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2 years ago

7(6m - 4) = -364 pls try and show work thanks!

Mathematics

2 answers:

mr Goodwill [35]2 years ago

5 0

Answer:

m = -8

Step-by-step explanation:

7 (6m - 4) = -364

7 (6m - 4) / 7 = -364 / 7 divide 7 from both sides

6m - 4 = -52 the 7s canceled out

6m - 4 + 4 = -52 add 4 to both sides

6m = -48 the 4s canceled out

6m / 6 = -48 / 6 divide both sides by 6

m = -8 the 6s canceled out

quester [9]2 years ago

3 0

Answer:

7(6m - 4) = -364

multiply 7 for both variables in the parenthesis

42m - 28 = -364

move 28 to other side

42m= -336

divide 42 on both sides

m= -8

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Katyanochek1 [597]

Answer: $-2+\sqrt{3}$

=========================================================

Work Shown:

Apply the following trig identity

$\tan(A - B) = \frac{\tan(A)-\tan(B)}{1+\tan(A)*\tan(B)}\\\\\tan(225 - 60) = \frac{\tan(225)-\tan(60)}{1+\tan(225)*\tan(60)}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+1*\sqrt{3}}\\\\\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\$

Now let's rationalize the denominator

$\tan(165) = \frac{1-\sqrt{3}}{1+\sqrt{3}}\\\\\tan(165) = \frac{(1-\sqrt{3})(1-\sqrt{3})}{(1+\sqrt{3})(1-\sqrt{3})}\\\\\tan(165) = \frac{(1-\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{(1)^2-2*1*\sqrt{3}+(\sqrt{3})^2}{(1)^2-(\sqrt{3})^2}\\\\\tan(165) = \frac{1-2\sqrt{3}+3}{1-3}\\\\\tan(165) = \frac{4-2\sqrt{3}}{-2}\\\\\tan(165) = -2+\sqrt{3}\\\\$

----------------------

As confirmation, you can use the idea that if x = y, then x-y = 0. We'll have x = tan(165) and y = -2+sqrt(3). When computing x-y, your calculator should get fairly close to 0, if not get 0 itself.

Or you can note how

$\tan(165) \approx -0.267949\\\\-2+\sqrt{3} \approx -0.267949$

which helps us see that they are the same thing.

Further confirmation comes from WolframAlpha (see attached image). They decided to write the answer as $\sqrt{3}-2$ but it's the same as above.

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