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goldfiish [28.3K]
1 year ago
10

Number 2 I need help on how to solve it

Mathematics
1 answer:
Hunter-Best [27]1 year ago
4 0

Let's simplify the following expression:

\text{ }\frac{\text{ \lparen2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{ }\frac{\text{ 6x}^5\text{y}^{-4}\text{z}^{-5}}{\text{ \lparen2x}^6\text{y}^3\text{z}^{-3})^2}

We get,

\text{ }\frac{(\text{2x}^4\text{ - y}^6\text{z}^{-4})^2}{\text{ 2x}^2\text{z}^3}\text{ }\cdot\text{  }\frac{6x^5y^{-4}z^{-5}}{(2x^6y^3z^{-3})^2}\text{ }\frac{(2x^4-y^6z^{-4})^2(6x^5y^{-4}z^{-5})}{(2x^2z^3)(2x^6y^3z^{-3})^2}\text{ }\frac{(4x^8\text{ - 4x}^4y^6z^{-4}\text{ + y}^{12}z^{-8})(6x^5y^{-4}z^{-5})}{(2x^2z^3)(4x^{12}y^6z^{-6})}\text{ }\frac{24x^{13}y^{-4}z^{-5}\text{ - 24x}^9y^2z^{-9}\text{ + 6x}^5y^8z^{-13}}{8x^{14}y^6z^{-3}}

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vova2212 [387]

Answer:

Part A)

This quadratic expression can be factored by using the difference of squares pattern.

Part B) (5x+y) and (5x-y)

Step-by-step explanation:

Given:

25x^2-y^2

above polynomial can be factorize by using difference of squares formula

a2-b2=(a+b)(a-b)

25x^2-y^2= (5x-y)(5x+y)

so for part A)

correct option is  This quadratic expression can be factored by using the difference of squares pattern.

Part B)

As factored above in part A,

the factors of given polynomial 25x^2-y^2 are (5x+y)(5x-y)

so for part B) correct options are (5x+y) and (5x-y) !

8 0
4 years ago
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Dahasolnce [82]

Answer:

I believe the answer is C.

Step-by-step explanation:

I put in 9 into the y

6(9)= 54

4(9)+18=

36+18= 54

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7 0
3 years ago
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Iteru [2.4K]

Answer:

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Step-by-step explanation:

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3 years ago
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bogdanovich [222]

Answer:

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Step-by-step explanation:

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3 years ago
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nasty-shy [4]
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