help mi wif tis..
To find $\displaystyle k$ we merely need to solve for $\displaystyle k$ when $\displaystyle y=90$ and $\displaystyle x=2$
$\displaystyle y=\frac{k}{x^2}$
$\displaystyle 90=\frac{k}{2^2}$
$\displaystyle 90=\frac{k}{4}$ multiply by 4
$\displaystyle 360=k$
3(b) simple...
$\displaystyle y=\frac{k}{x^2}$
$\displaystyle y=\frac{360}{(-10)^2}$
$\displaystyle y=\frac{360}{100}$
$\displaystyle y=\frac{36}{10}$
3(c) first, solve for x
$\displaystyle y=\frac{k}{x^2}$ reciprocate
$\displaystyle \frac{1}{y}=\frac{x^2}{k}$ multiply by $\displaystyle k$
$\displaystyle \frac{k}{y}=x^2$ find the square root for both sides
$\displaystyle \sqrt[\pm]{\frac{k}{y}}=x$ now substitute 40 for y
$\displaystyle \sqrt[\pm]{\frac{360}{40}}=x$ solve...
$\displaystyle \sqrt[\pm]{9}=x$ solve...
$\displaystyle \pm3=x$
The answer is $\displaystyle Q_ui_ck=2cool$Originally Posted by classicstrings
$\displaystyle Q_u=\sqrt{\frac{100C}{144l^2o}}=\frac{10\sqrt{C}}{ 12l\sqrt{o}}$
$\displaystyle i_c=\sqrt[3]{8o^9l^3}=2o^3l$
$\displaystyle k=6l\sqrt{\frac{C}{25o}}=\frac{6l\sqrt{C}}{5\sqrt{ o}}$
$\displaystyle Q_ui_ck=\frac{10\sqrt{C}}{12l\sqrt{o}}\times2o^3l
\times \frac{6l\sqrt{C}}{5\sqrt{o}}$
$\displaystyle =\frac{120o^3l^2\sqrt{C}\sqrt{C}}{60l\sqrt{o}\sqrt {o}}$
$\displaystyle =\frac{120o^3l^2C}{60lo}$
$\displaystyle =\frac{\not{60}\times2\not{o}\times o^2\not{l}\times lC}{\not{60}\not{l}\not{o}}$
$\displaystyle 2o^2lC\quad\rightarrow\quad2Co^2l\quad\rightarrow
\quad
2Cool$