If `f={(a, 0), (b, 2), (c, -3)}, g={(a, -1), (b, 1), (c, 2)}` then `2f-3g=`

If f={(a, 0), (b, -2), (c, 3)}, g={a, -2), (b, 0), (c,1)} then f//g=

If f={(a, 1), (b, -2), (c, 3)}, g={(a, -2), (b, 0), (c, 1)} then f^(2)+g^(2)=

If ={(a,1),(b,-2),(c,3)},g={(a,-2),(b,0),(c,1)} then f^(2)+g^(2)=

If f={(-2, 4), (0, 6), (2, 8)} and g={(-2, -1), (0, 3), (2, 5)} , then ((2f)/(3g)+(3g)/(2f))(0)=

If f={(1,a), (2, b), (1, b), (3,c), (1,c)}, g=(a, p), (b,r), (c,q), (c,p)} , then (gof)^(-1)=

If f={(-2,4),(0,6),(2,8)} and g={(-2,-1),(0,3),(25)}, then ((2f)/(3g)+(3g)/(2f))(0)=

Let A={1,2,3},B={a,b,c),C={p,q,r}. If f:A to B, g: B to C are defined by f={(1,a),(2,c),(3,b)},g={a,q),(b,r),(c,p)} then show that f^(-1)og^(-1)=(gof)^(-1) .

If f and g are two real functions defined by f={(a,1)(b,0),(c,8),(0,4)} and g={(a,5),(c,-3),(b,6)(0,2)} then dom ((sqrt(g))/(f)) is