We have been given two points. 
and 
. We are asked to find the point B such that it divides line segment AC so that the ratio of AB to BC is 4:1.
We will use segment formula to solve our given problem.
When a point P divides segment any segment internally in the ratio 
, then coordinates of point P are:
![[\right x=\frac{mx_2+nx_1}{m+n},y=\frac{my_2+ny_1}{m+n}\left]](https://tex.z-dn.net/?f=%5B%5Cright%20x%3D%5Cfrac%7Bmx_2%2Bnx_1%7D%7Bm%2Bn%7D%2Cy%3D%5Cfrac%7Bmy_2%2Bny_1%7D%7Bm%2Bn%7D%5Cleft%5D)
and 
.
Upon substituting our given information in above formula, we will get:
![[\right x=\frac{4(3)+1(3)}{4+1},y=\frac{4(9)+1(4)}{4+1}\left]](https://tex.z-dn.net/?f=%5B%5Cright%20x%3D%5Cfrac%7B4%283%29%2B1%283%29%7D%7B4%2B1%7D%2Cy%3D%5Cfrac%7B4%289%29%2B1%284%29%7D%7B4%2B1%7D%5Cleft%5D)
![[\right x=\frac{12+3}{5},y=\frac{36+4}{5}\left]](https://tex.z-dn.net/?f=%5B%5Cright%20x%3D%5Cfrac%7B12%2B3%7D%7B5%7D%2Cy%3D%5Cfrac%7B36%2B4%7D%7B5%7D%5Cleft%5D)
![[\right x=\frac{15}{5},y=\frac{40}{5}\left]](https://tex.z-dn.net/?f=%5B%5Cright%20x%3D%5Cfrac%7B15%7D%7B5%7D%2Cy%3D%5Cfrac%7B40%7D%7B5%7D%5Cleft%5D)
![[\right x=3,y=8\left]](https://tex.z-dn.net/?f=%5B%5Cright%20x%3D3%2Cy%3D8%5Cleft%5D)
Therefore, the coordinates of point B would be 
.
Taylor and her 3 friends.....so there is 4 people
total paid was 18.20
so the cost for each one of them is : 18.20/4 = 4.55 <== per person
Ooh, fun
what I would do is to make it a piecewise function where the absolute value becomse 0
because if you graphed y=x^2+x-12, some part of the garph would be under the line
with y=|x^2+x-12|, that part under the line is flipped up
so we need to find that flipping point which is at y=0
solve x^2+x-12=0
(x-3)(x+4)=0
at x=-4 and x=3 are the flipping points
we have 2 functions, the regular and flipped one
the regular, we will call f(x), it is f(x)=x^2+x-12
the flipped one, we call g(x), it is g(x)=-(x^2+x-12) or -x^2-x+12
so we do the integeral of f(x) from x=5 to x=-4, plus the integral of g(x) from x=-4 to x=3, plus the integral of f(x) from x=3 to x=5
A.
B.
sepearte the integrals
![\int\limits^{-5}_{-4} {x^2+x-12} \, dx = [\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-5}_{-4}=(\frac{-125}{3}+\frac{25}{2}+60)-(\frac{64}{3}+8+48)=\frac{23}{6}](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E%7B-5%7D_%7B-4%7D%20%7Bx%5E2%2Bx-12%7D%20%5C%2C%20dx%20%3D%20%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%2B%5Cfrac%7Bx%5E2%7D%7B2%7D-12x%5D%5E%7B-5%7D_%7B-4%7D%3D%28%5Cfrac%7B-125%7D%7B3%7D%2B%5Cfrac%7B25%7D%7B2%7D%2B60%29-%28%5Cfrac%7B64%7D%7B3%7D%2B8%2B48%29%3D%5Cfrac%7B23%7D%7B6%7D)
next one
![\int\limits^{-4}_3 {-x^2-x+12} \, dx=-1[\frac{x^3}{3}+\frac{x^2}{2}-12x]^{-4}_{3}=-1((-64/3)+8+48)-(9+(9/2)-36))=\frac{343}{6}](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E%7B-4%7D_3%20%7B-x%5E2-x%2B12%7D%20%5C%2C%20dx%3D-1%5B%5Cfrac%7Bx%5E3%7D%7B3%7D%2B%5Cfrac%7Bx%5E2%7D%7B2%7D-12x%5D%5E%7B-4%7D_%7B3%7D%3D-1%28%28-64%2F3%29%2B8%2B48%29-%289%2B%289%2F2%29-36%29%29%3D%5Cfrac%7B343%7D%7B6%7D)
the last one you can do yourself, it is
the sum is
so the area under the curve is
Answer:
C
Step-by-step explanation:
yeah-ya......... right?
