Isabella is playing with her yo-yo. The vertical distance Y (in cm ) between the yo-yo and her hand t seconds after she first spins it out is modeled by the following function. Here, t is in radians. Y(t)=40 cos((2pi)/(3)t)-71 How long does it take the yo-yo to fall all the way down from its peak, and then rise up to a vertical distance of -80cm ? Round your final answer to the nearest tenth of a second. seconds [Trigonometric identity reference sheet]

Given Function: We have the function Y(t) = 40 cos((2pi/3)t) - 71. We need to find t when Y(t) = -80.Set Equation: Set the equation -80 = 40 cos((2pi/3)t) - 71.Isolate Cosine Term: Add 71 to both sides to isolate the cosine term: -80 + 71 = 40 cos((2pi/3)t).Simplify Equation: Simplify the left side: -9 = 40 cos((2pi/3)t).Divide by 40: Divide both sides by 40 to solve for cos((2pi/3)t): -9/40 = cos((2pi/3)t).Calculate Value: Calculate the value of -9/40: -9/40 = -0.225.Inverse Cosine Function: Use the inverse cosine function to find (2pi/3)t: (2pi/3)t = cos^(-1)(-0.225).Calculate (2pi/3)t: Calculate (2pi/3)t using a calculator: (2pi/3)t ≈ 1.772 (radians).Solve for t: Solve for t by multiplying both sides by (3/(2pi)): t ≈ 1.772 * (3/(2pi)).Final Calculation: Calculate t: t ≈ 0.8 seconds.

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Isabella is playing with her yo-yo. The vertical distance
Y (in
cm ) between the yo-yo and her hand
t seconds after she first spins it out is modeled by the following function. Here,
t is in radians.

Y(t)=40 cos((2pi)/(3)t)-71
How long does it take the yo-yo to fall all the way down from its peak, and then rise up to a vertical distance of
-80cm ?
Round your final answer to the nearest tenth of a second.
seconds
[Trigonometric identity reference sheet]

Isabella is playing with her yo-yo. The vertical distance $Y$ (in $\mathrm{cm}$ ) between the yo-yo and her hand $t$ seconds after she first spins it out is modeled by the following function. Here, $t$ is in radians. $\newline$ $Y(t)=40 \cos \left(\frac{2 \pi}{3} t\right)-71$ $\newline$ How long does it take the yo-yo to fall all the way down from its peak, and then rise up to a vertical distance of $-80 \mathrm{~cm}$ ? $\newline$ Round your final answer to the nearest tenth of a second. $\newline$ $\square$ seconds

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Q. Isabella is playing with her yo-yo. The vertical distance

Y

(in

\mathrm{cm}

) between the yo-yo and her hand

t

seconds after she first spins it out is modeled by the following function. Here,

t

is in radians.

\newline

Y(t)=40 \cos \left(\frac{2 \pi}{3} t\right)-71

\newline

How long does it take the yo-yo to fall all the way down from its peak, and then rise up to a vertical distance of

-80 \mathrm{~cm}

\newline

Round your final answer to the nearest tenth of a second.

\newline

\square

seconds

Given Function: We have the function $Y(t) = 40 \cos\left(\frac{2\pi}{3}t\right) - 71$ . We need to find $t$ when $Y(t) = -80$ .
Set Equation: Set the equation $-80 = 40 \cos\left(\frac{2\pi}{3}t\right) - 71$ .
Isolate Cosine Term: Add $71$ to both sides to isolate the cosine term: $-80 + 71 = 40 \cos\left(\frac{2\pi}{3}t\right)$ .
Simplify Equation: Simplify the left side: $-9 = 40 \cos\left(\frac{2\pi}{3}t\right)$ .
Divide by $40$ : Divide both sides by $40$ to solve for $\cos\left(\frac{2\pi}{3}t\right)$ : $-\frac{9}{40} = \cos\left(\frac{2\pi}{3}t\right)$ .
Calculate Value: Calculate the value of $-\frac{9}{40}$ : $-\frac{9}{40} = -0.225$ .
Inverse Cosine Function: Use the inverse cosine function to find $(2\pi/3)t$ : $(2\pi/3)t = \cos^{-1}(-0.225)$ .
Calculate $(\frac{2\pi}{3})t$ : Calculate $(\frac{2\pi}{3})t$ using a calculator: $(\frac{2\pi}{3})t \approx 1.772$ (radians).
Solve for t: Solve for t by multiplying both sides by $\frac{3}{2\pi}$ : $t \approx 1.772 \times \left(\frac{3}{2\pi}\right)$ .
Final Calculation: Calculate $t$ : $t \approx 0.8$ seconds.