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Algebra 18 Online
OpenStudy (anonymous):

can anyone help me with this ratio and proportion problems

OpenStudy (anonymous):

If v is inversely proportional to x and y, and if v=20 when x=2 and y=8, find v when x =4 and y=10.

OpenStudy (anonymous):

@Nnesha

OpenStudy (anonymous):

@kropot72

OpenStudy (jhannybean):

You've been spamming this question constantly without giving anyone a chance to answer it, so no.

OpenStudy (igreen):

I believe we use the formula: \(y = \dfrac{k}{x}\)

OpenStudy (anonymous):

how about v?

OpenStudy (igreen):

I believe we replace 'x' with 'v'..

OpenStudy (igreen):

Maybe..It's a confusing question.. \(y = \dfrac{v}{x}\)

OpenStudy (jhannybean):

Just leave the variables as it is, and use the same methods of inverse proportions as \(y = \dfrac{k}{x}\).

OpenStudy (igreen):

I don't know how to fit in the 'v' though..

OpenStudy (igreen):

:D

OpenStudy (igreen):

Can you help out with this question? @Supreme_Kurt

OpenStudy (igreen):

Wait..maybe we find the constant and use those to set up a proportion..

OpenStudy (anonymous):

Just Kurt. And I'm a bit confused about what it means for a value to be inversely proportional to two values. Does it mean it's inversely proportional to their product?

OpenStudy (igreen):

The constant when x = 2 and y = 8 is 16. The constant when x = 4 and y = 10 is 40. v = 20 when the constant is 16 Find v when the constant is 40. Set up a proportion: \(\dfrac{20}{16} = \dfrac{x}{40}\) Idk if that's right..

OpenStudy (igreen):

Yay, hartnn! :D

OpenStudy (jhannybean):

Just kidding. I am not sure either.

hartnn (hartnn):

v= k/(xy)

OpenStudy (jhannybean):

Oh I was close....

OpenStudy (igreen):

Ohhhh

OpenStudy (igreen):

That explains it all :P

OpenStudy (anonymous):

so that IS what it meant, then? That for v to be inversely proportional to x and y, it means that v is inversely proportional to xy after all? I thought v had to be inversely proportional to x AND inversely proportional to y...

hartnn (hartnn):

both mean the same thing kurt :)

OpenStudy (igreen):

\(v = \dfrac{k}{xy}\) Plug in what we know: \(20 = \dfrac{k}{(8)(2)}\) Is this how we set it up..or..?

OpenStudy (igreen):

I've never learned this before xD

OpenStudy (anonymous):

No they don't. If it means v is inversely proportional to xy, then x is inversely proportional to y If it means v is inversely proportional to x AND y, then x is directly proportional to y.

hartnn (hartnn):

1. v = k/xy 2. v = k1/x, v = k2/y which indirectly implies v= k1k2/(xy) = k/(xy)

OpenStudy (anonymous):

oh wait... derp

hartnn (hartnn):

If it means v is inversely proportional to x AND y, then x is directly proportional to y. NOPE

hartnn (hartnn):

x can never be directly proportional to y in any cases here

hartnn (hartnn):

iGreen, thats correct. find k

OpenStudy (anonymous):

\[\Large vx = k_1\]\[\Large vy = k_2\] \[\Large \frac{vx}{vy}=\frac{k_1}{vy}\] \[\Large \frac{vx}{vy}=\frac{k_1}{k_2}\] \[\Large \frac{x}{y}=\frac{k_1}{k_2}=k\]

OpenStudy (jhannybean):

Are we proving inverse and direct proportionalities ? :D

OpenStudy (anonymous):

Kurt is confused :3 And we are dealing with poor Kurt's confusion D:

hartnn (hartnn):

\(\color{blue}{\text{Originally Posted by}}\) @hartnn 1. v = k/xy 2. v = k1/x, v = k2/y which indirectly implies v= k1k2/(xy) = k/(xy) <<<<<<MY MISTAKE, that should be v^2 \(\color{blue}{\text{End of Quote}}\) so you're right, kusrt :)

OpenStudy (anonymous):

dangit, hart you gave me a hart...attack. haha

hartnn (hartnn):

if v is inversely proportional to x and inversely proportional to y, then x and y are directly proportional. which is not the case here.

OpenStudy (anonymous):

brb guys, I think I've passed enough of my own confusion to OP. I'll just have dinner :3

OpenStudy (jhannybean):

Im confusing everyone here.

hartnn (hartnn):

\(\color{blue}{\text{Originally Posted by}}\) @iGreen \(v = \dfrac{k}{xy}\) Plug in what we know: \(20 = \dfrac{k}{(8)(2)}\) Is this how we set it up..or..? \(\color{blue}{\text{End of Quote}}\) continue, find k here

OpenStudy (igreen):

Okay, can you do that? @TheKinslayer

OpenStudy (anonymous):

so its the final formula then?

hartnn (hartnn):

yep, v= k/(xy)

OpenStudy (anonymous):

ok ty guys can you help me with this another problem out here

hartnn (hartnn):

what did you get the value of v finally ? just to check.... and yes

OpenStudy (anonymous):

wait im still working with it

OpenStudy (anonymous):

i got v=8

hartnn (hartnn):

\(\huge \checkmark \)

OpenStudy (anonymous):

who can i grand the best response?

OpenStudy (anonymous):

here is the other problem..

OpenStudy (anonymous):

The distance fallen by a body, starting from a position of rest in a vacuum near the easrth's surface, is proportional to the square of the time occupied in falling. If a body falls 256 ft. in 4sec, how far will it fall in 11 sec?

OpenStudy (anonymous):

Proportion this time is direct :) And when two values are in direct proportion, their QUOTIENT will be the same, no matter what :) Now, read through the problem again and tell me which two values are in direct proportion ^^

OpenStudy (igreen):

Btw, you should post this in a new question.

OpenStudy (anonymous):

ok

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