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Last updated at Dec. 24, 2018 by Teachoo
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Ex 9.5, 6 Using identities, evaluate. (i) γ71γ^2 γ71γ^2 =(70+1)^2 (π+π)^2=π^2+π^2+2ππ Putting π = 70 & π = 1 = (70)^2+(1)^2+2(70)(1) = 4900+1+140 = ππππ Ex 9.5, 6 Using identities, evaluate. (ii) γ99γ^2 γ99γ^2 =(100β1)^2 (πβπ)^2=π^2+π^2β2ππ Putting π = 100 & π = 1 = (100)^2+(1)^2β2(100)(1) = 10000+1β200 = 10001β200 = ππππ Ex 9.5, 6 Using identities, evaluate. (iii) γ102γ^2 γ102γ^2 =(100+2)^2 (π+π)^2=π^2+π^2+2ππ Putting π = 100 & π = 2 = (100)^2+(2)^2+2(100)(2) = 10000+4+400 = ππ404 Ex 9.5, 6 Using identities, evaluate. (iv) γ998γ^2 γ998γ^2 =(1000β2)^2 (πβπ)^2=π^2+π^2β2ππ Putting π = 1000 & π = 2 = (1000)^2+(2)^2β2(1000)(2) = 1000000+4β4000 = 1000004β4000 = ππππππ Ex 9.5, 6 Using identities, evaluate. (v) γ5.2γ^2 γ5.2γ^2 =(5+0.2)^2 (π+π)^2=π^2+π^2+2ππ Putting π = 5 & π = 0.2 (π+π)^2=π^2+π^2+2ππ Putting π = 5 & π = 0.2 = 27+ 4/100 = 27+0.04 = ππ.ππ Ex 9.5, 6 Using identities, evaluate. (vi) 297 Γ 303 297 Γ 303 = (300β3)Γ(300+3) (πβπ)(π+π)=π^2βπ^2 Putting π = 300 & π = 3 = (300)^2β(3)^2 = 90000β9 = πππππ Ex 9.5, 6 Using identities, evaluate. (vii) 78 Γ 82 78 Γ 82 = (80β2)Γ(80+2) (πβπ)(π+π)=π^2βπ^2 Putting π = 300 & π = 3 = (80)^2β(2)^2 = 6400β4 = ππππ Ex 9.5, 6 Using identities, evaluate. (viii) γ8.9γ^2 γ8.9γ^2 = (9β0.1)^2 (πβπ)^2=π^2+π^2β2ππ Putting π = 9 & π = 0.1 = (9)^2+(0.1)^2β2(9)(0.9) = 81+(1/10)^2β(2Γ9Γ1/10) = 81 + 1/100 β 18/10 = (81 Γ 100 + 1 β 180)/100 = (8100 + 1 β 18)/100 = (8101 β 18)/100 = 7921/100 = 79.21 Ex 9.5, 6 Using identities, evaluate. (ix) 1.05Γ9.5 1.05Γ9.5 = 105/100Γ95/10 = (105 Γ 95)/1000 = ((100 + 5) Γ (100 β 5))/1000 (π+π)(πβπ)=π^2βπ^2 Putting π = 100 & π = 5 = ((100)^2 β (5)^2)/1000 = (10000 β 25)/1000 = 9975/1000 = 9.975
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