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waste of time. it's like a 4-5 line question demoires would require like 10

10^x + 15^x/2^5.3^x + 2^x.2^5 =5^x(2^x+3^x) / 2^5 (2^x + 3^x) = 5^x / 2^5

then i'm pretty sure thats right but someone double check

well by replacing sinx and cosx with their values in t (cbf typing this up) you end up with: 2tb + a(1-t^2) = c(1+t^2) expand and rearange this you get: (c+a)t^2 -2bt +c -a = 0 there for for it to have real roots discriminant >o 4b^2 -4(c-a)(c+a)>=0 4b^2 + 4a^2 >=4c^2 b^2 + a^2 >= c^2 new...

hey technically you have to:P you posted a solution! xD

are you sure Q is the quantity of salt? or did you just assume that?

your way works too^^

is the answer 5^x / 32

isn't taht just simply because a^x is the inverse function of log_a(x) ?

oh sorry i keep forgetting, i'll find one. try this one: integrate between pi/2 -> 0 sqrt(1+sin2x) dx

i think there's already a thread on this pinned, but anyway, it's cambridge for sure!

woops got double angle the wrong way round=.=

cos^2x = 2cos^2(x/2) -cos^2x = -2cos^2(x/2) 1-cos^2x= 1-2cos^2(x/2) 1-cos^2x=cosx cos^2x + cosx -1=0 cosx = -1/2plus or minus sqrt(5)/2 try that and see if you get the answer=S

i) ok i formed the equation with roots 1/alpha etc so it becomes: rx^3 + qx^2 + px + 1= 0 then since the roots are 1/alpha + 1 /beta + 1/gamma and they are in arithmatic seq. then they can be written as : 1/beta - d , 1/beta , 1/beta + d then sum of roots = -q /r 1/beta -d + 1/beta + 1/beta...

ok, generally, when your given a question show that the integral between -1->1 of f(x) dx is = 0 you would use f(-x) = -f(x) to show that its an odd function and then you can say that the integral of an odd function between limits -a -> a is equal to 0. but, in the case of sinx it's not...

well, i don't know there could be a way to show that, but it's generally easier to use the graph and show it has point symmetry.

you cannot use f(-x) = -f(x) for sinx since you need to use the fact that it is an odd function to prove that. and even function is symmetrical about the y-axis an odd function has POINT symmetry about the origin

well i know as a fact that y= 2sinx is odd but to show it, you use the graph and show it has point symmetry about the origin. and the integral of an odd function with limits a->-a is 0

ahh yes forgot that its focal chord for pq=-1 not latus rectum

lol ima try 2003 tmr then xD