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Pie
3 years ago
15

Help me please answer neede asap

Mathematics
1 answer:
PolarNik [594]3 years ago
4 0
1 and 1 half oz (I DID IT YAY!!!!)
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Step-by-step explanation:

because 88 of 120 is about 73 %

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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}.

3 0
2 years ago
Line L1, L2 and L3 are in the same plane, L1 is perpendicular to L2, and L3 is parallel to L1. which of the following statement
exis [7]
Only Statement 2 is surely correct.
because there maybe chances that the line L1 and L3 lies above the line L2 and they can also fulfill the condition of perpendicularity so we can't be sure about statement 3 & statement 1 is clearly incorrect
7 0
4 years ago
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