A) \[-16\]
B) \[9\]
C) \[-9\]
D) \[16\]
Correct Answer: B
Solution :
Given \[p=\frac{2}{3}\] and \[q=\frac{3}{4}\] \[81{{p}^{2}}+16{{q}^{2}}-72pq\] \[=81{{\left( \frac{2}{3} \right)}^{2}}+16{{\left( \frac{3}{4} \right)}^{2}}-72\left( \frac{2}{3} \right)\left( \frac{3}{4} \right)\] \[=81\times \frac{4}{9}+16\times \frac{9}{16}-\frac{72}{2}\] \[=36+9-36=9\]You need to login to perform this action.
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