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worty [1.4K]
3 years ago
14

Algebraically prove that X^3+10/x^3+9=1+ 1/x^3+9 where x is not equal to -3

Mathematics
1 answer:
ehidna [41]3 years ago
7 0
<span>Algebraically prove that X^3+10/x^3+9=1+ 1/x^3+9 where x is not equal to -3
</span>\dfrac{x^3+10}{x^3+9}=1+\dfrac{1}{x^3+9} \\ \\ \\  \dfrac{x^3+10}{x^3+9}=\dfrac{(x^3+9)+1}{x^3+9}\\ \\ \\  \dfrac{x^3+10}{x^3+9}=\dfrac{x^3+10}{x^3+9}  \\ \\ \\x\in R \qquad x\in (-\infty, +\infty)<span>
</span>
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vlabodo [156]

Answer: 17.75

Step-by-step explanation:

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How to get quartiles:

First get the median:

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Then find the median of the first half of data(3.5, 10.4)

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Then find the median of the last half of data(21.7, 27.7)

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