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A ball's height above the ground is modeled by the equation h(t) = -16t^2 + 80t + 40 When does the ball hit the ground?
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set h(t)=0 and then solve for t
I tried, but I got 8(-2t^2+10t+5) = 0 what do I now? @baru
\(-2t^2 + 10t + 5=0\) this is a quadratic equation. you have to use the quadratic formula
\[t = { -10 \pm \sqrt{140}}\]
yes, simplify that, use a calculator
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also, have you done the quadratic formula correctly? \[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
over -4
? i dont get what u mean
oh, ok, yes your answer over -4
I have to go to class sorry. ty though.
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sure :)
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