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Ray Of Light [21]
3 years ago
14

Laws of indices please solve

Mathematics
2 answers:
lorasvet [3.4K]3 years ago
5 0

Answer:

here ,

\sqrt[3]{x {}^{3} y {}^{3}  }

= (x {}^{3} y {}^{3} ) {}^{ \frac{1}{3} }

therefore the anwer is xy.

was it hellful plz let me know..

saveliy_v [14]3 years ago
5 0

Answer:

<h2>xy</h2>

Solution,

\sqrt[3]{ {x}^{3}  {y}^{3} }  \\  =   {x}^{ \frac{3}{3} }  {y}^{ \frac{3}{3} }  \\  = xy

It is a root law of indices.

For example:

\: if \: {a}^{ \frac{m}{n} }

is an algebraic term, where m and n are the positive integers, then :

{a}^{ \frac{m}{n} }  =  \sqrt[n]{ {a}^{m} }

Hope this helps...

Good luck on your assignment...

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Suppose that $p$ and $q$ are positive numbers for which \[\log_9 p = \log_{12} q = \log_{16} (p + q).\] what is the value of $q/
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Given p,q>0 and \log_9p=\log_{12}q=\log_{16}(p+q)=x (say).

Then,

p=9^x\\ q=12^x\\ p+q=16^x

From the above 3 equations,

\frac{q}{p} =(\frac{12}{9} )^x\\ \frac{q}{p} =(\frac{4}{3} )^x\\ \frac{p+q}{p} =(\frac{16}{9} )^x\\ \frac{p+q}{p} =(\frac{4}{3} )^{2x}\\

From the equations, we get

\frac{p+q}{p}=(\frac{q}{p})^2\\ 1+\frac{q}{p}=(\frac{q}{p})^2\\ (\frac{q}{p})^2-\frac{q}{p}-1=0\\ \frac{q}{p}=\frac{1 \pm \sqrt{5}}{2}

Since p,q>0, the negative value is rejected.

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Divide (10x^3 + 19x^2 + x - 7) by (5x + 2)​
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Answer:

2x²  +  3x  - 1   - 5/(5x +2)

Step-by-step explanation:

               2x²  +  3x  - 1  

             __________________            

  5x + 2 |   10x³ + 19x² + x - 7

             |  -<u>10x³ -   4x²  </u>

             |              15x² + x

             |            <u>- 15x² - 6x </u>  

             |                     - 5x  - 7

             |                   <u>    5x   +2</u>

             |                             - 5

2x²  +  3x  - 1   - 5/(5x +2)

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