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
We use y=mx+c we know that c= 4 A’s this is the y intercept . Next we need to work out the gradient(m). For every two squares along we go three squares up. This means are gradient is 1.5. Finally we input are values into the equation y=1.5x+4
Answer:
3
Step-by-step explanation:
(You go from left to right)
(Dividend is 931, divisor is 7)
Step 1: Look for a number smaller than 9 that is divisible by 7. 7 is the only one which is smaller than 9 and is divisible by 7. Now that you know 7*1=7, you put 1 above 931 (Dividend)
Step 2: You subtract that number from the number in tens place. And bring the leftover down.
Step 3: You repeat Step 1 and 2 until either you get a remainder of 0 or the remainder in smaller than the divisor (in this case 7)
(Therefore 931/7=133, quotient=133, remainder=0)
^Hope this helps.