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MissTica
4 years ago
11

Halla la tasa de variación de cada funcion en el intervalo [-4,3] e indica si es positiva , negativa o nula A) f(x)=x2-2x+4 B) f

(x)= -3x + 2 En rl ejercico A es x elevado al 2 o cuadrado
Mathematics
1 answer:
masya89 [10]4 years ago
8 0

Answer:

A) \hspace{3}Rate\hspace{3}of\hspace{3}change=-5\hspace{3}Negative\\\\B)\hspace{3}Rate\hspace{3}of\hspace{3}change=-21\hspace{3}Negative  

Step-by-step explanation:

Given a function f(x), we called the rate of change to the number that represents the increase or decrease that the function experiences when increasing the independent variable from one value "x_1" to another "x_2".

The rate of change of f(x) between x_1 and x_2 can be calculated as follows:

Rate\hspace{3}of\hspace{3}change=f(x_2)-f(x_1)

For:

f(x)=x^2-2x+4

Let's find f(x_1) and f(x_2), where:

[x_1,x_2]=[-4,3]

f(x_1)=f(-4)=(-4)^2-2(4)+4=16-8+4=12\\f(x_2)=f(3)=(3)^2-2(3)+4=9-6+4=7

So:

Rate\hspace{3}of\hspace{3}change =7-12=-5\hspace{3}Negative

And for:

f(x)-3x+2

Let's find f(x_1) and f(x_2), where:

[x_1,x_2]=[-4,3]

f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7

So:

Rate\hspace{3}of\hspace{3}change =-7-14=-21\hspace{3}Negative

<em>Translation:</em>

Dada una función f(x), llamábamos tasa de variación al número que representa el aumento o disminución que experimenta la función al aumentar la variable independiente de un valor "x_1" a otro "x_2".

La tasa de variación de f(x) entre x_1 y x_2, puede ser calculada de la siguiente forma:

Tasa\hspace{3}de\hspace{3}variacion=f(x_2)-f(x_1)

Para:

f(x)=x^2-2x+4

Encontremos f(x_1) y f(x_2), donde:

[x_1,x_2]=[-4,3]

f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7

Entonces:

Tasa\hspace{3}de\hspace{3}variacion =7-12=-5\hspace{3}Negativa

Y para:

f(x)-3x+2

Encontremos f(x_1) y f(x_2), donde:

[x_1,x_2]=[-4,3]

f(x_1)=f(-4)=-3(-4)+2=12+2=14\\f(x_2)=f(3)=-3(3)+2=-9+2=-7

Entonces:

Tasa\hspace{3}de\hspace{3}variacion=-7-14=-21\hspace{3}Negativa

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Step-by-step explanation:

~~~~~~~~~~~~~~~~~

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Use the following pattern to answer the questions. 24.3, 24, 23.7, 23.4, ___,____. part A. ( what is the rule for the pattern?
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<em>Question 1: </em>

For this case we have the following pattern of numbers:

24.3, 24, 23.7, 23.4

It can be seen that subtracting 0.3 from the first number of the pattern gives 24.3-0.3 = 24\\

Equally if we subtract 0.3 successively we obtain:

24-0.3 = 23.7\\\\23.7-0.3 = 23.4\\\\23.4-0.3 = 23.1\\\\23.1-0.3 = 22.8\\

Thus, 0.3 must be subtracted from each term and the following numbers of the pattern shall be 23.1 and 22.8

Answer:

subtract 0.3 from each term

The next two numbers of the pattern are 23.1 and 22.8

<em>Question 2: </em>

For this case we have the following operation:

0.52 + 0.61 = 1.13\\\\1.13> 1\\

We note that the sum is greater than 1

The sum of two hundredth decimals does not necessarily have to be greater than 1.

Therefore, the correct option is: the sum is greater than 1 because the sum shades more than one hundredths grid.

Answer:

Option C

<em>Question 3: </em>

For this case we have to:

a + b + c = 2\\

Being a, b, c decimal numbers, one of them was 0.34. Suppose a = 0.34\\

Thus: 0.34 + b + c = 2\\

We choose values ​​for b and c, in order to obtain the number 2 as a result of the sum. We can choose infinite values ​​so that equality is fulfilled

Ifb = 0.96 and

c = 0.7\\

we have: 0.34 + 0.96 + 0.7 = 2\\

Thus, Holland can add to the sum b = 0.96 and c = 0.7\\

Answer:

b = 0.96\\\\c = 0.7\\

<em>Question 4: </em>

For this case we have the following subtraction:

3.5 - 2.46 = 1.16\\

This operation is wrong, subtracting 3.5-2.46 results in 1.04

Let's explain a model for Maysa:

We have the following expression

3.5 - 2.46\\

Rewriting each number in expanded form we have:

(3 + 0.5) - (2 + 0.46)\\

Using the distributive property we have:

(3 + 0.5) -2-0.46\\

Using the associative property we have:

(3-2) + (0.5-0.46)\\

Performing each operation:

1 + 0.04 = 1.04\\

Anwer:

The subtraction of Maysa is wrong, the correct answer is:

3.5 - 2.46 = 1.04

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3 years ago
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