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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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I don’t know how to find X or Y very urgent, need to figure this out how to solve plz help
Lena [83]

Answer:

x = 9

y = 9√3 = 15.6

Step-by-step explanation:

The triangle given is a right triangle, therefore:

✔️apply trigonometric ratio formula to find x:

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Thus:

Cos (60) = x/18

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