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madreJ [45]
1 year ago
13

PLEASE HELP!

Mathematics
1 answer:
Aleksandr-060686 [28]1 year ago
6 0

The equation of the function f(x) is f(x) = 2 sin(π/2(x + 6)) - 3

<h3>How to create the sine function?</h3>

A sine function is represented as:

f(x) = A sin(B(x + C)) + D

Where

A = Amplitude

Period = 2π/B

C = Phase shift

D = Vertical shift

The requirements in the question are:

  • Amplitude not equal to 1
  • Period not equal to 2π
  • Non-zero phase and vertical shifts

So, we can use the following assumptions

A = 2

Period = 4

C = 6

D = -3

So, we have:

f(x) = 2 sin(B(x + 6)) - 3

The value of B is

4 = 2π/B

This gives

B = π/2

So, we have:

f(x) = 2 sin(π/2(x + 6)) - 3

<h3>The amplitude, vertical shift, period of f(x)and the phase shift</h3>

Using the representations in (a), we have:

  • Amplitude = 2
  • Vertical shift = -3
  • Period = 4
  • Phase shift = 6

<h3>The graph of the function</h3>

See attachment for the graph of f(x)

<h3>The value of cos θ </h3>

Let θ = 3π

So, we have:

cos(3π)

This is calculated as:

cos(3π) = cos(2π + π)

Expand

cos(3π) = cos(2π) *cos(π) - sin(2π) *sin(π)

Evaluate

cos(3π) = -1

<h3>The value of sin θ </h3>

Let θ = 3π

So, we have:

sin(3π)

This is calculated as:

sin(3π) = sin(2π + π)

Expand

sin(3π) = sin(2π) *cos(π) + cos(2π) *sin(π)

Evaluate

sin(3π) = 0

<h3>The value of tan 2θ </h3>

Let θ = 3π

So, we have:

tan(2 * 3π)

tan(6π)

This is calculated as:

tan(6π) = tan(3π + 3π)

Evaluate

tan(3π) = 0

Read more about sinusoidal functions at:

brainly.com/question/10700288

#SPJ1

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What is the solution set for 2x+5y&gt;-1 and 4x-3&lt;-3?
bekas [8.4K]

PROBLEM ONE

•

Solving for x in 2x + 5y > -1.

•

Step 1 ) Subtract 5y from both sides.

2x + 5y > -1

2x + 5y - 5y > -1 - 5y

2x > -1 - 5y

Step 2 ) Divide both sides by 2.

2x > -1 - 5y

\displaystyle\frac{2x}{2} > \displaystyle\frac{-1 - 5y}{2}

\displaystyle\ x > \frac{-1 - 5y}{2}

So, the solution for x in 2x + 5y > -1 is...

\displaystyle\ x > \frac{-1 - 5y}{2}

•

Solving for y in 2x + 5y > -1.

•

Step 1 ) Subtract 2x from both sides.

2x + 5y > -1

2x - 2x + 5y > -1 - 2x

5y > -1 - 1x

Step 2 ) Divide both sides by 5.

5y > -1 - 1x

\displaystyle\frac{5x}{5} > \frac{-1 -1x}{5}

\displaystyle\ x > \frac{-1 -1x}{5}

So, the solution for y in 2x + 5y > -1 is...

\displaystyle\ x > \frac{-1 -1x}{5}

•

PROBLEM TWO

•

Solving for x in 4x - 3 < -3.

•

Step 1 ) Subtract 3 from both sides.

4x - 3 < -3

4x -3 - 3 < -3 - 3

4x < 0

Step 2 ) Divide both sides by x.

4x < 0

\displaystyle\frac{4x}{4}

x < 0

So, the solution for x in 4x - 3 < -3 is...

x < 0

•

•

- <em>Marlon Nunez</em>

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