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3 years ago

What is the domain of f(x) = cos(x)?

Mathematics

2 answers:

xxTIMURxx [149]3 years ago

5 0

Answer:

D) Set of all real numbers

Step-by-step explanation:

nika2105 [10]3 years ago

4 0

Option D: The set of all real numbers

Explanation:

Given that the function $f(x)=cos (x)$

We need to determine the domain of the function.

The domain of the function is the set of all input values for which the function is real and well defined.

In other words, the domain of the function is the set of all independent x - values of the function.

Thus, the function cos (x) is real and well defined in the interval $(-\infty, \infty)$

Hence, the function cos (x) is real and well defined for all real numbers.

Therefore, the domain of the function $f(x)=cos (x)$ is the set of all real numbers.

Hence, Option D is the correct answer.

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Determine whether the quadrilateral is a parallelogram. Justify you answer.

agasfer [191]

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It is a quadrilateral.

Step-by-step explanation:

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So when you add up all the angles, you will get 360°.

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3 years ago

What is the value of x in the equation shown? 2(X + 8) - 4x = 10x + 4

Anna007 [38]

Answer:

Option C: X=1

Step-by-step explanation:

2(X + 8) - 4x = 10x + 4

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2 years ago

Consider the two data sets below:

alina1380 [7]

Answer:

Option c) The data sets will have the same values of their interquartile range.

Explanation:

1. The values are in order: they are in increasing oder, from lowest to highest value.

2. Calculate the interquartile range.

Interquartile range, IQR, is the third quartile, Q3, less the first quartile Q1:

IQR = Q3 - Q1

To find the first and the third quartile, first find the median:

Data Set 1: 19, 25, 35, 38, 41, 49, 50, 52, 59

[19, 25, 35, 38], 41, [49, 50, 52, 59]

↑

median = 41

Data Set 2: 19, 25, 35, 38, 41, 49, 50, 52, 99

[19, 25, 35, 38] , 41, [49, 50, 52, 99]

↑

median = 41

Now find the median of each subset: the values below the median and the values above the median.

Data set 1: First quartile

[19, 25, 35, 38],

↑

Q1 = [25 + 35] / 2 = 30

Third quartile

[49, 50, 52, 59]

↑

Q3 = [50 + 52] / 2 = 51

IQR = Q3 - Q1 = 51 - 30 = 21

Data set 2: First quartile

[19, 25, 35, 38]

↑

Q1 = [25 + 35] / 2= 30

Third quartile

[49, 50, 52, 99]

↑

Q3 = [52 + 50]/2 = 51

IQR = 51 - 30 = 21

Thus, it is shown that the data sets have will have the same values for the interquartile range: IQR = 21. (option c)

This happens because replacing one extreme value (in this case the maximum value) by other extreme value does not affect the median.

An outlier will change the range because the range is the maximum value less the minimum value.

5 0

3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Nady [450]

Answer:

A) 2x³+11x²+8x-16

Step-by-step explanation:

When you multiply s(x) by t(x) you get something like this:

$s(x) \times t(x) = (2 {x}^{2} + 3x - 4) \times (x + 4) \\ = 2 {x}^{3} + 3 {x}^{2} - 4x + 8 {x}^{2} + 12x - 16 \\ = 2 {x}^{3} + 11 {x}^{2} + 8x - 16$

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3 years ago

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