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2 years ago

1. Solve x/5-3<5 A. x<2/5 B. x<10 C. x<40

Mathematics

2 answers:

RUDIKE [14]2 years ago

8 0

Add 3 to both sides of the equation
x/5-3+3<5+3
Solve
x/5<8
Now multiply both sides by 5
x<40
The answer is C

blsea [12.9K]2 years ago

5 0

x/5-3<5

add 3 to each side

x/5 < 8

multiply each side by 5

x<40

Choice C

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Step-by-step explanation:

We know that the volume of the Sphere is given by the formula

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$V=\pi r_1^2 \times h-\pi r_2^2 \times h$

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2 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

_______________

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

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}\\\\\\
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I hope this helps. =)

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

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