The ordered pair (-3 , 
) is on g(x) ⇒ 1st answer
Step-by-step explanation:
Let us revise the reflection across the axes
- If the function f(x) reflected across the x-axis, then its image is g(x) = - f(x) (change the sign of y)
- If the function f(x) reflected across the y-axis, then its image is g(x) = f(-x) (change the sign of x)
∵ 
∵ f(x) is reflected across the y-axis to create the function g(x)
- Change the sign of x
∴ 
To find the point that lies on g(x) substitute x in g(x) by the x-coordinate of the point if the answer equal to the y-coordinate of the point, then the point lies on it if not then the point does not lie on it
∵ The coordinates of the point are (-3 , )
∴ x = -3 and y =
- Substitute x by -3 in g(x)
∵
∴ 
∴ 
∵ 
∴ 
∴ 
- Divide up and down by 2
∴ 
∵ The value of g(x) equal to the y-coordinate of the point
∴ The point (-3 , ) lies on g(x)
The ordered pair (-3 , ) is on g(x)
Learn more:
You can learn more about the reflection in brainly.com/question/5017530
#LearnwithBrainly
Answer:
$150
Step-by-step explanation:
25x4=100
100+50=150
so she has $50 to spend on other things
i dont know if this is what its asking or how
Answer:
Step-by-step explanation:
6.
P(even)=4/8=1/2
=50 %
7.
P(8)=23/50
=46 %
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:
6x+21
Step-by-step explanation:
