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Determine which relation is a function. Question 13 options: a) {(3, 0), (– 2, – 2), (7, – 2), (– 2, 0)} b) c) y = 15x + 2 y = 1 5 x + 2 d)

User KenE
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1 Answer

6 votes

Answer:

x=3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5

Step-by-step explanation:Solving for x. Want to solve for y or solve for d instead?

1 Simplify 0-20−2 to -2−2.

3,-2,-27,-2-2,02y=1,5x+2d3,−2,−27,−2−2,02y=1,5x+2d

2 Simplify -2-2−2−2 to -4−4.

3,-2,-27,-4,02y=1,5x+2d3,−2,−27,−4,02y=1,5x+2d

3 Subtract 2d2d from both sides.

3-2d,-2-2d,-27-2d,-4-2d,02y-2d=1,5x3−2d,−2−2d,−27−2d,−4−2d,02y−2d=1,5x

4 Divide both sides by 1,51,5.

\frac{3-2d}{1},5,\frac{-2-2d}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x

​1

​3−2d

​​ ,5,

​1

​−2−2d

​​ ,5,

​1

​−27−2d

​​ ,5,

​1

​−4−2d

​​ ,5,

​1

​02y−2d

​​ ,5=x

5 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-4-2d}{1},5,\frac{02y-2d}{1},5=x

​1

​3−2d

​​ ,5,

​1

​−2(1+d)

​​ ,5,

​1

​−27−2d

​​ ,5,

​1

​−4−2d

​​ ,5,

​1

​02y−2d

​​ ,5=x

6 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{02y-2d}{1},5=x

​1

​3−2d

​​ ,5,

​1

​−2(1+d)

​​ ,5,

​1

​−27−2d

​​ ,5,

​1

​−2(2+d)

​​ ,5,

​1

​02y−2d

​​ ,5=x

7 Factor out the common term 22.

\frac{3-2d}{1},5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x

​1

​3−2d

​​ ,5,

​1

​−2(1+d)

​​ ,5,

​1

​−27−2d

​​ ,5,

​1

​−2(2+d)

​​ ,5,

​1

​2(y−d)

​​ ,5=x

8 Simplify \frac{3-2d}{1}

​1

​3−2d

​​ to (3-2d)(3−2d).

3-2d,5,\frac{-2(1+d)}{1},5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,

​1

​−2(1+d)

​​ ,5,

​1

​−27−2d

​​ ,5,

​1

​−2(2+d)

​​ ,5,

​1

​2(y−d)

​​ ,5=x

9 Simplify \frac{-2(1+d)}{1}

​1

​−2(1+d)

​​ to (-2(1+d))(−2(1+d)).

3-2d,5,-2(1+d),5,\frac{-27-2d}{1},5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,

​1

​−27−2d

​​ ,5,

​1

​−2(2+d)

​​ ,5,

​1

​2(y−d)

​​ ,5=x

10 Simplify \frac{-27-2d}{1}

​1

​−27−2d

​​ to (-27-2d)(−27−2d).

3-2d,5,-2(1+d),5,-27-2d,5,\frac{-2(2+d)}{1},5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,−27−2d,5,

​1

​−2(2+d)

​​ ,5,

​1

​2(y−d)

​​ ,5=x

11 Simplify \frac{-2(2+d)}{1}

​1

​−2(2+d)

​​ to (-2(2+d))(−2(2+d)).

3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,\frac{2(y-d)}{1},5=x3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,

​1

​2(y−d)

​​ ,5=x

12 Simplify \frac{2(y-d)}{1}

​1

​2(y−d)

​​ to (2(y-d))(2(y−d)).

3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5=x3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5=x

13 Switch sides.

x=3-2d,5,-2(1+d),5,-27-2d,5,-2(2+d),5,2(y-d),5x=3−2d,5,−2(1+d),5,−27−2d,5,−2(2+d),5,2(y−d),5

Done

User Klin
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