Mathematics: Are 7+², 2x like terms?

Part A)About 0.51% per year. Part B)About 0.30% per year. Part C)About 28.26%. Step-by-step explanation:We are given that the population of Americans age 55 and older as a percentange of the total population is approximated by the function: Where t is measured in years with t = 0 being the year 2000. Part A)Recall that the rate of change of a function at a point is given by its derivative. Thus, find the derivative of our function: Rewrite: We can use the chain rule. Recall that: Let: Then from...

Are 7+², 2x like terms? - id729125277, jazmine8194

Ответ:

miguelturner

21.01.2023

Mathematics

Part A)

About 0.51% per year.

Part B)

About 0.30% per year.

Part C)

About 28.26%.

Step-by-step explanation:

We are given that the population of Americans age 55 and older as a percentange of the total population is approximated by the function:

$f(t) = 10.72(0.9t+10)^{0.3}\text{ where } 0 \leq t \leq 20$

Where t is measured in years with t = 0 being the year 2000.

Part A)

Recall that the rate of change of a function at a point is given by its derivative. Thus, find the derivative of our function:

$\displaystyle f'(t) = \frac{d}{dt} \left[ 10.72\left(0.9t+10\right)^{0.3}\right]$

Rewrite:

$\displaystyle f'(t) = 10.72\frac{d}{dt} \left[(0.9t+10)^{0.3}\right]$

We can use the chain rule. Recall that:

$\displaystyle \frac{d}{dx} [u(v(x))] = u'(v(x)) \cdot v'(x)$

Let:

$\displaystyle u(t) = t^{0.3}\text{ and } v(t) = 0.9t+10 \text{ (so } u(v(t)) = (0.9t+10)^{0.3}\text{)}$

Then from the Power Rule:

$\displaystyle u'(t) = 0.3t^{-0.7}\text{ and } v'(t) = 0.9$

Thus:

$\displaystyle \frac{d}{dt}\left[(0.9t+10)^{0.3}\right]= 0.3(0.9t+10)^{-0.7}\cdot 0.9$

Substitute:

$\displaystyle f'(t) = 10.72\left( 0.3(0.9t+10)^{-0.7}\cdot 0.9 \right)$

And simplify:

$\displaystyle f'(t) = 2.8944(0.9t+10)^{-0.7}$

For 2002, t = 2. Then the rate at which the percentage is changing will be:

$\displaystyle f'(2) = 2.8944(0.9(2)+10)^{-0.7} = 0.5143...\approx 0.51$

Contextually, this means the percentage is increasing by about 0.51% per year.

Part B)

Evaluate f'(t) when t = 17. This yields:

$\displaystyle f'(17) = 2.8944(0.9(17)+10)^{-0.7} =0.3015...\approx 0.30$

Contextually, this means the percetange is increasing by about 0.30% per year.

Part C)

For this question, we will simply use the original function since it outputs the percentage of the American population 55 and older. Thus, evaluate f(t) when t = 17:

$\displaystyle f(17) = 10.72(0.9(17)+10)^{0.3}=28.2573...\approx 28.26$

So, about 28.26% of the American population in 2017 are age 55 and older.

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