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HACTEHA [7]
3 years ago
11

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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Answer:

6/-4   (right 6 points on x-axis & down 4 points on y- axis

Step-by-step explanation:

14- 8            6

---------- =    ------

5-9             - 4

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Which is the simplified form of the expression 12n-1/2(6n-4)
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The simplified for is 9n+2
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Select all equations that have two solutions. a x^2=16 b 4x^2=0 c x^2=-16 d 3x+2=14 e x^2-1=24 f (x+8)(x-8)=0
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a )

{x}^{2}  = 16

x = ± \: 4

Two solutions.

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b )

4 {x}^{2}  = 0

x = 0

One solution .

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c )

{x}^{2}  =  - 16

No  \:  \: solution.

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d )

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e )

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f )

(x + 8)(x - 8) = 0

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PLEASE HELP:
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Answer: (A) The image of JKL after a 90° counterclockwise about the origin is shown in figure 1. (B) The image of JKL after a reflection across the y-axis is shown in figure 2.

Explanation:

(A)

From the given figure it is noticed that the coordinate points are J(-4,1), K(-4,-2) and L(-3,-1).

If a shape rotate 90 degree counterclockwise about the origin, then,

(x,y)\rightarrow (-y,x)

J(-4,1)\rightarrow J'(-1,-4)

K(-4,-2)\rightarrow K'(2,-4)

L(-3,-1)\rightarrow L'(1,-3)

Therefore, the vertex of imare are  J'(-1,-4),  K'(2,-4) and  L'(1,-3). The graph is shown in figure (1).

(B)

If a figure reflect across the y-axis then,

(x,y)\rightarrow (-x,y)

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K(-4,-2)\rightarrow K''(2,-4)

L(-3,-1)\rightarrow L''(3,-1)

Therefore, the vertex of imare are  J''(4,1),  K''(2,-4) and  L''(3,-1). The graph is shown in figure (2).

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3/11*2

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