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vovikov84 [41]
4 years ago
6

How to write a square root as an exponent?

Mathematics
2 answers:
Doss [256]4 years ago
8 0
\sqrt{x} = x^{\frac{1}{2}}

A more general form is:
\sqrt[a]{x^{b}} = x^{\frac{b}{a}}
Nikolay [14]4 years ago
6 0
Hello : 
 a square root is : √a      a <span>≥ 0
</span><span>as an exponent : a ^(0.5)</span>
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3 years ago
Given cos(4x)+2<br><br> Find: Period, Shift, &amp; Range
Yuki888 [10]
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\bf \begin{array}{llll}&#10;% right side info&#10;\bullet \textit{ stretches or shrinks}\\&#10;\quad \textit{horizontally by amplitude } |{{  A}}|\\\\&#10;\bullet \textit{ flips it upside-down if }{{  A}}\textit{ is negative}\\\\&#10;\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\\\&#10;\end{array}

\bf \begin{array}{llll}&#10;&#10;&#10;\bullet \textit{vertical shift by }{{  D}}\\&#10;\qquad if\ {{  D}}\textit{ is negative, downwards}\\\\&#10;\qquad if\ {{  D}}\textit{ is positive, upwards}\\\\&#10;\bullet \textit{function period or frequency}\\&#10;\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\&#10;\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)&#10;\end{array}

now, with that template in mind, let's see

\bf \begin{array}{llccll}&#10;cos(&4x&+0)&+2\\&#10;&\uparrow &\uparrow &\uparrow \\&#10;&B&C&D&#10;\end{array} &#10;\\\\\\&#10;period\qquad \cfrac{2\pi }{B}\implies \cfrac{2\pi }{4}\implies \cfrac{\pi }{2}&#10;\\\\\\&#10;\textit{horizontal/phase shift}\qquad \cfrac{C}{B}\implies \cfrac{0}{4}\implies 0\impliedby none&#10;\\\\\\&#10;\textit{vertical shift}\qquad D=+2\impliedby \textit{2 units up}

now the range, how far up and down it goes on the y-axis

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the midline is at 0

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but with the midline at 2, goes up to 3 and down to 1, so the range is 3 ⩽ y ⩽ 1
4 0
4 years ago
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