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3 years ago

Need help of this!! Domain and range!!!

Mathematics

1 answer:

kvasek [131]3 years ago

8 0

Answer:

third one....domain starts negative (-7/3)

and range starts at -5

Step-by-step explanation:

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STALIN [3.7K]

Answer: should be B

Step-by-step explanation:

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2 years ago

What are the equations of the asymptotes of the graph of f(x)=(5/x+7)−8 ?

ludmilkaskok [199]

Answer:

$\mathrm{Vertical}:\:x=0,\:\mathrm{Horizontal}:\:y=-1$

The graph of the function is also attached.

Step-by-step explanation:

Considering the expression

$\:f\left(x\right)=\left(\frac{5}{x}+7\right)-8\:\:$

$\mathrm{Simplify}\:\frac{5}{x}+7-8:\quad \frac{5}{x}-1$

$\mathrm{Vertical\:asymptotes\:of\:}\frac{5}{x}-1:$

Go over every undefined point x = a and check if at least one of the following statement is true.

$\lim _{x\to a^-}f\left(x\right)=\pm \infty$

$\lim _{x\to a^+}f\left(x\right)=\pm \infty$

$\mathrm{Take\:the\:denominator\left(s\right)\:of\:}\frac{5}{x}-1\mathrm{\:and\:compare\:to\:zero}$

$x=0$

The following points are undefined

$x=0$

$\mathrm{The\:vertical\:asymptotes\:are:}$

$x=0$

$\mathrm{Horizontal\:Asymptotes\:of\:}\frac{5}{x}-1:$

$\mathrm{Check\:if\:at\:}x\to \pm \infty \mathrm{\:the\:function\:}y=\frac{5}{x}-1\mathrm{\:behaves\:as\:a\:line,\:}y=mx+b$

$\mathrm{Find\:an\:asymptote\:for\:}x\to -\infty \::$

$\lim _{x\to -\infty \:}\frac{f\left(x\right)}{x}=\lim _{x\to -\infty \:}\frac{\frac{5}{x}-1}{x}=0$

$\lim _{x\to -\infty \:}f\left(x\right)-mx=\lim _{x\to -\infty \:}\frac{5}{x}-1-0x=-1$

$y=-1$

$\mathrm{Find\:an\:asymptote\:for\:}x\to \infty:$

$\lim _{x\to \infty \:}\frac{f\left(x\right)}{x}=\lim _{x\to \infty \:}\frac{\frac{5}{x}-1}{x}=0$

$\lim _{x\to \infty \:}f\left(x\right)-mx=\lim _{x\to \infty \:}\frac{5}{x}-1-0x=-1$

$y=-1$

$\mathrm{The\:horizontal\:asymptote\:is:}$

$y=-1$

Therefore,

$\mathrm{Vertical}:\:x=0,\:\mathrm{Horizontal}:\:y=-1$

The graph of the function is also attached.

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4 years ago

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9 students have packed their lunches

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wolverine [178]

A and C is your answer

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3 years ago

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Factor completely. (x^2 – 4)(x^2 + 6x + 9)<br>

sertanlavr [38]

<h3>Answer: (x-2)(x+2)(x+3)(x+3)</h3>

This is the same as (x-2)(x+2)(x+3)^2. The order of the factors doesn't matter.

===============================================

Explanation:

x^2-4 factors to (x-2)(x+2) after using the difference of squares rule

x^2+6x+9 factors to (x+3)(x+3) after using the perfect square trinomial factoring rule

So overall, the original expression factors to (x-2)(x+2)(x+3)(x+3)

We can condense this into (x-2)(x+2)(x+3)^2 since (x+3)(x+3) is the same as (x+3)^2

-------------------

Side notes:

Difference of squares rule is a^2 - b^2 = (a-b)(a+b)
The perfect square trinomial factoring rule is a^2+2ab+b^2 = (a+b)^2

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3 years ago