The slopes of perpendicular lines are negative reciprocals.The slope of the given line is 4.The slope of the perpendicular is -1/4.
Now we need to find the equation of a line that has slope -1/4 and passes through the point (4, 20).
We use the point-slope form of the equation of a line, where m = slope, and the point is (x1, y1).
y - y1 = m(x - x1)
y - 20 = -1/4(x - 4)
-4y + 80 = x - 4
x + 4y = 84
Answer: 1 unit to the left, and 3 units down
Step-by-step explanation:
True
5^2 = 25
12^2 = 144
25+144=169
13^2=169
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:
<u>C) −2x − 2y + 17</u>
Step-by-step explanation (PEMDAS):
First, we do <u>3 (y + 5)</u> because of the parenthesis:
6x − 5y + 2 − 8x + 3y + 15
Since there are no exponents, multiplication ,or division, we will add and subtract. But we have to make sure we <u>combine like terms</u>. X with X's, Y with Y's and constants with constants. And ALWAYS go from left to right
6x − 5y + 2 − 8x + 3y + 15
<u>−2x</u> − 5y + 2 + 3y + 15
−2x <u>− 2y</u> + 2 + 15
<em>−2x − 2y</em><em> </em><em>+ </em><em><u>17</u></em><em>.</em>