f(x)= -x^2 + x^-1
f(-2)= - (-2)^2 + (-2)^-1 = -4 - 1/2 = -9/2
f(-1)= - (-1)^2 + (-1)^-1 = -1 -1 = -2
f(1) = - (1)^2 + (1)^-1 = -1 +1 = 0
The range is { -9/2, -2, 0}
Hi! :) How are you?
I'm doing well but I would like to hear about you!
Answer:
(x, y) = (40, 30)
Step-by-step explanation:
A graphing calculator can show you the solution to this system of equations is (x, y) = (40, 30). That is the point of intersection where the two lines cross.
__
An algebraic solution can be found by using the substitution method. An expression for y can be found using the second equation:
y = 110 -2x . . . . . . subtract 2x from both sides
Using this in the first equation gives ...
3x -4(110 -2x) = 0 . . . . substitute for y
11x = 440 . . . . . . . . . simplify, add 440
x = 40 . . . . . . . . . . divide by 11
y = 110 -2(40) = 30
The solution is (x, y) = (40, 30).
Answer:
Unfortunately it can't be solved.
Step-by-step explanation:
x is the all real number. which means it will have many solutions.
The result follows directly from properties of modular arithmetic:

That is,

means we can write
for some integer
. Then

and taken mod 12, the first term goes away, so

etc