d=(100b+n)/(n+t)
2006-10-06 15:27:27 · 9 answers · asked by Bob 3 in Science & Mathematics ➔ Mathematics
vovòzinha, it's because I am trying to determine something involving my "best answer" percentage.
2006-10-06 16:25:06 · update #1
Hi,
As there is no values for the variables here I'll just put the equation in another form.
=>d=(100b+n)/(n+t)
=>d(n+t) = 100b + n
=>dn + dt = 100b + n
=>(d-1)n = 100b -dt
=>n = (100b - dt)/(d-1)
Peace out.
2006-10-06 15:34:51 · answer #1 · answered by Pradyumna N 2 · 2⤊ 0⤋
Taking from my colleagues
n= (100b - dt) / (d -1)
for d = 1 "n" tends to infinity
for d = 0 n = 100b
for d > 1 or < 1 the equation holds as written.
Why????
2006-10-06 23:18:41 · answer #2 · answered by Vovó (Grandma) 7 · 0⤊ 0⤋
n=8
2006-10-06 22:28:52 · answer #3 · answered by tinkerbell 4 · 0⤊ 1⤋
d = (100b + n)/(n + t)
dn + dt = 100b + n
dn - n = 100b - dt
n(d - 1) = 100b - dt
n = (100b - dt)/(d - 1)
2006-10-07 00:30:37 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋
d(n+t) = 100b+n
dn + dt = 100b + n
dn - n = 100b - dt
n(d-1) = 100b - dt
n = (100b-dt)/(d-1)
2006-10-07 00:31:36 · answer #5 · answered by raghavan 1 · 0⤊ 0⤋
d(n+t) = 100b+n
dn + dt = 100b + n
dn - n = 100b - dt
n(d-1) = 100b - dt
n = (100b-dt)/(d-1)
2006-10-06 22:37:13 · answer #6 · answered by kezongputi 2 · 1⤊ 0⤋
n = d(n + t) - 100b
2006-10-06 22:35:33 · answer #7 · answered by ♥_mrs.smith 4 · 0⤊ 0⤋
dn+dt=100b+n
n(d-1)=100b-dt
n=(100b-dt)/(d-1)
2006-10-06 22:35:01 · answer #8 · answered by Helmut 7 · 1⤊ 0⤋
3 variables, 1 equation. it is indeterminate based on the information.
2006-10-06 22:31:56 · answer #9 · answered by David T 2 · 0⤊ 0⤋