
Mark M. answered 01/27/21
Mathematics Teacher - NCLB Highly Qualified
50 = -8t2 + 40t + 18
0 = -8t2 + 40t - 32
0 = t2 - 5t + 4
Can you solve for t and answer?
Sarai S.
asked 01/27/21An object is launched straight up into the air. It is launched from a height of H feet off the ground. Its height H in feet, at t seconds is given by the equation H= -8t^2 + 40 t +18. Find all times t that the object is at a height of 50 feet off the ground.
Mark M. answered 01/27/21
Mathematics Teacher - NCLB Highly Qualified
50 = -8t2 + 40t + 18
0 = -8t2 + 40t - 32
0 = t2 - 5t + 4
Can you solve for t and answer?
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