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Nady [450]
3 years ago
15

Which equation represents the data in

Mathematics
1 answer:
klasskru [66]3 years ago
4 0

Answer:

y = 2x - 4

Step-by-step explanation:

The equations are put in slope intercept form

Slope intercept form: y = mx + b

Where m = slope and b = y intersect

So in order to find the equation of the data represented by the table we will have to find the slope and y intercept

Let's begin!

First let's find the slope

We can find the slope by using the slope formula

m = (y2 - y1) / (x2 - x1) where the x and y values are derived from coordinates from the table

The points chosen may vary but I have chosen the points (0,-4) and (1,-2)

Now that we have chosen the points we will use to find the slope let's define the variables

remember coordinates are written like this: (x,y)

The x value of the second coordinate is 1 so x2 = 1

The x value of the first coordinate is 0

So x1 = 0

The y value of the second coordinate is -2 so y2 = -2

The y value of the first coordinate is -4

So y1 = -4

Now that we have defined each variable let's plug in the values into the formula

Formula: m = (y2 - y1) / (x2 - x1)

Variables: x2 = 1, x1 = 0, y2 = -2, y1 = -4

Substitute values

m = (-2 - (-4) / ( 1 - 0 )

Evaluate

The negative signs cancel out on top and it changes to +4

m = (-2 + 4)/(1-0)

Add top values

m = 2/(1-0)

Subtract bottom numbers

m = 2/1

Simplify fraction

m = 2

So we can conclude that the slope (m) = 2

Now let's find the y intercept or "b"

The y intercept is the value of y when x = 0

If you look at the table when x = 0 y = -4 meaning that the y intercept or "b" is -4

Now that we have found everything let's find the equation of the data represented by the table

The equation is in slope intercept form

y = mx + b

Define variables

m = 2 and b = -4

Substitute values

y = 2x - 4

The equation is y = 2x - 4

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step-by-step explanation:

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y = -1/2x + 4

Step-by-step explanation:

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It also crosses the y axis at 4,

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2 years ago
The Cartesian coordinates of a point are given. (a) (−5, 5) (i) Find polar coordinates (r, θ) of the point, where r > 0 and 0
Alex73 [517]

Answer:

a)

(i) The coordinates of the point in polar form is (5√2 , 7π/4)

(ii) The coordinates of the point in polar form is (-5√2 , 3π/4)

b)

(i) The coordinates of the point in polar form is (6 , π/3)

(ii) The coordinates of the point in polar form is (-6 , 4π/3)

Step-by-step explanation:

* Lets study the meaning of polar form

- To convert from Cartesian Coordinates (x,y) to Polar Coordinates (r,θ):

1. r = √( x2 + y2 )

2. θ = tan^-1 (y/x)

- In Cartesian coordinates there is exactly one set of coordinates for any

 given point

- In polar coordinates there is literally an infinite number of coordinates

 for a given point

- Example:

- The following four points are all coordinates for the same point.

# (5 , π/3) ⇒ 1st quadrant

# (5 , −5π/3) ⇒ 4th quadrant

# (−5 , 4π/3) ⇒ 3rd quadrant

# (−5 , −2π/3) ⇒ 2nd quadrant

- So we can find the points in polar form by using these rules:

 [r , θ + 2πn] , [−r , θ + (2n + 1) π] , where n is any integer

 (more than 1 turn)

* Lets solve the problem

(a)

∵ The point in the Cartesian plane is (-5 , 5)

∵ r = √x² + y²

∴ r = √[(5)² + (-5)²] = √[25 + 25] = √50 = ±5√2

∵ Ф = tan^-1 (y/x)

∴ Ф = tan^-1 (5/-5) = tan^-1 (-1)

- Tan is negative in the second and fourth quadrant

∵ 0 ≤ Ф < 2π

∴ Ф = 2π - tan^-1(1) ⇒ in fourth quadrant r > 0

∴ Ф = 2π - π/4 = 7π/4

OR

∴ Ф = π - tan^-1(1) ⇒ in second quadrant r < 0

∴ Ф = π - π/4 = 3π/4

(i) ∵ r > 0

∴ r = 5√2

∴ Ф = 7π/4 ⇒ 4th quadrant

∴ The coordinates of the point in polar form is (5√2 , 7π/4)

(ii) r < 0

∴ r = -5√2

∵ Ф = 3π/4 ⇒ 2nd quadrant

∴ The coordinates of the point in polar form is (-5√2 , 3π/4)

(b)

∵ The point in the Cartesian plane is (3 , 3√3)

∵ r = √x² + y²

∴ r = √[(3)² + (3√3)²] = √[9 + 27] = √36 = ±6

∵ Ф = tan^-1 (y/x)

∴ Ф = tan^-1 (3√3/3) = tan^-1 (√3)

- Tan is positive in the first and third quadrant

∵ 0 ≤ Ф < 2π

∴ Ф = tan^-1 (√3) ⇒ in first quadrant r > 0

∴ Ф = π/3

OR

∴ Ф = π + tan^-1 (√3) ⇒ in third quadrant r < 0

∴ Ф = π + π/3 = 4π/3

(i) ∵ r > 0

∴ r = 6

∴ Ф = π/3 ⇒ 1st quadrant

∴ The coordinates of the point in polar form is (6 , π/3)

(ii) r < 0

∴ r = -6

∵ Ф = 4π/3 ⇒ 3rd quadrant

∴ The coordinates of the point in polar form is (-6 , 4π/3)

6 0
3 years ago
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