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3 years ago

What is the perimeter of JKL?

Mathematics

1 answer:

Virty [35]3 years ago

8 0

Answer:

The perimeter of JKL is 60 units.

Step-by-step explanation:

The two tangents of a circle formed by a vertex outside the circle will always be the same distance, this means that JA = JB = 13, LA = LC = 7, and KB = KC = 10. JL = JA + LA = 20, KL = LC + KC = 17, and JK = JB + KB = 23. So the perimeter of triangle JKL is 20 + 17 + 23 = 60 units.

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dsp73

Answer:

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Step-by-step explanation:

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2 years ago

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denpristay [2]

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3 years ago

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Harman [31]

I assume the cone has equation $z=\sqrt{x^2+y^2}$ (i.e. the upper half of the infinite cone given by $z^2=x^2+y^2$ ). Take

$\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi$

The volume of the described region (call it $R$ ) is

$\displaystyle\iiint_R\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^{2\pi}\int_0^{\sqrt2}\int_{\pi/4}^\pi\rho^2\sin\varphi\,\mathrm d\varphi\,\mathrm d\rho\,\mathrm d\theta$

The limits on $\theta$ and $\rho$ should be obvious. The lower limit on $\varphi$ is obtained by first determining the intersection of the cone and sphere lies in the cylinder $x^2+y^2=1$ . The distance between the central axis of the cone and this intersection is 1. The sphere has radius $\sqrt2$ . Then $\varphi$ satisfies

$\sin\varphi=\dfrac1{\sqrt2}\implies\varphi=\dfrac\pi4$

(I've added a picture to better demonstrate this)

Computing the integral is trivial. We have

$\displaystyle2\pi\left(\int_0^{\sqrt2}\rho^2\,\mathrm d\rho\right)\left(\int_{\pi/4}^\pi\sin\varphi\,\mathrm d\varphi\right)=\boxed{\frac43(1+\sqrt2)\pi}$

4 0

2 years ago

As an estimation we are told 5 miles is 8 km.<br> Convert 40 miles to km.

sergeinik [125]

Answer:

5 miles=8 km

=)1 mile=8/5 km=1.6 km

=)40 miles=1.6×40=64 km

So, 40 miles is 64 km.

Step-by-step explanation:

7 0

3 years ago