Answer:
a^12+8
Step-by-step explanation:
let's check all the expressions
1. a^3+18
here 18 is not a perfect cube.
2. a^6+9
here 9 is not a perfect cube.
3. a^9+16
here 16 is not a perfect cube.
4. a^16+8
here 8 is a perfect cube, thus we can write above expression by using laws of exponents as
(a^4)^3+2^3
Thus, is a sum of cubes.
a^12+8
Answer:
Step-by-step explanation:
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Answer: a is the # of elements in A but NOT B, that is A ONLY
b is the # of elements in B but NOT A, that is B only
N is the # of elements in the INTERSECTION of A and B
a + N = 99
b + N = 25
a + b + N = 123
Subtracting the first two equations: a - b = 74
Subtracting the last two equations: -a = -98
a = 98
So then a - b = 74 and a = 98
98 - b = 74
-b = -24
b = 24
a = 98, b = 24 and a + b + N = 123
98 + 24 + N = 123
122 + N = 123
N = 1
98 are in A only
24 are in B only , which is the answer to the question
1 is in both A and B
Step-by-step explanation:
6/25 is the answer as a simplified fraction
