A buoy floating in the sea is bobbing in simple harmonic motion with period 1 second and amplitude 11 in . Its displacement d from sea level at time t=0 seconds is 0in, and initially it moves downward. (Note that downward is the negative direction.)

incomplete question

Complete question:

A buoy floating in the sea is bobbing in simple harmonic motion with period 1 second and amplitude 11 in . Its displacement d from sea level at time t=0 seconds is 0in, and initially it moves downward. (Note that downward is the negative direction.). Give the equation modelling displacement(d) as a function of time(t)

d = -11sin2πt

Step-by-step explanation:

Given,

Amplitude = 11 in

Period = 1 seconds.

Since, d at t = 0 in

Let displacement be d =asin(wt+∅)

So, a = amplitude = -11 in (it moves downward)

w = angular velocity

= 2π/period

= 2π/1

= 2πrad/sec

Therefore, at t = 0

So, since sin(∅) = 0

∅ = 0

So, inputing values of a,∅, and w we get

d = -11sin(2πt + 0)

d = -11sin2πt

x = 17

step-by-step explanation:

1st step: isolate x.

24 - 3x = -27

-24         -24

-3x   =   -51

2nd step: divide.

-3x     =   -51

  3             3

  x       =     17

hope this

1.3

step-by-step explanation:

3x2=6   9x+6

bring down the 6 on both sides

6+6=12

12= 9x

divide by 9

12/9 =

1.3 =x