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Sergeu [11.5K]
3 years ago
5

Which are counterexamples? Look at the picture below. Choose all that apply.

Mathematics
1 answer:
Flauer [41]3 years ago
8 0

Answer:

x^4/(2x^6)

Explanation:

In the first example the division results in a radical, not a polynomial.

The remaining examples are not counter-examples (they do result in a polynomial)

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I need help as soon as possible please
tatiyna

The centroid divides the median into a 2:1 ratio. So, 9x-9=4x+16. We solve and get x=5. Therefore,  OB is 36, and BE is half, 18.

6 0
3 years ago
Solucione la siguiente desigualdad. Haga la gráfica. 2 < I(x-10)/7I < 5
Arada [10]
Vaya a una pagina que se llama  y ponga la equacion donde dice inserte el problema y ahi le va a aparecer la grafica 
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8 0
3 years ago
Examples of data presentation in math
myrzilka [38]

Answer:

Data Presentation - Tables

Tables are a useful way to organize information using rows and columns. Tables are a versatile organization tool and can be used to communicate information on their own, or they can be used to accompany another data representation type (like a graph). Tables support a variety of parameters and can be used to keep track of frequencies, variable associations, and more.

For example, given below are the weights of 20 students in grade 10:

50, 45, 48, 39, 40, 48, 54, 50, 48, 48, \\ 50, 39, 41, 46, 44, 43, 54, 57, 60, 45.

50,45,48,39,40,48,54,50,48,48,

50,39,41,46,44,43,54,57,60,45.

To find the frequency of 4848 in this data, count the number of times that 4848 appears in the list. There are 44 students that have this weight.

The list above has information about the weight of 2020 students, and since the data has been arranged haphazardly, it is difficult to classify the students properly.

To make the information more clear, tabulate the given data.

\begin{array}{c}\\ \text{Weights in kg} & & & \text{Frequency} \\ 39 & & & 2 \\ 40 & & & 1 \\ 41 & & & 1 \\ 43 & & & 1 \\ 44 & & & 1 \\ 45 & & & 2 \\ 46 & & & 1 \\ 48 & & & 4 \\ 50 & & & 3 \\ 54 & & & 2 \\ 57 & & & 1 \\ 60 & & & 1 \end{array}

Weights in kg

39

40

41

43

44

45

46

48

50

54

57

60

Frequency

2

1

1

1

1

2

1

4

3

2

1

1

This table makes the data more easy to understand.

5 0
3 years ago
The number of fish in a small bay is modeled by the function F defined by F(t)=10 (t3 â 12t2 + 45t +100), where t is measured in
RSB [31]

The graph of the given function is attached showing characteristics of the function.

The correct responses are;

  • (a) F'(4) = -30 means that <u>on day 4, the number of fish in the bay is decreasing at a rate of 30 fish per day</u>.
  • (b) Over the interval 0 ≤ t ≤ 8 the absolute minimum number of fish in the bay is;<u> 1,000</u>.
  • (c) The values of <em>t</em> at which the rate of change is decreasing is; <u>3 ≤ t ≤ 8</u>
  • (d) The rate of change of the number of pelicans flying near is; P' =  <u>10·(3·c² - 24·c  + 45)</u>

Reasons:

The given function is; f(t) = 10·(t³ - 12·t² + 45·t + 100)

Where;

t = Number of days

0 ≤ t ≤ 8

(a) The derivative of the given function is presented as follows;

F'(t) = 10 × (3·t² - 2×12·t + 45) = 10·(3·t² - 24·t  + 45)

Therefore;

F'(4) = 10 × (3 × 4² - 24 × 4  + 45) = -30

Therefore F'(4) = -30 means that on day 4, the <u>number of fish in the bay is decreasing at 30 fish per day</u>.

(b) The absolute minimum is given as follows;

At a minimum or maximum value, F'(t) = 10·(3·t² - 24·t  + 45) = 0

Which gives;

3·(t - 5)·(t - 3) = 0

t = 5, or t = 3

At t = 5, we have;

f(5) = 10 × (5³ - 12 × 5² + 45 × 5 + 100) = 1,500

At t = 3, we have;

f(3) = 10 × (3³ - 12 × 3² + 45 × 3 + 100) = 1,540

Therefore, f(5) = 1,500 is a local maximum

However, at x = 0, we have;

f(0) = 10 × (0³ - 12 × 0² + 45 × 0 + 100) = 1000

At x = 8, we have;

f(8) = 10 × (8³ - 12 × 8² + 45 × 8 + 100) = 2,040

Therefore, the absolute minimum is given at t = 8, where f(t) = <u>1,000</u>

(c) The values of <em>t</em> at which the rate of change in the number of fish in the

bay is decreasing is between the local maximum at t = 3, and the local

minimum at t = 5, which gives;

The rate of change of the number of fish is decreasing for values of t in the range;

  • <u>3 ≤ t ≤ 5</u>

(d)  P = 10·(t³ - 12·t² + 45·t + 100)

P' =  10·(3·t² - 24·t  + 45)

At time t = c, we have;

  • P' =  10·(3·c² - 24·c  + 45)

<em>Based on a similar question online, we have;</em>

<em>(c) The interval over which the rate of change is decreasing</em>

<em>(d) The rate of change of the number of pelicans flying near the bay at t = c</em>

Learn more about differentiation of functions here:

brainly.com/question/1422315

5 0
2 years ago
PLS HELP!!!
satela [25.4K]
A=180-105
A=75

In the triangle
75+ 9x +12x =180
X=5
7 0
3 years ago
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