For this case, we must build a quotient that, when multiplied by the divisor, eliminates the terms of the dividend until it reaches the rest.
The attached figure shows the quotient given by:

Answer:
Quotient: 
See attached image
The cardinality of the union between two sets is given by

In fact, if an element is in both A and B, it is counted in both N(A) and N(B), so we're counting it twice. By subtracting the cardinality of the interserction, we're taking back one of these "extra-counted" element, and the count is correct. In your case, the numbers are

This means that the two sets cover, without repetitions, 43 of the 66 elements in the sets. So, there are

elements which are in neither A nor B
It’s probably 22 because 5 is on two sides and 11 is on two sides so 11 + 11+ 5+5 = 32
Answer:
543453
Step-by-step explanation:
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