it would be car d because of what the time was on the chart (i did the test and it was correct)
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
then she turns around and grabs those 4329.73 and put them in an account getting 8% APR I assume, so is annual compounding, for 7 years.

add both amounts, and that's her investment for the 11 years.
If

, then angle
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is obtuse and triangle with sides a, b, c is obtuse triangle.
In an arbitrary triangle can be only one obtuse angle, and the side which lies opposite to the largest angle is the largest. Then since <span>angle

is opposite the side of length a</span> you can conclude that a>c and a>b.