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Mashutka [201]
2 years ago
15

Helllp geometry! Pls!

Mathematics
1 answer:
creativ13 [48]2 years ago
7 0
He should use the Pythagorean Theorem to find the missing length.


Since KT is tangent to the circle and TL reaches the center of circle L, the measure of angle LTK is 90 degrees. This means that triangle LTK is a right triangle which means the Pythagorean Theorem can be used.

So,
TL²+(12)²=(13)²
=> TL²+144=169
=> TL²=25
=> TL = 5

Therefore, the radius of circle L is 5 feet.


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Can some 1 do that for me please​
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\displaystyle\iiint_R\mathrm dV=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta

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and

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r^2+z^2=72\implies z=\pm\sqrt{72-r^2}
(again, taking the positive root for the same reason)
z=r^2\implies 8=r^2\implies r=\sqrt8

\displaystyle\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr\,\mathrm d\theta
=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}\int_{\zeta=r^2}^{\zeta=\sqrt{72-r^2}}r\,\mathrm d\zeta\,\mathrm dr
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=\displaystyle2\pi\int_{r=0}^{r=\sqrt8}(r\sqrt{72-r^2}-r^3)\,\mathrm dr
=\dfrac{32(27\sqrt2-35)\pi}3
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