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Vlad1618 [11]
3 years ago
15

Which of the following is a polynomial?

Mathematics
1 answer:
Tcecarenko [31]3 years ago
3 0

Answer:

(D.) 3x^2 + 6x

Step-by-step explanation:

the other options aren't complete and some don't even create a parabola.

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The vertex of the parabola below is at the point (4, -1). Which of the equations below could be the one for this parabola?  
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Step-by-step explanation:

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Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0
eduard
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2787701

_______________


\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\
\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\
\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}


Square both sides:

\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\
\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\
\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}

\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\
\mathsf{17\,sin^2\,\theta=1}\\\\
\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}


Since \mathsf{sin\,\theta} is positive, you can discard the negative sign. So,

\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}


Substitute this value back into \mathsf{(i)} to find \mathsf{cos\,\theta:}

\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}


I hope this helps. =)


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

7 0
3 years ago
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