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3 years ago

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The population of Algebraville is 500,000 and growing at 6.25%. How many people will live in Algebraville in 10 years? 20?

1 answer:

otez555 [7]3 years ago

8 0

This would be exponential growth
A= 500,000(1+.0625)^10
=500,000(1.0625)^10
=500,000* <span>1.83353577126
the population after 10 years= </span><span>916,767
A=500,000(1+.0625)^20
=500,000(1.0625)^20
=500,000* 3.36185342451
the population after 20 years would be= </span><span>1,680,926</span>

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Read 2 more answers

Evaluate M^2 /p^2 for M = 10, N =-5p and P = -2

kondaur [170]

<u>Answer: </u>

The solution of $\bold{\frac{M^{2}}{P^{2}}}$ for M = 10, N = -5P and P = -2 is 25

<u>Solution: </u>

From question, given that the value of M is 10 and N is -5p and P is -2

We have to evaluate the value of $\frac{M^{2}}{P^{2}}$ ,

By substituting the values of M and N, we get

$\frac{M^{2}}{P^{2}} = \frac{10^{2}}{(-2)^{2}}$

Expanding $\bold{10^{2}}$ :

Here 10 is the base value and 2 is the exponent value. So the base term 10 is multiplied by itself two times.

$10^{2} = 10 \times 10 = 100$

Similarly expanding $\bold{(-2)^{2}}$ :

Here -2 is the base term and 2 is the exponent value. So the base term -2 is multiplied by itself two times.

$(-2)^{2} = -2 \times -2 = 4$

So the equation $\frac{M^{2}}{P^{2}} = \frac{10^{2}}{(-2)^{2}}$ becomes,

$\frac{M^{2}}{P^{2}} = \frac{100}{4}$

By dividing 100 by 4 , we get the result as 25

Hence the solution of $\bold{\frac{M^{2}}{P^{2}}}$ when M = 10 and P = -2 is 25

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