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masya89 [10]
2 years ago
9

What is the following quotient? 3 sqrt 8 / 4 sqrt 6

Mathematics
2 answers:
Stells [14]2 years ago
7 0

Answer:

The correct option is D) \frac{\sqrt{3}}{2}

Step-by-step explanation:

We need to find out the quotient of  \frac{3\sqrt{8}}{4\sqrt{6}}

=\frac{3\sqrt{8}}{4\sqrt{6}}

Rewrite the above expression;

=\frac{3\times2 \sqrt{2}}{4\sqrt{6}}

=\frac{6\sqrt{2}}{4\sqrt{6}}

Cancel out the common term from numerator and denominator,

=\frac{3\sqrt{2}}{2\sqrt{6}}

=\frac{3\sqrt{2}}{2\sqrt{2}\times \sqrt{3}}

=\frac{3}{2\sqrt{3}}

Multiply numerator and denominator with \sqrt{3}

=\frac{3\sqrt{3}}{2\sqrt{3}\sqrt{3}}

=\frac{3\sqrt{3}}{2\times 3}

=\frac{3\sqrt{3}}{6}

=\frac{\sqrt{3}}{2}

Therefore, the correct option is D) \frac{\sqrt{3}}{2}

miv72 [106K]2 years ago
6 0
For this case we have the following expression:
 3sqrt (8) / 4sqrt (6)
 Rewriting we have:
 3sqrt (2 * 4) / 4sqrt (6)
 3 * 2sqrt (2) / 4sqrt (6)
 6sqrt (2) / 4sqrt (6)
 3sqrt (2) / 2sqrt (6)
 3sqrt (2) / (2sqrt (2) sqrt (3))
 3 / (2sqrt (3))
 3sqrt (3) / (3 * 2)
 sqrt (3) / (2)
 Answer:
 
the following quotient is:
 D.sqrt 3/2
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It takes 8 minutes for Byron to fill the kiddie pool in the backyard using only a handheld hose. When his younger sister is impa
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8 0
1 year ago
Kali left school and traveled toward her friend’s house at an average speed of 40
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3 0
2 years ago
Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) |
Bas_tet [7]

Answer:

M=168k

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

Step-by-step explanation:

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

M=\int\int \rho dA

Now, we know the limits of the integrals of x and y, and also know that ρ = ky², so we will have:

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}

So the mass will be:

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!

3 0
2 years ago
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