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Mathematics 20 Online
OpenStudy (anonymous):

Help! I fan and give medals!! :) V varies inversely with T and V = 14 when T = 7. Which equation shows this relationship? A.v=2/t B.v=98/t C.v=2t D.v=98t

OpenStudy (anonymous):

@ParthKohli :) Sorry I'm such a bother....:)

OpenStudy (anonymous):

I have a feeling it might be a or b.:)

OpenStudy (whpalmer4):

So, if V varies inversely with T, that means that the relationship is of the form: \[V = \frac{k}{T}\]where \(k\) is the constant of variation. We know values for both \(V\) and \(T\) , so we can work out the value of \(k\) that makes the equation work.

OpenStudy (whpalmer4):

Your feeling is correct, as C and D are both examples of V varying directly with T:\[V = kT\]

OpenStudy (anonymous):

ok so can you help me work this out?

OpenStudy (whpalmer4):

So, for all the marbles, can you find the value of \(k\) and choose correctly between A and B?

OpenStudy (whpalmer4):

Hint: you can :-)

OpenStudy (anonymous):

B.? *crosses fingers*

OpenStudy (whpalmer4):

Here's the formula once again: \[V = \frac{k}{T}\] and the problem says "V = 14 when T = 7"

OpenStudy (whpalmer4):

No guessing allowed. You can get the answer with 100% certainty!

OpenStudy (anonymous):

oh hang on a sec.

OpenStudy (whpalmer4):

so plug V=14 and T = 7 into that equation and solve for \(k\)

OpenStudy (anonymous):

14 divieded by 7 is 2 so my answer is a?

OpenStudy (anonymous):

*divided I was multiplying!

OpenStudy (whpalmer4):

Slow down and think this through carefully. \[V = \frac{k}{T}\]\[V=14\]\[T=7\]\[14=\frac{k}{7}\]If we want to solve for the value of \(k\), what do we do?

OpenStudy (anonymous):

multiply?

OpenStudy (whpalmer4):

Multiply both sides by 7: \[14=\frac{k}{7}\]\[14*7 = 7*\frac{k}{7}\]\[98=k\] So our full equation to relate \(V\) and \(T\) at any value is...

OpenStudy (anonymous):

B.?

OpenStudy (whpalmer4):

As my old math teacher used to say, "is that an answer, or a prayer?" :-)

OpenStudy (whpalmer4):

your answer should be "it's B, of course!" :-)

OpenStudy (anonymous):

haha do you mind helping me on more?

OpenStudy (anonymous):

Your a great teacher:)

OpenStudy (whpalmer4):

I'm serious — you should be able to do these and be confident in your answers. Just takes a little practice. What's the next one?

OpenStudy (anonymous):

If x varies inversely with y and x = 8 when y = 10, find y when x = 6. A. B. C. D. If x varies inversely with y and x = 8 when y = 10, find y when x = 6. A. B. C. D.

OpenStudy (anonymous):

ahhh I typed it twice!!!

OpenStudy (anonymous):

A.y=3/20 B.y=4.8 C.y=7.5 D.y=40/3

OpenStudy (whpalmer4):

Okay, so this is exactly like the one we just did. Give it a shot and I'll guide you if you go astray

OpenStudy (whpalmer4):

"x varies inversely with y" means \[x = \frac{k}{y}\]

OpenStudy (whpalmer4):

plug in the known values of \(x\) and \(y\) to find the value of \(k\)

OpenStudy (anonymous):

ok so what was the formula again?

OpenStudy (anonymous):

urgh I suck at math so bad!

OpenStudy (anonymous):

6=k/10

OpenStudy (anonymous):

then I multiply each side by 10?

OpenStudy (whpalmer4):

okay, " x = 8 when y = 10" so plug those two into the formula and solve for \(k\)

OpenStudy (anonymous):

8=k/10

OpenStudy (whpalmer4):

formula is \[x=\frac{k}{y}\]

OpenStudy (whpalmer4):

right, so know we solve \[8 = k/10\] for \(k\) by multiplying each side by 10 as you suggested

OpenStudy (anonymous):

then I multiply each side by k/10 to get 80?

OpenStudy (whpalmer4):

close. multiply each side by 10, which will give us: \[8 = \frac{k}{10}\]\[8*10 = 10*\frac{k}{10}\]\[80=k\]

OpenStudy (anonymous):

then what?

OpenStudy (whpalmer4):

Now we plug the new value of \(x\) in to the equation along with \(k = 80\) to find the value of \(y\) when \(x = 6\) as requested

OpenStudy (whpalmer4):

we're just going one step further than we did in the previous problem.

OpenStudy (anonymous):

so then 6=k/10?

OpenStudy (anonymous):

then I multiply each side by 10?

OpenStudy (whpalmer4):

okay, we worked out the value of \(k\), right? what was it?

OpenStudy (anonymous):

oh duh 80.

OpenStudy (anonymous):

so 6=80/10

OpenStudy (whpalmer4):

so our formula is \[x = \frac{80}{y}\]

OpenStudy (anonymous):

Then I just plug in

OpenStudy (whpalmer4):

right. but be careful — it's asking for the value of \(y\) not \(x\) so we need to do a bit of algebra

OpenStudy (anonymous):

Wait so how do I do that? :)

OpenStudy (whpalmer4):

#1 secret to success in doing math problems is reading the problem carefully :-)

OpenStudy (anonymous):

6=80/k

OpenStudy (whpalmer4):

so we know that we want to find the value of \(y\) when \(x = 6\), and that our formula relating \(x\) and \(y\) is \[x = \frac{80}{y}\]We plug in the value of \(x = 6\): \[6 = \frac{80}{y}\] Now we need to solve for \(y\). What do you get if you multiply both sides of the equation by \(y\)?

OpenStudy (anonymous):

I have no idea

OpenStudy (anonymous):

6y and 80y?

OpenStudy (whpalmer4):

sure you do. \[6 * y = \frac{80}{\cancel{y}}*\cancel{y}\]\[6y=80\]

OpenStudy (whpalmer4):

Now can you solve that for the value of \(y\)?

OpenStudy (anonymous):

then divide each side by 6?

OpenStudy (whpalmer4):

Exactamundo!!!

OpenStudy (anonymous):

13.333333

OpenStudy (whpalmer4):

Right, but probably better to keep it as a fraction, given our answer choices

OpenStudy (anonymous):

Which is D.? :)

OpenStudy (whpalmer4):

80/6 = 40/3 so D is correct.

OpenStudy (anonymous):

Thanks so very much if I need help later could you possibly help my blonde head become a little smarter I already know I will. :)

OpenStudy (whpalmer4):

Sure, just shoot me a message and I'll respond next time I'm online

OpenStudy (anonymous):

Ok thanks I appreciate it alot!

OpenStudy (whpalmer4):

So to recap what we did: 1) write down the skeleton of the variation formula: x varies directly with y: \(x = ky\) x varies indirectly with y: \(x = \dfrac{k}{y}\) 2) Plug in known values of \(x, y\) to find \(k\) 3) (if needed) use completed formula with new value of \(x\) or \(y\) to find corresponding value of the other

OpenStudy (anonymous):

Thanks again!!!!

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