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charle [14.2K]
3 years ago
15

((81x^3 y^(-1/3))^1/4)/(x^2 y)^1/4 simplify

Mathematics
1 answer:
malfutka [58]3 years ago
7 0
\bf a^{\frac{{ n}}{{ m}}} \implies  \sqrt[{ m}]{a^{ n}} \qquad \qquad
\sqrt[{ m}]{a^{ n}}\implies a^{\frac{{ n}}{{ m}}}
\\\\\\
\left.\qquad \qquad \right.\textit{negative exponents}\\\\
a^{-{ n}} \implies \cfrac{1}{a^{ n}}
\qquad \qquad
\cfrac{1}{a^{ n}}\implies a^{-{ n}}
\qquad \qquad 
a^{{{  n}}}\implies \cfrac{1}{a^{-{{  n}}}}\\\\
-------------------------------\\\\

\bf \cfrac{\left( 81x^3y^{-\frac{1}{3}} \right)^{\frac{1}{4}}}{(x^2y)^{\frac{1}{4}}}\implies \cfrac{81^{\frac{1}{4}}x^{3\cdot {\frac{1}{4}}}y^{-\frac{1}{3}\cdot {\frac{1}{4}}}}{x^{2\cdot {\frac{1}{4}}}y^{\frac{1}{4}}}\implies
\cfrac{81^{\frac{1}{4}}x^{\frac{3}{4}}y^{-\frac{1}{12}}}{x^{\frac{2}{4}} y^{\frac{1}{4}}}

\bf \cfrac{81^{\frac{1}{4}}x^{\frac{3}{4}}x^{-\frac{2}{4}}}{ y^{\frac{1}{4}}y^{\frac{1}{12}}}\implies \cfrac{(3^4)^{\frac{1}{4}}x^{\frac{3}{4}-\frac{2}{4}}}{y^{\frac{1}{4}+\frac{1}{12}}}\implies \cfrac{3^{4\cdot \frac{1}{4}}x^{\frac{1}{4}}}{y^{\frac{4}{12}}}\implies \cfrac{3^{\frac{4}{4}}x^{\frac{1}{4}}}{y^{\frac{1}{3}}}
\\\\\\
\cfrac{3\sqrt[4]{x}}{\sqrt[3]{y}}
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Given tanj=5/12 find cosg
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Given tan g = 5/12,  find cos g:

                  5
tan g  =  ———
                 12

The tangent function is, by definition, the quotient between sine and cosine:

   sin g           5
————  =  ———
  cos g          12

Product of them extremes  =  product of the means

12 · sin g = 5 · cos g

Square both sides:

(12 · sin g)² = (5 · cos g)²

12² · sin² g = 5² · cos² g

144 · sin² g = 25 · cos² g

But  sin² g = 1 – cos² g.  Substitute it for  sin² g  into the equation above, and you have

144 · (1 – cos² g) = 25 · cos² g

Multiply out the brackets, and then isolate cos² g:

144 – 144 · cos² g = 25 · cos² g

144 = 25 · cos² g + 144 · cos² g

144 = 169 · cos² g

Divide both sides by 169:

                  144
cos² g  =  ———
                  169

                   12²
cos² g  =  ———
                   13²

cos² g = (12/13)²

Now, take the square root of both sides:

cos g = ± √(12/13)²
 
                    12
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The sign of  cos g  depends on which quadrant the angle  g  lies. As  tan g = 5/12,  which is positive, then  g  lies either in the 1st or the 3rd quadrant:

•  If g lies in the 1st quadrant, then

                                               5
cos g > 0     ⇒    cos g  =  ———          ✔
                                              13

•  If g lies in the 3rd quadrant, then

                                                   5
cos g < 0     ⇒    cos g  =  –  ———          ✔
                                                  13


I hope this helps. =)


Tags:  <em>trigonometric relation tangent cosine sine tan cos sin trig trigonometry</em>

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