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sergey [27]
3 years ago
15

Write 4x^3-6x+2x^5+3 in standard form

Mathematics
1 answer:
hodyreva [135]3 years ago
6 0

4x^3-6x+2x^5+3=2x^5+4x^3-6x+3

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Rewrite in simplest rational exponent form √x • 4√x. Show each step of your process.
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here, we have the equivalent (kind of) to x * 4x. 
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Im not sure what your teacher is asking you to communicate, but most simply would like to see evidence that you understand the material rather than simply looking up the answer... (i mean without reading an explanation xD)
4√x= √x*√x*√x*√xso √x • 4√x =5√x
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How do i do this question? someone help
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Read 2 more answers
Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.​
vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}

\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}

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2 years ago
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