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2 Responses to “I need help in MATH PLEASE! i’m desperate as you can see?”
1) Let x be the number of months. So you can say:
40 + 2x = 22 + 8x (Those two lines will intersect at x = 3)
18 = 6x
x = 3, so they’ll have the same number of people in 3 months.
2) Let x be the number of weeks. So you can say:
4w*1.25 = 400 + 1w
5w = w + 400 (These two will intersect at w=100)
4w = 400
w = 100, so you’ll need to do 100 loads for the costs to be equal.
let no of ppl at the end of nth month in morning class be p(n) & for evening class be Q(n).
p(n) = 2*n + 40
q(n) = 8*n + 22
now check for n=3…both are equal to 46. hence answer is 3rd month.
Problem 2:
Let x loads of loundary be the answer.
cost for case 1= (x/4)*(1.25)
Cost for case 2= 400 + (x/4)*(1.00)
Now for costs to be equal…
(x/4)*(1.25) = 400 + (x/4)*(1.00)
(x/4)*(0.25) = 400
x=6400
November 16th, 2008 at 8:26 am
belisle
1) Let x be the number of months. So you can say:
40 + 2x = 22 + 8x (Those two lines will intersect at x = 3)
18 = 6x
x = 3, so they’ll have the same number of people in 3 months.
2) Let x be the number of weeks. So you can say:
4w*1.25 = 400 + 1w
5w = w + 400 (These two will intersect at w=100)
4w = 400
w = 100, so you’ll need to do 100 loads for the costs to be equal.
November 19th, 2008 at 7:08 pm
isabel
hi jesayka…
Problem 1:
let no of ppl at the end of nth month in morning class be p(n) & for evening class be Q(n).
p(n) = 2*n + 40
q(n) = 8*n + 22
now check for n=3…both are equal to 46. hence answer is 3rd month.
Problem 2:
Let x loads of loundary be the answer.
cost for case 1= (x/4)*(1.25)
Cost for case 2= 400 + (x/4)*(1.00)
Now for costs to be equal…
(x/4)*(1.25) = 400 + (x/4)*(1.00)
(x/4)*(0.25) = 400
x=6400