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nexus9112 [7]
3 years ago
10

Find the sum (6a)/(a^(2)-6a+9)+(9)/(a^(2)-2a-3)

Mathematics
1 answer:
jeyben [28]3 years ago
6 0

Answer:

Step-by-step explanation:

\frac{6a}{a^2-6a+9}}+\frac{9}{a^2-2a-3)}\\\\=\frac{6a}{(a-3)^2}+\frac{9}{(a-3)(a+1)}\\\\\\

=\frac{6a(a+1)+9(a-3)}{(a-3)^2(a+1)}\\\\=\frac{6a^2+6a+9a-27}{(a-3)^2(a+1)}\\

=\frac{6a^2+15a-27}{(a-3)^2(a+1)}\\\\=\frac{3(2a^2+5a-9)}{(a-3)^2(a+1)}

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Answer:

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Answer:

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Step-by-step explanation:

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The side of the square is also the diameter of the circle.

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3 years ago
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Rewrite the limand as

(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = (1 - sin(<em>x</em>)) / (cos²(<em>x</em>) / sin²(<em>x</em>))

… = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / cos²(<em>x</em>)

Recall the Pythagorean identity,

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Then

(1 - sin(<em>x</em>)) / cot²(<em>x</em>) = ((1 - sin(<em>x</em>)) sin²(<em>x</em>)) / (1 - sin²(<em>x</em>))

Factorize the denominator; it's a difference of squares, so

1 - sin²(<em>x</em>) = (1 - sin(<em>x</em>)) (1 + sin(<em>x</em>))

Cancel the common factor of 1 - sin(<em>x</em>) in the numerator and denominator:

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Now the limand is continuous at <em>x</em> = <em>π</em>/2, so

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3 years ago
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Answer:

Number 2

Step-by-step explanation:

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