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3 years ago

The subject was greater than or less than

Mathematics

1 answer:

Llana [10]3 years ago

3 0

Answer:

greater than

Step-by-step explanation:

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Use the model to write the equation that represents the problem.

sdas [7]

5/2 • 4/1 (flip the second fraction to multiply)
20/2
10

A

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3 years ago

210 students were asked whether they want to go to a water park or a roller skating rink for a class party.

Brrunno [24]

Answer:

60.3

Step-by-step explanation:

hope this helps

4 0

3 years ago

Y/5+7=9 solve the problem by using addition or subtraction and multiply and divide

Elis [28]

Answer:

y=10

Step-by-step explanation:

y/5+7=9

y/5=9-7

y/5=2

y=2*5

y=10

3 0

2 years ago

What is the reference angle for cos 2pie/3

kirill [66]

Answer:

Step-by-step explanation:

If your mind works better in degrees than it does radians, as does mine, it will be beneficial to us to see what degree measure this angle is. Convert it to degrees using the fact that 180° = π:

$\frac{2\pi}{3}*\frac{180}{\pi}=120$ degrees

Now that we know that, we can plot that angle in a coordinate plane. The terminal side of the angle lands in quadrant 2. To find the reference angle, we subtract 120 from 180 and get that the reference angle is 60 degrees, which is the same as π/3.

5 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>

7 0

3 years ago