Question about Computers & Internet

Hi,

You are absolutely right.

The 3-D figure is of cylinder.

Cylinder's volume is given by (pie)*r*r*h.

Here, r is the radius = 6 in.

and h is the height = 10 in.

Volume = 3.14*6*6*10.
= 1130.4 cubic in.

Thanks for using Fixya.

Good Luck and have a nice day.

Posted on May 27, 2010

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Posted on Jan 02, 2017

The radius or the volume of the space?

Feb 07, 2015 | Miscellaneous

(pi)R squared (Fixya won't let me use the sign for Pi)

Area of a circle=pi*radius^2 or pi times radius times radius

example: suppose the radius of a circle is 10 inches, then the Area of the circle would be pi times 10 times 10 or 100*pi

pi represents the circumference (distance around) of a circle / diameter and for any circle is constantly the same number which is approximately 3.14

so in my example 100*pi=100*3.14 or 314 square inches

Area of a circle=pi*radius^2 or pi times radius times radius

example: suppose the radius of a circle is 10 inches, then the Area of the circle would be pi times 10 times 10 or 100*pi

pi represents the circumference (distance around) of a circle / diameter and for any circle is constantly the same number which is approximately 3.14

so in my example 100*pi=100*3.14 or 314 square inches

Jul 03, 2014 | Computers & Internet

Circumference of a Circle: 2 x pi x radius - OR - pi x diameter,

or if you need some sphere calculations?

Sphere Surface Area: 4 X pi X radius squared OR

pi X diameter squared Sphere Volume: 4/3 X pi X radius cubed OR

( pi X diameter cubed ) / 6

By the way.......... "pi" = 3.1415926536

or if you need some sphere calculations?

Sphere Surface Area: 4 X pi X radius squared OR

pi X diameter squared Sphere Volume: 4/3 X pi X radius cubed OR

( pi X diameter cubed ) / 6

By the way.......... "pi" = 3.1415926536

Apr 29, 2014 | Office Equipment & Supplies

To get a circle with a perimeter of 18 inches, you want a diameter of about 5.73 inches (just under 5 3/4 inches) or a radius of 2.86 inches (call it 2 7/8 inches). Get a compass, set its legs 2 and 7/8 inches apart, put one end at the center, and turn.

Oct 23, 2013 | Computers & Internet

(((a-10) + (b-10) + (c-10) + (d-10))/4) = 24

((a+b+c+d+e+e)/6) = 24

(((a-10) + (b-10) + (c-10) + (d-10))/4) = ((a+b+c+d+e+e)/6)

Solve for e

((a+b+c+d+e+e)/6) = 24

(((a-10) + (b-10) + (c-10) + (d-10))/4) = ((a+b+c+d+e+e)/6)

Solve for e

Nov 19, 2010 | Answer Racing Ion Motorcycle Jersey 2010

Hi

I'm studying Physics, not CS, but I've had a few brushes with OpenGL.

I'll use a Basic-like pseudocode syntax since I don't know what language you're using, and basic is very easy to read:

'Constants - Radius is radius of the circle

Const PI = 3.14159, Radius = 10

'Current angle, and angular velocity (dTheta / dt)

Dim Shared Theta as Single, Omega as Single

'Frame is called when you want to draw a new frame

Sub Frame(dt as Single)

'dt is number of seconds passed since last frame (typically less than one, since you want several frames per second)

'Initialize the frame (clear buffers, set up projection, etc)

InitFrame

'Increase angle

Theta = Theta + (Omega * dt)

'Calculate co-ordinates of point

X = Radius * Cos(Theta)

Y = Radius * Sin(Theta)

glBegin(GL_POINTS)

'Set the colour of the point

glColor(<r>, <g>, <b>, <a>)

'Draw the point

glVertex2f(X, Y)

glEnd()

'Display the frame

RenderFrame

End Sub

I'm studying Physics, not CS, but I've had a few brushes with OpenGL.

I'll use a Basic-like pseudocode syntax since I don't know what language you're using, and basic is very easy to read:

'Constants - Radius is radius of the circle

Const PI = 3.14159, Radius = 10

'Current angle, and angular velocity (dTheta / dt)

Dim Shared Theta as Single, Omega as Single

'Frame is called when you want to draw a new frame

Sub Frame(dt as Single)

'dt is number of seconds passed since last frame (typically less than one, since you want several frames per second)

'Initialize the frame (clear buffers, set up projection, etc)

InitFrame

'Increase angle

Theta = Theta + (Omega * dt)

'Calculate co-ordinates of point

X = Radius * Cos(Theta)

Y = Radius * Sin(Theta)

glBegin(GL_POINTS)

'Set the colour of the point

glColor(<r>, <g>, <b>, <a>)

'Draw the point

glVertex2f(X, Y)

glEnd()

'Display the frame

RenderFrame

End Sub

May 16, 2009 | Advanced Graphics Programming Using OpenGL...

Hello,

Sorry to say it but the question does not make sense. A circle is a figure that is perfectly defined by the knowledge of its center (or centre depending of the English you use) and its radius.

An ellipse is perfectly defined if you know its two foci (plural of focus) and the length each of its two axes (major and minor).

Is it the perimeter, the area, the parameter p of the ellipse you are lookong for.

Hope it will help you focus on your real the question.(The pun is intended)

Sorry to say it but the question does not make sense. A circle is a figure that is perfectly defined by the knowledge of its center (or centre depending of the English you use) and its radius.

An ellipse is perfectly defined if you know its two foci (plural of focus) and the length each of its two axes (major and minor).

Is it the perimeter, the area, the parameter p of the ellipse you are lookong for.

Hope it will help you focus on your real the question.(The pun is intended)

Feb 19, 2009 | The Learning Company Achieve! Math &...

I'm not sure what your ellipse looks like but the equation for the area of an ellipse is given by:

pi times the long axis times the short axis

pi = 3.1416

The length of the long axis is from the center of the ellipse to the outer edge of the ellipse

The length of the short axis is from the center of the ellipse to the outer edge of the ellipse

Hope this helps Loringh

pi times the long axis times the short axis

pi = 3.1416

The length of the long axis is from the center of the ellipse to the outer edge of the ellipse

The length of the short axis is from the center of the ellipse to the outer edge of the ellipse

Hope this helps Loringh

Nov 01, 2008 | The Learning Company Achieve! Math &...

assuming the question is what is the circle equation?

and if (-2,2) is the center of the circle

the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5

from that we can see that the slope of the line is 2/5

And from the fact of perpendicular line we can say that the slope

of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2

radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:

-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following

Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=

SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25

and if (-2,2) is the center of the circle

the equation should look like this: (x+2)^2+(Y-2)^2=R^2

And now only R is needed.

given 2x-5y+4=0 equation of line perpendicular

we can rearange the equation to be y=(2x+4)/5

from that we can see that the slope of the line is 2/5

And from the fact of perpendicular line we can say that the slope

of the radius line is -2/5.

The motivation now is to calculate the distance between the center of the circle to the cross point of the radius with the line perpendicular

For that we would calculate the radius line equation and compare it to the equation of line perpendicular

As mentioned earlier the slope of the radious line is -2/5.

So the equation is y=-2/5x+b and b can be calculated by using the center of the circle coordinates

2= - (2/5)*(-2)+b ------> b=2-4/5=1.2

radius equation is y=-(2/5)x+1.2

Now the cross point is calculated by comparing the equations:

-(2/5)x+1.2=(2x+4)/5 --> -2x+6=2x+4 --> 4x=2 --> x=1/2 --> y=1

So the cross point is (1/2,1).

The distance between the points is calculated by the following

Formula:

R=SQR(((1/2)-(-2))^2+(2-1)^2)=SQR(2.5^2+1^2)=SQR(6.25+1)=

SQR(7.25)

Therefore the circle eq is (x+2)^2+(Y-2)^2=7.25

Oct 26, 2008 | Casio FX-115ES Scientific Calculator

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