3 games for 45.90 what equation can be used to determine p of one game

2 answers:

8 0

To find the price of one game you..

Divide 45.90 by 3

Which equals = 15.30

Each of the games each costed 15.30

So your answer is: 15.30

DerKrebs [107]3 years ago

7 0

45.90 divided by 3
if you do this then you will get your answer for how much one costs. You might get a decimal. If you do then round to the nearest whole number. Of course while you answer the question you need to put your label. $

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Find the volume of a regular hexahedron if one of the diagonals of its faces is 8 root 2 inches.

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

D² = f² + a²
f² = 2a²

D² = 2a² + a² = 3a²
D = √3a² = √3 * √a² = √3 * a = a√3

Volume of a cube with side a is: V = a³
D = a√3
⇒ a = D/√3

V = a³ = (D/√3)³

We have:
D = 8√2 in

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

8 0

3 years ago

When is the additive relationship used? Select all that apply.

Mrac [35]

Answer:

your answer is for sure D but I think C is also the answer

Step-by-step explanation:

Additive relationship are two quantities can be expressed as related to each other through addition. It can be written as y = x + a, where y is related to x through the addition of a constant, a. The value for a may be positive or negative.

7 0

3 years ago

Read 2 more answers

The area of the circle is 144π m2<br><br> What is the diameter of the circle?

Over [174]

Answer:

Step-by-step explanation:

Remember that

Area of Circle = $\pi r^2$