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4 years ago

The square rout of 7

Mathematics

2 answers:

stich3 [128]4 years ago

7 0

Answer:

2.64575131106

its a irration number so it cant be simplified

____ [38]4 years ago

3 0

Answer:

2.6457513111

Step-by-step explanation:

when seven is put into the square root it doesn't come out perfectly even instead it comes out as a decimal

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1. Distribute 5/4

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2. Subtract three to collect like terms

y = 5/4x + 2

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Question: 69 = 9r -3<br> r =

In-s [12.5K]

Answer:

8 = r

Step-by-step explanation:

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3 years ago

Read 2 more answers

Rationalize the denominator and simplify.

cestrela7 [59]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3151018

——————————

$\mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}$

Multiply and divide by the conjugate of the denominator, which is $\mathsf{4\sqrt{6}+3\sqrt{2}:}$

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}$

Expand those products and eliminate the brackets:

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}$

$\mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}}$

$\mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}}$

$\mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

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3 years ago