If the operation Θ is defined for all real numbers a and b by the relation $$aΘ b =a^{2}\frac{b}{{3}}$$ then $$2Θ {3Θ(-1)} = ?$$
It is given that $$aΘ b =a^{2}\frac{b}{{3}}$$
Applying the same rule for $$2Θ {3Θ(-1)}$$
= 2 Θ $${\frac{3^2 \times (-1)}{3}}$$
= 2 Θ -3
= $$\frac{2^2 \times (-3)}{3} = -4$$
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