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3 years ago

A solution of the equation x^2/4=9 is

Mathematics

1 answer:

abruzzese [7]3 years ago

5 0

Answer:

x = ±6

Step-by-step explanation:

x^2/4=9

Multiply by 4 on each side

x^2/4 *4=9*4

x^2 = 36

Take the square root of each side

sqrt(x^2) = ±sqrt(36)

x = ±6

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Identify the graph of the equation and find (h k). x^2-2x-y^2-2y-36=0

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2 years ago

A school conducts 27 tests in 36 weeks. Assume the school conducts tests at a constant rate. What is the slope of the line that

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3 years ago

(7-4) + 17 / -14 + 2 x 5

Yuliya22 [10]

Answer:

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Step-by-step explanation:

(7-4) + 17 / -14 + 2 x 5

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165/14

8 0

3 years ago

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

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

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>

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2 years ago