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Alenkinab [10]
2 years ago
7

Find the value of x that maximized the volume of the following:

Mathematics
1 answer:
Shtirlitz [24]2 years ago
5 0

Check the picture below.

as you can see, the graph of the volume function comes from below goes up up up, reaches a U-turn then goes down down, U-turns again then back up to infinity.

the maximum is reached at the close up you see in the picture on the right-side.

Why we don't use a higher value from the graph since it's going to infinity?

well, "x" is constrained by the lengths of the box, specifically by the length of the smaller side, namely 5 - 2x, so whatever "x" is, it can't never zero out the smaller side, and that'd happen when x = 2.5, how so? well 5 - 2(2.5) = 0, so "x" whatever value is may be, must be less than 2.5, but more than 0, and within those constraints the maximum you see in the picture is obtained.

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Use the Distributive Property to rewrite the algebraic expression. 7(6m + 7)
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Step-by-step explanation:

first, distribute 7 to 6m (this means 7 • 6m)

7 • 6m = 42m

then distribute 7 to 7 (which is 7 • 7)

7 • 7 = 49

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3 years ago
HELP THIS IS DUE IN 15 MINUTES I DONT KNOW HOW TO SOLVE IT!!!
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Step-by-step explanation:

x/4 -y/12 = 0

--> x/4 = y/12

substitute into the first eq.

x/4 + y/12 = -5

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During an April rainstorm, it rained for six straight days. The amount of rainfall for each day was 0.8 inch, 2.15 inches, 2 inc
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Which represents r in terms of A and S?
ladessa [460]

Answer:

r=\frac{6\sqrt{10AS}}{S\sqrt{\pi }}

Step-by-step explanation:

Given the formula;

A=\frac{\pi r^2S}{360}

We want to solve the given formula for r.

Multiply both sides by \frac{360}{\pi S}

A\times \frac{360}{\pi S}=\frac{\pi r^2S}{360} \times \frac{360}{\pi S}

A\times \frac{360}{\pi S}=r^2

Take square root of both sides

r=\sqrt{\frac{360A}{\pi S}}

r=\frac{\sqrt{360A}}{\sqrt{\pi S}}

r=\frac{\sqrt{360A}}{\sqrt{\pi }\sqrt{S}}

r=\frac{\sqrt{360A}\times \sqrt{S}}{\sqrt{\pi }\sqrt{S} \times \sqrt{S}}

r=\frac{\sqrt{360AS}}{S\sqrt{\pi }}

r=\frac{6\sqrt{10AS}}{S\sqrt{\pi }}

5 0
3 years ago
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