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Sindrei [870]
3 years ago
12

Write an expression describing all the angles that are coterminal with 141°.

Mathematics
1 answer:
MA_775_DIABLO [31]3 years ago
7 0
<span>The coterminal angles of angle A are given by adding k times 360° to it, where k is an integer, positive or negative. Similarly, all angles that are coterminal with 141° are given by 141° + k * 360° where k is a positive or negative integer. That is 141° + 360°, 141 + 2*360° , 141+3*360° and so on and 141° - 360°, 141° -2*360°, 141° - 3*360° and so on. 9720)</span>
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Answer:

16x^{6} +56x^{3}y^{3} z^{4} +49y^{6} z^{8} \\

Step-by-step explanation:

To solve this type of problems first need to review some laws of exponents:

When you are multiplying the same base, you need to add the exponents.

x^{2} +x^{8}  = x^{2+8}  = x^{10}

When you are raising a base with power to another power, you should keep the base and multiply the exponents:

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Now for the expression (4x^{3} + 7y^{3} z^{4} )^2

Write the multiplying factors:

(4x^{3} + 7y^{3} z^{4} )(4x^{3} + 7y^{3} z^{4} )

Multiply the term 4x^{3}

(4x^{3} + 7y^{3} z^{4} )(4x^{3} + 7y^{3} z^{4} ) = 16x^{3+3} + 28x^3y^{3} z^{4}

Then multiply the term 7y^{3}z^{4}

(4x^{3} + 7y^{3} z^{4} )(4x^{3} + 7y^{3} z^{4} ) = 16x^{3+3} + 28x^3y^{3} z^{4}\\\\+ 28x^3y^{3} z^{4} + 49y^{3+3} z^{4+4}

Simplify the exponents:

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Add like terms:

= 16x^{6} + 56x^3y^{3} z^{4} + 49y^{6} z^{8}

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