Tough Problems :(

**justinwager** · September 13th 2008, 04:39 PM

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...

**Soroban** · September 13th 2008, 05:06 PM

Hello, justinwager!

WHO wrote these problem like this ??

$2(a)\;\;(2-2x)e^{-x} + (2x-x^2)e^{-x}(\text{-}1) \;=\;0$

We have: . $(2-2x)e^{-x} - (2x-x^2)e^{-x} \;=\;0$

Multiply by $e^x\!:\;\;(2-2x) - (2x - x^2) \;=\;0 \quad\Rightarrow\quad x^2-4x + 2 \;=\;0$

Quadratic Formula: . $x \;=\;\frac{4\pm\sqrt{8}}{2} \;=\;\frac{4 \pm2\sqrt{2}}{2} \;=\;\boxed{2 \pm \sqrt{2}}$

$2(b)\;\;(\ln x)^2 + x\cdot\ln x\cdot\frac{1}{x} \;=\;0$

We have: . $(\ln x)^2 + \ln x \;=\;0$

Factor: . $\ln x\,\bigg[\ln x + 1\bigg] \;=\;0$

Then: . $\begin{array}{ccccccc}<br /> \ln x \:=\:0 & \Rightarrow & x \:=\:e^0 & \Rightarrow &\boxed{x\:=\:1} \\ \\[-3mm]<br /> \ln x \:=\:\text{-}1 & \Rightarrow & x \:=\:e^{\text{-}1} & \Rightarrow & \boxed{x\:=\:\frac{1}{e}} \end{array}$

**galactus** · September 13th 2008, 05:22 PM

For the cone problem, try using similar triangles.

We get $\frac{H-h}{H}=\frac{r}{R}$

$h=\frac{H}{R}(R-r)$

Sub into cylinder volume formula:

$V={\pi}\frac{H}{R}(R-r)r^{2}={\pi}\frac{H}{R}(Rr^{2}-r^{3})$

There is a start. Can you do anything with that?.

**justinwager** · September 14th 2008, 12:13 PM

i still understand the cone question

**Jhevon** · September 14th 2008, 12:26 PM

Originally Posted by justinwager

i still understand the cone question

be specific as to what you don't get. this is an assignment. we should not be doing the problems for you. we can only guide you

**justinwager** · September 16th 2008, 09:38 PM

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!

**justinwager** · September 17th 2008, 10:16 AM

can some one please confirm? Thanks alot!

**justinwager** · September 17th 2008, 05:55 PM

for some reason I keep seeing 3) as incorrect that's why i really hoping for some confirmation

Thanks guys!

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