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Mathematics 17 Online
OpenStudy (anonymous):

r over 12 = 1 9/10 solve for r

OpenStudy (igreen):

\(\huge\frac{r}{12} = 1\frac{9}{10}\) Is this it?

OpenStudy (anonymous):

yeah

OpenStudy (anonymous):

Cross multiplication?

OpenStudy (igreen):

You multiply 12 to 1 9/10: \[12 \times \frac{9}{10} = 22 \frac{4}{5} \]

OpenStudy (anonymous):

^Cross multiplication

OpenStudy (anonymous):

114/5

OpenStudy (mathmale):

In your shoes I'd convert 1 9/10 to an improper fraction before doing that cross multiplication: 1 9/10 = 19/10

OpenStudy (igreen):

Oh

OpenStudy (mathmale):

Then,\[\frac{ r }{ 12 }=\frac{ 19 }{ 10 }\]

OpenStudy (anonymous):

still stuck

OpenStudy (mathmale):

What is the goal of this problem? Explain the goal in your own words. Have you done similar problems in the past?

OpenStudy (igreen):

\(\huge\frac{ r }{ 12 }=\frac{ 19 }{ 10 } = 10r = 228?\) @mathmale

OpenStudy (mathmale):

@iGreen: You're on the right track, except that that 2nd = sign should not be in there (use an arrow instead). Please, let qdbug have time to solve this problem himself / herself, OK?

OpenStudy (igreen):

Oh, right...

OpenStudy (anonymous):

im still stuck im sorry

OpenStudy (igreen):

Well since \(\Large\frac{ r }{ 12 }=\frac{ 19 }{ 10 } \LARGE\rightarrow 10r = 228\) Can you figure out \(10r = 228\)?

OpenStudy (mathmale):

First of all, @qdbug, please stop saying "I'm stuck." Instead, think of questions to ask that might produce the info you need to solve this problem, and share those questions, even if you don't think they're good questions. @gdbug, again I ask you to explain in your own words what the goal of the present problem is. Once you have that goal in mind, we can discuss how to get there.

OpenStudy (mathmale):

@qdbug: a question for you: what does "cross multiplication" mean?

OpenStudy (anonymous):

nvm my brother is helping me but ill giv u medal anyway tho

OpenStudy (mathmale):

If you're satisfied with your brother's explanation, good. If not, come back here, answer my questions, and then we'll continue in finding the solution.

OpenStudy (anonymous):

ok thx!

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