Answer:
It is the value of pie(22/7)
Answer:
Step-by-step explanation:
All we have to do is input the given values of x into the functions.
The first function:
f(x) = x^2 - 5x - 6
f(0) = 0^2 - 5(0) - 6 = 0 - 0 - 6 = -6
f(0) = -6
f(2) = 2^2 - 5(2) - 6 = 4 - 10 - 6 = -12
f(2) = -12
f(-1) = -1^2 - 5(-1) - 6 = 1 + 6 - 6 = 1
f(-1) = 1
f(6) = 6^2 -5(6) - 6 = 36 - 30 - 6 = 0
f(6) = 0
The second function:
f(x) = x^3 - x^2 - 12
f(0) = 0^3 - 0^2 - 12 = 0 - 0 - 12 = -12
f(0) = -12
f(2) = 2^3 - 2^2 - 12 = 8 - 4 - 12 = -8
f(2) = -8
f(-1) = -1^3 - (-1)^2 - 12 = -1 - 1 - 12 = -14
f(-1) = -14
f(6) = 6^3 - 6^2 - 12 = 216 - 36 - 12 = 168
f(6) = 168
The third function:
f(x) = 5 * 2^x
f(0) = 5 * 2^0 = 5 * 1 = 5
f(0) = 5
f(2) = 5 * 2^2 = 5 * 4 = 20
f(2) = 20
f(-1) = 5 * 2^-1 = 5 * 0.5 = 2.5
f(-1) = 2.5
f(6) = 5 * 2^6 = 5 * 64 = 320
f(6) = 320
I think right is correct ( - 1, 5 )
C is the answer. By only looking at the domain (the stuff on the right in the functions), you can see that the first equation has a hole at x=2, meaning it does not have a value here at x=2. When x<2, there are values for the 1st function. When x>2, there are values for the second function, as well as when x=2
Answer:Yes
Hope this helps!
Step-by-step explanation:
