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

25.6% of 250 +√ ? = 119

Solve: 12–[26–{2+5×(6–3)}]

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

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

$$ {\sqrt{5929}}+\sqrt{8464}=(?)^{2}$$

When x = 1, then value of |2x – 1| + |x – 3| will be -

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