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Kajal holds a ruler in some wavy water. The depth of the water
t seconds after she starts measuring it, in
cm, is given by

D(t)=50-23 sin(pi(t+0.23)).
After she starts measuring, when is the first time the depth of the waves is at its average value? Give an exact answer.
When
t= seconds

Kajal holds a ruler in some wavy water. The depth of the water $t$ seconds after she starts measuring it, in $\mathrm{cm}$ , is given by $\newline$ $D(t)=50-23 \sin (\pi(t+0.23)) .$ $\newline$ After she starts measuring, when is the first time the depth of the waves is at its average value? Give an exact answer. $\newline$ When $t= \square$ seconds

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Write linear functions: word problems

Full solution

Q. Kajal holds a ruler in some wavy water. The depth of the water

t

seconds after she starts measuring it, in

\mathrm{cm}

, is given by

\newline

D(t)=50-23 \sin (\pi(t+0.23)) .

\newline

After she starts measuring, when is the first time the depth of the waves is at its average value? Give an exact answer.

\newline

When

t= \square

seconds

Average Depth Value: The average value of the depth is given by the constant term in the equation, which is $50\,\text{cm}$ .
Finding Average Depth: To find when the depth is at its average value, we set $D(t)$ equal to $50$ cm. $50 = 50 - 23 \sin(\pi(t + 0.23))$
Isolating Trigonometric Function: Subtract $50$ from both sides to isolate the trigonometric function. $\newline$ $0 = -23 \sin(\pi(t + 0.23))$
Solving for Sine Function: Divide both sides by $-23$ to solve for the sine function. $\newline$ $0 = \sin(\pi(t + 0.23))$
Finding Average Value of Sine: The sine function is zero at its average value, which occurs at multiples of $\pi$ . So we need to find the smallest positive $t$ such that $\pi(t + 0.23)$ is a multiple of $\pi$ . $\pi(t + 0.23) = 0$
Solving for t: Divide both sides by $\pi$ to solve for $t$ . $t + 0.23 = 0$
Final Result: Subtract $0.23$ from both sides to find $t$ . $\newline$ $t = -0.23$

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