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mestny [16]
1 year ago
7

What is m? 49° 77° 98° 161°

Mathematics
2 answers:
Mariana [72]1 year ago
4 0

we know that

The measurement of <u>the external angle</u> is the semi-difference of the arcs it includes.

In this problem

21\°=\frac{1}{2}[arc\ RU-arc\ SU]

Solve for the measure of arc SU

42\°=[arc\ RU-arc\ SU]

arc\ SU=arc\ RU-42\°

arc\ SU=119\°-42\°=77\°

therefore

the answer is

The measure of the arc SU is 77\°

andrey2020 [161]1 year ago
4 0

Answer:

77

Step-by-step explanation:

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Answer:

M=168k

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Step-by-step explanation:

Let's begin with the mass definition in terms of density.

M=\int\int \rho dA

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M=\int^{9}_{1}\int^{4}_{1}ky^{2} dydx

Let's solve this integral:

M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx

M=k\int^{9}_{1}\frac{y^{3}}{3}|^{4}_{1}dx      

M=k\int^{9}_{1}21dx

M=21k\int^{9}_{1}dx=21k*x|^{9}_{1}

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M=21k*8=168k

Now we need to find the x-coordinate of the center of mass.

\bar{x}=\frac{1}{M}\int\int x*\rho dydx

\bar{x}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}x*ky^{2} dydx

\bar{x}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}x*y^{2} dydx

\bar{x}=\frac{1}{168}\int^{9}_{1}x*\frac{y^{3}}{3}|^{4}_{1}dx

\bar{x}=\frac{1}{168}\int^{9}_{1}x*21 dx

\bar{x}=\frac{21}{168}\frac{x^{2}}{2}|^{9}_{1}

\bar{x}=\frac{21}{168}*40=5

Now we need to find the y-coordinate of the center of mass.

\bar{y}=\frac{1}{M}\int\int y*\rho dydx

\bar{y}=\frac{1}{M}\int^{9}_{1}\int^{4}_{1}y*ky^{2} dydx

\bar{y}=\frac{k}{168k}\int^{9}_{1}\int^{4}_{1}y^{3} dydx

\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{y^{4}}{4}|^{4}_{1}dx

\bar{y}=\frac{1}{168}\int^{9}_{1}\frac{255}{4}dx

\bar{y}=\frac{255}{672}\int^{9}_{1}dx

\bar{y}=\frac{255}{672}8=\frac{2040}{672}

\bar{y}=\frac{85}{28}

Therefore the center of mass is:

(\bar{x},\bar{y})=(5,\frac{85}{28})

I hope it helps you!

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Answer:

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