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prealgebra intermediate

Problem

Simplify

(627 - 6 \frac{3}{4})^{2}

(A)

3/4

(B)

s q r t (3) /2

(C)

f r a c (3 s q r t (3), 4)

(D)

3/2

Solution

We have

627 = (3^{3})^{\frac{1}{6}} = 3^{\frac{1}{2}} = 3

and

6 \frac{3}{4} = \frac{27}{4} = \frac{3 3}{2}

, so the answer is

(3 - \frac{3 3}{2})^{2} = (- \frac{3}{2})^{2} = \frac{3}{4}

, which is

\frac{3}{4}

.

Final answer

A