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romanna [79]
3 years ago
8

Which function has a domain of all real numbers except x=π/2 ±nπ?

Mathematics
1 answer:
boyakko [2]3 years ago
4 0
<span>A. y=secx This problem deals with the various trig functions and is looking for those points where they are undefined. Since the only math operations involved is division, that will happen with the associated trig function attempts to divide by zero. So let's look at the functions that are a composite of sin and cos. sin and cos are defined for all real numbers and range in value from -1 to 1. sin is zero for all integral multiples of pi, and cos is zero for all integral multiples of pi plus pi over 2. So the functions that are undefined will be those that divide by cos. tan = sin/cos, which will be undefined for x = π/2 ±nπ cot = cos/sin, which will be undefined for x = ±nπ sec = 1/cos, which will be undefined for x = π/2 ±nπ csc = 1/sin, which will be undefined for x = ±nπ Now let's look at the options and pick the correct one. A. y=secx * There's a division by cos, so this is the correct choice. B. y=cosx * cos is defined over the entire domain, so this is a bad choice. C. y=1/sinx * The division is by sin, not cos. So this is a bad choice. D. y=cotx, * The division is by sin, not cos. So this is a bad choice.</span>
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Answer: 35,000,000
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3 years ago
Multiply and simplify
aleksandr82 [10.1K]

Answer and explanation:

1. 2 × 3/10 = 6/10 = 3/5

2. 8 × 4/8 = 32/8 = 4

3. 9 × 3/9 = 27/9 = 3

4. 12 × 2/11 = 24/11 = 2 2/11

5. 3 × 5/7 = 15/7 = 2 1/7

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7. 11 × 4/8 = 44/8 = 5 4/8 = 5 1/2

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4 0
3 years ago
Andrew claims the initial value and y - intercept are the same thing on a graph. Is he correct?
Agata [3.3K]

Answer:

We conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

Step-by-step explanation:

We know that the initial value on a graph is basically the out-put value y of the point where the line meets or crosses the y-axis.

In other words, the initial value is the y-value or output of the point at x = 0

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the initial value of the equation y = 2x+1 is: y = 1

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Thus, at y = 1, the line meets the y-axis.

Hence, the initial value of the line is: y = 1

Similarly, we know that the value of the y-intercept can be determined by setting x = 0 and determining the corresponding value of y.

For example,

Let the equation

y = 2x+1

substitute x = 0

y = 2(0)+1

y = 0+1

y = 1

Thus, the y-intercept of y = 2x+1 is y = 1.

Please check the attached graph of the equation y = 2x+1.

It is clear from the graph that at x = 0, the value of y = 1.

Therefore, the y-intercept of y = 2x+1 is y = 1.

Conclusion:

Therefore, we conclude that the initial value and y-intercept are the same thing on a graph.

Please check the attached graph of the equation y = 2x+1.

6 0
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Mars2501 [29]
I will go with B.

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