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2 years ago

Guys I need help with this problem pls!

Mathematics

2 answers:

Minchanka [31]2 years ago

8 0

Answer:

Most likely A, maybe b not c

Step-by-step explanation:

UkoKoshka [18]2 years ago

3 0

The answer is B

- hope this helped. :)

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77. the volume of a cube is increasing at a rate of <img src="https://tex.z-dn.net/?f=10%20%5Cmathrm%7B~cm%7D%5E%7B3%7D%20%2F%20

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Step-by-step explanation:

Given

$\displaystyle \frac{dV}{dt}=10\:\text{cm}^3/\text{min}\\ \\V=s^3\\\\SA=6s^2\\\\\frac{d(SA)}{dt}=?}\:;s=30\text{cm}$

Solution

(1) Find the rate of the cube's edge length with respect to time at s=30:

$\displaystyle V=s^3\\\\\frac{dV}{dt}=3s^2\frac{ds}{dt}\\ \\10=3(30)^2\frac{ds}{dt}\\ \\10=3(900)\frac{ds}{dt}\\\\10=2700\frac{ds}{dt}\\\\\frac{10}{2700}=\frac{ds}{dt}\\\\\frac{ds}{dt}=\frac{1}{270}\text{cm}/\text{min}$

(2) Find the rate of the cube's surface area with respect to time at s=30:

$\displaystyle SA=6s^2\\\\\frac{d(SA)}{dt}=12s\frac{ds}{dt}\\ \\\frac{d(SA)}{dt}=12(30)\biggr(\frac{1}{270}\biggr)\\\\\frac{d(SA)}{dt}=\frac{360}{270}\biggr\\\\\frac{d(SA)}{dt}=\frac{4}{3}\text{cm}^2/\text{min}$

Therefore, the surface area increases when the length of an edge is 30 cm at a rate of $\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$ .

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