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Ugo [173]
2 years ago
15

What is the range of the function y = 2sin x?

Mathematics
1 answer:
fiasKO [112]2 years ago
5 0

The standard form of equation for the sin function is y = Asin(Bx-C) where A is amplitude, Period is 2π/B, and the phase shift is equal to Bx-C set to zero. C is then obtained in this equation.

For given problem: A is equal to 2, period=2π/B=2π/1=2π, where B is equal to 1. Period then is equal to A. 2pi

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F(x) = 2x^2 – 5x – 3<br> g(x) = 2x^2 + 5x + 2<br> Find: (f/g) (x)
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Answer:

Since this means f of g so I will get the function g(x) and replace it with X in the function of f (x)

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What are the domain and range of f(x) = |x – 3 | + 6? Domain: {x | x is all real numbers} Range: {y | y ≥ 6} Domain: {x | x ≥ 3}
VARVARA [1.3K]

Answer:

Domain: {x | x is all real numbers} Range: {y | y ≥ 6}

Step-by-step explanation:

Domain is the set of all x values. No operations restrict the domain in this case. The domain is all real numbers.

Range is the set of all y values. Absolute value has a v shape starting at 0. Adding 6 raises this vertex to 6. The range is therefore all numbers greater than or equal to 6.

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Explain how to use the graph-and-check method to solve a linear system of two equations in two variables.
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4. Which of the following lines is perpendicular to y = -2X + 8?
Alenkinab [10]

For this case we have to by definition, if two lines are perpendicular then the product of its slopes is -1.

That is to say:

m_ {1} * m_ {2} = - 1

We have the following equation:

y = -2x + 8

So:

m_ {1} = - 2

Thus:

m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {- 2}\\m = \frac {1}{2}

Thus, a line perpendicular to the given line must have slope m = \frac {1} {2}.

Option A:

x + 2y = 8\\2y = -x + 8\\y = - \frac {1} {2} x + 4

It is not perpendicular!

Option B:

x-2y = 6\\2y = x-6\\y = \frac {1} {2} x-3

If it is perpendicular!

Option C:

2x + y = 4\\y = -2x + 4

It is not perpendicular!

Option D:

2x-y = 1\\y = 2x-1

It is not perpendicular!

The correct option is option B

ANswer:

Option B

5 0
3 years ago
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