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Alex learned 80 new vocabulary words for his English exam. The following function gives the number of words he remembers after
t days:

W(t)=80(1-0.1 t)^(3)
What is the instantaneous rate of change of the number of words Alex remembers after 5 days?
Choose 1 answer:
(A)
10 days
(B)
10 words per day
(C) -6 days
(D) -6 words per day

Alex learned $80$ new vocabulary words for his English exam. The following function gives the number of words he remembers after $t$ days: $\newline$ $W(t)=80(1-0.1 t)^{3}$ $\newline$ What is the instantaneous rate of change of the number of words Alex remembers after $5$ days? $\newline$ Choose $1$ answer: $\newline$ (A) $\mathbf{1 0}$ days $\newline$ (B) $\mathbf{1 0}$ words per day $\newline$ (C) $-6$ days $\newline$ (D) $-6$ words per day

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Write linear functions: word problems

Full solution

Q. Alex learned

80

new vocabulary words for his English exam. The following function gives the number of words he remembers after

t

days:

\newline

W(t)=80(1-0.1 t)^{3}

\newline

What is the instantaneous rate of change of the number of words Alex remembers after

5

days?

\newline

Choose

1

answer:

\newline

(A)

\mathbf{1 0}

days

\newline

(B)

\mathbf{1 0}

words per day

\newline

(C)

-6

days

\newline

(D)

-6

words per day

Take Derivative of $W(t)$ : To find the instantaneous rate of change, we need to take the derivative of $W(t)$ with respect to $t$ .
Apply Chain Rule: Differentiate $W(t) = 80(1 - 0.1t)^{3}$ . Using the chain rule, $\frac{d}{dt}[80(1 - 0.1t)^{3}] = 80 \times 3 \times (1 - 0.1t)^{2} \times (-0.1)$ .
Simplify Derivative: Simplify the derivative to get $W'(t) = -24 \times (1 - 0.1t)^{2}.$
Substitute $t = 5$ : Substitute $t = 5$ into $W'(t)$ to find the instantaneous rate of change after $5$ days. $\newline$ $W'(5) = -24 \times (1 - 0.1\times5)^{2}$ .
Calculate $W'(5)$ : Calculate $W'(5) = -24 \times (1 - 0.5)^{2}$ . $\newline$ $W'(5) = -24 \times (0.5)^{2}$ .
Final Result: $W'(5) = -24 \times 0.25$ . $\newline$ $W'(5) = -6$ .

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