Ex 2.5
Ex 2.5, 1 (ii)
Ex 2.5, 1 (iii) Important
Ex 2.5, 1 (iv)
Ex 2.5, 1 (v)
Ex 2.5, 2 (i)
Ex 2.5, 2 (ii) Important
Ex 2.5, 2 (iii)
Ex 2.5, 3 (i)
Ex 2.5, 3 (ii) Important
Ex 2.5, 3 (iii)
Ex 2.5, 4 (i)
Ex 2.5, 4 (ii)
Ex 2.5, 4 (iii) Important
Ex 2.5, 4 (iv)
Ex 2.5, 4 (v)
Ex 2.5, 4 (vi)
Ex 2.5, 5 (i)
Ex 2.5, 5 (ii) Important
Ex 2.5, 6 (i)
Ex 2.5, 6 (ii)
Ex 2.5, 6 (iii)
Ex 2.5, 6 (iv) Important
Ex 2.5, 7 (i) You are here
Ex 2.5, 7 (ii)
Ex 2.5, 7 (iii) Important
Ex 2.5, 8 (i)
Ex 2.5, 8 (ii)
Ex 2.5, 8 (iii) Important
Ex 2.5, 8 (iv) Important
Ex 2.5, 8 (v)
Ex 2.5, 9 (i)
Ex 2.5, 9 (ii)
Ex 2.5, 10 (i) Important
Ex 2.5, 10 (ii)
Ex 2.5, 11
Ex 2.5,12 Important Deleted for CBSE Board 2022 Exams
Ex 2.5,13 Deleted for CBSE Board 2022 Exams
Ex 2.5, 14 (i) Deleted for CBSE Board 2022 Exams
Ex 2.5, 14 (ii) Important Deleted for CBSE Board 2022 Exams
Ex 2.5, 15 (i)
Ex 2.5, 15 (ii) Important
Ex 2.5, 16 (i)
Ex 2.5, 16 (ii) Important
Last updated at Aug. 25, 2021 by Teachoo
Ex 2.5, 7 Evaluate the following using suitable identities: (i) (99)3 We write 99 = 100 – 1 (99)3 = (100 − 1)3 Using (a – b)3 = a3 – b3 – 3ab(a – b) Where a = 100 & b = 1 = (100)3 − (1)3 − 3(100) (1) (100 − 1) = 1000000 − 1 − 300(99) = 1000000 − 1 − 29700 = 970299