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andriy [413]
2 years ago
7

What is the quotient of 15p^-4q^-6/-20p^-12q^-3 in simplified form?

Mathematics
2 answers:
musickatia [10]2 years ago
3 0
\cfrac{15p^{-4}q^{-6}}{-20p^{-12}q^{-3}} =\cfrac{3p^{-4+12}}{-4q^{-3+6}} =- \cfrac{3p^{8}}{4q^{3}}
Novay_Z [31]2 years ago
3 0

Answer:

\dfrac{15p^{-4}q^{-6}}{-20p^{-12}q^{-3}}\Rightarrow \dfrac{3p^{8}}{-4q^{3}}

Step-by-step explanation:

We are given rational expression.

\text{Expression: }\dfrac{15p^{-4}q^{-6}}{-20p^{-12}q^{-3}}

We need to simplify and write as quotient.

Using exponent law simplify the expression

a^m\div a^n=a^{m-n}

a^m\times a^n=a^{m+n}

\Rightarrow \dfrac{3p^{-4+12}q^{-6+3}}{-4}

\Rightarrow \dfrac{3p^{8}q^{-3}}{-4}

\Rightarrow \dfrac{3p^{8}}{-4q^{3}}

Now we write quotient of simplest form.

Thus, simplified form is \Rightarrow \dfrac{3p^{8}}{-4q^{3}}


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kilograms of yellow potatoes bought is 8330 kg

<em><u>Solution:</u></em>

Given that James buys 245 kilograms of red potatoes

He buys yellow potatoes that weigh 34 as much as the red potatoes

To find: kilograms of yellow potatoes bought by James

Let "y" be the kilograms of yellow potatoes bought

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Which ordered pair is a solution of linear equations?
Bingel [31]

Answer:

\{\frac{2}{3}, \frac{1}{5}\}

ordered pair is the Solution to the given System.

Step-by-step explanation:

Given:

\frac{1}{2}x- \frac{3}{4}y= \frac{11}{60} ...........Equation( 1 )\\ \\\frac{2}{5}x+ \frac{1}{6}y= \frac{3}{10} ...........Equation( 2 )\\

To Check:

Which ordered pair is the Solution to the given System. ?

Solution:

If  \{\frac{2}{3}, \frac{1}{5}\}

is the Solution then it must satisfy the given equation. i.e .Left Hand Side (L.H.S) must be Equal to Right Hand Side (R.H.S) of both the given equation for

\{x=\frac{2}{3}, y=\frac{1}{5}\}

∴ For Equation ( 1 )

L.H.S=\frac{1}{2}\times \frac{2}{3}- \frac{3}{4}\times \frac{1}{5}\\\\\\=\frac{1}{3} -\frac{3}{20} \\\\=\frac{20-9}{20\times 3}\\\\ =\frac{11}{60} \\= R.H.S

∴ Left Hand Side = Right Hand Side

Hence,

\{x=\frac{2}{3}, y=\frac{1}{5}\}

The solution.

∴ For Equation ( 2 )

L.H.S=\frac{2}{5}\times \frac{2}{3}+ \frac{1}{6}\times \frac{1}{5}\\\\\\=\frac{4}{15} -\frac{1}{30} \\\\=\frac{120+15}{15\times 30}\\\\= \frac{135}{450}\\ \\\\=\frac{3}{30}\\ \\= R.H.S

∴ Left Hand Side = Right Hand Side

Hence ,

\{x=\frac{2}{3}, y=\frac{1}{5}\}

The solution.

For other Solutions  we do not have Left Hand Side equal to Right Hand Side.  

i.e  Left Hand Side ≠ Right Hand Side

Hence Not a Solution.

5 0
2 years ago
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