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2 years ago

HELP!!!

Mathematics

1 answer:

Naddika [18.5K]2 years ago

5 0

Check the picture below, so it reaches the maximum height at the vertex, let's check where that is

$h(t)=64t-16t^2+0 \\\\[-0.35em] ~\dotfill\\\\ \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+64}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 64}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (64)^2}{4(-16)}\right)\implies \stackrel{maximum~height}{(2~~,~~\stackrel{\downarrow }{64})}$

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Let the three dimensions(height, length, width) be x, y,z units respectively.

The surface area of the cuboid is given by

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For this case, tissue box is almost cuboid shaped.

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Learn more about surface area of cuboid here:

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2 years ago

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Step-by-step explanation:

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Step-by-step explanation:

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Answer:

Step-by-step explanation:

this is independent

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