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2 years ago

11

Read the instructions and then answer the questions

1 answer:

alexandr1967 [171]2 years ago

7 0

Answer:

2) $x - y = \frac{2}{3}$

$= > x = \frac{2}{3} + y = \frac{2 + 3y}{3}$

3) $3x + 2y = 6$

$= > 3x = 6 - 2y$

$= > x = \frac{6 - 2y}{3}$

4) $\frac{1}{4} x + y = 2$

$= > y = 2 - \frac{x}{4} = \frac{8 - x}{4}$

5) $4x - 3y = - 33$

$= > 3y = 4x + 33$

$= > y = \frac{4x + 33}{3}$

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estamos construyendo una carretera que enlace los puntos a = 12 ,21 y b =(17,23) otro punto se encuentra en c =(3,9) es posible

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is not possible

Step-by-step explanation:

<u><em>The question in English is</em></u>

we are building a road that links the points a = (12 ,21) and b =(17,23) another point is in c =(3,9) it is possible that a single road allows to join these three points?

we know that

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

step 1

Find the slope ab

we have

a = (12 ,21) and b =(17,23)

substitute

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step 2

Find the slope ac

we have

a = (12 ,21) and c =(3,9)

substitute

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simplify

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step 3

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Step-by-step explanation:

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The slope of the tangent line to $f$ at $x=-1$ is given by the derivative of $f$ at that point:

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{f(x)-f(-1)}{x-(-1)}=\lim_{x\to-1}\frac{2x^2-2}{x+1}$

Factorize the numerator:

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

We have $x$ approaching -1; in particular, this means $x\neq-1$ , so that

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

Then

$f'(-1)=\displaystyle\lim_{x\to-1}\frac{2x^2-2}{x+1}=\lim_{x\to-1}2(x-1)=2(-1-1)=-4$

and the tangent line's equation is

$y-f(-1)=f'(-1)(x-(-1))\implies y-4x-2$

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