2 years ago

Please I need the answer :-)

1 answer:

5 0

Answer:

Step-by-step explanation:

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mina [271]

Answer: 348

Step-by-step explanation:

5 0

3 years ago

Damm [24]

Answer: The answer should be B

6 0

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

Katarina [22]

Answer:

use desmos its a graphing caculator that gives you exaxt graphing answers

5 0

3 years ago

kirill [66]

360/12 = 30deg...so the arguments are multiples of 30deg.

5 0

3 years ago

Please I need the answer :-)​