Grade 128 grade maths
Using suitable algebraic identity, solve 10922 ejc jencjkewvhjkv j
Using suitable algebraic identity, solve 10922
ejc jencjkewvhjkv j
1 Answer
Harshit Singh
Dear Student
Use the algebraic identity: (a + b)² = a² + 2ab + b²
Now, 1092 = 1000 + 92
So, 10922= (1000 + 92)2
(1000 + 92)2= ( 1000 )² + 2 × 1000 × 92 + ( 92 )²
= 1000000 + 184000 + 8464
Thus, 10922= 1192464.
Thanks
Use the algebraic identity: (a + b)² = a² + 2ab + b²
Now, 1092 = 1000 + 92
So, 10922= (1000 + 92)2
(1000 + 92)2= ( 1000 )² + 2 × 1000 × 92 + ( 92 )²
= 1000000 + 184000 + 8464
Thus, 10922= 1192464.
Thanks
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