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:
C
Step-by-step explanation:
since -6 and -6 are factors of 36 that add up to equal -12, the factored form of the expression is (x-6)^2.
<span>We are asked if a number goes into 49 evenly.
First, let's explore what it means to "go in evenly."
A number (call it A) goes into another number (Call it B) if when we divide B by A there is no remainder. As an example, 5 goes into 10 evenly. This is because 10/5 = 2. There is no remainder. However, 6 does not go into 10 evenly. This is because 10/6 = 1 remainder 4. Since there is a remainder when we divide 10 by 6 we know 6 does not go evenly into 10.
If we go back to our example of 10 and 5. Since 5 goes evenly into 10 we call 5 a divisor (or factor) or 10. We call 10 a multiple of 5. Let's list all of the numbers that go evenly into 10: 1, 2 5, 10. These are the factors of 10.
Notice that 1 and the number itself are always factors of a number. What the question is asking us is whether there are any factors of 49. We know that 1 and 49 go evenly into 49, but in these types of questions we are looking for factors other than 1 or the number because those always work.
Do we have to test all of the numbers between 1 and 49? Do we have to divide 49 by all the number between 1 and 49 to find those that leave no remainder? Well, we could BUT you might have noticed that factors come in pairs so it's enough to check half the numbers, that is the numbers from 1 to 25.
When we divide 49 by one of these numbers and get a zero reminder the answer to our division problem is also a factor. If you do this you will see that 49 / 7 = 7 with no remainder. So 7 is a number that goes evenly into 49 (so is the other 7 but we don't have to write it twice).
You might have some familiarity with multiplication facts that allows you to get to the answer 7 (maybe you remember that 7 * 7 = 49) without testing the numbers one by one but testing them will always work.</span>
