Okay so I'm almost done my assignment and I'm completley lost to these (2) questions....I don't even understand where to begin...
Hello, justinwager!
WHO wrote these problem like this ??
$\displaystyle 2(a)\;\;(2-2x)e^{-x} + (2x-x^2)e^{-x}(\text{-}1) \;=\;0$
We have: .$\displaystyle (2-2x)e^{-x} - (2x-x^2)e^{-x} \;=\;0$
Multiply by $\displaystyle e^x\!:\;\;(2-2x) - (2x - x^2) \;=\;0 \quad\Rightarrow\quad x^2-4x + 2 \;=\;0$
Quadratic Formula: .$\displaystyle x \;=\;\frac{4\pm\sqrt{8}}{2} \;=\;\frac{4 \pm2\sqrt{2}}{2} \;=\;\boxed{2 \pm \sqrt{2}} $
$\displaystyle 2(b)\;\;(\ln x)^2 + x\cdot\ln x\cdot\frac{1}{x} \;=\;0$
We have: .$\displaystyle (\ln x)^2 + \ln x \;=\;0$
Factor: . $\displaystyle \ln x\,\bigg[\ln x + 1\bigg] \;=\;0$
Then: . $\displaystyle \begin{array}{ccccccc}
\ln x \:=\:0 & \Rightarrow & x \:=\:e^0 & \Rightarrow &\boxed{x\:=\:1} \\ \\[-3mm]
\ln x \:=\:\text{-}1 & \Rightarrow & x \:=\:e^{\text{-}1} & \Rightarrow & \boxed{x\:=\:\frac{1}{e}} \end{array}$
For the cone problem, try using similar triangles.
We get $\displaystyle \frac{H-h}{H}=\frac{r}{R}$
$\displaystyle h=\frac{H}{R}(R-r)$
Sub into cylinder volume formula:
$\displaystyle V={\pi}\frac{H}{R}(R-r)r^{2}={\pi}\frac{H}{R}(Rr^{2}-r^{3})$
There is a start. Can you do anything with that?.
okay so I did 2a) and got the right answer...Thanks for the help guys!
I scanned 2b) and 3 a) b) c) If you guys could verify to see if i got the correct answers I would really appreciate it!
Also Scroban I think you got the x = 1 and x = 1/e because you forgot the 2 in the beggining of the equastion!
Thanks for the help guys!