Given a coordinate point (x, y), the first value of the point represents the value on the x-axis while the second value represent the value on the y-axis.
1.) To express the values (-4, -1), (-1, 2), (1, -4), (2, -3), (4, 3) as a table, we have:
x y
-4 -1
-1 2
1 -4
2 -3
4 3
The values (-4, -1), (-1, 2), (1, -4), (2, -3), (4, 3) expressed as a graph have been attached as graph_1
To express the values (-4, -1), (-1, 2), (1, -4), (2, -3), (4, 3) as a mapping, we have two circles with one labelled x and the other one labelled y.
Inside the circle labelled x are the numbers -4, -1, 1, 2, 4 written vertically and inside the circle labelled y are the numbers -4, -3, -1, 2, 3 written vertically.
There are lines joining from the circle labelled x to the circle labelled y with line joining -4 in circle x to -1 in circle y, -1 in circle x to 2 in circle y, 1 in circle x to -4 in circle y, 2 in circle x to -3 in circle y, 4 in circle x to 3 in circle y.
The domain of the relation is the set of the x-values of the relation, i.e. domain is {-4, -1, 1, 2, 4}.
The range of the relation is the set of the y-values of the relation, i.e. range is {-4, -3, -1, 2, 3}
2.) To express the values (-2, 1), (-1, 0), (1, 2), (2, -4), (4, 3) as a table, we have:
x y
-2 1
-1 0
1 2
2 -4
4 3
The values (-2, 1), (-1, 0), (1, 2), (2, -4), (4, 3) expressed as a graph have been attached as graph_2
To express the values (-2, 1), (-1, 0), (1, 2), (2, -4), (4, 3) as a mapping, we have two circles with one labelled x and the other one labelled y.
Inside
the circle labelled x are the numbers -2, -1, 1, 2, 4 written
vertically and inside the circle labelled y are the numbers -4, 0, 1, 2, 3 written vertically.
There are lines joining from the circle labelled x to the circle labelled y with a line joining -2 in circle x to 1 in circle y, -1 in circle x to 0 in circle y, 1 in circle x to 2 in circle y, 2 in circle x to -4 in circle y, 4 in circle x to 3 in circle y.
The domain of the relation is the set of the x-values of the relation, i.e. domain is {-2, -1, 1, 2, 4}.
The range of the relation is the set of the y-values of the relation, i.e. range is {-4, 0, 1, 2, 3}
Answer: Here is the complete table, with the filled in values:
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Time (h) Distance (mi)
3 2
9 6
12 8
18 12
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Explanation:
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Let us begin by obtaining the "?" value; that is, the "distance" (in "mi.") ;
when the time (in "h") is "18" ;
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12/8 = 18/?
Note: "12/8 = (12÷4) / (8÷4) = 3/2 ;
Rewrite: 3/2 = 18/? ; cross-multiply: 3*? = 2 * 18 ;
3*? = 36 ;
Divide each side by "3" ;
The "?" = 36/3 = 12 ;
So, 12/8 = 18/12 ;
The value: "12" takes the place for the "?" in the table for "distance (in "mi.);
when the "time" (in "h") is "18".
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Now, let us obtain the "? " value for the "distance" (in "mi.");
when the "time" (in "h") is: "9" .
12/8 = 9/? ; Solve for "?" ;
We know (see aforementioned) that "12/8 = 3/2" ;
So, we can rewrite: 3/2 = 9/? ; Solve for "?" ;
Cross-multiply: 3* ? = 2* 9 ; 3* ? = 18 ;
Divide each side by "3" ;
to get: "6" for the "?" value.
When the time (in "h") is "9", the distance (in "mi.") is "6" .
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Now, to solve the final "?" value in the table given.
9/6 = ?/2 ; Note: We get the "6" from our "calculated value" (see above problem).
9/6 = (9÷3) / (6÷3) = 3/2 ;
So, we know that the "?" value is: "3" .
Alternately: 9/6 = ?/2 ;
Cross-multiply: 6*? = 2*9 ; 6 * ? = 18 ; Divide each side by "6" ;
to find the value for the "?" ;
"?" = 18/6 = "3" .
When the "distance" (in "mi.") is: "2" ; the time (in "h") is: "3" .
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Here is the complete table—with all the values filled in:
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<span>Time (h) Distance (mi)
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3 2
9 6
12 8
18 12
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