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3 years ago

7

What is the answer to this question 2x + x + x + 2y x 3y x y

1 answer:

nikklg [1K]3 years ago

3 0

Answer:

2 x (3 y^3 + 2)

Step-by-step explanation:

Simplify the following:

2 x + x + x + 2×3 y y x y

2 y×3 y = 2 y^2×3:

2 x + x + x + 2×3 y^2 x y

2×3 = 6:

2 x + x + x + 6 y^2 x y

6 y^2 x y = 6 y^(2 + 1) x:

2 x + x + x + 6 y^(2 + 1) x

2 + 1 = 3:

2 x + x + x + 6 y^3 x

Grouping like terms, 2 x + x + x + 6 y^3 x = 6 x y^3 + (2 x + x + x):

6 x y^3 + (2 x + x + x)

2 x + x + x = 4 x:

6 x y^3 + 4 x

Factor 2 x out of 6 x y^3 + 4 x:

Answer: 2 x (3 y^3 + 2)

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Find the value of tan(sin^-1(1/2))

suter [353]

If you know that $\sin\dfrac\pi3=\dfrac12$ , then you know right away

$\tan\left(\sin^{-1}\dfrac12\right)=\tan\dfrac\pi3=\dfrac1{\sqrt}3=\dfrac{\sqrt3}3$

###

Otherwise, you can derive the same result. Let $\theta=\sin^{-1}\dfrac12$ , so that $\sin\theta=\dfrac12$ . $\sin^{-1}$ is bounded, so we know $-\dfrac\pi2\le\theta\le\dfrac\pi2$ . For these values of $\theta$ , we always have $\cos\theta\ge0$ .

So, recalling the Pythagorean theorem, we find

$\cos^2\theta+\sin^2\theta=1\implies\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-\left(\dfrac12\right)^2}=\dfrac{\sqrt3}2$

Then

$\tan\theta=\tan\left(\sin^{-1}\dfrac12\right)=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac12}{\frac{\sqrt3}2}=\dfrac1{\sqrt3}=\dfrac{\sqrt3}3$

as expected.

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3 years ago

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irina1246 [14]

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Step-by-step explanation:

2(-4p-6)-5p²(-4p-6)-5p

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Irina18 [472]

<h3><u>QUES</u><u>TION</u><u>:</u><u> </u></h3>

what is x/9 = 9/27 help plz

<h3><u>ANSWER</u><u> </u><u>AND</u><u> </u><u>SOLU</u><u>TION</u><u>:</u><u> </u></h3>

SOLVING FOR X

<u>Reduce</u><u> </u><u>first</u><u> </u><u>the</u><u> </u><u>fraction</u><u> </u><u>with</u><u> </u><u>9</u>

$\large \sf = \dfrac{x}{9} = \dfrac{ \cancel9}{ \cancel{27}}$

<u> $\large \sf = \dfrac{x}{9} = \dfrac{1}{3}$ </u>

<u>Now</u><u> </u><u>simpli</u><u>fy</u><u> </u><u>using</u><u> </u><u>cross-multiply</u>

<u> $\large \sf = \dfrac{3x}{9} = 1$ </u>

<u> $\large \sf = 3x = 1 \times 9$ </u>

<u> $\large \sf = 3x = 9$ </u>

<u>Lastly</u><u> </u><u>divide</u><u> </u><u>both</u><u> </u><u>sides</u>

<u> $\large \sf = 3x \div 3 = 9 \div 3$ </u>

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<h3><u>FINAL</u><u> </u><u>ANSWER</u><u>:</u></h3>

$\huge \orange{ \underline{ \boxed{ \sf{x = 3}}}}$

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~kimtaetae92~

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NNADVOKAT [17]

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