Question : If $16 y^2-k=\left(4 y+\frac{3}{2}\right)\left(4 y-\frac{3}{2}\right)$, then the value of $k$ is:
Option 1: $\frac{9}{4}$
Option 2: $\frac{11}{4}$
Option 3: $\frac{6}{4}$
Option 4: $\frac{7}{4}$
Correct Answer: $\frac{9}{4}$
Solution : $16 y^2-k=\left(4 y+\frac{3}{2}\right)\left(4 y-\frac{3}{2}\right)$ $⇒(4y)^2-k=(4y)^2-(\frac{3}{2})^2$ $⇒k=(\frac{3}{2})^2$ $\therefore k = \frac{9}{4}$ Hence, the correct answer is $\frac{9}{4}$.
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Question : The value of $\left(5 \frac{1}{4} \div \frac{3}{7}\right.$ of $\left.\frac{1}{2}\right) \div\left(5 \frac{1}{9}-7 \frac{7}{8} \div 9 \frac{9}{20}\right) \times \frac{11}{21}+\left(2 \div 2\right.$ of $\left.\frac{1}{2}\right)$ is:
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Question : The value of $\left(5 \frac{1}{4} \div \frac{3}{7}\right.$ of $\left.\frac{1}{2}\right) \div\left(5 \frac{1}{9}-7 \frac{7}{8} \div 9 \frac{9}{20}\right) \times \frac{11}{21}-\left(5 \div 2\right.$ of $\left.\frac{1}{2}\right)$ is:
Question : If $\frac{x}{4 y}=\frac{3}{4}$ then, the value of $\frac{2 x+3 y}{x–2 y}$ is:
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