, how much would it cost to fertilize it if the fertilizer cost $5.25 and each bag of fertilizer covered 575 square yards?
2006-10-05 05:45:19 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
cost of fertiliser
=$75*207(1/9)(1/525)*5.25
=$15.75
2006-10-05 05:52:26 · answer #1 · answered by raj 7 · 0⤊ 0⤋
Multiply 207 by 75. Divide by 9, the number of square feet in a square yard, then divide that answer by 575 and multiply the result by $5.25, the cost a single bag. This is the process. You do the work.
2006-10-05 12:54:06 · answer #2 · answered by quietwalker 5 · 0⤊ 0⤋
Area of a rectangle
=length*breadth
=75*207 sq. feet
=15525 sq. feet
Now,1 square foot = 0.111 111 111 square yard
15 525 square foot = 1 725 square yard.
Each bag of fertilizer covered 575 square yards
1 725 square yard REQUIRES,
1725/575 Bags of fertilizer
=3 Bags of fertilizer
fertilizer cost $5.25
3 bags costs=3*$5.25 =15.75$
2006-10-05 13:46:40 · answer #3 · answered by Anonymous · 0⤊ 0⤋
75 ft by 207 ft
25 yds by 69 yds
1725 square yards
1725 / 575
3 bags of fertilizer
3 x $5.25
Total cost $15.75
2006-10-05 12:54:44 · answer #4 · answered by peter r 2 · 0⤊ 0⤋
75ft = 25 yards
207ft = 69 yards
area = 25 X 69 = 1725 square yards
575 square yards------$5.25
therefore 1725 square yards = (1725 X 5.25)/575
= $15.75
2006-10-05 13:07:40 · answer #5 · answered by mailfortarun 1 · 0⤊ 0⤋
[(207*75)/575]*5.25, in dollars
$141.75
bah.. sq. yards
divide by 9
$15.75
2006-10-05 12:48:33 · answer #6 · answered by bequalming 5 · 0⤊ 0⤋
75'x207'=25yds x 69 yds = 1725 yds^2
1725 yds^2/575 yds^2/bag=3 bags
3 bags * $5.25/bag= $15.75
2006-10-05 13:44:31 · answer #7 · answered by yupchagee 7 · 0⤊ 0⤋
$15.75, but if you go wal-mart, and buy bag for $4.25 and save $3.00. but sale end On Sunday.
2006-10-05 13:02:30 · answer #8 · answered by justmejimw 7 · 0⤊ 0⤋
$141.75 for 27 bags.
2006-10-05 12:47:18 · answer #9 · answered by jeffrcal 7 · 0⤊ 0⤋
hold on.. forgot to convert ft to yards..
edit..
$47.25
2006-10-05 12:46:53 · answer #10 · answered by mazter_ace 2 · 0⤊ 0⤋