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BARSIC [14]
3 years ago
5

What is the simplified form of the following expression 7(^3√2x)-3(^3√16x)-3(^3√8x)

Mathematics
2 answers:
enot [183]3 years ago
6 0
7 \sqrt[3]{2x} - 3 \sqrt[3]{16x} - 3 \sqrt[3]{8x}

7 \sqrt[3]{2x} - 3 \sqrt[3]{8 \times 2x} - 3 \sqrt[3]{4 \times 2x}

\sqrt[3]{2x} ( 7  - 3 \sqrt[3]{8} - 3 \sqrt[3]{4})

\sqrt[3]{2x} ( 7  - 6 - 3 \sqrt[3]{4})

\sqrt[3]{2x} ( 1 - 3 \sqrt[3]{4})

\sqrt[3]{2x} - 3 \sqrt[3]{8x}

\sqrt[3]{2x} - 6 \sqrt[3]{x}

svlad2 [7]3 years ago
5 0

Answer:

The simplified form of the expression is \sqrt[3]{2x}-6\sqrt[3]{x}

Step-by-step explanation:

Given : Expression 7\sqrt[3]{2x}-3\sqrt[3]{16x}-3\sqrt[3]{8x}

To Simplified : The expression

Solution :  

Step 1 - Write the expression

7\sqrt[3]{2x}-3\sqrt[3]{16x}-3\sqrt[3]{8x}

Step 2- Simplify the roots and re-write as

16=2^3\times2 and 8=2^3

7\sqrt[3]{2x}-3\times2\sqrt[3]{2x}-3\times2\sqrt[3]{x}

Step 3- Solve the multiplication

7\sqrt[3]{2x}-6\sqrt[3]{2x}-6\sqrt[3]{x}

Step 4- Taking \sqrt[3]{2x} common from first two terms

\sqrt[3]{2x}(7-6)-6\sqrt[3]{x}

\sqrt[3]{2x}-6\sqrt[3]{x}

Therefore, The simplified form of the expression is \sqrt[3]{2x}-6\sqrt[3]{x}

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Answer:

The magnetic field is B = 3.87 *10^{-6} \ T

Direction is inward

 

Step-by-step explanation:

The diagram for this question is shown on the first uploaded image  

From the question we are told that

      The radius of  BC is  r_{BC} =  30 \ cm  =  0.30 \ m

      The radius of  DA is r _{DA} =  20 \ cm =  0.20 \ m

      The length of CD is  r_{CD} =  10 cm =  0.1 \ m

      The length of AB is  r_{AB} =  10 cm =  0.1 \ m

      The current is  I  =  11.4A

The magnetic field is mathematically represented as

      B =  B_{BC} +  B_{DA} + B_{CD} + B_{AB}

       B =  \frac{\mu_o I}{4 \pi } [ \frac{\theta_{DA}}{r_{DA}} + \frac{\theta_{BC}}{r_{BC}}+ \frac{\theta_{CD}}{r_{CD}}+\frac{\theta_{AB}}{r_{AB}}]

        Where

                 \theta_{BC} = \theta_{DA} =  \frac{2\pi}{3}

Where  \frac{2 \pi}{3}  =  120^o

                \theta_{CD} = \theta_{AB}  =  0^o

so

        B =  \frac{\mu_o I}{4 \pi } [ \frac{\frac{2\pi}{3} }{0.20} - \frac{\frac{2\pi}{3}}{0.30}}+ \frac{0}{0.10}+\frac{0}{0.10}]

         B = 3.87 *10^{-6} \ T

The direction is into the page

    This because the magnitude of the magnetic field due to arc BC whose direction is outward is less than that of DA whose direction is inward

This is because according to Fleming's Left Hand Rule  the direction of current is perpendicular to the direction of magnetic field so since  current in arc BC and DA are moving in opposite direction their magnetic field will also be moving in opposite direction

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