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anonymous
 one year ago
1/a+1/b=1/c for c
anonymous
 one year ago
1/a+1/b=1/c for c

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How do I solve the formula for the specified variable?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{ 1 }{ a }+\frac{ 1 }{ b }=\frac{ 1 }{ c }\] this?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Few ways, but you can start off by getting the c in the numerator on the left side by multiplying both sides by c. \[c \times \left( \frac{ 1 }{ a } +\frac{ 1 }{ b }\right)=\frac{ 1 }{ c } \times c\] \[c \times \left( \frac{ 1 }{ a } +\frac{ 1 }{ b }\right)= 1\] see if you can finish it off

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do I do c times 1/a ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do you mean divide both sides? Both sides by c?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Well what I mean is we can just divide both sides by \[\left( \frac{ 1 }{ a }+\frac{ 1 }{ b } \right)\] so we have \[\huge c = \frac{ 1 }{ \left( \frac{ 1 }{ a } + \frac{ 1 }{ b } \right) }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1You can continue to simplify that if you like by using this for the denominator \[\huge \frac{ a }{ b } \pm \frac{ c }{ d } \implies \frac{ ad \pm bc }{ bd }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If I simplify with the last equation you gave me, where do the C and D variable substitutions come from

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Those I just used to specify two different fractions so you can see it easier \[\frac{ a }{ b } \implies \frac{ 1 }{ a }\]\[\frac{ c }{ d } \implies \frac{ 1 }{ b }\] don't let all the variables confuse you, it's just place holder from the "equation" above I could have very well used x/y+z/t

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01a +/ 1a  ab

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Oh, don't add the plus/ minus sign, I was using that to suggest it's the same whether it's + or , you can use either equation so you have \[\huge c= \frac{ 1 }{ \left( \frac{ b+a }{ ab } \right) }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1Now we just apply that little rule I showed you on your previous question to finish it off

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have no idea how to simplify this >.<

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.1It's ok, it takes practice to figure out this algebra, it can get confusing with the many rules so we have \[\large \frac{ 1 }{ \left( \frac{ a }{ b } \right) } = \frac{ b }{ a }\] so this implies \[c = \frac{ ab }{ b+a }\]
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