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$let eqn of cirlce be:$ $ $ x^2 +y^2=a^2 $since cylinder taken out has radius of c, let the lines y=c and y=-c be lines to represent the edges of the cylinder$ $when y=\pm c, x=\pm \sqrt{a^2-c^2} V_{remaing}=V_{revolution}-V_{cylinder} V_{revolution}=\pi \int_{-\sqrt{a^2 -c^2}}^{\sqrt{a^2...

well for a trig sub, let x=acos@... then u have to find what theta is in terms of a and c, since it is a definite integral ......so ur gonna get theta to equal the sin inverse of some fraction with a sqrt in it....... ill put up my solution up now... gimme time to play with latex

how would u integrate that with out a trig sub? u ca'nt use trig since u are given a and c, unless ut going to use sin^-1(a)?

Its in the terry lee

ohh thank God. i was going to kill myself :D

I AM OFFICIALLY STRESSING OUT. Q. find the modulus of 1+sqrt3 i... i keep getting fucking pi/3 and the answers say 2!??!?!?!??!??!??!?! WHAT THE FUCK!! someone help me out here.... and i only do 10 units

12 fo sho.... did jack all in years 7-11 and i don't regret it.... acing all me subs ever since i picked my game up in yr 12

u watch water go up the pipe and oil not go up.... something about cohession and adhession forces

x^2 -q=0 sum or roots of new equation is 0 and product is: -(a-b)(a-b) =-(a-b)^2= -sqrt(P+q-4abc+2sin\theta) *using the guassian distribution formula* then upon inspection, product of roots = -q thus x^2 -q=0

the heat is produced as a consequence of the resistance (collisions of electrons with the lattice)

"thanks"

can i tutor? im not that less experienced :P

maths is timeless:p

if they have vaginas and are drunk enough:drink:

case one: 2 couples 10C2 (pick the couples).... and i think thats it cause they will be chosen so that H1W1 and H2W2 will be playing, but there is only one posibility that will be H1W2 and H2W1... case 2: one couple 10C1 *18C2 case 3: no couoples: 20C4 ans:10C2+10C1*18C2+20C4

isn't it supposed to burn the oil? O.o.O (three eyed man is confused)

Write down the domain and range of y = pi/2 - sin^-1 (x/2) $DOMAIN:$ -1\leq \frac{x}{2}\leq 1 -2\leq x\leq 2 $RANGE:$ $when x= -2, y= \pi $when x= 2, y= 0 \therefore $RANGE:$ 0\leq y\leq \pi

Prove by induction that 9^(n+2) -4^(n) is divisable by 5 for all positive integers n . $Step one: test n=1$ LHS=9^3-4^1 =725,$which is divisible by five$ $hence n=1 holds$ $step two: assuming n=k holds true$ \therefore 9^{k+2}-4^k =5M $Step Three: proving true for n=k+1$ RTP:9^{k+1+2}-4^{k+1}...

|ex-a|= |-1||a-ex|= 1*|a-ex|

go with Nasa. it really doesn't matter... as long as you acknowledge it in ur biblio, then ur correct