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2 years ago

Which equation represents the data in

Mathematics

1 answer:

klasskru [66]2 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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3 years ago

Read 2 more answers

Consider the equation and the relation “(x, y) R (0, 2)”, where R is read as “has distance 1 of”. For example, “(0, 3) R (0, 2)”

Leviafan [203]

Answer:

The equation determine a relation between x and y

x = ± $\sqrt{1-(y-2)^{2}}$

y = ± $\sqrt{1-x^{2}}+2$

The domain is 1 ≤ y ≤ 3

The domain is -1 ≤ x ≤ 1

The graphs of these two function are half circle with center (0 , 2)

All of the points on the circle that have distance 1 from point (0 , 2)

Step-by-step explanation:

* Lets explain how to solve the problem

- The equation x² + (y - 2)² and the relation "(x , y) R (0, 2)", where

R is read as "has distance 1 of"

- This relation can also be read as “the point (x, y) is on the circle

of radius 1 with center (0, 2)”

- “(x, y) satisfies this equation , if and only if, (x, y) R (0, 2)”

* Lets solve the problem

- The equation of a circle of center (h , k) and radius r is

(x - h)² + (y - k)² = r²

∵ The center of the circle is (0 , 2)

∴ h = 0 and k = 2

∵ The radius is 1

∴ r = 1

∴ The equation is ⇒ (x - 0)² + (y - 2)² = 1²

∴ The equation is ⇒ x² + (y - 2)² = 1

∵ A circle represents the graph of a relation

∴ The equation determine a relation between x and y

* Lets prove that x=g(y)

- To do that find x in terms of y by separate x in side and all other

in the other side

∵ x² + (y - 2)² = 1

- Subtract (y - 2)² from both sides

∴ x² = 1 - (y - 2)²

- Take square root for both sides

∴ x = ± $\sqrt{1-(y-2)^{2}}$

∴ x = g(y)

* Lets prove that y=h(x)

- To do that find y in terms of x by separate y in side and all other

in the other side

∵ x² + (y - 2)² = 1

- Subtract x² from both sides

∴ (y - 2)² = 1 - x²

- Take square root for both sides

∴ y - 2 = ± $\sqrt{1-x^{2}}$

- Add 2 for both sides

∴ y = ± $\sqrt{1-x^{2}}+2$

∴ y = h(x)

- In the function x = ± $\sqrt{1-(y-2)^{2}}$

∵ $\sqrt{1-(y-2)^{2}}$ ≥ 0

∴ 1 - (y - 2)² ≥ 0

- Add (y - 2)² to both sides

∴ 1 ≥ (y - 2)²

- Take √ for both sides

∴ 1 ≥ y - 2 ≥ -1

- Add 2 for both sides

∴ 3 ≥ y ≥ 1

∴ The domain is 1 ≤ y ≤ 3

- In the function y = ± $\sqrt{1-x^{2}}+2$

∵ $\sqrt{1-x^{2}}$ ≥ 0

∴ 1 - x² ≥ 0

- Add x² for both sides

∴ 1 ≥ x²

- Take √ for both sides

∴ 1 ≥ x ≥ -1

∴ The domain is -1 ≤ x ≤ 1

* The graphs of these two function are half circle with center (0 , 2)

* All of the points on the circle that have distance 1 from point (0 , 2)

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scoundrel [369]

Step-by-step explanation:

An arithmetic sequence is given by relation as follows :

$A(n)=-6+(n-1)\dfrac{1}{5}$

For the first term, put n = 1. So,

$A(1)=-6+(1-1)\dfrac{1}{5}\\\\A(1)=-6$

For fourth term, put n = 4. So,

$A(4)=-6+(4-1)\dfrac{1}{5}\\\\A(4)=\dfrac{-27}{5}\\\\A(4)=-5\dfrac{2}{5}$

For tenth term, put n = 10. So,

$A(10)=-6+(10-1)\dfrac{1}{5}\\\\A(10)=-6+\dfrac{9}{5}\\\\A(10)=\dfrac{-21}{5}\\\\A(10)=-4\dfrac{1}{5}$

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Step-by-step explanation:

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