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For the two given equations I and II----
I. $$p=\frac{\sqrt{4}}{\sqrt{9}}$$ II. $$9q^{2}-12q+4=0$$
Give answer (A) if p is greater than q.
Give answer (B) if p is smaller than q.
Give answer (C) if p is equal to q.
Give answer (D) if p is either equal to or greater than q.
Give answer (E) if p is either equal to or smaller than q.
$$p = \frac{\sqrt{4}}{\sqrt{9}}$$$$p = \frac{2}{3}$$
$$9q^2-12q+4 = 0$$$${(3q-2)}^2 = 0$$$$q = \frac{2}{3}$$
p = q
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