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Eva8 [605]
3 years ago
6

What is the answer to this problem?

Mathematics
1 answer:
Vladimir [108]3 years ago
7 0
Volume of cone = (pi/3)r^2h
Volume of hemisphere = (1/2) (4pi/3)r^3

we are given r=6/2=3cm, h(cone)=8 cm,
so total volume
=(pi/3)r^2h + (1/2)(4pi/3)r^3
=(pi/3)*3^2*8 + (1/2)(4pi/3)*3^3
=24pi + 18pi
=42pi cm^3
=131.9 cm^3 (approx.)
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Write the given expression as an algebraic expression in x. tan(2 cos^-1(x))
miv72 [106K]
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\\\\\\
\textit{if we say }cos^{-1}\left( \frac{x}{1} \right)=\theta\textit{  that means }tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]\iff tan(2\theta)\\\\
-----------------------------\\\\
cos^{-1}\left( \frac{x}{1} \right)=\theta\implies cos(\theta)=\cfrac{x}{1}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}\\\\
-----------------------------\\\\


\bf \textit{again, using the pythagorean theorem to get the opposite side}
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c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{1^2-x^2}=b
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\pm\sqrt{1-x^2}=b\\\\
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tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\cdot \frac{\pm\sqrt{1-x^2}}{x}}{1-\left( \frac{\pm\sqrt{1-x^2}}{x} \right)^2}
\\\\\\

\bf tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{1-\frac{1-x}{x}}\implies 
tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{\frac{x-1+x}{x}}
\\\\\\
tan(2\theta)=\cfrac{\pm2\sqrt{1-x^2}}{x}\cdot \cfrac{x}{2x-1}
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tan(2\theta)=\cfrac{\pm 2\sqrt{1-x^2}}{2x-1}


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Step-by-step explanation:

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