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Rashid [163]
3 years ago
7

From the information given determine the quadrant in which the terminal point (determined by an angle with a measure of a) lies.

Input I, II, III, or IV.
If sin(θ)<0 and cos(θ)<0), then (x,y) lies in quadrant ___

If sin(θ)>0 and cos(θ)<0), then (x,y) lies in quadrant ____

If sin(θ)>0 and cos(θ)>0), then (x,y) lies in quadrant ____

If sin(θ)<0 and cos(θ)>0), then (x,y) lies in quadrant ____
​​

Mathematics
1 answer:
Otrada [13]3 years ago
6 0

Answer:

If sin(θ)<0 and cos(θ)<0), then (x,y) lies in quadrant III.

If sin(θ)>0 and cos(θ)<0), then (x,y) lies in quadrant II.

If sin(θ)>0 and cos(θ)>0), then (x,y) lies in quadrant I.

If sin(θ)<0 and cos(θ)>0), then (x,y) lies in quadrant IV.

Step-by-step explanation:

Sine and cosine values, relation with quadrant:

Quadrant 1: Sine > 0, cosine > 0.

Quadrant 2: Sine > 0, cosine < 0.

Quadrant 3: Sine < 0, cosine < 0.

Quadrant 4: Sine < 0, cosine > 0.

If sin(θ)<0 and cos(θ)<0), then (x,y) lies in quadrant

Both negative, so III.

If sin(θ)>0 and cos(θ)<0), then (x,y) lies in quadrant

Sine positive, cosine negative, so II.

If sin(θ)>0 and cos(θ)>0), then (x,y) lies in quadrant

Both positive, so I.

If sin(θ)<0 and cos(θ)>0), then (x,y) lies in quadrant

Sine negative, cosine positive, so IV.

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Rainbow [258]

By algebra properties we find the following relationships between each pair of algebraic expressions:

  1. First equation: Case 4
  2. Second equation: Case 1
  3. Third equation: Case 2
  4. Fourth equation: Case 5
  5. Fifth equation: Case 3

<h3>How to determine pairs of equivalent equations</h3>

In this we must determine the equivalent algebraic expression related to given expressions, this can be done by applying algebra properties on equations from the second column until equivalent expression is found. Now we proceed to find for each case:

First equation

(7 - 2 · x) + (3 · x - 11)

(7 - 11) + (- 2 · x + 3 · x)

- 4 + (- 2 + 3) · x

- 4 + (1) · x

- 4 + (5 - 4) · x

- 4 - 4 · x + 5 · x

- 4 · (x + 1) + 5 · x → Case 4

Second equation

- 7 + 6 · x - 4 · x + 3

(6 · x - 4 · x) + (- 7 + 3)

(6 - 4) · x - 4

2 · x - 4

2 · (x - 2) → Case 1

Third equation

9 · x - 2 · (3 · x - 3)

9 · x - 6 · x + 6

3 · x + 6

(2 + 1) · x + (14 - 8)

[1 - (- 2)] · x + (14 - 8)

(x + 14) - (8 - 2 · x) → Case 2

Fourth equation

- 3 · x + 6 + 4 · x

x + 6

(5 - 4) · x + (7 - 1)

(7 + 5 · x) + (- 4 · x - 1) → Case 5

Fifth equation

- 2 · x + 9 + 5 · x  + 6

3 · x + 15

3 · (x + 5) → Case 3

To learn more on algebraic equations: brainly.com/question/24875240

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Angles L and M are supplementary. What is the sum of their measures? The sum of the measures of angles L and M is
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Supplementary angles will always add up to 180. If you had a straight line, and you drew another line across that line, the two angles on either side of that line would add up to 180 because a flat line is equal to 180 degrees.
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2 years ago
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How many significant figures are in the number 12.001?
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What type of health screening would this patient most likely receive?
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Answer:

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Step-by-step explanation:

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8 0
3 years ago
Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units. What would be the perimeters of the
Dmitriy789 [7]

Answer:

The perimeter of other rectangles are 50, 28, 22 units

The area of all possible rectangles are same and equals to 24 unit²

Step-by-step explanation:

Given - Orbet used 24 square tiles to make a rectangle. The perimeter of the rectangle is 20 units.

To find -  What would be the perimeters of the other rectangles Orbet could make using only these 24 tiles? How do the areas of the rectangles compare?

Proof -

We know that

Are of Rectangle = Length × Breadth

And Perimeter of Rectangle = 2 (Length + Breadth )

Now,

Given that,

Orbet used 24 square tiles

And perimeter of the rectangle made = 20

So,

Possible Length of rectangle = 6

Possible breadth of Rectangle = 4

Or Vice-versa.

Total number of Rectangles possible = 4

Possibilities are -  1 × 24, 2 × 12, 3 × 8, 4 × 6

Case I :

Length of rectangle = 1

Breadth of rectangle = 24

∴ Perimeter of rectangle = 2(1 + 24) = 2(25) = 50 units

Area of rectangle = 1 × 24 = 24 unit²

Case II :

Length of rectangle = 2

Breadth of rectangle = 12

∴ Perimeter of rectangle = 2(2 + 12) = 2(14) = 28 units

Area of rectangle = 2 × 12 = 24 unit²

Case III :

Length of rectangle = 3

Breadth of rectangle = 8

∴ Perimeter of rectangle = 2(3 + 8) = 2(11) = 22 units

Area of rectangle = 3 × 8 = 24 unit²

Case IV :

Length of rectangle = 4

Breadth of rectangle = 6

∴ Perimeter of rectangle = 2(4 + 6) = 2(10) = 20 units

Area of rectangle = 4 × 6 = 24 unit²

∴ we get

The perimeter of other rectangles are 50, 28, 22 units

The area of all possible rectangles are same and equals to 24 unit²

5 0
3 years ago
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