Suppose x = √2 + √3, y = √5 + √1, and z = 2√2 + 1. ; then which of the following is true?Ax <y < zBz < y < xCy < x < zDx < z < yAnswer: A. x <y < z Read Explanation: Approximate the values:x=2+3≈1.414+1.732=3.146x = \sqrt{2} + \sqrt{3} \approx 1.414 + 1.732 = 3.146x=2+3≈1.414+1.732=3.146y=5+1=2.236+1=3.236y = \sqrt{5} + \sqrt{1} = 2.236 + 1 = 3.236y=5+1=2.236+1=3.236z=22+1≈2(1.414)+1=3.828z = 2\sqrt{2} + 1 \approx 2(1.414) + 1 = 3.828z=22+1≈2(1.414)+1=3.828Therefore,\boxed{z > y > x} Read more in App