5 – 10.525 = –0.025$$ $$(–0.025)^2 = 0.000625$$
10.5 $$10.5 – 10.525 = –0.025$$ $$(–0.025)^2 = 0.000625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = (0.475 \times 0.475)= 0.225625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = 0.225625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = 0.225625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = 0.225625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = 0.225625$$
11 $$11 – 10.525 = 0.475$$ $$(0.475)^2 = 0.225625$$
11.5 $$11.5 – 10.525 = 0.975$$ $$(0.975)^2 = (0.975 \times 0.975)= 0.950625$$
11.5 $$11.5 – 10.525 = 0.975$$ $$(0.975)^2 = 0.950625$$
11.5 $$11.5 – 10.525 = 0.975$$ $$(0.975)^2 = 0.950625$$
$$\displaystyle\sum X$$ 0 (basically) $$\sum = 9.7375$$
The first column in Table $$\PageIndex{1}$$ has the data, the second column has has deviations (each score minus the mean), the third column has deviations squared. The first row is the row's title, the second row is the symbols for that column, the rest of the rows are the scores until the bottom row, which is the sum of each of the rows.
Take the sum of the last column (9.7375) divided by the total number of data values minus one (20 – 1):
$\dfrac{9.7375}{20-1} = 0.5125 \nonumber$
The sample standard deviation s is the square root of $$\dfrac{SS}{df} \nonumber$$:
$s = \sqrt{0.5125} = 0.715891 \nonumber$
and this is rounded to two decimal places, $$s = 0.72$$. The standard deviation of the sample fo 20 fifth graders is 0.72 years.
Typically, you do the calculation for the standard deviation on your calculator or computer. When calculations are completed on devices, the intermediate results are not rounded so the results are more accurate. It's also darned easier. So why are spending time learning this outdated formula? So that you can see what's happening. We are finding the difference between each score and the mean to see how varied the distribution of data is around the center, dividing it by the sample size minus one to make it like an average, then square rooting it to get the final answer back into the units that we started with ( age in years).
• For the following problems, recall that value = mean + (#ofSTDEVs)(standard deviation).
• For a sample: $$x$$ = $$\bar{x}$$ + (#ofSTDEVs)(s)
• For a population: $$x$$ = $$\mu$$ + (#ofSTDEVs)$$\sigma$$
• For this example, use x = $$\bar{x}$$ + (#ofSTDEVs)(s) because the data is from a sample
1. Verify the mean and standard deviation on your own.
2. Find the value that is one standard deviation above the mean. Find ($$\bar{x}$$ + 1s).
3. Find the value that is two standard deviations below the mean. Find ($$\bar{x}$$ – 2s).
4. Find the values that are 1.5 standard deviations from (below and above) the mean.
Solution
1. You should get something close to 0.72 years, but anything from 0.70 to 0.74 shows that you have the general idea.
2. ($$\bar{x} + 1s) = 10.53 + (1)(0.72) = 11.25$$
3. $$(\bar{x} - 2s) = 10.53 – (2)(0.72) = 9.09$$
• $$(\bar{x} - 1.5s) = 10.53 – (1.5)(0.72) = 9.45$$
• $$(\bar{x} + 1.5s) = 10.53 + (1.5)(0.72) = 11.61$$
Notice that instead of dividing by $$n = 20$$, the calculation divided by $$n - 1 = 20 - 1 = 19$$ because the data is a sample. For the sample, we divide by the sample size minus one ($$n - 1$$). The sample variance is an estimate of the population variance. After countless replications, it turns out that when the formula division by only N (the size of the sample) is used on a sample to infer the population’s variance, it always under-estimates the variance of the population.
Which one has the bigger solution, the one with the smaller denominator or the larger denominator?
• $$\dfrac{10}{2}=$$
• $$\dfrac{10}{5}=$$
Smaller denominators make the resulting product larger. To solve our problem of using the population’s variance formula on a sample under-estimating the variance, we make the denominator of our equation smaller when calculating variance for a sample. In other words, based on the mathematics that lies behind these calculations, dividing by ($$n - 1$$) gives a better estimate of the population.
## What does it mean?
The deviations show how spread out the data are about the mean. From Table $$\PageIndex{1}$$, The data value 11.5 is farther from the mean than is the data value 11 which is indicated by the deviations 0.97 and 0.47. A positive deviation occurs when the data value (age, in this case) is greater than the mean, whereas a negative deviation occurs when the data value is less than the mean (that particular student is younger than the average age of the class) . The deviation is –1.525 for the data value nine. If you add the deviations, the sum is always zero, so you cannot simply add the deviations to get the spread of the data. By squaring the deviations, you make them positive numbers, and the sum will also be positive. The variance, then, is the average squared deviation. But the variance is a squared measure and does not have the same units as the data. No one knows what 9.7375 years squared means. Taking the square root solves the problem! The standard deviation measures the spread in the same units as the data.
The standard deviation, $$s$$ or $$\sigma$$, is either zero or larger than zero. When the standard deviation is zero, there is no spread; that is, all the data values are equal to each other. The standard deviation is small when the data are all concentrated close to the mean, and is larger when the data values show more variation from the mean. When the standard deviation is a lot larger than zero, the data values are very spread out about the mean; outliers can make $$s$$ or $$\sigma$$ very large.
##### Exercise $$\PageIndex{1}$$
Scenario: Using one baseball professional team as a sample for all professional baseball teams, the ages of each of the players are shown in Table $$\PageIndex{2}$$.
Table $$\PageIndex{2}$$- One Baseball Team's Ages
Data Deviations Deviations2
x (x – $$\bar{x}$$) (x – $$\bar{x})^2$$
21
21
22
23
24
24
25
25
28
29
29
31
32
33
33
34
35
36
36
36
36
38
38
38
40
$$\displaystyle\sum X$$ = 767 $$\displaystyle\sum X$$ should be 0 (basically) $$\displaystyle\sum X$$ = ?
If you get stuck after the table, don't forget that: $$s=\sqrt{\dfrac{\sum(X-\overline {X})^{2}}{N-1}}$$
All of your answers should be complete sentences, not just one word or one number. Behavioral statistics is about research, not math.
1. Who was the sample? Who could this sample represent (population)?
2. What was measured?
3. What is the mean? (Get in the practice of including the units of measurement when answering questions; a number is usually not a complete answer).
4. What is the standard deviation?
$s=\sqrt{\dfrac{\sum(X-\overline {X})^{2}}{N-1}}=\sqrt{\dfrac{S S}{d f}} \nonumber$
1. Find the value that is two standard deviations above the mean, and determine if there are any players that are more than two standard deviations above the mean.
1. The sample is 25 players from a professional baseball team. They were chosen to represent all professional baseball players (it says so in the scenario description!).
2. Age, in years, was measured.
3. The mean of the sample ($$\bar{X}$$ was 30.68 years.
4. The standard deviation was 6.09 years ( $$s = 6.09$$ ), although due to rounding differences you could get something from about 6.05 to 6.12. Don't worry too much if you don't get exactly 6.09; if you are close, then you did the formula correctly!
5. The age that is two standard deviations above the mean is 42.86 years, and none of the players are older than that.
$(\bar{x} + 2s = 30.68 + (2)(6.09) = 42.86 \nonumber$.
What standard deviation show us can seem unclear at first. Especially when you are unfamiliar (and maybe nervous) about using the formula. By graphing your data, you can get a better "feel" for what a standard deviation can show you. You will find that in symmetrical distributions, the standard deviation can be very helpful. Because numbers can be confusing, always graph your data.
## Summary
• $$s = \sqrt{\dfrac{\sum(x-\bar{x})^{2}}{n-1}}$$ is the formula for calculating the standard deviation of a sample. f(xμ)2N
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```Date: Dec 23, 2012 6:20 AM
Author: quasi
Subject: Re: convex polyhedra with all faces regular
achille wrote:>quasi wrote:>>>> Prove or disprove:>> >> For each positive integer n, there are only finitely many >> convex polyhedra, up to similarity, such that all faces are >> regular polygons (not necessarily of the same type) with at >> most n edges.>>Yes, it is finite.>>It is known that the strictly convex regular-faced polyhedra>comprises >> 2 infinite families (the prisms and antiprisms)> 5 Platonic solids,> 13 Archimedian solids>and 92 Johnson solids> >Let N(n) be the number of convex polyhedra with regular polygons>up to n sides as faces. One has:>> N(n) <= 2n+104>>Actually, it is pretty simple to prove N(n) < oo directly.>WOLOG, let us fix the sides of the regular polygons to has >length 1.>>Let's pick any convex polyhedron and one of its vertex v. >Let say's v is connected to k edges >e_0, e_1, e_2, ... e_k = e_0 >and a_i ( i = 1..k ) is the angle between e_(i-1) and e_i.>For this v, let >> A(v) := 2 pi - sum_{i=1..k} a_i>>Being a convex polyhedron, we have A(v) > 0. It is also easy>to see if we sum over all vertices of the convex polyhedron, >we get:> > sum_v A(v) = 4 pi >>If one build a convex polyhedron using regular polygons up to>n sides, it is easy to see 3 <= k <= 5 and there are only>finitely many possible choices of a_i:>> (1 - 2/3) pi, (1 - 2/4) pi, ... ( 1 - 2/n) pi>>This mean there are finitely many possible choices of >a_1,.., a_k which satisfy:>>(*) 2 pi - sum_{i=1..k} a_i > 0 >>Let M(n) be the smallest possible value of L.H.S of (*) for >given n. On any vertex v of any convex polyhedron build from >regular polygons up to n sides, A(v) >= M(n) and hence the >convex polyhedron has at most 4 pi / M(n) vertices.>>Since the number of vertices is bounded, there are finitely >many ways to connect them to build a polyhedron. Using Cauchy >theorem of convex polytopes, each way of connecting the >vertices to from a polyhedron corresponds to at most 1 convex>polyhedron in Euclidean space. (since the length of all edges >has been fixed to 1).>>As a result, there are only finitely many convex polyhedra one>can build using regular polygons up to n sides.When I get a chance, I'll try to digest the details, but at first glance, it looks very good. Thanks.quasi
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EULER'S DAY OFF
# Pi is the key to a beautiful, impossible formula that shows math is scarily perfect
Mathematics is beautiful.
Image: Quartz
By
In high school, I studied advanced mathematics. This has two uses: Firstly, I can tell people that I once studied advanced mathematics. Secondly, I know about Euler’s equation.
Euler’s equation should not exist. A science fiction author would never imagine a formula so mind-blowingly perfect; it’s far too neat to seem real in an imaginary world. And yet it’s true, and absolutely true in the way only mathematics can be. I may have forgotten how to derive Euler’s equation from first principles (along with nearly every other detail from my high school classes), but I will always remember the resulting formula. It is irrefutable proof of mathematical perfection, and a reminder that we will never be able to fully grasp the meaning and reasons behind math.
Lo, here is the equation:
𝑒𝑖𝜋 = -1
If it’s been a while since you studied math, perhaps this doesn’t look like much. But breaking it down reveals its beauty.
Firstly, in honor of pi day, take pi, or 𝜋: An irrational number, so-called because it goes on infinitely without repeating, 𝜋 was first discovered as the number that describes the relationship between a circle’s circumference and its diameter (circumference = 𝜋 x diameter.) In the centuries since, scientists have determined that 𝜋 also describes the way rivers wind and ripples of light in physics. Originally, though, 𝜋 belonged to the realm of mathematics that deals with shapes and sizes: geometry.
Another famous irrational number, 𝑒, comes from logarithms, which are a part of calculus—a totally different branch of mathematics. The full significance of 𝑒 takes a while to explain, but one key detail, forming the basis of its role in logarithms, is that the rate at which 𝑒x grows is 𝑒x. Like pi, e is the basis of many different formulas. Numerically, it equals 2.71828… going on continually, without repeating.
Then there’s 𝑖, which is an imaginary number. It’s a theoretical concept that can never practically exist. 𝑖 signifies the square root of -1, which is impossible. No two identical numbers can be multiplied together to get a negative, meaning there is no numerical square root of -1. The square root of 4 is 2, and you can have 2 apples. But you can never have √-1 apples.
And yet, take these three utterly different, complicated numbers and bring them together in Euler’s equation and you get a magically neat result: e to the power of (𝑖 multiplied by pi) equals -1. Or:
𝑒𝑖𝜋 = -1
You can also write this as:
𝑒𝑖𝜋 + 1 = 0
The number one, of course, is the first natural number, the first positive integer, and the most common lead in sets of data: Quite simply, it’s how we start counting. And the number zero is the only non-positive natural number, the smallest non-negative quantity, signifying nothing.
In other words, one short equation includes five of the most important numbers in all of mathematics.
It’s almost unnerving. Taken alone, numbers like e, 𝑖, and 𝜋 seem like the results of an imperfect human effort to understand the complexity of the world through mathematical relationships. Euler’s equation, though, shows there’s a unity behind these numbers. The sum total of human mathematic knowledge is no more than a tiny fraction of the complete, perfect system. And every number or equation we discover is a reflection of this abstract, inherent truth, rather than a human invention. Humans can hope to uncover mathematical truths, but we cannot create them.
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Compound angles help
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#1
I can't figure this compound angles question out for the life of me.
Question:
Determine for what range of values of θ between 0 and 2π, Sinθ > Sin3θ
A walkthrough of how to do it would be much appreciated. Thanks in advance.
0
3 years ago
#2
Have u tried drawing the graphs of both of these functions on the same axis?
0
3 years ago
#3
(Original post by TheGibbsy)
I can't figure this compound angles question out for the life of me.
Question:
Determine for what range of values of θ between 0 and 2π, Sinθ > Sin3θ.
Start by sketching y = sin θ and y = sin 3θ on the same set of axes. The former should be very easy, and the latter only requires you to know the effect of transformations of thie kind y = f(x) to y = f(3x).
You should now be able to see where these intervals are, from where the first curve is higher up than the second curve.
Now the tricky part! You need to find the points of intersection of the two curves, that is, you need to solve sin θ = sin 3θ. Any ideas?
0
#4
(Original post by Pangol)
Start by sketching y = sin θ and y = sin 3θ on the same set of axes. The former should be very easy, and the latter only requires you to know the effect of transformations of thie kind y = f(x) to y = f(3x).
You should now be able to see where these intervals are, from where the first curve is higher up than the second curve.
Now the tricky part! You need to find the points of intersection of the two curves, that is, you need to solve sin θ = sin 3θ. Any ideas?
Thanks for the pointer. Using Sin (A+B) i've changed the equation to:
Sin(2θ+θ) = Sinθ and from there using the rule,
Sin2θcosθ + Cos2θsinθ = sinθ
However, I've hit a wall again as to where to go from there.
0
3 years ago
#5
(Original post by TheGibbsy)
Thanks for the pointer. Using Sin (A+B) i've changed the equation to:
Sin(2θ+θ) = Sinθ and from there using the rule,
Sin2θcosθ + Cos2θsinθ = sinθ
However, I've hit a wall again as to where to go from there.
You need to express the LHS purely in terms of sin θ. Start by using the identites for sin 2θ and cos 2θ, and see if you can then get rid of any remaining cosines that you have.
1
#6
(Original post by Pangol)
You need to express the LHS purely in terms of sin θ. Start by using the identites for sin 2θ and cos 2θ, and see if you can then get rid of any remaining cosines that you have.
Thank you! I never even considered the Identities.
Substituting the identities in:
(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get
Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:
0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π
Thank you! You're a life saver!
0
3 years ago
#7
(Original post by TheGibbsy)
Thank you! I never even considered the Identities.
Substituting the identities in:
(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get
Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:
0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π
Thank you! You're a life saver!
For solving , you can save a lot of time if you consider when two sine functions are equal instead of using identities to change the equation. This method confuses students in my experience but it's very useful.
So clearly one solution is and there are other solutions if you consider the graph / CAST etc.
Let us know if you want to have a go at this method and if you need help with it.
0
3 years ago
#8
(Original post by TheGibbsy)
Thank you! I never even considered the Identities.
Substituting the identities in:
(1-2sin^2θ)sinθ + 2sinθcosθcosθ = sinθ - Expand and clean it up to get
Sinθ(2sin^2θ-1) = 0 and solve for Sinθ and 2sin^2θ resulting in:
0, π/4, 3π/4, π, 5π/4, 7π/4 and 2π
Thank you! You're a life saver!
Following on from Notnek's suggestion you could look at:
sin 3θ - sin θ = 0 then use the formula to turn this into a product
1
3 years ago
#9
(Original post by Muttley79)
Following on from Notnek's suggestion you could look at:
sin 3θ - sin θ = 0 then use the formula to turn this into a product
Yes this is also a nice quick way to do it.
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# Quants Questions : Time and Distance Set 4
Hello Aspirants. Welcome to Online Quantitative Aptitude in AffairsCloud.com. Here we are creating question sample in Time and Distance, which is common for all the IBPS,SBI exam and other competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!
1. A Bus covers a distance of 36km at a uniform speed.Had the speed been 6 kmps less, it would have taken one hour more for the journey.The original speed of the bus is
A)30kmps
B)12kmps
C)18mmps
D)20kmps
Explanation :
X^2 – 6x – 216 = 0
x^2-18x+12x-216 = 0
(x+12)(x-18) = 0
X = 18kmph
2. Two buses of same length are running in in the same direction with speed 50km/hr and 80km/ph.The latter completely cross the man in 18sec. The length of the each bus is
A)70m
B)75m
C)80m
D)150m
Explanation :
80 – 50 = 30
30×(5/18) = 25/3 m/s
Distance covered in 20sec = 18 ×(25/3) = 150
Length of each train = 150/2 = 75
3. Rahul started his journey on bike at 7.30pm at a speed of 8km/ph. After 30m, Lenin started his journey from the same place with the speed of 10km/ph. At what time did Lenin overtake Rahul ?
A)10 a.m
B)9 a.m
C)8.30 am
D)8 a.m
Explanation :
Relative speed = 10 – 8 = 2km/ph
Distance covered in 30 min = 8×(30/60) = 4km
Lenin take time to overtake Rahul = 4/2 = 2hr
8+2 = 10 a.m
4. A man can reach a certain place in 30hrs.If he reduces his speed by 1/10 th, he goes 9km less in that time. Find his speed ?
A)8kmph
B)7kmph
C)3kmph
D)1kmph
Explanation :
Let x be the speed
30x – (30*9/10)x = 9
30x – 27x = 9
3x = 9
X= 3 kmph
5. If Anil persons walks at 12kmph instead of 10kmph, he would have walked 18km more.The actual distance travelled by him is
A)100km
B)90km
C)80km
D)60km
Explanation :
x/10 = (x+18)/12
12x = 10x + 180
2x = 180
X = 90km
6. A lady travels equal distances with speed 4kmph,6kmph and 8kmph and takes a total time of 52min.The distance she travelled is
A)4km
B)4.2km
C)4.4km
D)4.8km
Explanation :
(x/4)+(x/6)+(x/8) = 52/60
(6x+4x+3x)/24 = 52/60
13x/24 = 52/60
X = (52×24)/(13×60) = 1.6km
Total distance = 3×1.6 = 4.8km
7. A man crosses a 500m long street in 8min.What is his speed in kmph
A)3.25kmph
B)3.5kmph
C)3.75kmph
D)4kmph
Explanation :
500/(8×60) = 1.04m/sec
1.04×(18/5) = 3.75kmph
8. A bus reached Mumbai from Hyderabad in 30min with an average speed of 50kmph. If the average speed is increased by 35kmph ,How long will it take to cover the same distance ?
A)17min
B)18min
C)19min
D)20min
Explanation :
Time = [ 50 × (30/60) ] / (50 +35)
= [150/60] / 85
= 17.6 / 60 hr = 17.6 min = 18min
9. An aeroplane covers a certain distance at a speed of 240kmph in 5hours. To cover the same distance in 1(2/3) hours it must travel at a speed of
A)700kmph
B)730kmph
C)720kmph
D)710kmph
Explanation :
D = 240×5 = 1200km
New speed = 1200/[5/3] = (1200×3) /5 = 720kmph
10. An athlete runs 240m race in 24 sec.Find his speed
A)30kmph
B)36kmph
C)40kmph
D)63kmph
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https://math.stackexchange.com/questions/3808159/are-the-ideals-of-a-ring-with-cyclic-additive-group-always-principal
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# Are the ideals of a ring with cyclic additive group always principal?
Note for me rings need not be unital or commutative.
Let $$R$$ be a ring with cyclic additive group $$(R, +, 0)$$ and let $$I$$ be an ideal in $$R$$. Is $$I$$ principal?
Here's my attempt, assuming $$R$$ has a $$1$$ and $$1$$ generates the additive group $$(R,+,0)$$:
Since $$(R,+,0)$$ is cyclic and $$(I,+,0)$$ is an additive subgroup of $$(R,+,0)$$, it is also cyclic and generated by some $$a \in R$$. Best guess is $$I = (a)$$.
By definition, as sets $$(I, +, 0 ) = (\langle a \rangle , +, 0) \subseteq (a)$$ . Also if $$x \in (a)$$ then $$x = \sum _i r_i a s_i$$ for some $$r_i, s_i$$. Hence ( using poor notation)
$$x = \sum_i r_i a (1+...+1) = \sum_i r_i (a+...+a) \\ = \sum_i (1+...+1) (a+...+a) = \sum_i ((a+...+a) +... +(a+...+a)) \in (\langle a \rangle, +, 0)$$.
By double inclusion we have the desired equality. $$\blacksquare$$
Firstly is this correct and also what about the case where $$R$$ is not unital or the case where $$R$$ is unital but $$1$$ doesn't generate the additive group?
Many thanks!
EDIT:
For future reference. It is argued here Does the unit generate the additive group in a unital ring with cyclic additive group? that the condition that $$1$$ generates the additive group is infact implied by $$R$$ being unital and is therefore not needed.
Not necessarily. Consider the ideal $$8\mathbb Z$$ within the ring $$4\mathbb Z$$.
Edit: Or, maybe this one is clearer: consider the ideal $$6\mathbb Z$$ within the ring $$2\mathbb Z$$.
• I'm a bit confused. $8\mathbb{Z}$ is principal in $4\mathbb{Z}$ isn't it? Aug 30, 2020 at 9:54
• What ring element would generate it? Aug 30, 2020 at 9:58
• (just to cross word limit) 8? Aug 30, 2020 at 9:59
• Within $4\mathbb Z$, the ideal $(8)=32\mathbb Z\neq 8\mathbb Z$. Aug 30, 2020 at 10:00
• Oh! Nice! I see it now, thanks! Aug 30, 2020 at 10:02
Well, an ideal is an additive subgroup of the given ring. If the additive structure of the ideal is cyclic, then each element of the ideal can be written as $$rg$$, where $$r\in R$$ and $$g$$ is a generator of the additive cyclic structure. Hence, the ideal is principal.
• How do you prove the claim "If the additive structure of the ideal is cyclic, then each element of the ideal can be written as $rg$..."? This is equivalent to saying if $I$ is cyclic then it is principal, which is my question. Many thanks! Aug 30, 2020 at 9:43
• @Bellem Hmmm... it seems to me that you are mixing notation. You are using concatenation both as addition and multiplication, no? Aug 30, 2020 at 9:47
• I have been sloppy, I will answer better. Aug 30, 2020 at 9:55
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# Simplify asymptotic notation
Premise for context: Reading a paper I stumbled upon the following expression $$n(1+p)^{O(\log_{\sigma}n)} + 2$$ which is claimed to be $$O(n)$$ when $$p \in O(1/\log_{\sigma}n)$$, under the (reasonable) assumption that $$n > \sigma$$. Substituting we get $$n\left(1+O\left(\frac{1}{\log_{\sigma}n}\right)\right)^{O(\log_{\sigma}n)} + 2$$
As we are dealing with asymptotic notation I would look only at the higher order terms (this passage actually turned out to be completely wrong, have a look at the comments for details) getting to
$$n\left(1+O\left(\frac{1}{\log_{\sigma}n}\right)^{O(\log_{\sigma}n)}\right) + 2$$
Let now $$x=\log_{\sigma}n$$ and discard the constant to get to $$O(n)+n \cdot O\left(\frac{1}{x}\right)^{O(x)}$$
end of premise, if you like more details feel free to ask them.
To verify the claim we need to check the value of $$O\left(\frac{1}{x}\right)^{O(x)}$$, which should turn out to be $$O(1)$$ to satisfy the initial claim.
Looking at the value of $$y^{-y}$$ which tends to 1 for growing values of $$y$$ in $$\mathbb{N}$$ (e.g. considering $$y^{-y} = e^{-y\cdot \log y }$$ and seeing that the value $$y\cdot \log y$$ goes to $$\infty$$).
This seems correct to me but I was wondering if my (not so formal) reasoning holds or not and whether there is a better line of reasoning to verify this.
• It's not true that $(1+f(n))^{g(n)} = O(1+f(n)^{g(n)})$. (For one thing, increasing $g(n)$ makes the left side increase, but makes the right side decrease when $f(n)<1$.) In this case (and often when dealing with functions in exponents), I recommend writing $(1+p)^{O(g(n))} = \exp( g(n)\log(1+p) )$ and finding an upper bound for $g(n)\log(1+p)$ as an intermediate step; here you should be able to prove that it's $O(1)$. Commented Sep 14, 2023 at 21:39
• Start by $1 + p \le {\rm e}^p$ for $p\ge 0$.
– Gary
Commented Sep 15, 2023 at 4:05
As $$n\to\infty$$, $$n(1+p)^{O(\log_{\sigma}n)}=\exp\left(\ln\left( n\right)+O(\log_{\sigma}n)\ln(1+p)\right)=n\exp(O(\log_{\sigma}n)\ln(1+p))$$ substitue as you did, but note that by Taylor's formula, $$\ln(1+O(1/\log_\sigma n))\sim O(1/\log_\sigma n)$$ as $$n\to\infty$$, then using the fact that $$e^{O(1)}=O(1)$$, i.e., it is bounded, $$n\exp(O(1))=O(n)$$ as $$n\to\infty$$. As you have noticed, discarding the constant $$2$$ which is $$O(1)$$ doesn't matter here.
• Writing $\exp(O(1))\sim 1+O(1)$ does not make much sense. Just note that $\exp(O(1))$ is a bounded quantity, i.e., it is $O(1)$.
• @Gary. You are right, $e^{O(1)}$ is just a number. I will edit that.
• Thanks a lot. In place of Taylor's formula I think we can also use the simpler obesrvation $log(1+x) \le x$ s.t. $x > -1$ (which holds since $x = O(1/\log_{\sigma}n) \ge 0$ in our case) in this case as we are looking for an upper bound right? Commented Sep 15, 2023 at 7:48
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# What Exactly Is A Gaussian Blur?
Blurring is a commonly used visual effect when digitally editing photos and videos. One of the most common blurs used in these fields is the Gaussian blur. You may have used this tool thousands of times without ever giving it greater thought. After all, it does a nice job and does indeed make things blurrier.
Of course, we often like to dig deeper here at Hackaday, so here’s our crash course on what’s going on when you run a Gaussian blur operation.
## It’s Math! It’s All Math.
Digital images are really just lots of numbers, so we can work with them mathematically. Each pixel that makes up a typical digital color image has three values- its intensity in red, green and blue. Of course, greyscale images consist of just a single value per pixel, representing its intensity on a scale from black to white, with greys in between.
Regardless of the image, whether color or greyscale, the basic principle of a Gaussian blur remains the same. Each pixel in the image we wish to blur is considered independently, and its value changed depending on its own value, and those of its surroundings, based on a filter matrix called a kernel.
The kernel consists of a rectangular array of numbers that follow a Gaussian distribution, AKA a normal distribution, or a bell curve.
Our rectangular kernel consists of values that are higher in the middle and drop off towards the outer edges of the square array, like the height of a bell curve in two dimensions. The kernel corresponds to the number of pixels we consider when blurring each individual pixel. Larger kernels spread the blur around a wider region, as each pixel is modified by more of its surrounding pixels.
For each pixel to be subject to the blur operation, a rectangular section equal to the size of the kernel is taken around the pixel of interest itself. These surrounding pixel values are used to calculate a weighted average for the original pixel’s new value based on the Gaussian distribution in the kernel itself.
Thanks to the distribution, the central pixel’s original value has the highest weight, so it doesn’t obliterate the image entirely. Immediately neighboring pixels having the next highest influence on the new pixel, and so on. This local averaging smoothes out the pixel values, and that’s the blur.
Edge cases are straightforward too. Where an edge pixel is sampled, the otherwise non-existent surrounding pixels are either given the same value of their nearest neighbor, or given a value matching up with their mirror opposite pixel in the sampled area.
The same calculation is run for each pixel in the original image to be blurred, with the final output image made up of the pixel values calculated through the process. For grayscale images, it’s that simple. Color images can be done the same way, with the blur calculated separately for the red, green, and blue values of each pixel. Alternatively, you can specify the pixel values in some other color space and smooth them there.
Here we see an original image, and a version filtered with a Gaussian blur of kernel size three and kernel size ten. Note the increased blur as the kernel size increases. More pixels incorporated in the averaging results in more smoothing.
Of course, larger images require more calculations to deal with the greater number of pixels, and larger kernel sizes sample more surrounding pixels for each pixel of interest, and can thus take much longer to calculate. However, on modern computers, even blurring high-resolution images with huge kernel sizes can be done in the blink of an eye. Typically, however, it’s uncommon to use a kernel size larger than around 50 or so as things are usually already pretty blurry by that point.
The Gaussian blur is a great example of simple mathematics put to a powerful use in image processing. Now you know how it works on a fundamental level!
## 29 thoughts on “What Exactly Is A Gaussian Blur?”
1. quinor says:
That’s a great paper. I’ve been using a hillbilly version of that algorithm for my hanging plotter project but this one is significantly nicer! How did you stumble upon it?
1. I don’t remember … It was years ago where I implemented an BW filter for laser.im-pro.at ….
1. Steven Clark says:
Also the classic Sharpen filter is a RREF-style reversal of a gaussian blur kernel. Meanwhile Unsharp Mask is a gaussian blur applied in negative which is why it’s easier to parametrize: there’s no solving step.
2. snarkysparky says:
now how about inverting an applied gaussian blur to recover the prior image.
1. RustyHydrogen says:
Ah, the fun worlds of deconvolution, and blind deconvolution.
3. grounded says:
And if you apply the same blur sequentially? Are two 3×3 blurs the same as a single 5×5?
I guess I should do some math…
1. Marcus says:
Yep, doing the math helps here, to some degree:
the convolution of a Gaussian with a Gaussian is still a Gaussian (without that, there wouldn’t be the central limit theorem, i.e. the reason why sums of independent identically distributed random variables add up to Gaussian distribution for large numbers of summands). So, yes, if we take a Gaussian with some sigma² and convolve it with itself, we get a new Gaussian with twice the sigma² (Variance).
This means that if we decided the original discrete 3×3 kernel was “large” enough to cover the “interesting parts of the Gaussian because the stuff that was “cut off” from the (infinitely extending) Gaussian function was small enough for our tastes to not matter.
(I’m ***very*** much at odds with the article: “larger kernels spread further”: No; you can put the same 3×3 in a 11×11 matrix and fill in with the smaller values from the actual Gaussian function. What the author means is that a higher-variance Gaussian spreads further.)
But 5 is not at least twice 3… so, you’d get a worse approximation to an actual Gaussian filter with this finite kernel.
However 5×5 is indeed the size you get when you, in the discrete domain, convolve two 3×3 kernels. So, by doing it that way, you get a different Gaussian kernel, but it’s a worse approximation to the Gaussian of twice the variance of the Gaussian in the 3×3 kernel.
1. Lasse says:
Also the Gaussian is separable, so you can do the 2D blur by doing a 1D horizontally and then vertically.
4. Marcus says:
Is it just me that’s bothered by the logo in the post header *not* being the result of the sharp hackaday logo being filtered with a symmetric kernel,i.e. especially *not* with a Gaussian? That blur isn’t symmetrical at all!
1. Chris Cox says:
Nope, not just you. That kinda jumped out at me as well.
5. Marcus says:
Honestly, I like when HaD does such articles!
I must still voice my disagreement on a few things: the first two are fundamental, the third I think is a bit of an overgeneralization but not a fundamentally wrong statement; my apologies.
> The kernel consists of a rectangular array of numbers that follow a Gaussian distribution,
These numbers don’t follow a Gaussian distribution. They are deterministic; they are samples taken from the *Gaussian function*, which happens to be the probability density of the Gaussian distribution.
> Larger kernels spread the blur around a wider region,
This is wrong – the size of the kernel doesn’t define the spread, just the number of pixels taken into account.
The spread is defined by the variance of the Gaussian function that gets sampled for the kernel. You can have a very narrow blur “spread” in a large filter, if you need the accuracy, or a very wide (=high variance) blur in a small filter.
The question is really just how much of the actual continuous function you cut off when you decide on your filter size.
But these are two different properties of the filter, and you’re conflating them!
> as each pixel is modified by more of its surrounding pixels.
That’s right – a larger Kernel takes more pixels into account.
> Edge cases are straightforward too. Where an edge pixel is sampled, the otherwise non-existent surrounding pixels are either given the same value of their nearest neighbor, or given a value matching up with their mirror opposite pixel in the sampled area.
These are **two** options, but they are not necessarily the right ones for any particular application. In fact, that choice can become very awkward for the pixels that are not at the very image border – the actual border pixel get their “weight” increased very much, so that their values overshadow that of their neighbors.
So, not a great generalization to make!
6. echodelta says:
Now for a math routine to put shake and jiggle into stable video and stills for enhanced viewing.
1. Nick says:
Oh god, don’t give them ideas! We already have fake film lines and blemishes, and fake artefacts and glitches. Overuse of those is bad enough.
7. Ian Dobbie says:
I would question the large kernels taking significant processing time. Applying the blur is a convolution and almost any sensible algorithm will actually compute this as a multiplication in Fourier space. Take your kernel and image, FFT both, multiple them together and then do in the inverse FFT. Almost certainly massively faster than a direct real space convolution where you generate a new image, multiple every pixel by the kernel and sum the the relevant pixels into the results image.
1. ROFL. Nope. Gaussian kernels are separable, and can use the central limit theorem, plus some linear and constant time algorithms, to do the work much more quickly than you could do a 2D FFT (much less the multiply and reverse FFT). Plus the separability makes the operation much more cache friendly than an FFT. (see also: bicubic and sinc based resampling kernels)
If you are doing a convolution that isn’t separable, then an FFT (or other frequency space transform) might be a good idea. But please talk to algorithms and performance experts first, and measure, measure, measure.
1. intheoryonly says:
The source you linked to only considers theoretical number of operations. In practice, memory access order and caches are far more important. The crossover point where FFT-based convolution is faster than separable convolution will only happen for much larger inputs.
2. Yeah, that completely ignores better algorithms and cache effects. It’s using the number of operations for a direct convolution – not separable, not box filters (certainly not the constant time algorithms for box filters).
Next time, you probably should consult with someone who actually knows algorithms and performance.
1. snarkysparky says:
These people at Analog Devices are idiots. Shouldn’t consult them at all.
“”This chapter presents two important DSP techniques, the overlap-add method, and FFT
convolution. The overlap-add method is used to break long signals into smaller segments for
easier processing. FFT convolution uses the overlap-add method together with the Fast Fourier
Transform, allowing signals to be convolved by multiplying their frequency spectra. For filter
kernels longer than about 64 points, FFT convolution is faster than standard convolution, while
producing exactly the same result. “”
https://www.analog.com/media/en/technical-documentation/dsp-book/dsp_book_Ch18.pdf
Mr Algorithm guru would you be so kind as to point me to where I can gain understanding about this from the REAL experts?
2. Yes sparky, you are in way over your head. FFTs are good in some places, and not in many others. In real world image processing, an FFT for gaussian blur would be among the slowest algorithms (only exceeded by brute force convolution). You are confusing theory (written by people who don’t know the practice and who are talking entirely in generalities) with actual practice. I’m talking about actual code in major applications that has been highly optimized, measured, copied by other applications, and hits the limits for throughput on a variety of architectures and inputs (image size, kernel size, etc.).
How can you gain experience? Listen to the experts and stop trying to claim you know more than they do by quoting other people who know even less.
Here’s a hint: click on the link in my name above.
3. snarkysparky says:
Ok smartie Chris. You win. All i did was google fft vs convolution and the first 63455 articles all went with fft as faster once past a very minimal kernel size. I looked for something agreeing with you but couldn’t find it. One thing i have always noticed is that when someone makes the boneheaded claim about the difference between theory and practice they don’t actually know the theory. What they have in brain is a conglomeration of anecdotes that support their view.
You thought you could “one up” the experts by looking things up on Google. But you got the theory (and ignored all the warning signs that it was just theory), with zero practice and zero experience.
Again, you probably should click on the link in my name here to learn who it is that you are addressing. Yes, I know the theory, and I lecture on it a few times a year at the larger universities here in the SF Bay Area. I also spent many years on the practice: measuring everything, improving the algorithms until they were limited by the speed of RAM instead of the CPU, squeezing out a bit more performance with significant cache optimization, then re-optimizing for each new major chip architecture. There is a reason why the chip and motherboard makers consult with me about their latest designs and how their changes will affect real world throughput…
5. Hallo Chris,
I really appreciate your input to this topic. I’m now myself working in a research institute and my experience is not really good. We have an high incentive to write publications and there the quality is often not optimal. Furthermore when doing research I takes a lot of effort to filter out bad papers and find the practical problems.
I hob if somebody searches about FFT for convolution this will show up to save them a lot of time to go throw with the implementation and find in the end that it will not be faster for real applications.
Thanks Chris!
@sparky here is one link about chis you should open : http://www.photoshophalloffame.com/chris-cox
8. Chris Maple says:
The article would be greatly improved if the actual formula for pixel weighting was provided.
9. brandon says:
you guys are gaussian blurring and im still reticulating splines :(
10. Alan says:
If you have any interest in stitching images together (to make a panorama) then it helps to know about Gaussian blur.
Blurring to different kernel sizes, then subtracting (Difference of Gaussians), is one of the steps in keypoint detection algorithms like SIFT and SURF.
https://en.wikipedia.org/wiki/Image_stitching#Keypoint_detection
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ex2-sol10
# ex2-sol10 - AMS 341 (Spring, 2010) Exam 2 - Solution notes...
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Unformatted text preview: AMS 341 (Spring, 2010) Exam 2 - Solution notes Estie Arkin Mean 73.925, median 76, top quartile 87, high 99, low 12. 1. (10 points) I am planning a vacation once the spring semester is over. There are many tasks that have to be completed before I can go. Fortunately, I can get help from family and friends. The following are the tasks that would have to be undertaken before the vacation: Task Predecessors Time (hours) A- 2 B- 3 C A 1 D B 8 E B,C 7 F D,E 5 (a). Draw a project network. 1 2 3 4 5 6 A-2 B-3 C-1 dummy-0 D-8 E-7 F-5 Common mistakes: several start nodes, undirected edges, wrong predecessors. (b). What is my critical path? You may find the path either by computing the total float for each node, or by inspection. (Your answer should be a list all critical activities.) Tasks B,D,F (or nodes 1,3,5,6). 2. (10 points) Consider the following (minimum) Balanced Transportation problem: Find an initial BFS for the problem using the min cost method: 1 100 100 100 75 75 125 25 1 2 3 4 5 6 7 8 9 1 1 25 75 100 25 75 Common mistake: not enough basic variables. You should have 4 + 3 − 1 = 6. 3. (15 points) Plans are being made for the energy systems for a new building. The three possible sources of energy are electricity, natural gas and a solar heating unit. The energy requirements are: 20 units of electricity, 10 units for water heating and 30 units for space heating. The size of the roof limits the solar heater to 30 units. There is no limit on the amount of electricity or natural gas bought. Electricity needs can only be met by buying electricity at a cost of \$ 50 per unit. Other energy needs can be met by any sources with the following unit costs: Electricity Natural Gas Solar heater Water heating \$ 90 \$ 60 \$ 30 Space heating \$ 80 \$ 50 \$ 40 Formulate a Balanced Transportation Problem to minimize the...
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# ou have been given the following information for Moore’s HoneyBee Corp.: Net sales = \$44,000,000.Gross profit = \$19,400,000.Other operating expenses = \$3,400,000.Addition to retained earnings = \$8,328,000.Dividends paid to preferred and common stockholders = \$2,100,000.Depreciation expense = \$2,000,000. The firm’s tax rate is 21 percent. The firm's interest expense is all tax deductible.Calculate the cost of goods sold and the interest expense for Moore’s HoneyBee Corp. (Round your answers to the nearest dollar amount.) Cost of goods sold Interest expense
Question
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ou have been given the following information for Moore’s HoneyBee Corp.:
1. Net sales = \$44,000,000.
2. Gross profit = \$19,400,000.
3. Other operating expenses = \$3,400,000.
4. Addition to retained earnings = \$8,328,000.
5. Dividends paid to preferred and common stockholders = \$2,100,000.
6. Depreciation expense = \$2,000,000.
The firm’s tax rate is 21 percent. The firm's interest expense is all tax deductible.
Calculate the cost of goods sold and the interest expense for Moore’s HoneyBee Corp. (Round your answers to the nearest dollar amount.)
Cost of goods sold Interest expense
check_circle
Step 1
Recall the mathematical equation:
Gross Profit = Net Sales - Cost of Goods sold
Hence, 19,400,000 = 44,000,000 - Cost of Goods sold
Hence, Cost of Goods sold = 44,000,000 - 19,400,000 = \$ 24,600,000
Step 2
Recall the fundamental equation of accounting of net income:
Net income = Addition to retained earnings + Dividends paid to preferred and common stockholders =(Gross Profit - Other Operating expenses - Depreciaton - Interest) x (1 - Tax rat...
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This site is being phased out.
# Calculus III -- Spring 2014 -- final exam
Name:_________________________
10 problems, 10 points each
• Justify every step you make with as thorough explanation as possible.
• Unless requested, no decimal representation of the answers is necessary.
• Start every problem at the top of the page.
$\bullet$ 1. (1) Sketch the parametric curve $x=\cos t,y=\sin 2t$. (2) The curve intersects itself. Find the angle of this intersection.
$\bullet$ 2. Sketch the graph of a function of two variables $z=f(x,y)$ the derivatives of which have the following signs: $f_x>0, f_{xx}>0, f_y<0, f_{yy}<0$.
$\bullet$ 3. The graph of a function of two variables $z=f(x,y)$ is given below along with four points on the graph. Sketch the gradient for each on a separate $xy$-plane:
$\bullet$ 4. Give the definition of the curvature. Give examples of curves with various curvatures.
$\bullet$ 5. A vector field is sketched below. Suppose $C$ is the clockwise oriented square centered at the origin. Is $\int _{C}\mathbf{F}\cdot d\mathbf{r}$ positive, negative or $0$? Explain.
$\bullet$ 6. This is the formula of Green's Theorem: $$\oint_{C} (L\, dx + M\, dy) = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, dA.$$ Explain its parts and the conditions that have to be satisfied.
$\bullet$ 7. Prove that the vector field $F(x,y,z)=z\mathbf{j} - y\mathbf{k}$ is not conservative.
$\bullet$ 8. Make a sketch of contour (level) curves for the function below:
$\bullet$ 9. Sketch the vector field $$F(x,y)=\dfrac{1}{x^{2}+y^{2}}(y\mathbf{i}-x\mathbf{j}).$$
$\bullet$ 10. Find the work done by force field $$F(x,y)=< xy,y^{2}>.$$ in moving an object along the parabola $x=t,y=t^{2},0\leq t\leq1.$
$\bullet$ 11. Use a Riemann sum with $8$ terms to estimate the value of the integral $$\iiint_{D}(x+y+z)dV,$$ over the cube $D=[0,1]\times[0,1]\times[0,1]$. Choose your own sample points.
$\bullet$ 12. (1) Represent the cylinder of radius $1$ and height $1$ centered on the $z$-axis as a parametric surface. (2) Find the tangent plane to the cylinder at the point $(\sqrt{2}/2,\sqrt{2}/2,1/2)$. (3) Compute the flux of the vector field $F=< 2,1,1 >$ across the part of the cylinder that lies in the first octant.
$\bullet$ Extra credit problem. (5 points, no partial credit) Suppose you are towing a trailer-home. During the first few minutes, every time you look at the rear view mirror you can see only the lower part of the home. Later, every time you look you can see only the top part. Discuss the profile of the road.
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https://www.esaral.com/q/find-the-coordinates-of-the-foci-the-vertices-the-length-of-major-axis-the-minor-axis-the-eccentricity-and-the-length-81740
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Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length
Question:
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $16 x^{2}+y^{2}=16$
Solution:
The given equation is $16 x^{2}+y^{2}=16$.
It can be written as
$16 x^{2}+y^{2}=16$
Or, $\frac{x^{2}}{1}+\frac{y^{2}}{16}=1$
Or, $\frac{x^{2}}{1^{2}}+\frac{y^{2}}{4^{2}}=1$ $\ldots(1)$
Here, the denominator of $\frac{y^{2}}{4^{2}}$ is greater than the denominator of $\frac{x^{2}}{1^{2}}$.
Therefore, the major axis is along the $y$-axis, while the minor axis is along the $x$-axis.
On comparing equation (1) with $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$, we obtain = 1 and a = 4.
$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{16-1}=\sqrt{15}$
Therefore,
The coordinates of the foci are $(0, \pm \sqrt{15})$.
The coordinates of the vertices are $(0, \pm 4)$.
Length of major axis = 2a = 8
Length of minor axis = 2b = 2
Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{15}}{4}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 1}{4}=\frac{1}{2}$
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CC-MAIN-2024-38
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latest
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