using identities evaluate 1.05 ×9.5
dinkupal:
(x+a)(x-a)
kk bt how
1.05×9.5
yupzzz
Answers
Answered by
4
hey friend here is your answer
1.05*9.5
1.05*9.5*10
(1+0.05)*(1-0.05)*10
[(1)2 -(0.05)2*10]
[1-0.0025]*10
0.9975*10
9.975
If it is helpful then mark my answer please
1.05*9.5
1.05*9.5*10
(1+0.05)*(1-0.05)*10
[(1)2 -(0.05)2*10]
[1-0.0025]*10
0.9975*10
9.975
If it is helpful then mark my answer please
thanks
Answered by
3
=(0.5 + 1) (0.5 +9)
=(0.5)^2+(9+1)×0.5 + (9×1)
(using (x+a)(x+b)=x^2+(a+b)x+ (ab) identity)
=0.25 + 5 + 9
=14.25
=(0.5)^2+(9+1)×0.5 + (9×1)
(using (x+a)(x+b)=x^2+(a+b)x+ (ab) identity)
=0.25 + 5 + 9
=14.25
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