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kvv77 [185]
3 years ago
14

Which expression is equivalent to (3a2b−3)−1

Mathematics
1 answer:
RoseWind [281]3 years ago
8 0
2(3ab-2) or also 6ab-4
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Find the values of x and y that makes these triangles congruent by HL.
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What is HL?  holey logic? hawk a loogy? half lies? honey lemon? heavenly lampshade? hand lotion?

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as AB = DE, the only options given where x = y - 1 are the the second and third.

as AC and AD are equal from ASA

3x - 2 = 2y + 1

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Simplify (5+√5) (5–√5)​
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The angle \theta_1θ
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Answer:

sin\theta_1 = \dfrac{\sqrt{217}}{19}

Step-by-step explanation:

It is given that:

cos\theta_1 = -\dfrac{12}{19}

And we have to find the value of sin\theta_1 = ?

As per trigonometric identities, the relation between sin\theta\ and \ cos\theta can represented as:

sin^2\theta + cos^2\theta = 1

Putting \theta_1 in place of \theta Because we are given

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Putting value of cosine:

cos\theta_1 = -\dfrac{12}{19}

sin^2\theta_1 + (\dfrac{12}{19})^2 = 1\\\Rightarrow sin^2\theta_1 + \dfrac{144}{361} = 1\\\Rightarrow sin^2\theta_1 = 1-\dfrac{144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{361-144}{361}\\\Rightarrow sin^2\theta_1 = \dfrac{217}{361}\\\Rightarrow sin\theta_1 = +\sqrt{\dfrac{217}{361}}, -\sqrt{\dfrac{217}{361}}\\\Rightarrow sin\theta_1 = +\dfrac{\sqrt{217}}{19}, -\dfrac{\sqrt{217}}{19}

It is given that \theta_1 is in 2nd quadrant and value of sine is always positive in 2nd quadrant. So, the answer is.

\Rightarrow sin\theta_1 = \dfrac{\sqrt{217}}{19}

8 0
3 years ago
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