C 15 x 12 =$180 plus $8 shipping is $188.00
Let there be x quarts of the 20% salt solution, and (25 - x) quarts of the 80% salt solution.
Then the total amount of salt in the 20% portion would be 0.2x, while that from the 80% portion would be (0.8)(25 - x). This would have to total up to (0.44)(25) in the final solution, so the equation is:
0.2x + 0.8(25 - x) = (0.44)(25)
-0.6x + 20 = 11
x = 15 quarts of the 20% salt solution
25 - x = 10 quarts of the 80% salt solution
-12c to the fourth power +18c to the third power! :)
Answer:
One possible answer would be they sold 80 student tickets and 30 adult tickets
Step-by-step explanation:
set up two inequalities, one to show number of tickets that can be sold, and the other is how much money they need to make
5x+10y G than or = 700
x+y L than or = 110
The first inequality shows that each student (x) ticket sells for $5, and each adult (y) ticket sells for $10, and the amound has to be greater or equal to 700.
The second inequality shows that both student and adult tickets sold have to be less than or equal to 110.
First find either x or y, in this case finding y was easier
y is g than or = to 30
This means that they sold at least 30 adult tickets= $300
With y, plug into the inequality x+y L than or = to 110 to find x
x is less than or = to 80
This means that they sold at most 80 student tickets = $400
Hope this helps!
Mel should use the least common multiple to solve the problem
<u>Solution:</u>
Given, Mel has to put the greatest number of bolts and nuts in each box so each box has the same number of bolts and the same number of nuts.
We have to find that should Mel use the greatest common factor or the least common multiple to solve the problem?
He should use least common multiple.
Let us see an example, suppose 12 bolts and nuts are to be fit in 6 boxes.
Then, if we took H.C.F of 12 and 6, it is 6, which means 6 bolts and nuts in each box, but, after filling 2 boxes with 6 bolts and nuts, there will be nothing left, which is wrong as remaining boxes are empty.
So the remaining method to choose is L.C.M.
Hence, he should use L.C.M method.
