An object is launched straight up into
the air at an initial velocity of 32 feet
per second. it is launched from a height
of 3 feet off the ground. its height h, in
feet, at t seconds is given by the
equation h = -8t^2 + 32t + 3. find all
times t that the object is at a height of
27 feet off the ground.

  t ∈ {1, 3}

Step-by-step explanation:

You want to find t such that ...

  h = 27

  27 = -8t^2 +32t +3 . . . . . . substitute the expression for h

  24 = -8t^2 +32t . . . . . . . . . subtract 3

  -3 = t^2 -4t . . . . . . . . . . . . . divide by -8

  1 = t^2 -4t +4 = (t -2)^2 . . . . add 4 to complete the square

  ±√1 = t -2 . . . . . . . . . . . . . . take the square root

  t = 2 ± 1 . . . . . . . . . . . . . . . . add 2

  t = 1 or 3

The object is 27 ft off the ground at t = 1 and again at t = 3.

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18

first one because it can't have a fraction in it

step-by-step explanation: