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kipiarov [429]
3 years ago
9

What is the fully simplified answer to sqrt(7/18) + sqrt(5/8) - sqrt(7/2)

Mathematics
1 answer:
NeX [460]3 years ago
6 0

Answer:

=\frac{-4\sqrt{7}+3\sqrt{5}}{6\sqrt{2}}

Step-by-step explanation:

Given expression as:

=\sqrt{\frac{7}{18}}+\sqrt{\frac{5}{8}}-\sqrt{\frac{7}{2} }

We need to simplify the given expression.

Solution:

We have:

=\sqrt{\frac{7}{18}}+\sqrt{\frac{5}{8}}-\sqrt{\frac{7}{2} }

Rewrite the expression as:

=\frac{\sqrt{7}}{3\sqrt{2}} - \frac{\sqrt{7}}{\sqrt{2}} +\frac{\sqrt{5}}{2\sqrt{2}}

\frac{\sqrt{7}}{\sqrt{2} } is a common factor to the first two terms.

Using distributive property we can factor out \frac{\sqrt{7}}{\sqrt{2} } from the first two terms.

=\frac{\sqrt{7}}{\sqrt{2}}(\frac{1}{3} -1)  +\frac{\sqrt{5}}{2\sqrt{2}}

=\frac{\sqrt{7}}{\sqrt{2}}(\frac{1-3}{3})  +\frac{\sqrt{5}}{2\sqrt{2}}

=\frac{\sqrt{7}}{\sqrt{2}}(\frac{-2}{3})  +\frac{\sqrt{5}}{2\sqrt{2}}

=-\frac{2\sqrt{7}}{3\sqrt{2}} +\frac{\sqrt{5}}{2\sqrt{2}}

\sqrt{2} is common factor, so we can factor \sqrt{2} from the above expression.

=\frac{1}{\sqrt{2} }( -\frac{2\sqrt{7}}{3} +\frac{\sqrt{5}}{2})

=\frac{1}{\sqrt{2} }( \frac{-2\times 2\sqrt{7}+3\times \sqrt{5}}{6})

=\frac{1}{\sqrt{2} }( \frac{-4\sqrt{7}+3\sqrt{5}}{6})

=\frac{-4\sqrt{7}+3\sqrt{5}}{6\sqrt{2}}

Therefore, we get simplified answer as.

=\frac{-4\sqrt{7}+3\sqrt{5}}{6\sqrt{2}}

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Answer:

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What is the solution to -2(3x - 9) + 5x = -10?
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x = 28 <===
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Can someone please help me solve these two problems and explain them to me? About the one with proportions though... I've always
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Answers:

A) -\frac{11}{4}

C) m_{1}=\frac{Fr^{2}}{G m_{2}}

Step-by-step explanation:

<u>Part 1:</u>

We have the followig equation:

\frac{x-1}{3}-\frac{x+4}{5}=\frac{4x-1}{8}

Calculating the least common multiple (l.c.m) in the denominator in the left side of the equation, being l.c.m=15:

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Dividing numerator and denominator by 11:

x=-\frac{11}{4} Hence, the correct option is A

<u>Part 2:</u>

We have the followig equation:

F=\frac{G m_{1} m_{2}}{r^{2}}

Operating with cross product:

Fr^{2}=G m_{1} m_{2}

Isolating m_{1}:

m_{1}=\frac{Fr^{2}}{G m_{2}} Hence, the correct option is C

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Step-by-step explanation:

you jus subtract :)  

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