Recent content by santa

ok the constant of Hooke's_law F=-KX k positive, negative, real, complex. or not

thanks but a have another Definition of directly proportional - can k be negative? In almost all textbooks, "directly proportional" is defined by saying that a is directly proportional to b if and only if a = kb for some constant k. That's perfectly sensible, but taking the...

if y\propto x at z constant and y\propto z at x constant then y\propto xz why not y^2\propto xz thank you

OK but the solution where

find a and r a+ar+ar^2=7 a^3+a^3r^3+a^3r^6=73

There are two pipes, Pipe A and Pipe B. Pipe A filled a tank in for minutes less than B does. If both pipes are open the tank is filled in 24 minutes. Find the time A will take if B is closed

good work but let $ \sqrt[3]{x^2+2}=a, \ \sqrt[3]{4x^2+3x-2}=b, \ \sqrt[3]{3x^2+x+5}=c, \ \sqrt[3]{2x^2+2x-5}=d. these may be help

prove that \frac{1}{1^2.3^3.5^2}-\frac{1}{3^2.5^3.7^2}+\frac{1}{5^2.7^3.9^2}-...=\frac{1}{9}-\frac{\pi}{2^6}-\frac{\pi^3}{2^9}

solve in R (x^2+2)^{1/3}+(4x^2+3x-2)^{1/3}=(3x^2+x+5)^{1/3}+(2x^2+2x-5)^{1/3}

thanks for all from these http://mathworld.wolfram.com/PiFormulas.html let S=1- \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - ... = \frac{\pi}{4}. thus \frac{1}{3} S= \frac{1}{3} - \frac{1}{9}+ \frac{1}{15} - \frac{1}{21} + ... = \frac{\pi}{12}. series = S...

prove the sum of this series =pi/3 1 +\frac{ 1}{5} -\frac{ 1}{7} - \frac{1}{11 }+\frac{1}{13} + \frac{1}{17} - ...

If \frac{x}{y}=\frac{3}{5} then (A) x<y ( B) x> y (C)x=y (D)noon of these

\frac{1}{1^4+1^2+1}+\frac{1}{2^4+2^2+1}+...+\frac{1}{2007^4+2007^2+1}

\tan^2{(1^\circ)}+\tan^2{(3^\circ)}+\tan^2{(5^\circ)}+\ldots+\tan^2{(89^\circ)}

solve in R x+y+z=2 2^{x+y^2}+2^{y+z^2}+2^{z+x^2}=6\sqrt[9]{2}