" (iv) "2m^(3)-13m^(2)+22m-8

Add (i) 5m(3-m) and 6m^2-13m-13m (ii) 4y(3y^2+5y-7) and 2(y^3-4y^2+5)

Add (i)5m(3-m) and 6m^(2)-13m-13m (ii) 4y(3y^(2)+5y-7) and 2(y^(3)-4y^(2)+5)

If m is a real number such that m^(2)+1=3m the value of (2m^(5)-5m^(4)+2m^(3)-8m^(2))/(1+m^(2))

Solve the following quadratic equations by factorisation method. (i) x^(2) +8x+15=0 (ii) 5m^(2) - 22m - 15= 0

Simplify the following: 10m^(2) - 9m + 7m - 3m^(2) - 5m -8

If pi is taken as 22//7 , the distance (in metres) covered by a wheel of diameter 35 cm, in one revolution, is (a) 2.2m (b) 1.1m (c) 9.625m (d) 96.25m