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  1. Chapter 9 Class 8 Algebraic Expressions and Identities
  2. Serial order wise

Transcript

Ex 9.5, 7 Using ๐‘Ž^2โˆ’๐‘^2=(๐‘Ž+๐‘)(๐‘Žโˆ’๐‘) , find (i) ใ€–51ใ€—^2โˆ’ใ€–49ใ€—^2 ใ€–51ใ€—^2โˆ’ใ€–49ใ€—^2 ๐‘Ž^2โˆ’๐‘^2=(๐‘Ž+๐‘)(๐‘Žโˆ’๐‘) Putting ๐‘Ž = 51 & ๐‘ = 49 = (51+49)ร—(51โˆ’49) = 100ร—2 = ๐Ÿ๐ŸŽ๐ŸŽ Ex 9.5, 7 Using ๐‘Ž^2โˆ’๐‘^2=(๐‘Ž+๐‘)(๐‘Žโˆ’๐‘) , find (ii) (1.02)^2โˆ’(0.98)^2 (1.02)^2โˆ’(0.98)^2 ๐‘Ž^2โˆ’๐‘^2=(๐‘Ž+๐‘)(๐‘Žโˆ’๐‘) Putting ๐‘Ž = 1.02 & ๐‘ = 0.98 = (1.02+0.98)ร—(1.02โˆ’0.98) = 2ร—0.04 = 2ร—4/100 = 8/100 = ๐ŸŽ.๐ŸŽ๐Ÿ– Ex 9.5, 7 Using ๐‘Ž^2โˆ’๐‘^2=(๐‘Ž+๐‘)(๐‘Žโˆ’๐‘) , find (iii) ใ€–153ใ€—^2โˆ’ใ€–147ใ€—^2 ใ€–153ใ€—^2โˆ’ใ€–147ใ€—^2 ๐‘Ž^2โˆ’๐‘^2=(๐‘Ž+๐‘)(๐‘Žโˆ’๐‘) Putting ๐‘Ž = 153 & ๐‘ = 147 ๐‘Ž^2โˆ’๐‘^2=(๐‘Ž+๐‘)(๐‘Žโˆ’๐‘) Putting ๐‘Ž = 153 & ๐‘ = 147 Ex 9.5, 7 Using ๐‘Ž^2โˆ’๐‘^2=(๐‘Ž+๐‘)(๐‘Žโˆ’๐‘) , find (iv) ใ€–12.1ใ€—^2โˆ’ใ€–7.9ใ€—^2 ใ€–12.1ใ€—^2โˆ’ใ€–7.9ใ€—^2 ๐‘Ž^2โˆ’๐‘^2=(๐‘Ž+๐‘)(๐‘Žโˆ’๐‘) Putting ๐‘Ž = 12.1 & ๐‘ = 7.9 = (12.1+7.9)ร—(12.1โˆ’7.9) = 20ร—4.2 = 20ร—42/10 = 2ร—42 = ๐Ÿ–๐Ÿ’

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.