write the following as percent
3/4
x^10 + y^15z^25 sum and diff of two odd powers
[(23/36)+(1/40)]/16
8
√
3
−
4
x^3-5x factorize
(23/36)+(1/40)
440x$10^{-3}$ = 0,44
Determine the solution of the initial boundary-value problem:
u_tt =c^ 2 u_xx
0 < x < l, t > 0
u(x, 0) = f(x)
0 <= x <= l
u_{...
[(23/16)+(1/40)/23-7]
x+y=4
x-y=2
