2 years ago

Find the value of x for the quadrilateral with exterior angles of 6x+12, 15x-5, 10x+13, and 12x-4.

Mathematics

1 answer:

vagabundo [1.1K]2 years ago

6 0

The answer for the question would’ve been 90

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$p(x)=\dfrac{1}{x+1}$

$q(x)=\dfrac{1}{x-1}$

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We have,

$p(x)=\dfrac{1}{x+1}$

$q(x)=\dfrac{1}{x-1}$

Now,

$p(x)-q(x)=\dfrac{1}{x+1}-\dfrac{1}{x-1}$

$p(x)-q(x)=\dfrac{(x-1)-(x+1)}{(x+1)(x-1)}$

$p(x)-q(x)=\dfrac{x-1-x-1}{x^2-1^2}$ $[\because a^2-b^2=(a-b)(a+b)]$

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