In order to solve this we'll start by assigning variables to hamburgers and cheeseburgers, since these are what we're trying to find. Lets say x = hamburgers and y = cheeseburgers. So we know two things, we know that x+y= 763 (hamburgers plus cheeseburgers sold equals 763, and we know that y= x+63 (cheeseburgers sold equals 63 more than hamburgers sold). Now we have a system of equations. This can be solved most easily by rearranging each equation to each y, and then set them equal to each other:
x+y=763 -> y=763-x, and we already have y=x+63. Set them equal to each other:
x+63 = 763-x (add x to both sides) -> 2x+63 = 763 (subtract 63 from both sides) -> 2x = 700 (divide both sides by 2) x = 350. So we solved for x, which is hamburgers sold, which is what the question asks for, so your answer is 350 hamburgers were sold on Saturday
As a group they were paid $1000 after putting in 4+6+5+5 = 20 worker-hours.
$1000/(20 worker-hours) = $50/(worker-hour)
Alemu received 4*$50 = $200
Tulu received 6*$50 = $300
Kassa and Tegitu each received 5*$50 = $250
Answer:
sleeping the floor and vacuuming
Answer:0.91
Step-by-step explanation:
3.64/4=0.91
unit price=cost of one item/thing
Answer:
Step-by-step explanation:
Given that there are 3 sets such that there are 100 elements in A1, 1000 in A2, and 10,000 in A3
a) If A1 ⊆ A2 and A2 ⊆ A3
then union will contain the same number of elements as that of A3
i.e. 
b) If the sets are pairwise disjoint.
union will contain the sum of elements of each set

c) If there are two elements common to each pair of sets and one element in all three sets
We subtract common elements pairwise and add common element in 3
i.e. 