Aball is launched into the sky at 18 feet per second from a 18 meter tall building. the equation for the ball’s height, h, at time t seconds is h = -6t2 + 18t + 18. when will the ball strike the ground?

I believe that you meant:

h(t)=-16t^2+18t+18  (because -6t^2 would mean the acceleration due to gravity would be -12ft/s^2, which certainly is not the case on Earth :P )

Anyway, the ball will hit the ground when h(t)=0 so:

-16t^2+18t+18=0

16t^2-18t-18=0

16t^2-18t=18

t^2-9t/8=9/8

t^2-9t/8+81/256=9/8+81/256

(t-9/16)^2=369/256

t-9/16=±√(369/256)

t=9/16±√(369/256)

t=(9±√369)/16, since t>0

t=(9±3√41)/16 seconds

t≈1.763 seconds  (to nearest thousandth of a second)

0=6x+2y+10 is equivalent to the simplified form: y= -3x - 5• parallel lines always have the same slope, different y intercept knowing this, we can say the slope, or m of the line parallel to y= -3x - 5 is -3.m= -3 hope this could !