If the sum of the squares of three consecutive natural numbers is 110, then the smallest of these natural numbers is:

Evaluate: $${-(4-6)^2-3(-2)+।-6।\over 18-9÷3×5}$$

Simplify the following expression.

 25-(4÷2)+11.5+9.5

If $$ {(x-2)^{2}+(y-{1\over2})^{2}}=0$$, then the value of $$ {x\over y}$$ is

25.6% of 250 +√ ? = 119

Solve: 12–[26–{2+5×(6–3)}]

  $$25\% \ of (50\% \ of \ 30\% \ of \ 150)\over 40\% of \ 2250$$ is equal to:

If p = 8.15, q = 9.06 and r = −17.21, then the value of p³+q³+r³-3pqr