A rocket is launched with an initial upward velocity of 320 feet per second from an initial height of 15 feet. The function h (t) = -16t^2 + 320t + 15 models the height of the rocket, in feet, t seconds after it was launched. For how many seconds is the height of the rocket greater than or equal to 1,039 feet?

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

do your own work and don"t cheat.

step-by-step explanation:

(0.2)+(-8)=0.3+-3+0.3m is the equation i guess? ?

step-by-step explanation:

step-by-step explanation: