I'll come celebrate with you.

HAPPY BIRTHDAY OMI! =D 

x+17=18

   -17 -17

x    =    1    

Happy Birthday !!!    I am going to celebrate  FOR you !....

LOL

36!/[36 - 12]! x 12!=1,251,677,700

happy b-day Omi

x^(-6)  =  1/x^6        NOT  1/6x

\(\dfrac{12}{5^{-2}}\\ =12\cdot 5^2\\ = 12\cdot 25\\ = 300\\ \color{blue}{\cdot\;\;\;\;\;\cdot\\ \;\smile}\)

\(\dfrac{\text{miles}}{\text{miles per gallon}}\\ =\dfrac{\text{miles}}{\dfrac{\text{miles}}{\text{gallon}}}\\ =\dfrac{1}{\dfrac{1}{\text{gallon}}}\\ =\text{gallon}\)

This is correct.

\(\sqrt{1000000}\\ =\sqrt{10^6}\\ =10^{6\div2}\\ =10^3\)

\(\sqrt{5x-4}-\sqrt x = 2\\ 5x-4+x-2(\sqrt{5x^2-4x})=4\\ 6x - 2\sqrt{5x^2-4x}=8\\ 3x-4=\sqrt{5x^2-4x}\\ 5x^2-4x=9x^2+16-24x\\ 4x^2-20x+16=0\\ x^2 - 5x + 4 = 0\\ (x-1)(x-4)=0\\ \color{blue}x=1\text{(rej.) OR }x=4\\ \color{red}x=4\)

D operators work well for the first equation.

\(\displaystyle y'' - 2y' + y = (x^{2}+1)e^{x}\)

so the equation becomes

\(\displaystyle D^{2}y-2Dy + y =(x^{2}+1)e^{x}\),

\(\displaystyle (D^{2}-2D+1)y=(x^{2}+1)e^{x}\),

\(\displaystyle y = \frac{1}{(D-1)^{2}}e^{x}(x^{2}+1)=e^{x}\frac{1}{D^{2}}(x^{2}+1)=e^{x}\big(\frac{x^{4}}{12}+\frac{x^{2}}{2}+Ax+B\big)\),

where \(\displaystyle A \text{ and }B\) are arbitrary constants.

Tiggsy

702 is not a fraction.

Hi Max,

I was about to ask you what it was but now I know - at least I know what it was supposed to be,  :)

\(360\div 3\\ = 120\)

Whopps it should be a birthday cake :(

Happy birthday!!
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LOL Thanks Omi,

I thought that you were born in 1967, but you have been around for much longer than that!

Do you have a pet dinosaur too ???

I have Red, he is a bit put out that I took his photo off my icon, my new icon is 'HAPPY' he is one of RED's friends but still RED is jealous..  I might put RED back again soon.  Some of the kids are complaining too, they like me using RED as my icon..... ..   

Hello.

\(126\div 7\\ =18 \)

Hallo Melody

Thanks Alan,

I might just do that - does sound rather like cheating though   

You will need to become pretty familiar with Bessel functions to answer Guest #3's new question Melody!  I suggest you just enter it into WolframAlpha.

Hi guest/s

Thanks for your participation here :)     

I shall look at your solution and your new question too :)

I might take a while to get too it but I will  :))