Answer: h(x)
Explanation:
1) f(x) is not a function because there is an ambiguity for x = 4.
x = 4 belongs to both intervals - 2 ≤ x ≤ 4 and x ≥ 4
Then you find two different possible images for x: 0 and -(2)^2 = - 4.
That makes that f(x) be not a function.
2) similar thing happens with g(x)
as per the given relation the value of g(x) for x = 2 is 4 and 4+1 = 5. Which makes that g(x) be not a function.
3) j(x) is not a function because the image of x = -4 is -3(-4) = 12 and 3.
4) h(x) is a function, because there is not any ambiguity in its definition, for every x in its domain there is only one image h(x).
Answer:
17/28
if you're solving for cosine
Step-by-step explanation:
Can u say this in English please
In −3z+6b−7, the first term is -3z. It's a variable term, since z is assumed to be a variable. The coefficient of this variable, z, is -3.
Ypu would subtract Y from the other points Y
Then you would subtract X from the ither points X.
Then you would put the Y value over the X
Then simiplify
