The range is the set of all y-coordinates.
R = {2, 6, 8}
if indeed two functions are inverse of each other, then their composite will render a result of "x", namely, if g(x) is indeed an inverse of f(x), then
![\bf (g\circ f)(x)=x\implies g(~~f(x)~~)=x \\\\\\ \begin{cases} f(x) = 3x\\ g(x)=\cfrac{1}{3}x \end{cases}\qquad \qquad g(~~f(x)~~)=\cfrac{1}{3}[f(x)]\implies g(~~f(x)~~)=\cfrac{1}{3}(3x)](https://tex.z-dn.net/?f=%5Cbf%20%28g%5Ccirc%20f%29%28x%29%3Dx%5Cimplies%20g%28~~f%28x%29~~%29%3Dx%20%5C%5C%5C%5C%5C%5C%20%5Cbegin%7Bcases%7D%20f%28x%29%20%3D%203x%5C%5C%20g%28x%29%3D%5Ccfrac%7B1%7D%7B3%7Dx%20%5Cend%7Bcases%7D%5Cqquad%20%5Cqquad%20g%28~~f%28x%29~~%29%3D%5Ccfrac%7B1%7D%7B3%7D%5Bf%28x%29%5D%5Cimplies%20g%28~~f%28x%29~~%29%3D%5Ccfrac%7B1%7D%7B3%7D%283x%29)
The smaller number is 63 I think.
I think this because 150/2 is 75
75-12=63
the small number would be 63
the larger number just in case u need it btw is 87 I think
hope this helps
Happy New Year from MrBillDoesMath!
Answer:
Part A:
point \ coordinates \ quadrant:
I (1,2) 1
L (-4,2) 2
K (-4,0) does not belong to any quadrant
J (3,-1) 4
M (4,-4) 4
Part B:
Point K = (-4,0) lies on the x-axis so reflecting it about the x-axis gives the same point. In other words, the "reflected" point has coordinates (-4,0)
Part C:
L = (-4,2). Reflecting it about the y-axis gives the point (4,2)
Thank you,
MrB
A) True
The domain of a quadratic function in standard form is always all real numbers, meaning you can substitute any real number for x. The range of a function is the set of all real values of y that you can get by plugging real numbers into x.
