HELP ME PLEASE Find the times (to the nearest hundredth of a second) that the weight is halfway to its maximum negative position over the interval . Solve algebraically, and show your work and final answer in the response box. Hint: Use the amplitude to determine what y(t) must be when the weight is halfway to its maximum negative position. Graph the equation and explain how it confirms your solution(s).

  0.20, 0.36 seconds

Step-by-step explanation:

We have already seen that the equation for y(t) can be written as ...

  y(t) = √29·sin(4πt +arctan(5/2))

The sine function will have a value of -1/2 for the angles 7π/6 and 11π/6. Then the weight will be halfway from its equilibrium position to the maximum negative position when ...

  4πt +arctan(5/2) = 7π/6 or 11π/6

  t = (7π/6 -arctan(5/2))/(4π) ≈ 0.196946 . . . seconds

and

  t = (11π/6 -arctan(5/2))/(4π) ≈ 0.363613 . . . seconds

The weight will be halfway from equilibrium to the maximum negative position at approximately 0.20 seconds and 0.36 seconds and every half-second thereafter.

b

step-by-step explanation:

a difference is the result of subtraction. while each of these expressions has subtraction in it, a difference will be the result when subtraction is the last operation applied in the order of operations for the expression. this means subtraction must be outside of the parenthesis. only b has this.

answer: he is 272195

step-by-step explanation: all you have to do is multiple 3535 and 77 and it equals 272195