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torisob [31]
3 years ago
12

If a town starts with population of 63,500 that declines by 700 people each year, construct an equation to model its population

size over time. How long would it take for the population to drop to 53,000?
Mathematics
1 answer:
ankoles [38]3 years ago
6 0

The equation would be 63500-700x=P, where 63500 is the original population, -700 is the amount the population decreases each year, x is the number of years, and P is the total population.


To answer the second part of the question, plug in 53000 for P and solve for x:

63500-700x=53000

<em>*Subtract 63500 from both sides*</em>

-700x=-10500

<em>*Divide both sides by -700*</em>

x=15


It would take 15 years for the population to drop to 53000.


Hope this helps!!

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2 years ago
A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula
Ilia_Sergeevich [38]

Answer: A. A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}

              B. A'(5) = 1.76 cm/s

Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

A=\pi.r^{2}

So to find the rate of the area:

\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}

\frac{dA}{dr} =2.\pi.r

Using r(t)=3-\frac{363}{(t+11)^{2}}

\frac{dr}{dt}=\frac{726}{(t+11)^{3}}

Then

\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]

\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}

Multipying and simplifying:

\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}

The rate at which the area is increasing is given by expression A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}.

B. At t = 5, rate is:

A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}

A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}

A'(5)=\frac{2408693760\pi}{4294967296}

A'(5)=1.76268

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

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