A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the time that the rocket will hit the ground, to the nearest 100th of second. y=-16x^2+224x+121 y=−16x 2 +224x+121

54

step-by-step explanation:

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116

the answer is: -22 also a

hope i

-lavira: )

given the equation is 8+9z=-6, the first step would be to subtract the 8 from both sides:

8+9z-8=-6-8

since 8 minus 8 = 0, this would cancel out the 8. and since -6 minus 8 equals -14, the equation would look like this

9z = -14