1. 5 \sqrt{27} = 3 \sqrt{75}

2. \sqrt{8} + \sqrt{32} = \sqrt{40}

3. 2 \sqrt{\frac{2}{3}} = 3 \sqrt{\frac{3}{2}}

Brenda Jordan
2022-11-17
Answered

True or false? Give your reasoning either way.

1. 5 \sqrt{27} = 3 \sqrt{75}

2. \sqrt{8} + \sqrt{32} = \sqrt{40}

3. 2 \sqrt{\frac{2}{3}} = 3 \sqrt{\frac{3}{2}}

1. 5 \sqrt{27} = 3 \sqrt{75}

2. \sqrt{8} + \sqrt{32} = \sqrt{40}

3. 2 \sqrt{\frac{2}{3}} = 3 \sqrt{\frac{3}{2}}

You can still ask an expert for help

Kaeden Lara

Answered 2022-11-18
Author has **23** answers

1) $3\sqrt{75}=3\sqrt{[(25)\ast 3]}=5\ast 3\sqrt{3}=5\sqrt{[(3)2\ast (3)]}=5\sqrt{(9\ast 3)}=5\sqrt{27}$ TRUE

2) $\sqrt{8}+\sqrt{32}=2\sqrt{2}+4\sqrt{2}=6\sqrt{2}(\text{}as\text{}8=(2)2\ast 2\text{}and\text{}32=(4)2\ast 2\text{}Also\text{}\sqrt{40}=\sqrt{[(2)2\ast 2\ast 5]}=2(\sqrt{2})(\sqrt{5})$ FALSE

3) FALSE The statement is apparently false on 1st scrutiny itself . 3/2 is 1.5 and 2/3 is 0.6666666666......

Therefore 3/2 > 2/3 so that $(3/2{)}^{1/2}>(2/3{)}^{1/2}$ so that $3(3/2{)}^{1/2}>3(2/3{)}^{1/2}>2(2/3{)}^{1/2}$ . Hemce the statement is FALSE

2) $\sqrt{8}+\sqrt{32}=2\sqrt{2}+4\sqrt{2}=6\sqrt{2}(\text{}as\text{}8=(2)2\ast 2\text{}and\text{}32=(4)2\ast 2\text{}Also\text{}\sqrt{40}=\sqrt{[(2)2\ast 2\ast 5]}=2(\sqrt{2})(\sqrt{5})$ FALSE

3) FALSE The statement is apparently false on 1st scrutiny itself . 3/2 is 1.5 and 2/3 is 0.6666666666......

Therefore 3/2 > 2/3 so that $(3/2{)}^{1/2}>(2/3{)}^{1/2}$ so that $3(3/2{)}^{1/2}>3(2/3{)}^{1/2}>2(2/3{)}^{1/2}$ . Hemce the statement is FALSE

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