1+2
solve the following inequality: $\frac{f}{2}$ +_25>45
if solution of recurrence relation aSn+bSn-1+cSn-2 = 6 is 3n+ 4n+ 2 , find a,b,c
x + 5% = 68.99
1/4+1/5+1/8
3b-3a-2b
2+2
5x+2y = 1250
3.14(8+4/3*4.25*4.25*4.25)
3a+2c-2c
=
