If i don't get this question right i fail the anyone who thinks they know the answer .

the height, h, in feet of a golf ball above the ground after being hit into the air is given by the equation, h = -16t 2 + 64t, where t is the number of seconds elapsed since the ball was hit. how many seconds does it take for the golf ball to hit the ground?

it will take a0 seconds for the ball to hit the ground.

When h is zero it hit the ground.so put h=0 and solve for t.The height, h, in feet of a golf ball above the ground after being hit into the air is given by the equation, h = -16t^2 + 64t, where t is the number of seconds elapsed since the ball was hit. How many seconds does it take for the golf ball to hit the ground?  The height, h, in feet of a golf ball above the ground after being hit into the air is given by the equation, h = -16t^2 + 64t, where t is the number of seconds elapsed since the ball was hit. How many seconds does it take for the golf ball to hit the ground?

h = -16t^2 + 64tIt hits the ground when h=0.
-16t^2 + 64t = 0
Solve for tThe ball is on the ground whenever the height is 0:

Obviously, is the time when the ball is thrown, so ignore that. It thus takes the ball seconds to hit the ground after it's been launched. -16t^2+64=0 > 16t(t-4)=0> t=0, t=4t=0
  is the time when the ball is thrown, so ignore that. It thus takes the ball thus t=4 seconds to hit the ground after the ball is launched. So with that being said it took 4 seconds for the ball to hit the ground or 0 seconds. 0-4 Hope this helps!

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

ac ≅ ab (given)< cax ≅ < bax   (given)ax ≅ ax (common side)

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

ac ≅ ab (given)< cay ≅ < bay (since < cax ≅ < bax is given and xy is the extension of line ax)ay ≅ ay (common side)

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)