Solving Inequalities: h(t) > 25

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To solve the inequality h(t) = 30t - 5t^2 > 25, the function is rearranged to h(t) - 25 > 0, leading to -5(t^2 - 6t + 5) > 0. Factoring gives t^2 - 6t + 5 > 0, which results in critical points at t = 1 and t = 5. The correct intervals where h(t) exceeds 25 are between these points, specifically for 1 < t < 5. Multiplying by -1 reverses the inequality, confirming the solution's validity.
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Homework Statement



I have a function h(t) = 30t - 5t^2.
I need to find the interval for which h is > 25.

Homework Equations





The Attempt at a Solution



h(t) = 30t - 5t2 - 25 > 0
-5(t2 - 6t + 5) > 0
iff t2 - 6t + 5 > 0
Then the answer is t > 5 and t < 1.

But it is actually between those points.
Is it because I factored out a negative?
 
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In your last step, you get rid of the -5 by multiplying each side by -1/5.
This effectively reverses the inequality sign.

You may want to think of it another way...
You could add 5(t2 - 6t + 5) to both sides, which would give you
0 > 5(t2 - 6t + 5)
 
oicic ty
 
Not being a "texter," I'll assume that means "Oh, I see, I see. Thank you"

You're welcome!
(or, should I say, "yw"?)
 
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