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Last updated at Dec. 24, 2018 by Teachoo
Transcript
Ex 9.5, 8 Using (π₯+π)(π₯+π)=π₯^2+(π+π)π₯+ππ , find (i) 103 Γ 104 103 Γ 104 = (100+3) Γ(100+4) (π₯+π)(π₯+π)=π₯^2+(π+π)π₯+ππ Putting π₯ = 100 , π = 3 & π = 4 (π₯+π)(π₯+π)=π₯^2+(π+π)π₯+ππ Putting π₯ = 100 , π = 3 & π = 4 Ex 9.5, 8 Using (π₯+π)(π₯+π)=π₯^2+(π+π)π₯+ππ , find (ii) 5.1 Γ 5.2 5.1 Γ 5.2 = (5+0.1)Γ(5+0.2) (π₯+π)(π₯+π)=π₯^2+(π+π)π₯+ππ Putting π₯ = 5 , π = 0.1 & π = 0.2 = (5)^2+(0.1+0.2)(5)+(0.1)(0.2) = 25+(0.3)5+(0.1Γ0.2) = 25+3/10Γ5+(1/10Γ2/10) = 25+(15/10)+2/100 = 25+1.5+0.02 = ππ.ππ Ex 9.5, 8 Using (π₯+π)(π₯+π)=π₯^2+(π+π)π₯+ππ , find (iii) 103 Γ 98 103 Γ98 = (100+3)Γ(100β2) = (100+3)Γ[100+(β2)] (π₯+π)(π₯+π)=π₯^2+(π+π)π₯+ππ Putting π₯ = 100 , π = 3 & π = β2 = (100)^2+(3+2)(100)+(3)(β2) = 10000+100β6 = 10000+94 = πππππ Ex 9.5, 8 Using (π₯+π)(π₯+π)=π₯^2+(π+π)π₯+ππ , find (iv) 9.7 Γ 9.8 9.7 Γ9.8 = (10β0.3)Γ(10β0.2) = (10+(β0.3))Γ[10+(β0.2)] (π₯+π)(π₯+π)=π₯^2+(π+π)π₯+ππ Putting π₯ = 10 , π = β0.3 & π = β0.2 = (10)^2+(β0.3β0.2)(10)+(β0.3)(β0.2) = 100+(β0.5Γ10)+(0.3Γ0.2) = 100β(5/10Γ10)+(3/10Γ2/10) = 100β5+6/100 = 95+0.06 = ππ.ππ
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