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irina [24]
3 years ago
13

URGENT....HELP QUICKLY

Mathematics
2 answers:
Ber [7]3 years ago
6 0

Answer:

0.10 is the first slider

0.60 is the second slider

0.20 is the third slider

0.10 is the final slider

<em>I just took the test and this is the correct answer! Good luck! </em>

ankoles [38]3 years ago
3 0

Answer: Here is your answer, 0.10, 0.60, 0.20, and 0.1

Step-by-step explanation:

You add up the f values and then divide them each by the value you got, which is 200. So in this case 20/200 = 0.1, and so forth. So you drag the slider up to 0.10, and do the same for each. I will send a picture with the exact answer.

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The population of a town in 2014 was 78,918 people with an annual rate of increase of about 1.7%. Which type of function can rep
skelet666 [1.2K]

Answer:

B.  Exponential.

Step-by-step explanation:

Each year you can find the estimated population for the following year by multiplying by  1 + 1.7%  = 1.017.

The population estimate in  x years time =  78,918 (1.017)^x.

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3 years ago
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What is true regarding secants and chords
Elenna [48]

Answer:

Chords and secants intersect the perimeter of a circle twice. Chords are segments entirely within a circle, while secants are lines or rays that extend through a circle.

Step-by-step explanation:

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Consider the sequence {an}={3n+13n−3n3n+1}. Graph this sequence and use your graph to help you answer the following questions.
Fantom [35]

Part 1: You can simplify a_n to

\dfrac{3n+1}{3n}-\dfrac{3n}{3n+1} = \dfrac1{3n}+\dfrac1{3n+1}

Presumably, the sequence starts at <em>n</em> = 1. It's easy to see that the sequence is strictly decreasing, since larger values of <em>n</em> make either fraction smaller.

(a) So, the sequence is bounded above by its first value,

|a_n| \le a_1 = \dfrac13+\dfrac14 = \boxed{\dfrac7{12}}

(b) And because both fractions in a_n converge to 0, while remaining positive for any natural number <em>n</em>, the sequence is bounded below by 0,

|a_n| \ge \boxed{0}

(c) Finally, a_n is bounded above and below, so it is a bounded sequence.

Part 2: Yes, a_n is monotonic and strictly decreasing.

Part 3:

(a) I assume the choices are between convergent and divergent. Any monotonic and bounded sequence is convergent.

(b) Since a_n is decreasing and bounded below by 0, its limit as <em>n</em> goes to infinity is 0.

Part 4:

(a) We have

\displaystyle \lim_{n\to\infty} \frac{10n^2+1}{n^2+n} = \lim_{n\to\infty}10+\frac1{n^2}}{1+\frac1n} = 10

and the (-1)ⁿ makes this limit alternate between -10 and 10. So the sequence is bounded but clearly not monotonic, and hence divergent.

(b) Taking the limit gives

\displaystyle\lim_{n\to\infty}\frac{10n^3+1}{n^2+n} = \lim_{n\to\infty}\frac{10+\frac1{n^3}}{\frac1n+\frac1{n^2}} = \infty

so the sequence is unbounded and divergent. It should also be easy to see or establish that the sequence is strictly increasing and thus monotonic.

For the next three, I'm guessing the options here are something to the effect of "does", "may", or "does not".

(c) may : the sequence in (a) demonstrates that a bounded sequence need not converge

(d) does not : a monotonic sequence has to be bounded in order to converge, otherwise it grows to ± infinity.

(e) does : this is true and is known as the monotone convergence theorem.

5 0
3 years ago
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jeka57 [31]

Answer:

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5 0
3 years ago
What equation represents a parabola that opens UP with the vertex at 2,4
lubasha [3.4K]

Answer:

  y = (x -2)^2 +4

Step-by-step explanation:

The vertex form of the equation of a parabola is ...

  y = a(x -h)^2 +k

for vertex (h, k) and vertical scale factor "a". When a > 0, the parabola opens upward.

One equation for your parabola could be ...

  y = (x -2)^2 +4

__

In standard form this one is ...

  y = x^2 -4x +8

3 0
3 years ago
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