Answer:C
Step-by-step explanation: In order for it to become the second figure it needs to be enlarged and reflected. It needs to be enlarged first so it will end up placed in the right spot.
First solve for g(1)
Plug in 1 for the equation g(x)
g(1) = -(1) + 4
g(1) = 3
Now plug 3 in for f(x)
f(3) = 3(3)^2 + 4(3) + 1
f(3) = 3(9) + 12 + 1
f(3) = 27 + 12 + 1
f(3) = 40
Solution: f(g(1)) = 40
Answer:
3 2/3
Step-by-step explanation:
12 1/2 - 8 5/6
25/2 - 53/6
(75-53)/6
22/6
11/3
3 2/3
Hello from MrBillDoesMath!
Answer:
Choice C, sqrt(3-x)
Discussion:
Have no fear, MBill is here!
The quotient is
sqrt (9-x^2) / sqrt( 3+x) = => as sqrt(a)/sqrt(b) = sqrt(a/b), b <> 0
sqrt ( (9-x^2)/( 3+x) ) = => as 9 - x^2 = (3-x) * (3+x)
sqrt ( (3-x)(3+x) /(3+x) ) = => cancel (3+x) from num. and denom.
sqrt( 3-x)
This is Choice C
Note the domain restriction -3 < x < = 3 guarantees that division by zero can't happen.
Thank you,
MrB
Answer:
16) 2
17) -5
18) doesn't exist
19) doesn't exist
20) doesn't exist
21) 3
22) 4
23) 6
Step-by-step explanation:
16) as you move towards -9, the function adopts the value 2
17) as one moves towards x = -6 , from both sides (right and left) the function goes to the value -5
18) As one moves towards x = -4 (from the right and from the left, the functions seems to diverge towards + ∞. So normally the convention for limits stipulates: Undefined or Doesn't exist
19) f(-4) doesn't exist (for same reasons as above (there is a singularity here)
20) As one moves towards 2 from the right, the function gets towards the value 3, while approaching from the left the function goes towards the value 5. So formally we say that the limit doesn't exist (from the left and from the right limits don't agree)
21) f(2) is the well defined value of 3
22) approaching x= 4 from the right and from the left both lead towards the value 4.
23) f(4) is 6
