evaluate b^2 - 9(b - 1)^2 , if b = 1.1
Answers
Answered by
11
FORMULA USED,
{(a+b)^2 = a^2 + b^2 + 2ab}
SOLUTION,
b^2 - 9(b^2 + 1^2 + 2*b*1)
b^2 - 9 (b^2 + 1 + 2b)
b^2 - 9b^2 + 9 + 18b
On putting the values,
1.1^2 - 9(1.1)^2 + 9 + 18*1.1
1.21 - 10.89 + 9 + 19.8
30.01 - 10.89
19.12 (ANS)
PLS MARK IT THE BRAINLIESR
{(a+b)^2 = a^2 + b^2 + 2ab}
SOLUTION,
b^2 - 9(b^2 + 1^2 + 2*b*1)
b^2 - 9 (b^2 + 1 + 2b)
b^2 - 9b^2 + 9 + 18b
On putting the values,
1.1^2 - 9(1.1)^2 + 9 + 18*1.1
1.21 - 10.89 + 9 + 19.8
30.01 - 10.89
19.12 (ANS)
PLS MARK IT THE BRAINLIESR
Jack183739:
I AM SORRY
it must be -2ab, I.e., it will be -18*1.1.......pls evaluate this way
Answered by
0
Answer:
formula used
{(a+b)^2= a^2+b^2+2ab}
b^2-9(b^2+1^2+2*b*1)
b^2-9(b^2+1+2b)
b^2-9b^2+9+18b
on putting the values
1*1^2-9(1*1)^2+9+18*1*1
1*21-10.89+9+19.8
30.01-10.89
19.12
Similar questions