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3 years ago

3/5 (10+5x)−1/3(6x+3)=9 ok answer fast

Mathematics

1 answer:

Anton [14]3 years ago

6 0

Answer: x = 4

Step-by-step explanation:

$\dfrac{3}{5}(10 + 5x) - \dfrac{1}{3}(6x + 3)=9$

$\dfrac{3}{5}(10 + 5x)(15) - \dfrac{1}{3}(6x + 3)(15)=9(15)$ multiplied by common denominator

3(10 + 5x)(3) - 1(6x + 3)(5) = 9(15) reduced all fractions

90 + 45x - 30x - 15 = 135 distributed

15x + 75 = 135 simplified (added like terms)

15x = 60 subtracted 75 from both sides

x = 4 divided 15 from both sides

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What is the maximum percent of net spendable income that should be set aside for housing

stepladder [879]

Answer:

Answer;

-38 %

Step-by-step explanation:

Explanation;

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3 years ago

Read 2 more answers

Find m∠IUV if m∠IUV=x+49, m∠TUI=x+63, and m∠TUV=106∘.? I'll mark you brainliest!

8090 [49]

Answer:

Step-by-step explanation:

You have to do IVU+TUI=TUV

so... x+49+x+63=106

5 0

3 years ago

Plz help i will mark you branlyest also plz answer both questions

Margaret [11]

Answer:

D(1, -2)

E(3, -4)

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3 years ago

You want to buy a camera. The retail price is $75. The camera is on sale for 15% off . What is the sale price of the camera?

ludmilkaskok [199]

Answer:

$63.75

Step-by-step explanation:

15% off

Get percentage of retail price

1-0.15 = .85

0.85 = 85%

The sale price is 85% of the retail price

Multiply

0.85 * 75

$63.75

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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

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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: trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus

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3 years ago