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Aleonysh [2.5K]
3 years ago
11

Which of the following function types exhibit the end behavior f(x)-->0 as x --> -infinity?

Mathematics
1 answer:
ipn [44]3 years ago
5 0
Let's consider the functions one by one.

i) y=x^n,

x --> -infinity means that x is a very very small negative number. To model it in our mind let's think of -10^{15}. This number to an even power becomes 10^{30}, 10^{60}... etc.

Indeed, we can see that the smaller the x, the greater the value x^n. In fact, as x --> -infinity,  f(x)-->+infinity.

ii)
y=x, 

this clearly means that the behaviors of x and y are identical.

As x --> -infinity, y --> -infinity as well.


iii) y=|x|,

We can think of x --> -infinity, again, as a very very small number, like -10^{15}. For this value of x, y is |-10^{15}|, that is 10^{15}.

Indeed, the smaller the x, the greater the y. As in the function in part (i), as x --> -infinity,  f(x)-->+infinity.

iv) 

y=1/x

consider the values of y for the following values of x:

for x=-10, -100, -1000, y is respectively -0.1, -0.01, -0.001.

We can see that the smaller the x, the closer y gets to 0.

Thus, <span>f(x)-->0 as x --> -infinity
</span>
v) n'th root of x in not defined for negative x, when n is even.

vi) y=b^x, b>0

Here note that b cannot be equal to 1, otherwise the function is not exponential.

Let b=5, consider the values of y for the following values of x:

for x=-10, -100, -1000, y is respectively \displaystyle{ \frac{1}{5^{10}},  \frac{1}{5^{100}}, \frac{1}{5^{1000}}.

That is, the smaller the value of x, the closer y gets to 0.

Thus, <span>f(x)-->0 as x --> -infinity</span>
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Andrews [41]

Answer:

\huge\boxed{2 \sqrt{10}}

Step-by-step explanation:

When taking a square root and making it a radical, we need to list out it's factors and find which one we can take the square root of.

The factors of 40 are:

1, 2, 4, 5, 8, 10, 20, 40.

Out of these, we need to look for one that we can find the square root of.

We know that the square root of 4 is 2.

Therefore, we can make \sqrt{40} into \sqrt{4 \cdot 10}.

The square root of 4 is 2. Therefore, we can put that outside the radical.

2 \sqrt{10}.

Hope this helped!

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2 years ago
Write a quadratic equation ax²+ bx + C = 0, whose coefficients are a =-9, b = 0, c = -2
aksik [14]
<h3>ax² + bx + c = 0</h3>

<em>Let's write -9 where we see A</em><em>:</em>

<h3>-9x² + bx + c = 0</h3>

<em>Let's</em><em> </em><em>write</em><em> </em><em>0</em><em> </em><em>where</em><em> </em><em>we</em><em> </em><em>see</em><em> </em><em>B</em><em>:</em>

<h3>-9x² + 0.x + c = 0</h3>

<em>(</em><em>Since B = 0, when it is multiplied by x, it becomes 0 again</em><em>)</em>

<h3>-9x² + c = 0</h3>

<em>Let's</em><em> </em><em>write</em><em> </em><em>-2</em><em> </em><em>where</em><em> </em><em>we</em><em> </em><em>see</em><em> </em><em>C</em><em>:</em>

<h3>-9x² + -2 = 0</h3>

<em>Now we can move on to solving our equation</em><em>:</em><em>)</em>

<em>Let's put the known and the unknown on different sides:</em>

<em>(</em><em>-2 goes to the opposite side positively</em><em>)</em>

<h3>-9x² = 2</h3>

<em>(</em><em>i</em><em>t goes as a division because it is in the case of multiplying -9 across</em><em>)</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

<h3>x² = 2/-9</h3>

<em>I could not find the rest of it, but I did not want to delete it for trying very hard. Sorry. It felt like we should take the square root, but I couldn't find it, maybe this can help you a little bit.</em>

<em>Please do not report</em><em>:</em><em>(</em>

<em>I hope I got it right, I'm trying to improve my English a little :)</em>

<h3><em>Greetings from Turke</em><em>y</em><em>:</em><em>)</em></h3>

<h3><em><u>#XBadeX</u></em></h3>
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2 years ago
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Alex787 [66]

Answer:

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Step-by-step explanation:

I multiplied 120 by 0.1 to find how much she would be marking up the price of each apple. I got the answer of $12. So then I added that $12 to $120 and got the new price of $132.

I checked my work I swear it's correct

hope this helps chu <3

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Answer:

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Step-by-step explanation:

bruh, it's right there

8 0
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