Mathematics, 24.12.2019 21:31 fatherbamboo

Use the limit theorem and the properties of limits to find the limit.
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Use the limit theorem and the properties of limits to find the limit. picture below

Answers: 2

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Answer from: bre2795

We have

$\dfrac{-6x^3+7x+7}{8x^3-8x+5}=\dfrac{-6+\frac7{x^2}+\frac7{x^3}}{8-\frac8{x^2}+\frac5{x^3}}$

As $x\to-\infty$ , all of terms with powers of in their denominators will converge to 0, leaving you with

$\displaystyle\lim_{x\to-\infty}\frac{-6x^3+7x+7}{8x^3-8x+5}=\lim_{x\to-\infty}\frac{-6}8=-\frac34$

making the answer B.

Answer from: cruzhazeL999

The value of limit is $\dfrac{-3}{4}$

Option B is correct.

Step-by-step explanation:

Given:

$L=\lim_{x\rightarrow -\infty}\left ( \dfrac{-6x^3+7x+7}{8x^3-8x+5}\right )$

Here we have rational function whose limit is minus infinity.

Divide numerator and denominator by highest degree of polynomial (x³)

$L=\lim_{x\rightarrow -\infty}\left ( \dfrac{-6+7/x^2+7/x^3}{8-8/x^2+5/x^3}\right )$

Apply limit and we get

$L=\left ( \dfrac{-6+\frac{7}{\infty}+\frac{7}{-\infty}}{8-\frac{8}{\infty}+\frac{5}{-\infty}}\right )$

$L=\left ( \dfrac{-6+0+0}{8-0+0}\right )$

$L=\dfrac{-6}{8}\Rightarrow -\dfrac{3}{4}$

Hence, The value of limit is $\dfrac{-3}{4}$

Answer from: Quest

first anwser choice.

step-by-step explanation:

the greater/less or equal to symbol (> or < with the line at the bottom) is a closed/filled in circle.

when multiplying or dividing by a negative on both sides you have to flip the inequality sign. this rule only applies to multiplying or dividing by a negative number in inequality equations. this rule is really important.

-5c + 2 _< _27

subtract from both sides by 2

-5c _< _ 27

divide both sides by -5

c _< _ 5.4

flip the inequality sign

c _> _5.4

the only possible answer that would match the simplified inequality is the first answer choice

5c+2< 27 i really do not understand this.

Answer from: Quest

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf =

pf = 6,25 u

ver más en -

step-by-step explanation:

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Use the limit theorem and the properties of limits to find the limit.
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