Mathematics
7 Online
OpenStudy (amtran_bus):
I am ashamed I need help on this easy question
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OpenStudy (amtran_bus):
OpenStudy (amtran_bus):
I need help setting this up.
OpenStudy (anonymous):
you should !
OpenStudy (amtran_bus):
Your a true friend @Luis_Rivera
OpenStudy (anonymous):
i am going to bet you use distance equals rate times time
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OpenStudy (amtran_bus):
D=R*T
Well, I know that rate 1=100 and you are looking for D
OpenStudy (anonymous):
the time going was \(\frac{D}{100}\) and the time returning was \(\frac{D}{75}\) and the total time is \(7\)
OpenStudy (amtran_bus):
You can believe it or not, but I was setting it up the same way in the equation box.
OpenStudy (anonymous):
when i said distance equals rate times time, i means it and all its cousins
\[D=RT,\frac{D}{R}=T, \frac{D}{T}=R\]
OpenStudy (anonymous):
you got it now right?
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OpenStudy (amtran_bus):
Do you cross multiply?
OpenStudy (amtran_bus):
That is, 100D+75D=7?
OpenStudy (anonymous):
if you mean "do i add" then yes
OpenStudy (anonymous):
oh no!
OpenStudy (anonymous):
\[\frac{D}{100}+\frac{D}{75}=\frac{100D+75D}{100\times 75}=7\]
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OpenStudy (amtran_bus):
\[\frac{ 175D }{ 7500 }=7\]
OpenStudy (amtran_bus):
YEEHAWW!
OpenStudy (amtran_bus):
Thanks!
OpenStudy (amtran_bus):
@satellite73 , I really think this is right. Will you double check real fast?
OpenStudy (amtran_bus):
Thanks again