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is it all ages?

light, L, is proportional (by constant k) to rectangle area + half of semicircle area L/k = xy + πx2/16 and x + 2y + (πx)/2 = 20 [sub in for y, differentiate, solve for max] hopefully gives x = 10/(1+3π/8)

what did you put as your limits? where does the curve cut the y-axis?

there's still a 1/2 chance of getting H. it's just very unlikely that you got the string of H's in the first place as for your weird example, if there is a 9/10 chance that "the US loses the War in Iraq" and a 1/10 chance that "women have mustache", and they're independent events, then there...

what's wrong with "AND = multiply"? I thought that was pretty intuitive say, you're rolling a dice and if you don't roll a 6 you stop rolling there's a 1/6 chance of rolling 6 on the first roll if you made it to the second roll, there would be a 1/6 chance again of rolling 6 so the chance of...

hint: pythagoras.. (and circle geo properties)

I guess it could be partially content based, and have a problem set given as the assessment (like what is in maths competitions) still probably wouldn't work, but it would be cool if school was like that (I think so at least)

motors hum, equipment switches on and off, valves open and close.. but the truth rings clear we have a destination WE HAVE A DESTINATION

in my opinion the courses should focus on general problem solving/thought processes, rather than being content/topic based (it would probably be pretty hard to structure it that way though). same goes for pretty much any subject :)

just put each in polar (mod/arg) form

infinity + 1

my notation is a bit messy

you're just putting an integral that you can't identify how to solve into a form you can maybe if you give an example, it could help us explain clearly why it works

so you're not able to put a limit in the integal?, ie. +anyone who hasn't read through the solutions, feel free to have a go, or at least think about the original question

(a-b)2 >= 0 a2 + b2 >= 2ab a/b + b/a >= 2 (dividing by ab which is >0 as a,b >0) similarly, a2/b2 + b2/a2 > 2 (a+b)^2(1/a^2+1/b^2) = (a2 + 2ab + b2)(1/a2 + 1/b2) = 1 + a2/b2 + 2b/a + 2a/b + b2/a2 + 1 = 2 + 2[a/b + b/a] + [b2/a2 + a2/b2] >= 2 + 2(2) + 2 = 8 .'. (a+b)^2(1/a^2+1/b^2)>=8 a,b>0...

oh right.. well you can just include 0; that doesn't affect the limiting sum not too sure about the 1 though.. unless you make the limit 1- (value of x as x->1), such that it is just below 1, but the sum still approaches the value for x=1.. or something

is there anything wrong with

find the limiting sum of 1 - x2 + x4 - x6 + x8 ... given that 0 < or = x < 1 hence find the limiting sum of 1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 ...

haha Rkiuyto, where is this question (the tan2 one) from? solution is here.. Art of Problem Solving Forum

"In how many ways can n things be shared between two people?" Each of the n objects can be given to one of 2 people (2n ways) but neither of the 2 can recieve every thing (-2 of the ways)