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torisob [31]
4 years ago
10

Identify the domain and range of each relationship. Represent the relation with a mapping diagram. Is the relation a function?

Mathematics
1 answer:
Ivenika [448]4 years ago
4 0
Remember, to be a function, all elements in the domain should correspond to one element in the range. 
The domain is the x coordinate, range is the y coordinate.
The domain we can solve for as the set {4.2, 5, 7}. The range is the set {0, 1.5, 2.2, 4.8}.
But we see that an element in the domain corresponds to more than one element in the range, thus making it not a function.
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8.18

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Write 2 subtraction equations to show how to find 15-7
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So 15-7 is 8

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6 0
4 years ago
Please help with the questions in the image
Pavel [41]

Answer:   f(x) = (x - 3)²

<u>Step-by-step explanation:</u>

Use the vertex formula: y = a(x - h)² + k

and input the vertex (h, k) = (3, 0) and the given point (x, y) = (4, 2) to solve for a

2 = a(4 - 3)² + 0

2 = a(1)²

2 = a

Now, input (h, k) = (3, 0) and a = 1 into the vertex formula:

y = 1(x - 3)² + 0    →     y = (x - 3)²  

****************************************************************************

Answer:   g(x) = -(x + 1.25)² + 2.5625

<u>Step-by-step explanation:</u>

Use the vertex formula: y = a(x - h)² + k

and input the given points for (x, y) to create a system of equations, then solve for a, h, and k.

EQ1: 2 = a(-2 - h)² + k  

       2 = a(4 + 4h + h²) + k

       2 - 4a - 4ah - ah² = k


EQ2: 1 = a(0 - h)² + k

        1 = ah² + k

        1 - ah² = k


EQ3: -2.5 = a(1 - h)² + k

        -2.5 = a(1 - 2h + h²) + k

        -2.5 -a + 2h + ah² = k

<u>Substitute</u> - set EQ1 = EQ2 and EQ2 = EQ3 to eliminate k

EQ1 = EQ2:     2 - 4a - 4ah - ah² = 1 - ah²

                                 1 - 4a - 4ah = 0    

EQ2 = EQ3:     1 - ah² = -2.5 - a + 2ah - ah²

                                3.5 + a - 2ah = 0  

<u>Elimination:</u>  - now solve the system for "a"

 1 - 4a - 4ah = 0    →    1(1 - 4a - 4ah = 0)    →     1 - 4a - 4ah = 0

3.5 + a - 2ah = 0   →   -2(3.5 + a - 2ah = 0) →  <u> -7 - 2a + 4ah = 0 </u>

                                                                         -6 - 6a           = 0

                                                                              -6a            = 6

<h2>                                                     a     = -1</h2>

Next, replace "a" with -1 into either of the equations to solve for "h"

                                         1 - 4a - 4ah = 0

                                     1 - 4(-1) - 4(-1)h = 0

                                            1 + 4 + 4h = 0

                                                 5 + 4h = 0

                                                       4h = -5

<h2>                                     h = -1.25</h2>

Now, replace "a" with -1 and "h" with -1.25 into any of the original equations (EQ1, EQ2, or EQ3) to solve for k:

1 - ah² = k

1 - (-1)(-1.25)² = k

1 - (-1)(1.5625) = k

1 + 1.5625 = k

<h2> 2.5625 = k</h2>

Now, input (h, k) = (-1.25, 2.5625) and a = -1 into the vertex formula:

y = -1(x - (-1.25))² + 2.5625    →     y = -(x + 1.25)² + 2.5625

4 0
4 years ago
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