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.
That would be 0.24(0.4). First, estimate the result: 0.24 is about 1/4, and (1/4)(0.4) is about 0.1.
0.24(0.4) = 0.096 (which agrees with the estimated answer, 0.1).
Answer:
y=2x-3
when two lines are parallel, they have the same slope.
in the formula y=mx+b, m=slope of the line
we know 2 will be the slope for the line passing through (4,5)
next plug in (4,5) into the formula with 2 as the slope
4=x 5=y
next 5=2(4)+b
and you end up with
y=2x-3 when you simplify
x = number of adult tickets
y = number of children tickets
9x + 6y = 1053
x+y = 127 so x = 127 - y
replace x = 127 - y into 9x + 6y = 1053
9(127 - y) + 6y = 1053
1143 -9y + 6y = 1053
-3y = 1053 -1143
-3y = -90
y = 30
x = 127 - 30
x = 97
answer: 97 adults and 30 children
