Maths question

(a)
Let h & k are constant

Suppose the equation is a = H(b)^0.5 + K/(c^2)

For a=3 when b=c=1,
3 = H + K ---(Eqn. 1)
For a=5/2 when b=4 and c=2
(5/2) = 2H + 0.25K
=> Mutlipy 4 to both side => 10 = 8H + K ---(Eqn. 2)

(Eqn. 2) - (Eqn. 1)
=> 7 = 7H => H = 1
=> K = 2

The equation is a = (b)^0.5 + 2/(c^2)

(b)
when b=16 and c=3
a
= (16)^0.5 + 2/(3^2)
= 4 + 2/9
= 38/9

(a). let k be the constant
a=kb^1/2+K/(c^2)
3=k+K------1
5/2=2k+K/4-----2
k=3-K-----3
sub 3 into 2
10=24-8K+K
K=2
k=1
so a=b^1/2+2/(c^2)

(b). a=16^1/2+2/3^2
=4+2/9
=38/9