Answer:
Both x=6 and x=22/3 are extraneous. There are no actual solutions to the given equation.
The equation given was: 2|x-7|=x-8.
Step-by-step explanation:
2|x-7|=x-8
Distribute the 2. We can do this because 2=|2|.
|2x-14|=x-8
|2x-14|=|-(2x-14)| holds since |1|=|-1|.
This means we have the following two equations to solve:
2x-14=x-8 or -(2x-14)=x-8
Let's first solve:
2x-14=x-8
Subtract x on both sides:
x-14=-8
Add 14 on both sides:
x=-8+14
Simplify:
x=6
Let's solve:
-(2x-14)=x-8
Distribute:
-2x+14=x-8
Add 2x on both sides:
14=3x-8
Add 8 on both sides:
22=3x
Divide both sides by 3:
22/3=x
So let's check both x=6 and x=22.3.
2|x-7|=x-8 for x=6:
2|6-7|=6-8
2|-1|=-2
2(1)=-2
2=-2 is not true so x=6 is extraneous.
2|x-7|=x-8 for x=22/3:
2|22/3-7|=22/3-8
2|22/3-21/3|=22/3-24/3
2|1/3|=-2/3
2/3=-2/3 is not true so x=22/3 is extraneous.
Answer:
Equation of a circle:
radius = the sq. rt of (x - h)^2 + (y - k)^2
Plug in the values for the center of a circle and the radius:
15 = the sq. rt of (x + 9)^2 + (y + 4)^2
(x + 9)^2 + (y + 4)^2 = 225
The answer is A or the first one.
Answer:
x≠-1,1
Step-by-step explanation:
f(g(x)) is a composition where g(x) is is substituted for x in f(x).
Recall, f(x) is 2/x. So we write 2/|x|-1. This places x in the denominator and 0 cannot be in the denominator x. Any value of x that makes the denominator 0 will not be in the domain.
|x|-1=0
|x|=1
x=1,-1
Answer:
Ur answer will be C. -20 + 4
Step-by-step explanation:
So, when u do -20 - (-4), u get -16. So, u just add them to see which one gets -16. Also, it would be C. -20 + 4 because a negative plus a positive is a negative.