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3 years ago

PWEASE HELP ME!! I need it ASAP

Mathematics

1 answer:

dem82 [27]3 years ago

5 0

Answer:

V=6/13.5

Step-by-step explanation:

8-8.5(v)=6v+2-v

We move all terms to the left:

8-8.5(v)-(6v+2-v)=0

We add all the numbers together, and all the variables

-8.5v-(5v+2)+8=0

We get rid of parentheses

-8.5v-5v-2+8=0

We add all the numbers together, and all the variables

-13.5v+6=0

We move all terms containing v to the left, all other terms to the right

-13.5v=-6

v=-6/-13.5

v=6/13.5

<em>I hope this helps. </em>

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vesna_86 [32]

Factorize the denominator:

$\dfrac{x^2-4}{x^3+x^2-4x-4}=\dfrac{x^2-4}{x^2(x+1)-4(x+1)}=\dfrac{x^2-4}{(x^2-4)(x+1)}$

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We're then left with

$\dfrac1{x+1}$

which is undefined when $x=-1$ , so this is the site of a vertical asymptote.

As $x$ gets arbitrarily large in magnitude, we find

$\displaystyle\lim_{x\to-\infty}g(x)=\lim_{x\to+\infty}g(x)=0$

since the degree of the denominator (3) is greater than the degree of the numerator (2). So $y=0$ is a horizontal asymptote.

Intercepts occur where $g(x)=0$ ( $x$ -intercepts) and the value of $g(x)$ when $x=0$ ( $y$ -intercept). There are no $x$ -intercepts because $\dfrac1{x+1}$ is never 0. On the other hand,

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The domain of $g(x)$ is the set of values that $x$ can take on for which $g(x)$ exists. We've already shown that $x$ can't be -2, 2, or -1, so the domain is the set

$\{x\in\mathbb R\mid x\neq-2,x\neq-1,x\neq2\}$