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Nikitich [7]
3 years ago
13

What is the value of x in the equaton

itle=" {25}x^{2} = 16" alt=" {25}x^{2} = 16" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
lisov135 [29]3 years ago
5 0
25x^2=16\ \ \ |:25\\\\x^2=\dfrac{16}{25}\to x=\pm\sqrt{\dfrac{16}{25}}\\\\x=-\dfrac{\sqrt{16}}{\sqrt{25}}\ \vee\ x=\dfrac{\sqrt{16}}{\sqrt{25}}\\\\\boxed{x=-\dfrac{4}{5}\ \vee\ x=\dfrac{4}{5}}
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Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= \pi r^2h

According to the problem,

\pi r^2 h=25

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The metal cost for the base is =$(2.00× \pi r^2)

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C'=4\pi r- \frac{62.5}{ r^2}

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C''=4\pi + \frac{125}{ r^3}

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4\pi r- \frac{62.5}{ r^2}=0

\Rightarrow 4\pi r=\frac{62.5}{ r^2}

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\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0

When r=1.71 cm, the metal cost will be minimum.

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