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wariber [46]
2 years ago
13

Please help me I need someone to help me with this ​

Mathematics
2 answers:
dlinn [17]2 years ago
8 0

Answer:

-2

Step-by-step explanation:

We'll have to solve the problem first, the strategy used here is to divide the first term of the dividend by the first term of the divisor (x^4 / x) which is x^3 and then we add it as the first term in the quotient, and then multiply x^3 by the divisor ( x^3 * (x-3)) which is  x^4 - 3x^3, then we subtract the x^4 - 3x^3 from the dividend ( ( x^4 + 2x^3 - 10x^2 - 4x - 35 ) - ( x^4 - 3x^3 ) ) which is equal to 5x^3 - 10x^2 - 4x - 35, and we do the last steps over and over until the remainder equals -2, in case of confusion, here are the other steps:

5x^3 / x = 5x^2

the current quotient = x^3 + 5x^2

5x^2 * x-3 = 5x^3-15x^2

( 5x^3 - 10x^2 - 4x - 35 )-( 5x^3-15x^2 ) = 5x^2 - 4x - 35

5x^2 / x = 5x

the current quotient = x^3 + 5x^2 + 5x

5x * x-3 = 5x^2-15x

( 5x^2 - 4x - 35 ) - ( 5x^2-15x ) = 11x - 35

11x / x = 11

the final quotient = x^3 + 5x^2 + 5x + 11

11 * x-3 = 11x - 33

(11x - 35) - (11x - 33) = -2

So, the remainder will be -2

Digiron [165]2 years ago
6 0

Answer:

-2

Step-by-step explanation:

essay with the title stolen

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jek_recluse [69]

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It follows that

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Similarly, you would find

Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}

You can solve the second system in the same fashion. You would end up with

P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}

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