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Answer :
We are given that the function for the total weight when [tex]$x$[/tex] pairs of plates are added to the 4 lb barbell is
[tex]$$
f(x) = 20x + 4.
$$[/tex]
Since there are 6 pairs of weight plates available, the possible number of pairs [tex]$x$[/tex] can be any whole number from [tex]$0$[/tex] to [tex]$6$[/tex]. Let’s compute [tex]$f(x)$[/tex] for each value of [tex]$x$[/tex]:
1. For [tex]$x = 0$[/tex]:
[tex]$$
f(0) = 20(0) + 4 = 4.
$$[/tex]
2. For [tex]$x = 1$[/tex]:
[tex]$$
f(1) = 20(1) + 4 = 20 + 4 = 24.
$$[/tex]
3. For [tex]$x = 2$[/tex]:
[tex]$$
f(2) = 20(2) + 4 = 40 + 4 = 44.
$$[/tex]
4. For [tex]$x = 3$[/tex]:
[tex]$$
f(3) = 20(3) + 4 = 60 + 4 = 64.
$$[/tex]
5. For [tex]$x = 4$[/tex]:
[tex]$$
f(4) = 20(4) + 4 = 80 + 4 = 84.
$$[/tex]
6. For [tex]$x = 5$[/tex]:
[tex]$$
f(5) = 20(5) + 4 = 100 + 4 = 104.
$$[/tex]
7. For [tex]$x = 6$[/tex]:
[tex]$$
f(6) = 20(6) + 4 = 120 + 4 = 124.
$$[/tex]
Thus, the range of the function, which is the set of all possible total weights, is
[tex]$$
\{4, 24, 44, 64, 84, 104, 124\}.
$$[/tex]
This corresponds to option B.
[tex]$$
f(x) = 20x + 4.
$$[/tex]
Since there are 6 pairs of weight plates available, the possible number of pairs [tex]$x$[/tex] can be any whole number from [tex]$0$[/tex] to [tex]$6$[/tex]. Let’s compute [tex]$f(x)$[/tex] for each value of [tex]$x$[/tex]:
1. For [tex]$x = 0$[/tex]:
[tex]$$
f(0) = 20(0) + 4 = 4.
$$[/tex]
2. For [tex]$x = 1$[/tex]:
[tex]$$
f(1) = 20(1) + 4 = 20 + 4 = 24.
$$[/tex]
3. For [tex]$x = 2$[/tex]:
[tex]$$
f(2) = 20(2) + 4 = 40 + 4 = 44.
$$[/tex]
4. For [tex]$x = 3$[/tex]:
[tex]$$
f(3) = 20(3) + 4 = 60 + 4 = 64.
$$[/tex]
5. For [tex]$x = 4$[/tex]:
[tex]$$
f(4) = 20(4) + 4 = 80 + 4 = 84.
$$[/tex]
6. For [tex]$x = 5$[/tex]:
[tex]$$
f(5) = 20(5) + 4 = 100 + 4 = 104.
$$[/tex]
7. For [tex]$x = 6$[/tex]:
[tex]$$
f(6) = 20(6) + 4 = 120 + 4 = 124.
$$[/tex]
Thus, the range of the function, which is the set of all possible total weights, is
[tex]$$
\{4, 24, 44, 64, 84, 104, 124\}.
$$[/tex]
This corresponds to option B.
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