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How does $h(t) = 9^t$ change over the interval from $t = 0$ to $t = 1$ ? $\newline$ Choices: $\newline$ (A) $h(t)$ decreases by $9$ $\newline$ (B) $h(t)$ increases by $800\%$ $\newline$ (C) $h(t)$ increases by $900\%$ $\newline$ (D) $h(t)$ increases by $t = 0$ $0$

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Exponential functions over unit intervals

Full solution

Q. How does

h(t) = 9^t

change over the interval from

t = 0

to

t = 1

?

\newline

Choices:

\newline

(A)

h(t)

decreases by

9

\newline

(B)

h(t)

increases by

800\%

\newline

(C)

h(t)

increases by

900\%

\newline

(D)

h(t)

increases by

t = 0

0

Calculate $h(0)$ : Calculate $h(0)$ by substituting $t = 0$ into $h(t) = 9^t$ . $\newline$ $h(0) = 9^0$ $\newline$ $h(0) = 1$
Calculate $h(1)$ : Calculate $h(1)$ by substituting $t = 1$ into $h(t) = 9^t$ . $\newline$ $h(1) = 9^1$ $\newline$ $h(1) = 9$
Determine increase or decrease: Determine if $h(t)$ increases or decreases from $t = 0$ to $t = 1$ . Since $h(0) = 1$ and $h(1) = 9$ , $h(t)$ increases.
Calculate percentage increase: Calculate the percentage increase from $t = 0$ to $t = 1$ .
% increase = $\left[\frac{h(1) - h(0)}{h(0)}\right] \times 100\%$
% increase = $\left[\frac{9 - 1}{1}\right] \times 100\%$
% increase = $\left[\frac{8}{1}\right] \times 100\%$
% increase = $8 \times 100\%$
% increase = $800\%$

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