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I understand it but then I don’t I think you you do y=mx + b could I get some help question 11 only

I understand it but then I don’t I think you you do y=mx + b could I get some help-example-1

1 Answer

5 votes

Given:


h(x)=-10t^2+20t+80

Where h(x) is the height of an object t second after launched

We will find the time (t) when hitting the ground.

So, the height will be = 0

substitute with h = 0 then solve the equation to find t


\begin{gathered} -10t^2+20t+80=0\rightarrow(/-10) \\ t^2-2t-8=0 \\ (t-4)(t+2)=0 \\ t-4=0\rightarrow t=4 \\ t+2=0\rightarrow t=-2 \end{gathered}

The negative result is illogical (there is no negative time)

So, the answer will be option D. 4

User JamesFrost
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