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Ex 9.5
Ex 9.5, 1 (ii) Important
Ex 9.5, 1 (iii)
Ex 9.5, 1 (iv)
Ex 9.5, 1 (v) Important
Ex 9.5, 1 (vi) Important
Ex 9.5, 1 (vii)
Ex 9.5, 1 (viii)
Ex 9.5, 1 (ix) Important
Ex 9.5, 1 (x)
Ex 9.5, 2 (i)
Ex 9.5, 2 (ii)
Ex 9.5, 2 (iii)
Ex 9.5, 2 (iv) Important
Ex 9.5, 2 (v)
Ex 9.5, 2 (vi) Important
Ex 9.5, 2 (vii) Important
Ex 9.5, 3 (i)
Ex 9.5, 3 (ii) Important
Ex 9.5, 3 (iii)
Ex 9.5, 3 (iv) Important
Ex 9.5, 3 (v) Important
Ex 9.5, 3 (vi)
Ex 9.5, 4 (i)
Ex 9.5, 4 (ii)
Ex 9.5, 4 (iii) Important
Ex 9.5, 4 (iv)
Ex 9.5, 4 (v) Important
Ex 9.5, 4 (vi)
Ex 9.5, 4 (vii) Important
Ex 9.5, 5 (i)
Ex 9.5, 5 (ii) You are here
Ex 9.5, 5 (iii) Important
Ex 9.5, 5 (iv)
Ex 9.5, 5 (v) Important
Ex 9.5, 6 (i)
Ex 9.5, 6 (ii) Important
Ex 9.5, 6 (iii)
Ex 9.5, 6 (iv)
Ex 9.5, 6 (v) Important
Ex 9.5, 6 (vi)
Ex 9.5, 6 (vii) Important
Ex 9.5, 6 (viii)
Ex 9.5, 6 (ix) Important
Ex 9.5, 7 (i)
Ex 9.5, 7 (ii) Important
Ex 9.5, 7 (iii)
Ex 9.5, 7 (iv) Important
Ex 9.5, 8 (i)
Ex 9.5, 8 (ii)
Ex 9.5, 8 (iii) Important
Ex 9.5, 8 (iv) Important
Last updated at March 22, 2023 by Teachoo
Ex 9.5, 5 Show that. (ii) (9πβ5π)^2+180ππ=(9π+5π)^2 Solving LHS (9πβ5π)^2+180ππ (πβπ)^2=π^2+π^2β2ππ Putting π = 9π & π = 5π (πβπ)^2=π^2+π^2β2ππ Putting π = 9π & π = 5π Solving RHS (9π+5π)^2 (π+π)^2=π^2+π^2+2ππ Putting π = 9π & π = 5π = (9π)^2+(5π)^2+2(9π)(5π) = 81π^2+25π^2+90ππ Thus LHS = RHS Hence proved