(1.02^12-1)/0.02
Draw the graph of the given function for the interval mentioned.
i. y = cos x/2 x ∈ [- π,π]
1,02^12-1/0,02
(e+f) (e-f) -2e (4f-e)
(2-b)(-3a)-((4a+1)(2)-5(6ab-2a+3))
(8^8×4^5)^2
3x-1
2x+y=14
12 + x = 58
R1=2+R2=4+R3=6
