ER = RF + b(RM)
How would you re-arrange this equation to find b
Any help would be greatly appreciated
2007-05-12 08:35:00 · 9 answers · asked by ZS 1 in Science & Mathematics ➔ Mathematics
ER-RF=b(RM)
b=(ER-RF)/RM
2007-05-12 08:38:42 · answer #1 · answered by ~*tigger*~ ** 7 · 1⤊ 1⤋
It all depends on what E, R, F, and M are! Assuming they are ordinary variables then this might help:
ER = RF + b(RM)
ER - RF = b(RM)
So, b = (ER - RF)/RM
= R(E - F)/RM
So b = (E- F)/M
If they are not variables in their own way then:
b = (ER - RF)/RM
Hope this helps.
2007-05-12 09:44:11 · answer #2 · answered by Anonymous · 0⤊ 0⤋
To rearrange an equation, you have to do the same to both sides of the equation until you can cancel the entire of one side to leave just b.
ER = RF + b(RM)
minus RF from both sides
ER - RF = RF + b(RM) - RF
RF - RF cancels to 0.
ER - RF = b(RM)
Divide both sides by (RM)
(ER - RF) / (RM) =(b(RM))/(RM)
b(RM)/(RM) cancels to give b
(ER - RF) / (RM) = b
b = (ER - RF) / (RM)
2007-05-12 10:02:15 · answer #3 · answered by Anonymous · 1⤊ 0⤋
ER = RF + b(RM)
ER - RF = b(RM)
(ER - RF)/RM = b
Simplyfing:
ER/RM minus RF/RM = b
b = E/M - R/M
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2007-05-12 16:07:29 · answer #4 · answered by aeiou 7 · 0⤊ 0⤋
ER-RF=b(RM)
b=ER-RF/RM
2007-05-12 23:39:35 · answer #5 · answered by Anonymous · 0⤊ 0⤋
subtract RF from both sides:
ER - RF = b(RM)
divide by RM
(ER - RF) / RM = b
2007-05-12 08:39:37 · answer #6 · answered by Anonymous · 2⤊ 1⤋
ER = RF + b(RM)
b(RM=ER-RF
2007-05-12 08:59:52 · answer #7 · answered by Anonymous · 0⤊ 1⤋
ER=RF+b(RM)
ER-RF=RF-RF+b(RM)
ER-RF=b(RM)
ER-RF/RM=b(RM)/RM
ER-RF=b
--------
RM
2007-05-12 08:50:58 · answer #8 · answered by Dave aka Spider Monkey 7 · 0⤊ 2⤋
b(RM) = ER - RF
b(RM) = R(E - F)
b = R.(E - F) / (RM)
b = (E - F) / M
2007-05-12 20:25:05 · answer #9 · answered by Como 7 · 0⤊ 0⤋