I don't know how to solve this at all. please help

Mathematics

1 answer:

dybincka [34]3 years ago

3 0

Answer:

5. 97

Step-by-step explanation:

Well, we know that all we have to do is add QRP and PRS. so 71+25 is 96 (but put 97). For the other one you can do 2x-5=x+1. Then you just solve for x. For the y value you do the same thing, 3y=5y+3. So, x=2 but y=1.5 but its not an answer choice... I know the second one didn't help you, sorry. :(

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Naily [24]

The answer is 114sqrt{6} in³

A regular hexahedron is actually a cube.

Diagonal of a cube D is a hypotenuse of a right triangle which other two legs are face diagonal (f) and length of a side (a):
D² = f² + a²

Face diagonal is a hypotenuse of a right triangle which sides are a and a:
f² = a² + a² =2a²

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f² = 2a²

D² = 2a² + a² = 3a²
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V = a³ = (D/√3)³

We have:
D = 8√2 in

$V= (\frac{D}{ \sqrt{3} } )^{3} =( \frac{8 \sqrt{2} }{ \sqrt{3} } )^{3}=( \frac{8 \sqrt{2} * \sqrt{3} }{ \sqrt{3}* \sqrt{3}} )^{3} =( \frac{8 \sqrt{2*3} }{ \sqrt{3*3}} )^{3} =( \frac{8 \sqrt{6} }{ \sqrt{3^{2} }} )^{3} =( \frac{8 \sqrt{6} }{ 3} )^{3} = \\ \\ = \frac{ 8^{3} *( \sqrt{6} )^{3}}{3^{3}} = \frac{512* \sqrt{6 ^{2} }* \sqrt{6} }{27} = \frac{512*6* \sqrt{6} }{27} = 114sqrt{6}$

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I don't know how to solve this at all. please help​

I don't know how to solve this at all. please help