Lines intersected by a transversal can never be parallel lines. True or false

Mathematics

1 answer:

gayaneshka [121]2 years ago

5 0

Answer : True

When lines intersect they can't be parallel.!

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m<A = 84

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48+48+A=180

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We can then use the isosceles triangle similarity theorem and reason that since this is an isosceles triangle and one of the side lengths (AB) is 9, the other (AC) would also be 9.

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scoray [572]

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Step-by-step explanation:

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Rationalize the denominator.

Maslowich

If you're using the app, try seeing this answer through your browser: brainly.com/question/3166412

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$\mathsf{\dfrac{\sqrt{a}+2\sqrt{y}}{\sqrt{a}-2\sqrt{y}}}$

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

$\mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}\\\\\\ \mathsf{=\dfrac{\big(\sqrt{a}+2\sqrt{y}\big)^2}{\big(\sqrt{a}-2\sqrt{y}\big)\cdot \big(\sqrt{a}+2\sqrt{y}\big)}}$

The numerator is a square of a sum, and the denominator is the product of a difference and a sum (product of two conjugates). Then, expand the common products:

$\mathsf{=\dfrac{\big(\sqrt{a}\big)^2+2\cdot \sqrt{a}\cdot 2\sqrt{y}+\big(2\sqrt{y}\big)^2}{\big(\sqrt{a}\big)^2-\big(2\sqrt{y}\big)^2}}\\\\\\ \mathsf{=\dfrac{a+4\cdot \sqrt{a}\cdot \sqrt{y}+2^2\cdot \big(\sqrt{y}\big)^2}{a-2^2\cdot \big(\sqrt{y}\big)^2}}$

$\mathsf{=\dfrac{a+4\sqrt{ay}+4y}{a-4y}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

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