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3 years ago

Factor 64a^2b^36-1

1 answer:

3 0

$64a^2b^{36} -1$

$= (8ab^{18})^{2} - (1)^2$

$=(8ab^{18} - 1)(8ab^{18} + 1)$

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Answer: (8ab¹⁸ - 1)(8ab¹⁸ + 1) (Answer 3)
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To model this situation, we are going to use the decay formula: $A=Pe^{rt}$
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$A$ is the final pupolation
$P$ is the initial population
$e$ is the Euler's constant
$r$ is the decay rate
$t$ is the time in years

A. We know for our problem that the initial population is 1,250, so $P=1250$ ; we also know that after a year the population is 1000, so $A=1000$ and $t=1$ . Lets replace those values in our formula to find $r$ :
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Now that we have $r$ , we can write a function to model this scenario:
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B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.

C.
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- The function doe snot have a x-intercept
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0:
$A(0)=1250e^{(-0.2231)(0)$
$A_{0}=1250e^{0}$
$A_{0}=1250$
- Over the interval [0,10], the function will have a minimum at t=10:
$A(10)=1250e^{(-0.2231)(10)$
$A_{10}=134.28$

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula: $m= \frac{A(0)-A(10)}{10-0}$
where
$m$ is the rate of change
$A(10)$ is the function evaluated at 10
$A(0)$ is the function evaluated at 0
We know from previous calculations that $A(10)=134.28$ and $A(0)=1250$ , so lets replace those values in our formula to find $m$ :
$m= \frac{134.28-1250}{10-0}$
$m= \frac{-1115.72}{10}$
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We can conclude that the rate of change of the function over the interval [0,10] is -111.572.

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