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Hollywood_chrissy
 3 years ago
Can somebody please explain th3 attached to me. is there a proof or something???
Hollywood_chrissy
 3 years ago
Can somebody please explain th3 attached to me. is there a proof or something???

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koalamon
 3 years ago
Best ResponseYou've already chosen the best response.0just think about it, it will come to you.

Hollywood_chrissy
 3 years ago
Best ResponseYou've already chosen the best response.0@koalamon now about a hint

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0hint: in this equation:\[\frac{a+b}{ab}=\frac{c+d}{cd}\]divide the numerator and denominator of the lefthandside by 'b' and divide the numerator and denominator of the righthandside by 'd'.

Hollywood_chrissy
 3 years ago
Best ResponseYou've already chosen the best response.0@asnaseer, even after your hint, i am still running into difficulty

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1Isn't this Componendo & Dividendo? ;)

Hollywood_chrissy
 3 years ago
Best ResponseYou've already chosen the best response.0what is componedo and dividendo???

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1It is a theorem in which the denominator is added to the numerator & numerator is added with negative denominator e. g. if \[\frac{x}{y}=\frac{a}{b}\] then by componendo & dividendo we have \[\frac{( x + y)}{( x y ) }\]= \[\frac{( a +b )}{( a  b )}\] (Proof) \[\frac{3}{2}\] =\[\frac{ 6}{4}\] then by componendo & dividendo \[\frac{3+ 2}{32}\] = \[\frac{6 +4}{6 4}\] OR 5/ 1 = 10/ 2 that is true

Hollywood_chrissy
 3 years ago
Best ResponseYou've already chosen the best response.0@DLS do you understand asnaseer's hint???

DLS
 3 years ago
Best ResponseYou've already chosen the best response.1I think the theorem is better, he just said if u do the same thing/operation with a digit on one side,u do that on other too so overall effect nullifies

Hollywood_chrissy
 3 years ago
Best ResponseYou've already chosen the best response.0you helped me greatly thanks

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0This is what I meant  starting with this:\[\frac{a+b}{ab}=\frac{c+d}{cd}\]lets first divide the numerator and denominator of the lefthandside by 'b' to give:\[\frac{\frac{a}{b}+1}{\frac{a}{b}1}=\frac{c+d}{cd}\]then lets divide the numerator and denominator of the righthandside by 'd' to give:\[\frac{\frac{a}{b}+1}{\frac{a}{b}1}=\frac{\frac{c}{d}+1}{\frac{c}{d}1}\]now note that you are given that:\[\frac{a}{b}=\frac{c}{d}\]so the equality of both sides follows from this relation.

asnaseer
 3 years ago
Best ResponseYou've already chosen the best response.0I don't think you can just assume the "Componendo and dividendo" theorem directly as this question is effectively asking you to prove it.
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