Question

Which relationships have the same constant of proportionality between yyy and xxx as the equation y=\dfrac{1}{2}xy= 2 1 ​ xy, equals, start fraction, 1, divided by, 2, end fraction, x? Choose 3 answers: Choose 3 answers:

Answer 1

A,B, and E

Explanation:

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Answer 2

A & B

Explanation:

See attachment for complete question

Given

`y = (1)/(2)x`

Analyzing the given options

Option A:

`6y = 3x`

Divide both sides by 6

`(6y)/(6) = (3x)/(6)`

`y = (3x)/(6)`

`y = (1)/(2)x`

This is true for the given expression:

Option B:

From the graph:

x = 2, when y = 1

x = 4, when y = 2

Solving for the slope:

`m = (y_2 - y_1)/(x_2 - x_1)`

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

`m = (1)/(2)`

The equation is then calculated as:

`y - y_1 = m(x - x_1)`

`y - 1 = (1)/(2)(x - 2)`

`y - 1 = (1)/(2)x - 1`

Add 1 to both sides

`y - 1 +1 = (1)/(2)x - 1 + 1`

`y = (1)/(2)x`

This is also true for the given expression.

Option C:

From the graph:

x = 2, when y = 4

x = 4, when y = 8

Solving for the slope:

`m = (y_2 - y_1)/(x_2 - x_1)`

`m = (8 - 4)/(4 - 2)`

`m = (4)/(2)`

`m =2`

The equation is then calculated as:

`y - y_1 = m(x - x_1)`

`y - 4 = 2(x - 2)`

`y - 4 = 2x - 4`

Add 4 to both sides

`y - 4 + 4= 2x - 4 + 4`

`y = 2x`

This isn't true for the given expression.

Option (D):

From the table:

x = 2, when y = 1

x = 4 when y = 3

Solving for the slope:

`m = (y_2 - y_1)/(x_2 - x_1)`

`m = (3 - 1)/(4 - 2)`

`m = (2)/(2)`

`m = 1`

The equation is then calculated as:

`y - y_1 = m(x - x_1)`

`y - 1 = 1 * (x - 2)`

`y - 1 = x - 2`

`y = x - 2 +1`

`y = x - 1`

This isn't true for the given expression.