how to find the value of i for the following problem:
ln (1+i)^10 = Ln2
ln (1_i)^10 = Ln2
4y=2x-4
1130=250+350 0+560+ (x-80) find the x
Y= - 2x+1
3^12/13
Evaluate the integral: $\int_{0}^{\frac{\pi}{2}}{\log\left( 1 - {a}^{2} {\sin\left( x \right)}^{2} \right)}dx$
(x+3)(x+3)(3x-2)
2099188.39=A-((A/1.12)*.01)
