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Q.) (0.1x-0.5x)^2 (Solve by using II identity)
identity is:- a^2-2ab+b^2
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Answered by
1
Answer:
(0.1x-0.5x)^2. (a^2-2ab+b^2)
a=0.1x,b=0.5x
((0.1x)^2-2×0.1x×0.5x+(0.5x)^2)
here 0.1×0.5=0.05 and 0.05×2=0.10
0.1x^2- 0.1x×0.0025x^2
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Answer:
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