Aball is launched into the sky at 272 feet per second from the roof of a skyscraper 1,344 feet tall. the equation for the ball’s height h at time t seconds is h = -16t^2 + 272t + 1344. when will the ball strike the ground?

Hello @Wpisd3039,

How are you doing? In this case,
The ball will strike the ground when h=0.-16t²+272t+1344=0t₁,₂=(-272+-√(272²+4*16*1344))/2*(-16)t₁,₂=(-272+-400)/(-32)t₁=(-272+400)/(-32)=-4 or t₂=(-272-400)/(-32)=21Time cannot be negative value so t₂ is solution. The ball will strike the ground in 21 seconds.

I hope I helped, Contact me if you need more assistance.

Thank you,
Darian D.

Answercounterclockwise rotation about the origin by 180 degrees followed by reflection about the y-axisexplanationthe given figure has verticesrecall that for counterclockwise rotation about the origin, when we rotate figure pqrs counterclockwise through an angle of 180° about the origin, the coordinates will becomenext, we reflect in the y-axis by negating the x-coordinates of the resulting figure to obtain,therefore the correct answer is c