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:120°
Step-by-step explanation:
sum of exterior angles=360
In a triangle we have 3 exterior angles
Size for each exterior angle=360/3
Size for each exterior angle=120°
Answer:
A. -¾ + 0 = -¾
B. -¾ - ¾ = -(¾ + ¾)
C. ¾ - ¾ = ¾ + (-¾)
E. -¾ + ¾ = ¾ + (-¾)
F. -¾ + ¾ = 0
Step-by-step explanation:
Let's check each equation to determine whether they are true or false.
If what we have in the both sides are equal, then the equation is true, if they're not, them it is false.
✔️-¾ + 0 = -¾
Add everything on your left together
-¾ = -¾ (TRUE)
✔️-¾ - ¾ = -(¾ + ¾)
Add everything on both sides together respectively
(-3 - 3)/4 = -(3 + 3)/4
-6/4 = -6/4 (TRUE)
✔️¾ - ¾ = ¾ + (-¾)
0 = ¾ - ¾ (+ × - = -)
0 = 0 (TRUE)
✔️-¾ + ¾ = ¾ - (-¾)
0 = ¾ + ¾ (- × - = +)
0 = 6/4 (FALSE)
✔️-¾ + ¾ = ¾ + (-¾)
0 = ¾ - ¾ (+ × - = -)
0 = 0 (TRUE)
✔️-¾ + ¾ = 0
0 = 0 (TRUE)
yaaaa I am here what about u
Answer:
A = 3; B = -2; C = 2
Step-by-step explanation:
Original line:
3x - 2y = 1
Slope of original line:
-2y = -3x + 1
y = 3/2 x - 1/2
slope = m = 3/2
Parallel lines have equal slopes, so we need the equation of the line with slope 3/2 that passes through the point (-6, -10)
y = mx + b
y = (3/2)x + b
Substitute -6 for x and -10 for y and solve for b.
-10 = (3/2)(-6) + b
-10 + 9 = b
b = -1
Equation: y = (3/2)x - 1
2y = 3x - 2
3x - 2y = 2