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Mathematics 7 Online
OpenStudy (amtran_bus):

I am ashamed I need help on this easy question

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

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?

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):

\[\frac{D}{100}+\frac{D}{75}=\frac{100D+75D}{100\times 75}=7\]

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

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