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GREYUIT [131]
3 years ago
14

Which of the following is false for f(x) = (10x^3-10x^2-10x)/(2x^5-2x)

Mathematics
1 answer:
Trava [24]3 years ago
4 0
1) Factor and simplify:

f(x) = [10x (x^2 - x - 1) ] / [ 2x (x^4 - 1) ] =5 (x^2 - x - 1) / (x^4 - 1)

2) Calculate limits

A) limit of f(x) when x -> 0 = 5(-1)/(-1) = 5

=> y-axis is not an asymptote and A is TRUE.

B) Lim of f(x) when x -> +/- ∞ =

5 * (x^2 / x^4 - x / x^4  - 1/ x^4 ) / (x^4 / x^4 - 1 /x^4) = 0 / ∞ = 0

=> x-axis is an asymptote and B is TRUE

C) Lim of f(x) when x -> - 1 =

5 * (x^2 - x - 1) / (x^4 - 1) = 5 * (1 + 1 - 1) / (1 - 1) = 5 / 0 = ∞

=> x = - 1 is an asymptote and C. is FALSE.

Now you have the answer.

If you want you can verify that the last option is TRUE.

I also enclose a picture showing the asymptotes which may help you.

Answer: Option C. is the false one.

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A square fits exactly inside a circle with each of its vertices being on the circumference of the circle.
swat32

Answer:

The side of the square is 5.971 cm

Step-by-step explanation:

If the square fits the circle exactly, then its diagonal is equal to the diameter of the circle. Since the side of the square has a length of x cm, then it's diagonal have the length of x\sqrt2 cm. Using the circle's area we can find the diagonal of the square, as shown below:

area = \pi*r^2\\56 = pi*r^2\\r^2 = \frac{56}{\pi}\\r = \sqrt{\frac{56}{\pi}}\\r = 4.222

Since the diameter of the circle is the same as the diagonal of the square, then:

x\sqrt{2} = 2*r \\x = \frac{2*r}{\sqrt{2}}\\x = \frac{2*4.222}{\sqrt{2}}\\x = 5.971

The side of the square is 5.971 cm

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Step-by-step explanation:

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gogolik [260]
<span>DE=EF  hope I helped with your studies :)

</span>
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Consider the two functions shown here. What is the rate of change of each function?
Katarina [22]
The rate of change is the same as the slope.

Let's find the slope of function 1 using the Rise Over Run rule.
The rise is 2 and run is 1. So your rate of change is 2/1 or 2 for function 1

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y = 1/2x + 7
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27 / 3 =9

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