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iren2701 [21]
3 years ago
13

3x+6/x^2-x-6+2x/x^2+x-12

Mathematics
1 answer:
Tanzania [10]3 years ago
6 0
\dfrac{3x+6}{x^2-x-6} +  \dfrac{2x}{x^2+x-12}

= \dfrac{3(x+2)}{(x + 2)(x - 3)} +  \dfrac{2x}{ (x - 3)(x + 4)}

= \dfrac{3}{(x - 3)} +  \dfrac{2x}{ (x -3)(x + 4)}

= \dfrac{3(x+4)}{(x - 3)(x + 4)} +  \dfrac{2x}{ (x -3 )(x + 4)}

= \dfrac{3(x+4) + 2x }{(x - 3)(x + 4)}

= \dfrac{3x+12 + 2x }{(x - 3)(x + 4)}

= \dfrac{5x+12}{(x - 3)(x + 4)}



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We have to use the function A(t)= a(1 +/- r)^t
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A=2400(.9425)^8
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