Please help. I'll give you a medal and fan.
@xapproachesinfinity
Multiply both sides by hmn so that way we can get rid of the fraction
\(\sf\LARGE hmn (\frac{1}{h} = hmn(\frac{1}{m} + \frac{1}{n})\)
what do you get? :P
\(\sf\LARGE hmn \frac{1}{h} = hmn(\frac{1}{m} + \frac{1}{n})\)***
\[h=\frac{ 1 }{ m }+\frac{ 1 }{ n }\]
wrong
\(\sf\LARGE hmn \frac{1}{h} = hmn(\frac{1}{m} + \frac{1}{n})\) we then get \(\sf\Large mn = hn + hm\) factor out h from the right hand side
I need you to cooperate with me...
Sorry, I'm not the best at literal equations...
that is not an excuse...
can you factor out h from hn + hm? hint: ab + ac = a (b+c)
we are solving for h, so how can we have `h` in the equation? so once again, wrong
also, if you will only post your final answer, then also include your work along with it :)
\[h=mn=n+m\]
uhhh, also, there is no need to delete your replies
You seem very critical, sorry :)
\(\color{#0cbb34}{\text{Originally Posted by}}\) @shaleiah \[h=mn=n+m\] \(\color{#0cbb34}{\text{End of Quote}}\) this is incorrect also
\(\color{#0cbb34}{\text{Originally Posted by}}\) @TheSmartOne can you factor out h from hn + hm? hint: ab + ac = a (b+c) \(\color{#0cbb34}{\text{End of Quote}}\) please do this
I may seem very critical because you are not cooperating. We would have gotten the answer a long time ago if you did what I told you to do :P
\[hn+hm=h(n+m)\]
\(\sf\Large mn = h(n+m)\) now fivide both sides by (n+m) and that will be your answer :)
thanks for cooperating now :)
\[h=\frac{ mn }{ n+m}\]
and that is your answer! :D
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