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3 years ago

If two pyramids have the same height, what must be true of the pyramids for them to also have the same volume?

Mathematics

2 answers:

Alik [6]3 years ago

8 0

The area of the bases must be the same. I just got it correct on the quiz.

victus00 [196]3 years ago

4 0

Answer:

<u>The answer is the option</u>

The areas of the bases must be the same

Step-by-step explanation:

we know that

The volume of the pyramid is equal to

$V=\frac{1}{3}Bh$

where

B is the area of the base of the pyramid

h is the height of the pyramid

In this problem we have

<u>Pyramid N 1</u>

$h1=h\ units$

$B=B1\ units^{2}$

Substitute

$V1=\frac{1}{3}B1h$

<u>Pyramid N 2</u>

$h2=h\ units$

$B=B2\ units^{2}$

Substitute

$V2=\frac{1}{3}B2h$

Remember that

the two pyramids have the same volume

$V1=V2$

$\frac{1}{3}B1h=\frac{1}{3}B2h$

$B1=B2$

therefore

The areas of the bases must be the same

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2 years ago

Find the distance between the points (5, -5) and (-7,4).

abruzzese [7]

Answer:

<h2>The answer is 15 units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{ ({x1 - x2})^{2} + ({y1 - y2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(5, -5) and (-7,4)

The distance between them is

$d = \sqrt{ ({5 + 7})^{2} + ({ - 5 - 4})^{2} } \\ = \sqrt{ {12}^{2} + ( { - 9})^{2} } \\ = \sqrt{144 + 81} \\ = \sqrt{225} \\ = 15 \: \: \: \: \: \:$

We have the final answer as

<h3>15 units</h3>

Hope this helps you

4 0

3 years ago

Find the coordinates of T' after the given translation.

Karo-lina-s [1.5K]

I believe it’s C because the original coordinates where (1, -3) all you will have to do it’s subtract -6 by 1 and add 2 and -3

8 0

3 years ago

Please help with these problems <br> thank you so much!<br><br> 9. 10. 11. 12. 13. 14.

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4 0

4 years ago

Find the surface area of the solid generated by revolving the region bounded by the graphs of y = x2, y = 0, x = 0, and x = 2 ab

Nikitich [7]

Answer:

See explanation

Step-by-step explanation:

The surface area of the solid generated by revolving the region bounded by the graphs can be calculated using formula

$SA=2\pi \int\limits^a_b f(x)\sqrt{1+f'^2(x)} \, dx$

If $f(x)=x^2,$ then

$f'(x)=2x$

and

$b=0\\ \\a=2$

Therefore,

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+(2x)^2} \, dx=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx$

Apply substitution

$x=\dfrac{1}{2}\tan u\\ \\dx=\dfrac{1}{2}\cdot \dfrac{1}{\cos ^2 u}du$

Then

$SA=2\pi \int\limits^2_0 x^2\sqrt{1+4x^2} \, dx=2\pi \int\limits^{\arctan(4)}_0 \dfrac{1}{4}\tan^2u\sqrt{1+\tan^2u} \, \dfrac{1}{2}\dfrac{1}{\cos^2u}du=\\ \\=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0 \tan^2u\sec^3udu=\dfrac{\pi}{4}\int\limits^{\arctan(4)}_0(\sec^3u+\sec^5u)du$

Now

$\int\limits^{\arctan(4)}_0 \sec^3udu=2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17})\\ \\ \int\limits^{\arctan(4)}_0 \sec^5udu=\dfrac{1}{8}(-(2\sqrt{17}+\dfrac{1}{2}\ln(4+\sqrt{17})))+17\sqrt{17}+\dfrac{3}{4}(2\sqrt{17}+\dfrac{1}{2}\ln (4+\sqrt{17}))$

Hence,

$SA=\pi \dfrac{-\ln(4+\sqrt{17})+132\sqrt{17}}{32}$

3 0

3 years ago