Melody

Then you can addd interest :)))   That is the icing on the cake :))

$$\begin{array}{rlll}
\frac{x-2}{2x+2}-\frac{1}{3x^2-6x-9}-\frac{x-30}{18-6x}&=&3\\\\
\frac{x-2}{2(x+1)}-\frac{1}{3(x^2-2x-3)}-\frac{x-30}{6(3-x)}&=&3\\\\
\frac{x-2}{2(x+1)}-\frac{1}{3(x-3)(x+1)}+\frac{x-30}{6(x-3)}&=&3\\\\
\frac{3(x-3)(x-2)}{3(x-3)*2(x+1)}-\frac{2*1}{2*3(x-3)(x+1)}+\frac{(x+1)(x-30)}{6(x+1)(x-3)}&=&3\\\\
3(x-3)(x-2)-2*1+(x+1)(x-30)&=&3*6(x+1)(x-3)\\\\
3(x^2-5x+6)-2+(x^2-29x-30)&=&18(x^2-2x-3)\\\\
3x^2-15x+18-2+x^2-29x-30&=&18x^2-36x-54\\\\
3x^2+x^2-15x-29x+18-2-30&=&18x^2-36x-54\\\\
4x^2-44x-14&=&18x^2-36x-54\\\\
0&=&14x^2+8x-40\\\\
0&=&7x^2+4x-20\\\\
\end{array}$$

$$\\$Now I need 2 numbers that multiply to give 7*-20=-140$\\
$and add to give +4$\\
$Obviously one will be negative and one will be positive,$\\
$The bigger one will be positive$\\
$And the absolute value difference will be 4$\\
140=7*20\\
140=7*2*2*5\\
140=14*10\\
$The numbers that we want are +14 and -10$\\
$replace 4x with 14x-10x$\\
0&=&7x^2+4x-20\\
7x^2+14x-10x-20=0\\
$Now factor in pairs$\\
7x(x+2)-10(x-2)=0\\
(7x-10)(x-2)&=&0\\
7x-10=0\quad or \quad x-2=0\\
7x=10\quad or \quad x=2\\
x=1\frac{3}{7}\quad or \quad x=2\\$$

Ofcourse you could have just used the quadratic equation to get the final answer but if you wanted to factor it than is the best way to go about it.   IMHO     

Thanks Alan for spotting my error :)

Yes, i can see that Zac  

$${\frac{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{3.5}}\right)}{\left({\mathtt{2.44}}{\mathtt{\,\times\,}}{\mathtt{1.22}}\right)}} = {\mathtt{5.878\: \!796\: \!022\: \!574\: \!576\: \!7}}$$

minimum of 6 

$$\\250(1/2)^{10/2}\\\\
=250*(\frac{1}{2})^5\\\\
=250*\frac{1}{32}\\\\
=\frac{125}{16}\\\\
=7\frac{13}{16}\\\\$$

@@ End of Day Wrap  Fri 19/6/15   Sydney, Australia Time   9:35pm     ♪ ♫

Hi all,

It was very quiet until a short time ago when some fun hard ones came along.  Anyway, there were some really good questions and answers.  Today our honorable answerers were CPhill, Civonamzuk, Syllogist, Headingnorth, Alan, Radix and Heureka.  Thanks all  

Let me think about this

What is the lowest common multiple of 30 and 95?

The prime factors of 30 = 3*2*5

The prime factors of 95 are  5*19

so the lowest common multiple is  3*2*5*19

$${\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{19}} = {\mathtt{570}}$$

Syllogist is right :))

$$\\X^6-X^3=2\\\\
Let \;Y=X^3\\\\
Y^2-Y=2\\\\
$Completing the square$\\\\
Y^2-Y+\frac{1}{4}=2+\frac{1}{4}\\\\
(Y-\frac{1}{2})^2=\frac{9}{4}\\\\
Y-\frac{1}{2}=\pm\frac{3}{2}\\\\
Y-\frac{1}{2}=\frac{3}{2}\qquad OR \qquad Y-\frac{1}{2}=-\frac{3}{2}\\\\
Y=2\qquad \qquad OR \qquad \qquad Y=-1\\\\
X^3=2\qquad \qquad OR \qquad \qquad X^3=-1\\\\
X=\sqrt[3]{2}\qquad \qquad OR \qquad \qquad X=-1\\\\$$

These are just the real solutions :)

Let me think about the complex solutions

I think they are

$$\\x=\sqrt[3]{2}\;\;e^{\frac{2\pi i}{3}},
\qquad x=\sqrt[3]{2}\;\;e^{\frac{-2\pi i}{3}}, \qquad
x=e^{\frac{-\pi i}{3}}, \quad and \quad x= e^{\frac{\pi i}{3}}$$

I hope I didn't s***w that one up :/ 

Thanks Syllogist.  You're right of course, I just want to try and explain.  

$$\\\int\;\frac{(cos^{-1}x)^2 }{(1-x^2)^{1/2}}\;dx\\\\
$What I notice is that $\;\; \frac{d}{dx}cos^{-1}x=\frac{-1}{\sqrt{1-x^2}}\\\\
$So I think the answer will be $ some constant* (cos^{-1}x)^3 \;\;+c\\\\
$Now if i differentiated this (without the constant multiple) I get$\\\\
\frac{d}{dx}(cos^{-1})^3x=3(cos^{-1})^2x\times \frac{-1}{\sqrt{1-x^2}}\\\\
$I need to get rid of the - and the 3 so I divide by -3 (that will be the constant multiple$\\\\
\int\;\frac{(cos^{-1}x)^2 }{(1-x^2)^{1/2}}\;dx=\frac{-1}{3}cos^{-1}x+c$$

Just like Syllogist said:))

NOTE: I can never remember that inverse trig differential, I have to work it out from scratch every time.  

(It doesn't take long to work out)

Thanks Chris

I had to think quite hard about this - maybe I shouldn't admit that   :))

He has a maximum of   $100,000 for fuel

Fuel costs $400 per gallon so he has enough money for

100,000 / 400 = 250 gallons

Now the distance is 600 miles so it will be alright so long has he gets at least 

$$\frac{600\;miles}{250\;gallons}=\frac{2400\;miles}{1000\;gallons}=2.4miles/gallon$$

Just like Chris said