the number of elements in the union of the A sets is:5(30)−rAwhere r is the number of repeats.Likewise the number of elements in the B sets is:3n−rB
Each element in the union (in S) is repeated 10 times in A, which means if x was the real number of elements in A (not counting repeats) then 9 out of those 10 should be thrown away, or 9x. Likewise on the B side, 8x of those elements should be thrown away. so now we have:150−9x=3n−8x⟺150−x=3n⟺50−x3=n
Now, to figure out what x is, we need to use the fact that the union of a group of sets contains every member of each set. if every element in S is repeated 10 times, that means every element in the union of the A's is repeated 10 times. This means that:150 /10=15is the number of elements in the the A's without repeats counted (same for the Bs as well).So now we have:50−15 /3=n⟺n=45
Answer: m = 0
Step-by-step explanation:
Answer:
f(-3) = -1
General Formulas and Concepts:
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Algebra I</u>
- Functions
- Function Notation
Step-by-step explanation:
<u>Step 1: Define</u>
<em>Identify</em>
f(x) = -1x - 4
<u>Step 2: Evaluate</u>
- Substitute in <em>x</em> [Function f(x)]: f(-3) = -1(-3) - 4
- Multiply: f(-3) = 3 - 4
- Subtract: f(-3) = -1
Answer:
What to know about the lines is the exact point is 0 what you need to know to find that is answer 1, 2 and 3
Step-by-step explanation:
Answer:
165 ways
Step-by-step explanation:
Selection deals with combination
There are a total of 11 from which 3 are to be selected
11C3 = 11!/3!(11-3)!
= 11!/(3!x8!)
=(11x10x9x8!)/(3x2x8!)
=11x10x9/6
=11x5x3 = 165 ways
