An object is launched from a platform.
Its height (in meters),
x seconds after the launch is modeled by
h(x)=−5(x+1)(x−9)h, left parenthesis, x, right parenthesis, equals, minus, 5, left parenthesis, x, plus, 1, right parenthesis, left parenthesis, x, minus, 9, right parenthesis
What is the height of the object at the time of launch?

100

Step-by-step explanation:

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the correct answer is x/3+10≥32

hope this !

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204