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2 years ago

Complete the two column proof Given: 22= 24,m_2 = 110°

Mathematics

1 answer:

Savatey [412]2 years ago

6 0

Step-by-step explanation:

I took 3 indicated as 5 and its adjacent angle to be 3

<2 = <4

As <2 =<3( corresponding angle)

And <3 = <4 ( Vertically opp.angle)

hence <2 = <4

<2 =>110

so , <4 = 110

So, <4 =>110

As <4 and <5 form linear pair

So <4 + <5 =>180

<5 = 180 -110 =>70

As i took <5 as replacing angle to <3

So According to Question fig

<3 =>70

Hence proved

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Consider the set whose elements are the graphs having vertex set {1, 2, 3, 4}, and consider the relation on that set, where two

Damm [24]

Answer:

Step-by-step explanation:

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The number of edges is 0 forms a disconnected graph which makes an equivalent class.

The graphs of 1 edge makes an equivalent class.

The graphs of 2 edges makes an equivalent class.

The graphs of 3 edges makes an equivalent class.

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2 years ago

What are the solutions of 4x2 + x = -3? (5 points) and <br> x2 - 7x = -12 (5 points)

Yuki888 [10]

4x^2 + x + 3 = 0

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7 0

3 years ago

Read 2 more answers

Help me plz fot 27 28 and 29 tysm

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3 years ago

A swimmer does 3,560 J of work in 55 s. What is the swimmer's power output? Round your answer to two significant

Vesna [10]

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64.73 Watts

Step-by-step explanation:

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2 years ago

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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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2 years ago