# Thread: Proportions

1. ## Urgent..

help mi wif tis..

2. Originally Posted by xiaoz
help mi wif tis..
your equation involves three variables and you are given two of them in every part. Simply plug in the values and you will get the answer.

Keep Smiling
Malay

3. To find $k$ we merely need to solve for $k$ when $y=90$ and $x=2$

$y=\frac{k}{x^2}$

$90=\frac{k}{2^2}$

$90=\frac{k}{4}$ multiply by 4

$360=k$

3(b) simple...

$y=\frac{k}{x^2}$

$y=\frac{360}{(-10)^2}$

$y=\frac{360}{100}$

$y=\frac{36}{10}$

3(c) first, solve for x

$y=\frac{k}{x^2}$ reciprocate

$\frac{1}{y}=\frac{x^2}{k}$ multiply by $k$

$\frac{k}{y}=x^2$ find the square root for both sides

$\sqrt[\pm]{\frac{k}{y}}=x$ now substitute 40 for y

$\sqrt[\pm]{\frac{360}{40}}=x$ solve...

$\sqrt[\pm]{9}=x$ solve...

$\pm3=x$

4. Zzz.. so diff.. ur question.. my question more easy.. XD

5. Originally Posted by xiaoz
Zzz.. so diff.. ur question.. my question more easy.. XD
I assume you mean the question in my signature. The only person (that I know of) to actually solve it is PH, although it's grown since then...

6. Originally Posted by Quick
I assume you mean the question in my signature. The only person (that I know of) to actually solve it is PH, although it's grown since then...
I solved the signature a long time ago, I just haven't gotten the point of it yet.

-Dan

7. Originally Posted by topsquark
I solved the signature a long time ago, I just haven't gotten the point of it yet.

-Dan
Originally the answer was $Q_ui_ck=co^2l$ whitch can be written as $Q_ui_ck=cool$ now the answer is different by 1 number...

8. Wait is the answer still cool? If so that is cool.

9. Originally Posted by classicstrings
Wait is the answer still cool? If so that is cool.
The answer is $Q_ui_ck=2cool$

$Q_u=\sqrt{\frac{100C}{144l^2o}}=\frac{10\sqrt{C}}{ 12l\sqrt{o}}$

$i_c=\sqrt[3]{8o^9l^3}=2o^3l$

$k=6l\sqrt{\frac{C}{25o}}=\frac{6l\sqrt{C}}{5\sqrt{ o}}$

$Q_ui_ck=\frac{10\sqrt{C}}{12l\sqrt{o}}\times2o^3l
\times \frac{6l\sqrt{C}}{5\sqrt{o}}$

$=\frac{120o^3l^2\sqrt{C}\sqrt{C}}{60l\sqrt{o}\sqrt {o}}$

$=\frac{120o^3l^2C}{60lo}$

$=\frac{\not{60}\times2\not{o}\times o^2\not{l}\times lC}{\not{60}\not{l}\not{o}}$

$2o^2lC\quad\rightarrow\quad2Co^2l\quad\rightarrow
\quad
2Cool$

10. Ah ha...

Is there a thread which shows how to work that out? Thanks I'm interested.

11. I edited it into my last post.

THIS IS MY 200TH POST! (and it says quick is 2cool)

Quick is:

The 11th highest poster!
6th highest reputation holder!

12. Originally Posted by Quick
The answer is $Q_ui_ck=2cool$

$Q_u=\sqrt{\frac{100C}{144l^2o}}=\frac{10\sqrt{C}}{ 12l\sqrt{o}}$

$i_c=\sqrt[3]{8o^9l^3}=2o^3l$

$k=\frac{\sqrt{\frac{C}{25o}}}{6l}=\frac{6l\sqrt{C} }{5\sqrt{o}}$

$Q_ui_ck=\frac{10\sqrt{C}}{12l\sqrt{o}}\times2o^3l
\times \frac{6l\sqrt{C}}{5\sqrt{o}}$

$=\frac{120o^3l^2\sqrt{C}\sqrt{C}}{60l\sqrt{o}\sqrt {o}}$

$=\frac{120o^3l^2C}{60lo}$

$=\frac{\not{60}\times2\not{o}\times o^2\not{l}\times lC}{\not{60}\not{l}\not{o}}$

$2o^2lC\quad\rightarrow\quad2Co^2l\quad\rightarrow
\quad
2Cool$
Wow that's very nice. I'm going to show it to my friends.

13. Originally Posted by Quick
The answer is $Q_ui_ck=2cool$

$Q_u=\sqrt{\frac{100C}{144l^2o}}=\frac{10\sqrt{C}}{ 12l\sqrt{o}}$

$i_c=\sqrt[3]{8o^9l^3}=2o^3l$

$k=\frac{\sqrt{\frac{C}{25o}}}{6l}=\frac{6l\sqrt{C} }{5\sqrt{o}}$

$Q_ui_ck=\frac{10\sqrt{C}}{12l\sqrt{o}}\times2o^3l
\times \frac{6l\sqrt{C}}{5\sqrt{o}}$

$=\frac{120o^3l^2\sqrt{C}\sqrt{C}}{60l\sqrt{o}\sqrt {o}}$

$=\frac{120o^3l^2C}{60lo}$

$=\frac{\not{60}\times2\not{o}\times o^2\not{l}\times lC}{\not{60}\not{l}\not{o}}$

$2o^2lC\quad\rightarrow\quad2Co^2l\quad\rightarrow
\quad
2Cool$
NOW it makes sense. The previous formula didn't.

-Dan