Maths

CHECK THAT; first term should be [ 1 + ( x^2 / y^2 ) ]^( 3/2 )

Math is easy.
Fuck your British "s".
It's Mathematics. Not Mathsematics.

Anywho, it's a puzzle.
You have your end solution and you have your tools. How do you make your end look like your beginning.

I've honestly lost a lot of my math skills. It's all about transformations.
So look at your cheat sheet. Look at how you turn one thing into another.
The key to math is looking at your end first, not your beginning.
Look at the end, then look at the ends of all your equations. Look at what looks most similar.
Then look at what that started from. And that started from.

It's all a puzzle. And you have all the moves before you

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447KB, 720x528px

>>17148629

1 + (x^2 / y^2) = (y^2 / y^2) + (x^2 / y^2)
= ([x^2 + y^2] / y^2) = (1 / y^2) * (x^2 + y^2)

and (a * b)^c = a^c * b^c

Also holy fuck you're stupid if you're over 18 and don't know this shit.

Okay, think I have it figured out.

Take the first 1 in [ 1 + ( x^2 / y^2 ) ]^( 3/2 ), rewrite it as y^2 / y^2, and you get [ ( y^2 / y^2 ) + ( x^2 / y^2 ) ]^( 3/2 ).

Add the first two terms together and you get [ ( x^2 + y^2 ) / y^2 ]^( 3/2 ).

Multiply ( x^2 + y^2 ) / y^2 by 1/(y^2) • (y^2)/1 and you cancel out the y^2 in the denominator, leaving you with [ ( 1 / y^2 ) • ( y^2 + x^2 ) ]^( 3/2 ), and the last step's obvious.

Geez. Math takes more creativity than poetry ever did.

>>17148817
>>17148819

Spoke too soon; you guys replied while I was typing out >>17148825. Thanks for the help anyway.

>>17148767
It might be simpler to look at this as

1+(a/b) = (1/b)*(b+a)

Do the addition for 1+(a/b)

>>17148826
I was like you.
I always asked questions, then as I was asking them figured it out.

Math is beautiful.
It's about seeing the solutions in problems when you are already given the answers.

Connect the two and you're made.