solve
2007-01-14 04:26:10 · 8 answers · asked by Julie 1 in Science & Mathematics ➔ Mathematics
-0.5(30000) + 0.15p = 0.02(30000 + p)
-15000 + .15p = 600 + .02p
.13p -15000 = 600
.13p = 15600
p = 120000
2007-01-14 04:33:41 · answer #1 · answered by achue500 3 · 2⤊ 0⤋
-0.5(30,000) + 0.15p= 0.02(30,000 + p)
-15000 + 0.15p = 600 + 0.02p
-1500000 + 15p = 60000 + 2p
13p = 1 560 000
p = 120000
2007-01-14 12:35:45 · answer #2 · answered by Tom :: Athier than Thou 6 · 2⤊ 1⤋
-0.5(30,000)+0.15p= 0.02(30,000+p)
multiply and stuff: 0.15p - 15000 = 0.02p + 600
0.13 p = 15000 + 600 = 15600
p = 15600/0.13
p = 120,000
2007-01-14 12:33:52 · answer #3 · answered by car of boat 4 · 2⤊ 0⤋
I assume you need to solve for "p" correct?
First: use distribution > multiply 0.02 with both terms in the parenthesis...
- 0.5(30,000) + 0.15p = 0.02(30,000 + p)
- 15,000 + 0.15p = (0.02)(30,000) + 0.02p
- 15,000 + 0.15p = 600 + 0.02p
Sec: add 15,000 to both sides...
- 15,000 + 15,000 + 0.15p = 600 + 0.02p + 15,000
0.15p = 15,600 + 0.02p
Third: subtract 0.02p from both sides...
0.15p -0.02p = 15,600 + 0.02p - 0.02p
0.13p = 15,600
Fourth: solve for "p" by isolating it on one side > divide both sides by 0.13
0.13p/0.13 = 15,600/0.13
p = 120,000
2007-01-14 12:34:33 · answer #4 · answered by ♪♥Annie♥♪ 6 · 1⤊ 1⤋
Use distributive property:
-15,000+0.15p=600+0.02p
Subtract 0.02p from both sides:
0.13p-15,000=600
Add 15,000 from both sides:
0.13p=15,600
Divide 0.13 from both sides:
p=120,000
2007-01-14 13:00:34 · answer #5 · answered by Anonymous · 0⤊ 0⤋
This is mostly a calculator exercise, but:
multiplying through:
.15p -15,000 = .02p + 600
Collect terms:
.13p = 15,600
multiply by 100/13
p = 120,000
2007-01-14 12:33:57 · answer #6 · answered by John T 6 · 2⤊ 0⤋
-.05(30,000) + .15p = .02(30,000 + p)
work math in parenthesis
-15,000 + .15p = 600 + .02p
add 15,000 and subtract .02p from both sides
.13p = 15,600
p = 120,000
2007-01-14 12:43:30 · answer #7 · answered by TJ 1 · 1⤊ 0⤋
-15000+0.15p=600+0.02p
0.13p=15600
p=120000
2007-01-14 12:35:15 · answer #8 · answered by raj 7 · 2⤊ 1⤋