evaluate using identity (a) (0.99)2 (b)1053 3.
Answers
Answered by
3
(0.99)^2 = ( 1 - 0.01)^2
= (1)^2 + (0.01)^2 - 2 × 1 × 0.01 [ (a - b)^2= a^2 + b^2 - 2ab ]
= 1 + 0.0001 - 0.02
= 1.0001 - 0.02 = 0.9801
= (1)^2 + (0.01)^2 - 2 × 1 × 0.01 [ (a - b)^2= a^2 + b^2 - 2ab ]
= 1 + 0.0001 - 0.02
= 1.0001 - 0.02 = 0.9801
Answered by
2
l think you wanna asked this
I couldn't understand what was it power or multiple
I couldn't understand what was it power or multiple
Attachments:
Similar questions