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Yuki888 [10]
3 years ago
12

Simplify EXPLAIN your answer.

Mathematics
1 answer:
krek1111 [17]3 years ago
8 0
\frac{x-5}{ x^{3} +27} +  \frac{2}{ x^{2} -9} =  \frac{x-5}{ x^{3}+27 } +  \frac{2}{(x+3)(x-3)}  \\  \\ long-division.. \\  (x^{3} +27)/(x+3) =  x^{2} -3x+9 \\  \\  \frac{x-5}{(x+3)( x^{2} -3x+9)} +  \frac{2}{(x+3)(x-3)}   \\  \\  \frac{x-5}{(x+3)( x^{2} -3x+9)}* \frac{x-3}{x-3}  +  \frac{2}{(x+3)(x-3)} * \frac{x^{2} -3x+9}{ x^{2} -3x+9} =  \frac{(x-5)(x-3)+2( x^{2} -3x+9)}{(x+3)(x-3)( x^{2} -3x+9)}

\frac{(x-5)(x-3)+2( x^{2} -3x+9)}{(x+3)(x-3)( x^{2} -3x+9)} = \frac{ x^{2} -8x+15+2x^{2} -6x+18}{(x+3)(x-3)( x^{2} -3x+9)}  \\  \\ \frac{ 3x^{2} -14x+33}{(x+3)(x-3)( x^{2} -3x+9)}
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Step-by-step explanation:  Given that five members of gymnastics team and 7 members of weight lifting team are travelling in two vans  to the rival school for competition. We are to find the number of ways in which they can ride in two vans such that no van contains more than eight passengers and there is at least one member of each team in each van.

Let 'g' and 'w' denote a member of gymnastics team and weight lifting team respectively.

Then, the possibilities are

For 1 g, we can choose 3w, 4w, 5w, 6w.

For 2g, we can choose 2w, 3w, 4w, 5w, 6w.

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For 4g, we can choose, 1w, 2w, 3w, 4w.

Therefore, there are (4 + 5 + 5 + 4) = 18 ways.

Since, here the position of the vans does not matter, so the options are double here. Hence, we must divide it by 2.

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