Answer:
see explanation
Step-by-step explanation:
Inequalities of the type | x | ≤ a always have solutions of the form
- a ≤ x ≤ a
Given
| x - 1 | ≤ 1 then
- 1 ≤ x - 1 ≤ 1 ( add 1 to each of the 3 intervals )
0 ≤ x ≤ 2
Since we require an integer solution then x = 0, 1, 2
Answer:
2.
Domain: all real values of x
Range: f(x)》-4
3.
Domain: all real values except x = 0
Range: All real values except f(x) = 0
Step-by-step explanation:
2.
Domain: all real values of x
Range: f(x)》-4
Because this is a concave up quadratic with a minimum turning point at (0,-4)
3.
Domain: all real values except x = 0
Range: All real values except f(x) = 0
Since x is in the denominator, it can't be 0
For f(x) = 0, x has to be infinity
Answer:
Step-by-step explanation:
make y=0
0=-2x+3 (Switch the 3 sides)
-3=-2x (Divide everything by -2, cant have a negative x)
2/3=x or 0.66=x
Answer:
B. Solve each equation for a variable
Is (-1,13) a solution of the system: Yes it is x =1, y= 13
Step-by-step explanation:
-4x=30-2y
- 15 = - 2y + 11x
-4x=30-2y
-2y+4y = -30-------------------i
- 2y + 11x= - 15 ---------------ii
make x the subject of formula in equ i
-4x=30-2y
x= (30-2y)/-4
insert x= (30-2y)/-4 in equ ii
- 2y + 11x= - 15
-2y+ 11((30-2y)/-4)= -15
-2y+(330-22y)/4 = -15
-8y+330-22y/4 = -15
-8y-22y+330/4= -15
cross multiply
-30y+330 = 4x-15 = -60
-30y = -60-330
-30y =--390
y= -390/-30 = 13
if y = 13
x = (30-2y)/-4 = (30-2x13)/-4 = (30-26)/4 = 4/4 = 1
x=1 , y=13
B I’m pretty sure dmdndndnsmndmddmd
