What is the circumference of the circle that

Mathematics

1 answer:

saul85 [17]2 years ago

6 0

The answer is D 25 units

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Determine whether each equation is True or False. In case you find a "False" equation, explain why is False.

elixir [45]

Answer:

(1) TRUE.

(2) FALSE.

(3) FALSE.

(4) TRUE.

(5) FALSE.

Step-by-step explanation:

(1) $\sqrt{32} = 2^{\frac{5}{2} }$

$2^{\frac{5}{2} } = (\sqrt{2} )^5 = (\sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2}) = 4\sqrt{2}\\\\\sqrt{32} = \sqrt{16 \ \times \ 2}\ = \ \sqrt{16} \ \times \ \sqrt{2} \ = \ 4\sqrt{2}$

Thus, the equation is TRUE.

(2) $16^{\frac{3}{8} } = 8^2$

$16^{\frac{3}{8} } =(2^4)^{\frac{3}{8} } = 2^\frac{3}{2} }= (\sqrt{2} )^3 = (\sqrt{2} \ \times \ \sqrt{2} \ \times \ \sqrt{2}) = 2\sqrt{2} \\\\8^2 = 64$

Thus, the equation is FALSE.

(3) $4^{\frac{1}{2} } = \sqrt[4]{64}$

$4^{\frac{1}{2} }= \sqrt{4} = 2\\\\\sqrt[4]{64} = (64)^{\frac{1}{4} } = (2^6)^{\frac{1}{4} }= 2^{\frac{6}{4} } = 2^{\frac{3}{2} }=(\sqrt{2} )^3 = (\sqrt{2} \times \sqrt{2} \times \sqrt{2} ) = 2\sqrt{2}$

Thus, the equation is FALSE.

(4) $2^8 = (\sqrt[3]{16} )^6$

$2^8 = 256\\\\ (\sqrt[3]{16} )^6 = (16)^{\frac{6}{3} } = (2^4)^{\frac{6}{3} } = (2)^{\frac{24}{3} } = 2^8 = 256$

Thus, the equation is TRUE.

(5) $(\sqrt{64} )^{\frac{1}{3} } = 8^{\frac{1}{6} }\\\\$

$8^{\frac{1}{6} } = (2^3)^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} } = \sqrt{2} \\\\(\sqrt{64} )^{\frac{1}{3} } = (2^6)^{\frac{1}{3} } = 2^{\frac{6}{3} } = 2^2 = 4$