without actual multiplication find the square of
(1) 98
(2) 1003
(3) 999
(4) 10.8
please answer
Answers
Answer:
Step-by-step explanation:
(1) 98² = (100-2)² = 100²- 2x100x2+2²
= 10000-400+4=9604
(2) 1003² = (1000+3)² = 1000²+2x1000x3+3²= 1000000+6000+9 =1006009
(3) 999² = (1000-1)² = 1000²-2x1000x1+1² = 1000000-2000+1
=998001
(4) 10.8²= (10+0.8)² = 10²+2x10x(0.8)+(0.8)² = 100+16+0.64= 116.64
N. B.= Here I've used two formulas:
(i) (a+b)² = a²+2ab+b²
(ii) (a-b)² = a²-2ab+b²
Hope, it will help you. Please mark me as a brainliest.
Answer:
1 (98)^2
(100-2)^2
(100)^2+(2)^2-2×100×2
10000+4-400
10004-400
9604 ANSWER
2 1003²=(1000+3)²=1000000+6000+9=1006009
3 The value of 999×999 is
=(1000−1)(1000−1)
=(1000−1)
2
We know that suitable identities (x−y)
2 The value of 999×999 is
=(1000-1) (1000-1)
=(1000-1)^2
We know that suitable identities ( x- y)^2 = x^2-2xy + y^2
Thus ( 1000 - 1 ) ^2 = (1000)^2 - 2(1000)(1) + (1)^2
= 1000000 - 2000+ 1
= 998001
4 116.64
Step-by-step explanation: