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(22)2+2(2−1)+((210)12)\sqrt{(2^2)^2+2(2^{-1})+((2^{10})^{\frac{1}{2}})} = 

A6

B$\sqrt20$

C$\sqrt48$

D7

Answer:

D. 7

Read Explanation:

The value of the expression is 7.

We simplify each term inside the square root one by one:

1. Simplify the first term:
(22)2=22×2=24=16(2^2)^2 = 2^{2 \times 2} = 2^4 = 16

2. Simplify the second term:
2(2−1)=21×2−1=21−1=20=12(2^{-1}) = 2^1 \times 2^{-1} = 2^{1-1} = 2^0 = 1

3. Simplify the third term:
((210)12)=210×12=25=32((2^{10})^{\frac{1}{2}}) = 2^{10 \times \frac{1}{2}} = 2^5 = 32

4. Add the terms together inside the square root:
16+1+32=49\sqrt{16 + 1 + 32} = \sqrt{49}

5. Find the square root:
49=7\sqrt{49} = 7


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