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stira [4]
3 years ago
10

Can someone help me find the equivalent expressions to the picture below? I’m having trouble

Mathematics
1 answer:
miss Akunina [59]3 years ago
6 0

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is \frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}.

Now we will solve this expression with the help of law of exponents.

\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}

           =2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}

           =2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}

           =2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}

           =2^{\frac{2}{9}}\times 3^{-\frac{2}{3} } [Option 2]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2 [Option 1]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2

                =(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }

                =\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} } [Option 3]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }

               =\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}} [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

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Answer:

A. \sqrt{-64} = 8i

B. 8i + 5i = 13i

C.  8i - 5i = 3i

D. 8i x 5i = -40

Step-by-step explanation:

A. The square root of any negative number would lead to a complex number. Complex numbers are number which consist a complex part denoted by i.

-1 = i^{2}

\sqrt{-1} = \sqrt{i^{2} } = i

Example: 1. What is the square root of -64?

square root of -64 = \sqrt{-64}

                         = \sqrt{-1 *64}

                         = \sqrt{-1} x \sqrt{64}

                         = i x 8

                         = 8i

\sqrt{-64} = 8i

2. find the square root of -25.

\sqrt{-25} = \sqrt{-1*25}

          = 5i

B. To add two complex numbers, they are considered as algebraic expressions.

Example, the sum of 8i and 5i can be determined as;

                 8i + 5i = 13i

C. To add two complex numbers, they are considered as algebraic expressions.

Example, the subtraction of 8i and 5i can be determined as;

                 8i - 5i = 3i

D. To multiply two complex numbers, the complex part is considered.

Example, determine the product of 8i and 5i.

8i x 5i = 8 x 5 x i x i

          = 40i^{2}

         = -40                (∵      i^{2} = -1)

8i x 5i = -40

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