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3 years ago

Maximum volume of a pyramid that can fit inside a cube with a side length of 12 cm?

Mathematics

1 answer:

Luden [163]3 years ago

4 0

Hello, did you ever finish this test?

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<img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B7%20%7B%7D%5E%7B7%7D%20%7D" id="TexFormula1" title=" \sqrt{7 {}^{7} }" alt=" \sqr

Mrrafil [7]

Answer:

$\boxed{\tt\blue{=343\sqrt{7}}}$

$\tt\red{=907.4927}$

Step-by-step explanation:

$\sqrt{7 {}^{7} }$

$\sf=343\sqrt{7}$

<u>OR</u>

Decimal:

$\tt=907.4927$

7 0

2 years ago

Read 2 more answers

Please help me asap.

KonstantinChe [14]

Answer:

Step-by-step explanation:

You can tell that the angles and length of the triangles are equal from the marks. For example, angle J has 2 lines on its angle so i looked for the other angle with 2 lines on it which happens to be angle S.

Comment if u dont understand

3 0

2 years ago

Please help! I mark brainliest :3

vladimir1956 [14]

no for a
yes for B
no for c
that what I think It is opinion could be wrong

3 0

2 years ago

RIP OpenStudy ;(

klemol [59]

First note that $\frac{2^n+1}{2^{n+1}} = \frac{2^n}{2^{n+1}} + \frac{1}{2^{n+1}} = \frac{1}{2} + \frac{1}{2^{n+1}}$

If you take limit, then you have $\lim_{n \to \infty}( \frac{1}{2} + \frac{1}{2^{n+1}})= \lim_{n \to \infty}( \frac{1}{2}) +\lim_{n \to \infty}(\frac{1}{2^{n+1}})=\frac{1}{2} +0= \frac{1}{2}$

3 0

2 years ago

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What is the soulution of x^2 + 64 = 0

lys-0071 [83]

Solve out
x^2 + 64 = 0
x^2 = -64 ( we are going to use imaginary numbers to solve)
x = $i \sqrt{64}$
x = 8i, -8i final answers

3 0

2 years ago

Read 2 more answers