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A function $h(t)$ increases by $9$ over every unit interval in $t$ and $h(0) = 0$ . $\newline$ Which could be a function rule for $h(t)$ ? $\newline$ Choices: $\newline$ (A) $h(t) = 9t$ $\newline$ (B) $h(t) = -\frac{t}{9}$ $\newline$ (C) $h(t) = t - 9$ $\newline$ (D) $h(t) = 9^t$

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Dilations of functions

Full solution

Q. A function

h(t)

increases by

9

over every unit interval in

t

and

h(0) = 0

.

\newline

Which could be a function rule for

h(t)

?

\newline

Choices:

\newline

(A)

h(t) = 9t

\newline

(B)

h(t) = -\frac{t}{9}

\newline

(C)

h(t) = t - 9

\newline

(D)

h(t) = 9^t

Rate of Change Analysis: $h(t)$ increases by $9$ for each unit interval in $t$ . This means for every increase in $t$ by $1$ , $h(t)$ increases by $9$ . So, the rate of change is $9$ .
Initial Value Consideration: Since $h(0) = 0$ , the function starts at the origin. This means the function has no initial value other than $0$ .
Linear Function Identification: A linear function with a rate of change of $9$ and starting at the origin is $h(t) = 9t$ . This matches choice $(A)$ .
Option Comparison: Check the other options to ensure they do not describe the function correctly. $\newline$ (B) $h(t) = -\frac{t}{9}$ would decrease as $t$ increases, which is not correct. $\newline$ (C) $h(t) = t - 9$ does not start at the origin, so it's not correct. $\newline$ (D) $h(t) = 9^t$ is exponential, not linear, so it's not correct.

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Question

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Tess tried to solve the differential equation

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\frac{d y}{d x}=2 x y-6 x

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1

\quad \frac{d y}{d x}=(y-3) 2 x

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2

\quad \int \frac{1}{y-3} d y=\int 2 x d x

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(A) Tess's work is correct.

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4

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\newline

7

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Aditya tried to solve the differential equation

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\newline

\frac{d y}{d x}=6 x-30

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2

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2

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2

of the following expressions represents the distance between the sandbag and the target?

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2

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\newline

The function represents a parabola in the

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\newline

1

\newline

f(x) = -(x+4)^{2}+9

\newline

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f(x)

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