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Let u = 16-x^2 \frac{du}{dx}=-2x dx = \frac{du}{-2x} \int_{0}^{4}x\sqrt[]{u}.\frac{du}{-2x} \frac{-1}{2}.[\frac{2u^{\tfrac{3}{2}}}{3}]_{16}^{0} And just minus u should get the answer from there.

I will try to beat Spi to this question.

How about you teach me now before i procrastinate tomorrow evening? :)

I hope you keep your word.

Yes vectors.... Msn with me tomorrow night please.

School assessments.

Maybe your teacher was right. SC is not good enough to determine whether or not you're suited for extension.

^ yea pretty cut.

Indeed. How about i rephrase it: Imma just take it easy till Christmas. xD

Imma just take it easy till new years.

Yea lol im sensing trolling. Who gets embarrassed with a 97/100 ? LMFAO

Wow wtf? Beast. And i thought the person who topped the state for physics was a guy o.O

Tomorrow you will know

I like to chew gum before and during the test.

Lol normal. He/she does 4u. 2U don't count.

Yea you've digged yourself a very big hole that will be very hard to climb out of.

I'm still buying that Ruse will still keep its 1st position.

I'm still buying that Ruse will still keep its number 1 position this year though.

Ruse prolly most likely notched up most of the state ranks anyways haha.

Spoiler much lol