First, we sketch a picture to get a sense of the problem. g(x)=x is a diagonal line through (0,0) with slope = = 1. Since we are interested in the area between x = -4 and x = 8, we find the points on the line at these values. These are (-4, -4) and (8,8).
f(x) is a parabola. It's lowest point occurs when x = 0. It is the point (0,7). At x = -4 and x=8 it has the values 11.8 and 26.2 respectively. That is, it contains the points (-4, 11.8) and (8,26.2).
From these we make a rough sketch (see attachment). This is a sketch and mine is very incorrect when it comes to scale but what matters here is which of the curves is on top, which is below and whether they intersect anywhere in the interval, so my rough sketch is good enough. From the sketch we see that f(x) is always above (greater than) g(x).
To find the area between the curves over the given interval we integrate their difference and since f(x) is strictly greater than g(x) we subtract as follows: f(x) - g(x). The limits of integration are the values -4 and 8 (the x-values between which we are looking for the area.
Now let's integrate:

The integral yields:
^{3} }{3} +7(8)- \frac{ (8)^{2} }{2}) -(\frac{.3 (-4)^{3} }{3} +7(-4)- \frac{ (-4)^{2} }{2}) = 117.6](https://tex.z-dn.net/?f=%20%5Btex%5D%28%5Cfrac%7B.3%20%288%29%5E%7B3%7D%20%7D%7B3%7D%20%2B7%288%29-%20%5Cfrac%7B%20%288%29%5E%7B2%7D%20%7D%7B2%7D%29%20-%28%5Cfrac%7B.3%20%28-4%29%5E%7B3%7D%20%7D%7B3%7D%20%2B7%28-4%29-%20%5Cfrac%7B%20%28-4%29%5E%7B2%7D%20%7D%7B2%7D%29%20%3D%20117.6)
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We evaluate this for 8 and for -4 subtracting the second FROM the first to get:
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Short answer: C. 9
Why?
Box and whisker plots are confusing, the line in the middle is the median, but we don't need to do that now. The dots on the lines outside of the box show the largest and the shortest number. The plotted point on the way left is your answer, the line plotted is 9.
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Answer:
how you doing
Step-by-step explanation:
Answer:
100 yards or 300 feet
Step-by-step explanation:
Answer:
D. <C'D'A'
Step-by-step explanation:
Given that figure ABCD is transformed to figure A′B′C′D′, therefore the angles of ABCD would be congruent to the corresponding angles of A′B′C′D′.
<CDA corresponds to <C'D'A'. Therefore:
<CDA = <C'D'A'