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:
- 439
Step-by-step explanation:
- s ( x ) = 2 - x²
- t ( x ) = 3x
- s ( t ( -7 ) ) = ?
--------------------------
- t (-7) = 3 * ( -7 ) = -21
- s ( t ( -7 ) ) = s ( -21 ) = 2 - ( -21 )² = 2 - 441 = - 439
- s ( t ( -7 ) ) = -439
Answer:
. ........
Step-by-step explanation:
may i know were is the question??
lol
Answer:
x1 = -4; x2 = 10
Step-by-step explanation:
1) Expand the module as two separate equations:
x - 3 = 7
x - 3 = -7
2) Solve the equations:
x = 10
x = -4
=> x1 = -4; x2 = 10