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velikii [3]
3 years ago
10

The radius, r, of a sphere is increasing at 0.03 meters per second. The rate of increase is constant.

Mathematics
1 answer:
Vladimir79 [104]3 years ago
3 0
A)
Given: dr/dt = .03 m/s and r = 18 m
Volume of a sphere = (4/3)*PI*r^3
Take the derivative of both sides, so
dv/dt = 4*PI*r^2*dr/dt
Plug in the givens and you have the rate of increase of the volume
B)
Given: V = 288 m^3 and I'm going to assume that r and dr/dt are the same as in part A
Surface Area = 4*PI*r^2
Take the derivative on both sides, so
dA/dt = 8*PI*r*dr/dt
Again, plug in the givens and you'll have the rate of increase in the surface area
C)
This part, it's been awhile since I've done related rates, so it may take me awhile, but perhaps the next person can answer it before I finish.

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Answer:

Choice C is correct

Step-by-step explanation:

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The second equation can be written as;

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The second step is to make N the subject of the formula in both equations.

Solving for N from this equation \frac{M}{N}=10^{4}, yields;

N=\frac{M}{10^{4}}

Solving for N from the second equation \frac{P}{N}=10^{5}, yields;

N=\frac{P}{10^{5}}

Therefore;

\frac{M}{10^{4}}=\frac{P}{10^{5} }\\\\P=10M

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<u>SOLUTION:</u>

Given, a cellular family phone plan cost $49 per minute plus five cents per minute of long distance service. We have to write an equation for the monthly payment b when M minutes of long distance service are used.

Now, given that,

\begin{array}{l}{\text { Monthly payment = } \$ 49 \text { per minute plus five cents per minute of long distance }} \\\\ {\Rightarrow \$ 49 \times \text { number of minutes }+\$ 0.05 \times \text { number of minutes of long distance }} \\\\ {\Rightarrow \mathrm{b}=49 \times \mathrm{M}+0.05 \times \mathrm{M} \rightarrow \mathrm{b}=49 \mathrm{M}+0.05 \mathrm{M} \rightarrow \mathrm{b}=49.05 \mathrm{M}}\end{array}

<em>Equation:</em>

An equation is a statement that the values of two mathematical expressions are equal <em>(indicated by the sign =)</em>

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