bloodhound

I am currently doing the first one by hand... tell you the result later. seem to get a wrong complementary function... anyway ill try it later again. its 3.43 am. have to get up at 7:30 for lectures

I am currently doing the first one by hand... tell you the result later. seem to get a wrong complementary function... anyway ill try it later again. its 3.43 am. have to get up at 7:30 for lectures

derivative of x'/2 wrt t is just x''/2

derivative of x'/2 wrt t is just x''/2

for the second one i got from maple [math]x(t)=-\frac{26}{121}-\frac{4}{11}t+Ae^{11t}[/math] [math]y(t)=\frac{45}{121}-\frac{1}{11}t-\frac{4}{5}Ae^{11t}[/math] note again A is same in x and y

for the second one i got from maple [math]x(t)=-\frac{26}{121}-\frac{4}{11}t+Ae^{11t}[/math] [math]y(t)=\frac{45}{121}-\frac{1}{11}t-\frac{4}{5}Ae^{11t}[/math] note again A is same in x and y

i couldnt be bothered to do it myself. but i just banged the problem into maple and here are the sultions. the for the first [math]y(t)=e^{t}(B\sin(2t) + A\cos(2t))+\frac{7}{10}\sin(t)+\frac{11}{10}\cos(t)[/math] [math]x(t)=\frac{e^{t}}{2}(B\sin(2t)+B\cos(2t)+A\cos(2t)-A\sin(2t))+\frac{7}{10}\cos(t)-\frac{1}{10}\sin(t)[/math] note: A and B are same constants in x andy since the two diff equations are coupled. edit: why isnt x(t) showing up properly? anyway, i am sure you can understand the solution from the latex

i couldnt be bothered to do it myself. but i just banged the problem into maple and here are the sultions. the for the first [math]y(t)=e^{t}(B\sin(2t) + A\cos(2t))+\frac{7}{10}\sin(t)+\frac{11}{10}\cos(t)[/math] [math]x(t)=\frac{e^{t}}{2}(B\sin(2t)+B\cos(2t)+A\cos(2t)-A\sin(2t))+\frac{7}{10}\cos(t)-\frac{1}{10}\sin(t)[/math] note: A and B are same constants in x andy since the two diff equations are coupled. edit: why isnt x(t) showing up properly? anyway, i am sure you can understand the solution from the latex

what do u mean getting shot results in slow deaths. all that is required is two shots at max.

what do u mean getting shot results in slow deaths. all that is required is two shots at max.

example: sigma = 1 2 3 4 5 6 4 1 6 2 3 5 write it down in dijoint cycles you get sigma = (142)(365) now the order of sigma will be lcm(3,3)=3

example: sigma = 1 2 3 4 5 6 4 1 6 2 3 5 write it down in dijoint cycles you get sigma = (142)(365) now the order of sigma will be lcm(3,3)=3

if u write that permutation in disjoint cycles, the order will be the lowest common multiple of the order of the disjoing cycles. which is quite easy to find ,

if u write that permutation in disjoint cycles, the order will be the lowest common multiple of the order of the disjoing cycles. which is quite easy to find ,

the order is the least n such that sigma^n = sigma

the order is the least n such that sigma^n = sigma

jsatan? is that coventry in UK ur located at?

jsatan? is that coventry in UK ur located at?

what made u think it was OUF?

what made u think it was OUF?

If foxes were really vermin, then surely there are better ways to put them down than make hounds chase them while you are on horseback and watch the dogs tear the fox apart. Anyone who has "fun" , whether it be upper. middle or lower class, by killing animals in such a bloody way must have something wrong with them and should be put down themselves

If foxes were really vermin, then surely there are better ways to put them down than make hounds chase them while you are on horseback and watch the dogs tear the fox apart. Anyone who has "fun" , whether it be upper. middle or lower class, by killing animals in such a bloody way must have something wrong with them and should be put down themselves

if i read that correctly then it says the minimum of that integral where x and y are continuously differentiable functions from [0,1] to R. Unfortunately i havent done calculus of variations yet, and i cant solve that problem sucks'';;;;;;,,,,.....

if i read that correctly then it says the minimum of that integral where x and y are continuously differentiable functions from [0,1] to R. Unfortunately i havent done calculus of variations yet, and i cant solve that problem sucks'';;;;;;,,,,.....

I would say its more spherical, as spherical is defined in precise mathematical terms, whereas the concept of colour is different to everyone. i.e if an alien came about, it would not be seeing the same colour, but it would see the same shape.