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X^2/4 - y^2/19 = 1 has major axis of length ___?

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Error 86:01


did you perform a scan axis calibration after changing all these parts? If not then you should do that before doing anything else.

Sep 01, 2014 | HP Designjet Z6100 InkJet Printer

1 Answer

How to calculate the volume of a oval shaped spher


The volume of a solid of revolution can be calculated by making use of some methods of integral calculus. For that you will need the exact equation of the the surface that limits the solid. Nothing you can do on a calculator.
However for the spheroids there exit formulas you can use on a calculator.
  • For an ellipsoid
If a, b, c are the lengths of the semiaxes (length of semiaxis=half the length of the axis)
V=(4/3)PI*a*b*c

  • For an oblate spheroid
An oblate spheroid is the solid of revolution formed by the rotation of an ellipse about its minor axis (here an image of the cross section from Wikipedia).

e8cb930e-aad9-48e2-b8a8-58c916a2eda7.png

V=(4/3)PI*(a^2)*b

  • For a prolate spheroid
A prolate spheroid is the solid of revolution formed by the rotation of an ellipse about its major axis

V=(4/3)PI*a*(b^2)

Feb 27, 2014 | Office Equipment & Supplies

2 Answers

What operations can a lathe perform?


a lathe can cut, sand, drill, knurl, deface, face, turn, and deform

Jan 23, 2013 | Tools & Hardware - Others

1 Answer

Given a length of 30 cm on the x axis and a length of 10 cm on the y axis and a angle of 90 at the x/y axis intersection is there a short cut to calculate interior angles and hypot. of the triangle?


Yes, there is shortcut because this is right triangle, so you can use Pythagorean theorem (see picture).
  1. Length of hypotenuse is square root of sum of squares of lengths of other two sides of triangle, which is equal to square root of 30^2+10^2=31.6 cm.
  2. Sin(a)=longer cathetus/hypotenuse=0.949 so a=arcsin(0.949)=71.6 degrees
  3. Finally b=90-a=18.4 degrees.
elessaelle.png

If this was helpful please rate 4 thumbs :)

Sep 05, 2011 | Texas Instruments TI-30XA Calculator

1 Answer

Find the equation of the hyperbola which satisfies the given equation center(1,-2) transverse axis to the x-axis transverse 6 and conjugate axis 12


I am not quite sure how the major axis of your hyperbola is directed and i do not know if the lengths you give are measures of the major and minor axes or the measures of the semi-major and semi-minor axes. So I am giving you the equations and the graphs so that you can decide for yourself what is appropriate for your problem.

Major axis parallel to the X-axis
Equation and graph

k24674_37.jpgk24674_38.jpg
Center is at x=1 and y=-2, semi-major axis length is a=6, and semi-minor axis length is b=12


Major axis is parallel to the y-axis
Equation
k24674_39.jpgk24674_40.jpg

Center is at x=1 and y=-2, semi-major axis length is a=12, and semi-minor axis length is b=6.

I trust you can customize the equations to fit your need.

Mar 24, 2011 | Office Equipment & Supplies

1 Answer

Two vectors, 10 and 15 units long, make 60 degrees with each other. Find the magnitude and direction of their resultant with respect to the longer vectors.


A(the longer vector=15units) will have the components on x and y axis. (x=15, y=0)
B(the shortest vector=10units) will have the components(x=5,y=5/sqrt(3))
R = A+B (the sum vector) will have the components (x=20, y=5*sqrt(3)), thus the length of R is 5*sqrt(19) and the angle with respect to A is 23.41deg or arctan(5*sqrt(3)/20) rad. = arctan(sqrt(3)/4)~0.40rad.

Jul 11, 2009 | Darda Race Track Physics _

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Radius of an ellipse


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)

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

1 Answer

Image problem


IF I understand the problem from the limited info in your question, the problem is a distorted picture on whatever surface [probably a wall, or screen] onto which the image is projected.

With all lenses, the axis [centerline of the lens projected] MUST be absolutely PERPENDICULAR [in both the vertical and horizontal planes] to the perfectly plane [flat] surface onto which the image is projected.

IF for example, the projector is mounted in a rear corner of the room, and the camera is aimed downward and to one side to project the image on the center of the opposite wall, or a screen located there, the image will be distorted. It will be larger on one edge [fartherist from the lens] and smaller on the opposite [nearest] edge.

If it is non-perpendicular in only one axis, then two opposite edges will be different lengths, and if out of perpendicular in two axis then no edges will be the same length.

The solution to this problem is that the center axis of the lens must be perpendicular to the surface projected on, in both vertical and horizontal axis.

Feb 14, 2009 | Sharp BQC-XGE630U/1 Projector Lamp for...

1 Answer

What is the forular/working out of- ellipse with a long axis 2.500mm and short axis .850 (double the 0.425 radius) the area is 1.669 thus the area of the part circle is 0.835. what is the formular


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

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

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