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fiasKO [112]
2 years ago
5

The graph of a sine function is shown. Plot the line that represents the midline of this periodic function.

Mathematics
2 answers:
BARSIC [14]2 years ago
6 0

Answer: The midline should be a horizontal one, at y=4

Step-by-step explanation: See above answer.

marysya [2.9K]2 years ago
6 0

Answer:

y=4 (see pic for proof)

Step-by-step explanation:

Have a great day :)

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1) Is 3 for K

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Parallel / Perpendicular Practice
deff fn [24]

The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

  1. Neither
  2. ║
  3. Neither
  4. ⊥
  5. ║
  6. Neither
  7. Neither
  8. Neither

Reason:

The slope and intercept form is the form y = m·x + c

Where;

m = The slope

Two equations are parallel if their slopes are equal

Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; m_1 = -\dfrac{1}{m_2}

1. The given equations are in the slope and intercept form

\ y = 3 \cdot x + 1

The slope, m₁ = 3

y = \dfrac{1}{3} \cdot x + 1

The slope, m₂ = \dfrac{1}{3}

Therefore, the equations are <u>neither</u> parallel or perpendicular

  • Neither

2. y = 5·x - 3

10·x - 2·y = 7

The second equation can be rewritten in the slope and intercept form as follows;

y = 5 \cdot x -\dfrac{7}{2}

Therefore, the two equations are <u>parallel</u>

  • ║

3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

y = 2 -\dfrac{1}{2}  \cdot x

Slope, m₁ = -\dfrac{1}{2}

y = \dfrac{1}{2}  \cdot x - 2

Slope, m₂ = \dfrac{1}{2}

The slopes

Therefore, m₁ ≠ m₂

m_1 \neq -\dfrac{1}{m_2}

The lines are <u>Neither</u> parallel nor perpendicular

  • <u>Neither</u>

4. The given equations are;

2·y - x = 2

y = \dfrac{1}{2} \cdot   x +1

m₁ = \dfrac{1}{2}

y = -2·x + 4

m₂ = -2

Therefore;

m_1 \neq -\dfrac{1}{m_2}

Therefore, the lines are <u>perpendicular</u>

  • ⊥

5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, y = \dfrac{3}{4} \cdot x + 3

Second equation, y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}

Therefore, m₁ = m₂, the lines are <u>parallel</u>

  • ║

6. The given equations are;

8·x - 4·y = 16

Which gives; y = 2·x - 4

5·y - 10 = 3, therefore, y = \dfrac{13}{5}

Therefore, the two equations are <u>neither</u> parallel nor perpendicular

  • <u>Neither</u>

7. The equations are;

2·x + 6·y = -3

Which gives y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}

12·y = 4·x + 20

Which gives

y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}

m₁ ≠ m₂

m_1 \neq -\dfrac{1}{m_2}

  • <u>Neither</u>

8. 2·x - 5·y = -3

Which gives; y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}

5·x + 27 = 6

x = -\dfrac{21}{5}

  • Therefore, the slopes are not equal, or perpendicular, the correct option is <u>Neither</u>

Learn more here:

brainly.com/question/16732089

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Anyone wanna help with this?????
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The answer is c hope it’s right
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