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Max wrote an algorithm that searches for a specific term within a large set of terms. The following function gives the length of the search, in number of steps, over a set with
n terms:

S(n)=1.6*ln(0.9 n)
What is the instantaneous rate of change of the search length for a set of 10 terms?
Choose 1 answer:
(A) 0.16 steps per term
(B) 0.16 steps per second
(C) 3.5 steps per term
(D) 3.5 steps per second

Max wrote an algorithm that searches for a specific term within a large set of terms. The following function gives the length of the search, in number of steps, over a set with $n$ terms: $\newline$ $S(n)=1.6 \cdot \ln (0.9 n)$ $\newline$ What is the instantaneous rate of change of the search length for a set of $10$ terms? $\newline$ Choose $1$ answer: $\newline$ (A) $0$ . $16$ steps per term $\newline$ (B) $0$ . $16$ steps per second $\newline$ (C) $3$ . $5$ steps per term $\newline$ (D) $3$ . $5$ steps per second

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Volume of cubes and rectangular prisms: word problems

Full solution

Q. Max wrote an algorithm that searches for a specific term within a large set of terms. The following function gives the length of the search, in number of steps, over a set with

n

terms:

\newline

S(n)=1.6 \cdot \ln (0.9 n)

\newline

What is the instantaneous rate of change of the search length for a set of

10

terms?

\newline

Choose

1

answer:

\newline

(A)

0

.

16

steps per term

\newline

(B)

0

.

16

steps per second

\newline

(C)

3

.

5

steps per term

\newline

(D)

3

.

5

steps per second

Differentiate $S(n)$ : To find the instantaneous rate of change, we need to differentiate the function $S(n)$ with respect to $n$ . $\newline$ $S(n) = 1.6 \cdot \ln(0.9n)$ $\newline$ Let's find $S'(n)$ , the derivative of $S(n)$ .
Apply Chain Rule: Using the chain rule, the derivative of $\ln(0.9n)$ is $\frac{1}{(0.9n)}$ times the derivative of $0.9n$ , which is $0.9$ . So, $S'(n) = 1.6 \times \left(\frac{1}{(0.9n)}\right) \times 0.9$
Simplify $S'(n)$ : Simplify the expression to find $S'(n)$ . $S'(n) = 1.6 \times (\frac{0.9}{0.9n})$
Evaluate at $n=10$ : Cancel out the $0.9$ in the numerator and denominator. $\newline$ $S'(n) = 1.6 \times (1/n)$
Calculate $S'(10)$ : Now we need to evaluate $S'(n)$ at $n = 10$ to find the instantaneous rate of change when there are $10$ terms. $\newline$ $S'(10) = 1.6 \times (\frac{1}{10})$
Final Result: Calculate the value of $S'(10)$ . $S'(10) = 1.6 \times 0.1$
Final Result: Calculate the value of $S'(10)$ . $\newline$ $S'(10) = 1.6 \times 0.1S'(10) = 0.16$ $\newline$ This is the instantaneous rate of change of the search length when there are $10$ terms.

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