can anyone help me with this ratio and proportion problems
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.
@Nnesha
@kropot72
You've been spamming this question constantly without giving anyone a chance to answer it, so no.
I believe we use the formula: \(y = \dfrac{k}{x}\)
how about v?
I believe we replace 'x' with 'v'..
Maybe..It's a confusing question.. \(y = \dfrac{v}{x}\)
Just leave the variables as it is, and use the same methods of inverse proportions as \(y = \dfrac{k}{x}\).
I don't know how to fit in the 'v' though..
:D
Can you help out with this question? @Supreme_Kurt
Wait..maybe we find the constant and use those to set up a proportion..
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?
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..
Yay, hartnn! :D
Just kidding. I am not sure either.
v= k/(xy)
Oh I was close....
Ohhhh
That explains it all :P
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...
both mean the same thing kurt :)
\(v = \dfrac{k}{xy}\) Plug in what we know: \(20 = \dfrac{k}{(8)(2)}\) Is this how we set it up..or..?
I've never learned this before xD
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.
1. v = k/xy 2. v = k1/x, v = k2/y which indirectly implies v= k1k2/(xy) = k/(xy)
oh wait... derp
If it means v is inversely proportional to x AND y, then x is directly proportional to y. NOPE
x can never be directly proportional to y in any cases here
iGreen, thats correct. find k
\[\Large vx = k_1\]\[\Large vy = k_2\] \[\Large \frac{vx}{vy}=\frac{k_1}{vy}\] \[\Large \frac{vx}{vy}=\frac{k_1}{k_2}\] \[\Large \frac{x}{y}=\frac{k_1}{k_2}=k\]
Are we proving inverse and direct proportionalities ? :D
Kurt is confused :3 And we are dealing with poor Kurt's confusion D:
\(\color{blue}{\text{Originally Posted by}}\) @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 :)
dangit, hart you gave me a hart...attack. haha
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.
brb guys, I think I've passed enough of my own confusion to OP. I'll just have dinner :3
Im confusing everyone here.
\(\color{blue}{\text{Originally Posted by}}\) @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
Okay, can you do that? @TheKinslayer
so its the final formula then?
yep, v= k/(xy)
ok ty guys can you help me with this another problem out here
what did you get the value of v finally ? just to check.... and yes
wait im still working with it
i got v=8
\(\huge \checkmark \)
who can i grand the best response?
here is the other problem..
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?
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 ^^
Btw, you should post this in a new question.
ok
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