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Algebra 17 Online
OpenStudy (drmoss):

Suppose c and d vary inversely, and d=2 when c=17. a)write an equation that models the variation b)find d when c=68

OpenStudy (imstuck):

The inverse variation equation is this one: \[c=\frac{ k }{ d }\], where k is the constant of variation. Solve it like this, first for k

OpenStudy (drmoss):

so for a it is: 17=k/2

OpenStudy (imstuck):

\[c=\frac{ k }{ d }\] fill in the values they gave you: \[17=\frac{ k }{ 2 }\] and solve for k first.

OpenStudy (imstuck):

k= ???

OpenStudy (drmoss):

k = 34

OpenStudy (imstuck):

Yes! Now use that k value to solve for d when c = 68. Any ideas on how to set it up?

OpenStudy (imstuck):

The same equation is used, but now you have a k value and a c value to use to help find d.

OpenStudy (drmoss):

sorry my brain just died lol i couldnt think of anything for a minute

OpenStudy (imstuck):

It's ok! This is kinda daunting, I get it!

OpenStudy (drmoss):

68=34/d?

OpenStudy (drmoss):

so d = 2

OpenStudy (imstuck):

\[c = \frac{ k }{ d }\] and we have k to be 34 and c to be 68 \[68=\frac{ 34 }{ d }\] and 68d = 34

OpenStudy (imstuck):

No, d doesn't equal 2...watch:

OpenStudy (drmoss):

oh its 1/2

OpenStudy (imstuck):

\[d=\frac{ 34 }{ 68 }\] \[d=\frac{ 1 }{ 2 }\]

OpenStudy (imstuck):

Good! See how that works?! Good job!!

OpenStudy (drmoss):

so.... a) 68=34/d

OpenStudy (drmoss):

b) d = 1/2

OpenStudy (imstuck):

yes, that's correct

OpenStudy (drmoss):

ok thx

OpenStudy (imstuck):

Any time!

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