Answer: 484
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Work Shown:
First let's compute f(2). We replace every x with 2 and then use PEMDAS to simplify
f(x) = -x^4 + 5x - 4x^2
f(2) = -(2)^4 + 5(2) - 4(2)^2
f(2) = -16 + 5(2) - 4(4)
f(2) = -16 + 10 - 16
f(2) = -6 - 16
f(2) = -22
Then we square this result to find the value of ![[ f(2) ]^2](https://tex.z-dn.net/?f=%5B%20f%282%29%20%5D%5E2)
![f(2) = -22\\\\\left[ f(2) \right]^2 = [ -22 ]^2\\\\\left[ f(2) \right]^2 = 484](https://tex.z-dn.net/?f=f%282%29%20%3D%20-22%5C%5C%5C%5C%5Cleft%5B%20f%282%29%20%5Cright%5D%5E2%20%3D%20%5B%20-22%20%5D%5E2%5C%5C%5C%5C%5Cleft%5B%20f%282%29%20%5Cright%5D%5E2%20%3D%20484)
n(A-B) denotes elements which are in A but not in B
n(Au B) denotes elements in A and B
n(AnB) denotes elements that are common in A and B
Now I will add one more set
n(B-A) which denotes elements in B but not in A
So, n(AuB) = n(A-B) + n( B-A) +n(AnB)
70 = 18 +n(B-A) + 25
70 = 43 + n(B-A)
n(B-A) = 70-43
n(B-A) = 27
So, n(B) = n( B-A) + n( AnB)
= 27+25
= 52
Answer:
D
Step-by-step explanation:
It is on the line so it is a solution, hope this helps
Answer: Answer should be C
Step-by-step explanation: I had this question on my math test and it was C for me
