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Last updated at Dec. 24, 2018 by Teachoo

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Ex 9.3, 3 Find the product. (i) (๐^2 ) ร (2๐^22 ) ร (4๐^26 ) (๐^2 ) ร (2๐^22 ) ร (4๐^26 ) = ๐^2ร2ร๐^22ร4ร๐^26 = (2ร4)ร(๐^2ร๐^22ร๐^26 ) = (2ร4)ร(๐^(2 + 22 + 26) ) ( ๐ฅ^๐ ร๐ฅ^๐ร๐ฅ^๐ = ๐ฅ^(๐ + ๐ + ๐)) = 8ร๐^50 = ๐๐^๐๐ Ex 9.3, 3 Find the product. (ii) (2/3 ๐ฅ๐ฆ)ร((โ9)/10 ๐ฅ^2 ๐ฆ^2 ) (2/3 ๐ฅ๐ฆ)ร((โ9)/10 ๐ฅ^2 ๐ฆ^2 ) = 2/3ร๐ฅร๐ฆร(โ9)/10ร๐ฅ^2ร๐ฆ^2 = (2/3ร(โ9)/10)ร(๐ฅร๐ฅ^2 )ร(๐ฆร๐ฆ^2 ) = (2/10ร(โ9)/3)ร๐ฅ^3ร๐ฆ^3 = (1/5รโ3)ร๐ฅ^3ร๐ฆ^3 = (โ๐)/๐ ๐^๐ ๐^๐ Ex 9.3, 3 Find the product. (iii) ((โ10)/3 ๐๐^3 )ร(6/5 ๐^3 ๐) ((โ10)/3 ๐๐^3 )ร(6/5 ๐^3 ๐) = (โ10)/3ร๐ร๐^3ร6/5ร๐^3ร๐ = ((โ10)/3ร6/5)ร(๐ร๐^3 )ร(๐^3ร๐) = ((โ10)/5ร6/3)ร๐^4ร๐^4 = (โ2ร2)ร๐^4ร๐^4 = โ๐๐^๐ ๐^๐ Ex 9.3, 3 Find the product. (iv) ๐ฅร๐ฅ^2ร๐ฅ^3ร๐ฅ^4 ๐ฅร๐ฅ^2ร๐ฅ^3ร๐ฅ^4 = ๐ฅ^(1 + 2 + 3 + 4) = ๐^๐๐ (๐ฅ^๐ ร๐ฅ^๐ = ๐ฅ^(๐ + ๐))

About the Author

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.