Evaluate
$$\begin{aligned} \sqrt{248+\sqrt{64}} \end{aligned}$$

Answer: Option D

Explanation:

$$\begin{aligned} = \sqrt{41 - \sqrt{21 + \sqrt{19 - 3}}} \end{aligned}$$

$$\begin{aligned} = \sqrt{41 - \sqrt{21 + \sqrt{16}}} \end{aligned}$$

$$\begin{aligned} = \sqrt{41 - \sqrt{21 + 4}} \end{aligned}$$

$$\begin{aligned} = \sqrt{41 - \sqrt{25}} \end{aligned}$$

$$\begin{aligned} = \sqrt{41 - \sqrt{25}} \end{aligned}$$

$$\begin{aligned} = \sqrt{41 - 5} \end{aligned}$$

$$\begin{aligned} = \sqrt{36} = 6 \end{aligned}$$

Answer: Option A

Answer: Option A

Explanation:

$$\begin{aligned} 3600 = 2^3 \times 5^2 \times 3^2 \times 2 \end{aligned}$$
To make it a perfect cube it must be divided by
$$\begin{aligned} 5^2 \times 3^2 \times 2 = 450 \end{aligned}$$

Answer: Option D

Answer: Option C