3 years ago

ALGEBRA

1 answer:

7 0

Answer:

11,7,5,3

Step-by-step explanation:

thats all thank you

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Firlakuza [10]

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3 years ago

pychu [463]

20/3
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Anettt [7]

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14/-7

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3 years ago

Marta_Voda [28]

Oh ok so square root of 2 is irrational since it goes on and on. square root of 18 is an irrational number too

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3 years ago

UkoKoshka [18]

Since the limit becomes the undetermined form

$\displaystyle \lim_{x\to 1} \dfrac{x^3-2x^2+3x-2}{2x^4-3x+1} \to \dfrac{0}{0}$

it means that both polynomials have a root at $x=1$ . So, we can fact both numerator and denominator:

$x^3-2x^2+3x-2 = (x-1)(x^2-x+2)$

$2x^4-3x+1 = (x-1)(2x^3+2x^2+2x-1)$

So, the fraction becomes

$\dfrac{(x-1)(x^2-x+2)}{(x-1)(2x^3+2x^2+2x-1)} = \dfrac{x^2-x+2}{2x^3+2x^2+2x-1}$

Now, as x approaches 1, you have no problems anymore:

$\displaystyle \lim_{x\to 1} \dfrac{x^2-x+2}{2x^3+2x^2+2x-1} \to \dfrac{2}{5}$

4 0

3 years ago