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3 years ago

Name a point that is NOT coplanar with G, A, and B

Mathematics

1 answer:

GenaCL600 [577]3 years ago

5 0

The point that is not coplanar with G, A, and B are E, F, G, H and J.

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A pyramid has a base with a demension 5 inches by 7 inches .The height of the pyramid is unknown .If the volume of the pyramid i

Ulleksa [173]

Answer:

9 inches

Step-by-step explanation:

The volume of a pyramid is given as:

$V = \frac{l* w*h}{3}$

where l = length of base of pyramid

w = width of base of pyramid

h = height of pyramid

The volume of the cylinder is already known ( $105 in^3$ ) and the other two dimensions are known as 5 in and 7 in.

From the formula of the volume of the pyramid, the height of the pyramid is given as:

$h = \frac{3V}{l*h}$

Hence, the height of the pyramid is:

$h = \frac{3 * 105}{5 * 7}\\ \\h = \frac{315}{35} = 9in$

The pyramid's height is 9 inches.

7 0

3 years ago

Read 2 more answers

Which of the following is the cube root of 125 ?

atroni [7]

Answer:

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

This is algebra please help

Ghella [55]

Answer:

1/8 and 1/16

Step-by-step explanation:

The negative exponent just means to right the negative inverse.

For example:

4^-2 means 4^2 but in a fraction

the answer would be 1/16

4 0

3 years ago

Find the area of the helicoid (or spiral ramp) with vector equation r(u, v) = ucos(v) i + usin(v) j + v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 9π

Natasha2012 [34]

Let $H$ denote the helicoid parameterized by

$\mathbb r(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+v\,\mathbf k$

for $0\le u\le1$ and $0\le v\le9\pi$ . The surface area is given by the surface integral,

$\displaystyle\iint_H\mathrm dS=\iint_H\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv$

We have

$\mathbf r_u=\dfrac{\partial\mathbf r(u,v)}{\partial u}=\cos v\,\mathbf i+\sin v\,\mathbf j$
$\mathbf r_v=\dfrac{\partial\mathbf r(u,v)}{\partial v}=-u\sin v\,\mathbf i+u\cos v\,\mathbf j+\mathbf k$
$\implies\mathbf r_u\times\mathbf r_v=\sin v\,\mathbf i-\cos v\,\mathbf j+u\,\mathbf k$
$\implies\|\mathbf r_u\times\mathbf r_v\|=\sqrt{1+u^2}$

So the area of $H$ is

$\displaystyle\iint_H\mathrm dS=\int_{v=0}^{v=9\pi}\int_{u=0}^{u=1}\sqrt{1+u^2}\,\mathrm du\,\mathrm dv$
$=\dfrac{9(\sqrt2+\sinh^{-1}(1))\pi}2$

5 0

3 years ago

How do i solve this

Montano1993 [528]

The answer to this problem is 3 is less than or equal to 17

You solve the problem by subtracting 2 from 5 because you have to solve for X. 5 minus 2 is 3, so you would get 3 if you had the problem.

Hope this helps!

-Dub-

6 0

3 years ago