M+C=152 This is the correct answer because the to variables added together makes 152
Hey there!
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The answer to your question is 
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<u>Formulas:</u>
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<em>Area of semi-circle = </em>
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<em>Area of rectangle = </em>
<em></em>
<em>Area of triangle = </em>
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<u>Now solve:</u>
<em>semi-circle</em>
<em>(We will use 3.14 as approx. for pi)</em>
<em />
<em>rectangle</em>
<em>triangle</em>
<em />
<em />
<em />
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<em>Now add them all together:</em>
<em />
<em />
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Have a terrificly amazing day!
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If tan(<span>θ) is negative, then </span><span>θ must be either in Q-II or else in Q-IV.
Fortunately, the question tells us that it's in Q-II.
If you draw a circle on the x- and y-axes, then draw a right triangle
in Q-II, then mark the legs 3 and -2, then the hypotenuse of the
triangle ... also the radius of the circle ... is √13 .
Look for the angle whose tangent is -3/2.
tangent = (opposite) / (adjacent)
So the side opp</span>osite is the 3 and the side adjacent is the -2.
For that same angle, cosine = (adjacent) / (hypotenuse) .
The adjacent side is still the -2, and the hypotenuse is √13 .
So the cosine of the same angle is
- 2 / √13 .
To rationalize the denominator (get that square root out of there),
multiply top and bottom by √13 . Then you have
(- 2 / √13) · (√13 / √13)
= - 2 √13 / 13 .
