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Setting cos(x)=i/sqrt(6) works. Not sure for the real value(s) tho.
is this asking what is the probality that x is in A given that it is in B, or is it asking what is the probability that a divides b
for lines I get:
1+((n-2)(n-1)/2)
dont know if it makes any difference but you dont need all of the sin(x) to be positive. you could have an even number of them be negative
From dy/dx = e^2x + 1/e^2x + 2x +2 i simply get y=(1/2)(e^2x- 1/e^2x)+x^2+2x+3
dont know how to write it like above.
((4)(z^-2)+(-8)(z^-1)+4)/((4+w-2w/Q)(z^-2)+(2w-8)(z^-1)+(4+w+2w/Q))
that is my answer to the first question. i used w instead od omega
there is a way to tell if a number is prime or not. it just isnt very practical for large numbers. if x divides (x-1)! then it is not a prime. so all prime numbers,p, do not divide (p-1)!
lucy=lana+6
jo=lucy+5=lana+6+5=lana+11
41=lana+lana+6+lana+11
3*lana=24
lana=8
i wanted an equation with zeroes at 350 and 550. this could be (x-350)(x-550). i needed a maximum at 500 so i added the (x-650) to give this by symmetry. i then included another (x-550) in order to keep the shape right. then i multplyed by a constant to get it to go through the points
that face is meant to be an = then a (. it replaced them
f(x)=((350-x)(x-650)(x-550)^2)/562500
i get 2*(pi^2)log(2)
I made the substitution z=exp(ix). it then reduces to a complex integration using a semicircle with radius one as the contour. i then construced the the power series and used the coeficient of 1/z as the residue.
not sure if there are any other singular points in the contour apart from z=0
If 1/((x(1- (ln^2(x))^0.5)) can be writtn as 1/((x(1- ln(x))) then the integral is
-ln(1-ln(x)).
put three into (cuberoot(112-3*x))/2. then put this answer back in. and so on.
I am not sure if this counts as trial and improvement
I dont think that this can be expressed in terms of elementary functions. You can express sin in terms of exponentials and i know that exp(x^2) cant be integrated.
1)letters, aplhabet
2)Days week
3)?
4)dalmations
5)cards deck
6)wonders world
suppose that ab and ba have different orders the one of then must have an order greater then the other. let the order of ab be greater than ba. let n be the order of ab. so (ab)^n=1.
so (ab)(ab)......(ab)(ab)=1 (with n lots of ab)
a(ba)(ba)......(ba)(ba)b=1 since the order of ba is less than that of ab some combination of these ba must equal one. this gives:
(ab)(ab)......(ab)(ab)=1 (with less than n lots of ab)
this gives the order of ab less than n. contradiction. therfore the order of ab = the order of ba
You can prove that the ratio is sqrt((1-cos(108))/(1-cos(36))) by using the cosine rule to get expressions for the lentghs of th e two pentagons
The ratio of the two pentagons is equal to sqrt((1-cos(108))/(1-cos(36)))
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