Solve for x. 2/x+2 + 1/5 = 6/x+5

Mathematics

1 answer:

olganol [36]3 years ago

5 0

Answer:

Exact Form:

x=−107

Decimal Form:

x=−1.428571

Mixed Number Form:

x=−137

Step-by-step explanation:

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Simplify 81p6q2÷3p2q5 HELPPP

goldfiish [28.3K]

Answer:

$\displaystyle \frac{81\, p^{6}\, q^{2}}{3\, p^{2} \, q^{5}} = \frac{27\, p^{4}}{q^{3}}$ .

Step-by-step explanation:

Make use of the fact that for any $x \ne 0$ and integer $n$ :

$\displaystyle \frac{1}{x^{n}} = x^{-n}$ .

For example, in this question:

$\displaystyle \frac{1}{p^{2}} = p^{-2}$ .

$\displaystyle \frac{1}{q^{5}} = q^{-5}$ .

Thus, the original expression would be equivalent to:

$\begin{aligned}& \frac{81\, p^{6}\, q^{2}}{3\, p^{2} \, q^{5}} \\ =\; & \frac{81}{3}\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5}) \\ =\; & 27\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5})\end{aligned}$ .

Also make use of the fact that for any $x \ne 0$ , integer $m$ , and integer $n$ :

$x^{m}\, x^{n} = x^{m + n}$ .

Thus:

$\begin{aligned}& 27\, (p^{6}\, p^{-2})\, (q^{2}\, q^{-5}) \\=\; & 27\, p^{6 + (-2)}\, q^{2 + (-5)} \\ =\; & 27\, p^{4}\, q^{-3} \\ =\; & \frac{27\, p^{4}}{q^{3}}\end{aligned}$ .