How does h(t) = 2^t change over the interval from t = 7 to t = 8? Choices: (A)h(t) decreases by a factor of 2 (B)h(t) decreases by 2 (C)h(t) increases by a factor of 2 (D)h(t) increases by 200%

Calculate h(t): Calculate h(t) when t = 7. h(7) = 2^7 h(7) = 128Calculate h(t): Calculate h(t) when t = 8. h(8) = 2^8 h(8) = 256Compare h(7) and h(8): Compare h(7) and h(8) to determine the change. h(8) is twice h(7), so h(t) increases by a factor of 2 from t = 7 to t = 8.

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How does $h(t) = 2^t$ change over the interval from $t = 7$ to $t = 8$ ? $\newline$ Choices: $\newline$ (A) $h(t)$ decreases by a factor of $2$ $\newline$ (B) $h(t)$ decreases by $2$ $\newline$ (C) $h(t)$ increases by a factor of $2$ $\newline$ (D) $h(t)$ increases by $t = 7$ $0$

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Math Problems

Grade 8

Interpreting Linear Expressions

Full solution

Q. How does

h(t) = 2^t

change over the interval from

t = 7

t = 8

\newline

Choices:

\newline

(A)

h(t)

decreases by a factor of

2

\newline

(B)

h(t)

decreases by

2

\newline

(C)

h(t)

increases by a factor of

2

\newline

(D)

h(t)

increases by

t = 7

0

Calculate $h(t)$ : Calculate $h(t)$ when $t = 7$ . $\newline$ $h(7) = 2^7$ $\newline$ $h(7) = 128$
Calculate $h(t)$ : Calculate $h(t)$ when $t = 8$ .
$h(8) = 2^8$
$h(8) = 256$
Compare $h(7)$ and $h(8)$ : Compare $h(7)$ and $h(8)$ to determine the change. $h(8)$ is twice $h(7)$ , so $h(t)$ increases by a factor of $2$ from $t = 7$ to $t = 8$ .