Evaluate 6j+23+kl−3, when j=3, k=7, and l=33. Enter your answer in the box.
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Answer:
Step-by-step explanation:
Given : universal set in this diagram is the set of integers from 1 to 15.
Solution :
The intersection of odd integer,multiples of 3 and Factors of 15 are 3,15
The intersection of odd integer and Factors of 15 are 1,5
The intersection of odd integer,multiples of 3 is 9
The remaining multiples of 3 are 6,12
The remaining odd integers are 7,11,13
Now the remaining integers are 2,4,8,10,14 and these integers must be placed in the boxes outside the circles Since they does not belong any intersection or odd integer or factor of 15 .
Refer the attached figure for the answer.
