# Math Help - Tough Problems :(

1. ## Tough Problems :(

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...

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

3. 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?.

4. i still understand the cone question

5. 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

6. 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!

7. can some one please confirm? Thanks alot!

8. for some reason I keep seeing 3) as incorrect that's why i really hoping for some confirmation Thanks guys!