Question

Find the values of the following
(a) 912− 232
(b) 101 × 99
(c) 9992​

Answer 1

Step-by-step explanation:

1) 912-232

= 680

2) 101 × 99

=. 9,999

3) 9992

=. 998001

Answer 2

Answer:

a) $91^{2}$ - $23^{2}$ = $68^{2}$

b) 101 * 99 = 9999

c) $999^{2}$ = 998001

Step-by-step explanation:

I am assuming you copy-pasted these questions and put it on Brainly, and the powers of the questions have been turned into 2s.

a) According to the laws of exponents, if the number's power are the same and these to powers are being subtracted, the answer will be the number's difference + its power. So accordingly, $91^{2} - 23^{2} = (91 - 28)^{2} = 68^{2} = 4624$

b) 99 * 101 = (100 - 1) * (100 + 1) = $(100)^{2}$  - $1^{2}$ = 10000 -1   = 9999                                 [ $a^{2} - b^{2} = (a - b) * (a + b)$ ]

c) $999^{2}$ = $(1000 - 1)^{2}$ = $(1000^{2} - 2 * 100 * 1 + 1^{2} )$ = 100000 - 2000 + 1 = 998001    [$a^{2} - b^{2} = a^{2} - 2*1000*1 + 1^{2}$ ]

Hope it helps! And please mark me as the brainliest! :D