P(t)P(t)P, left parenthesis, t, right parenthesis models the distance of a swinging pendulum (in \text{cm}cmstart text, c, m, end text) from the place it was released ttt seconds after it starts to swing. Here, ttt is entered in radians. P(t) = -5\cos\left(2\pi t\right) + 5P(t)=−5cos(2πt)+5P, left parenthesis, t, right parenthesis, equals, minus, 5, cosine, left parenthesis, 2, pi, t, right parenthesis, plus, 5 What is the first time the pendulum reaches 3.5\text{ cm}3.5 cm3, point, 5, start text, space, c, m, end text from the place it was released? Round your final answer to the nearest hundredth of a second.

Step-by-step explanation:

Given the distance of a swinging pendulum modeled by the equation P(t)=−5cos(2πt)+5, to get the first time the pendulum reaches 3.5cm, we will have to substitute P(t)= 3.5cm into the given equation and calculate the value of 't' as shown;

3.5 = −5cos(2πt)+5

Subtracting 5 from both sides;

3.5-5 = −5cos(2πt)+5-5

-1.5 = −5cos(2πt)

Dividing both sides by -5 will give;

The first time the pendulum reaches 3.5cm is after 0.20secs

the entire expression should be subtracted from 30. to subtract the quantity, distribute the negative sign, which changes the signs on the terms. so, the –2 changes to positive and gets added to 30 instead of subtracted from it.

hope this ! ^w^

huh? im pretty sure it is in 199

step-by-step explanation: