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The drama club is selling candy for a school fundraiser. Each club member starts with 140 pieces of candy to sell. The linear function [tex]$c(x) = -10x + 140$[/tex] models the number of pieces of candy remaining, [tex]$c(x)$[/tex], after a given number of days, [tex][tex]$x$[/tex][/tex]. What is the range of the function?

A. [tex][140, \infty][/tex]
B. [tex](-\infty, 140][/tex]
C. [tex](-\infty, \infty)[/tex]
D. [tex][0, 140][/tex]


Sagot :

Let's analyze the linear function [tex]\( c(x) = -10x + 140 \)[/tex] which models the number of pieces of candy remaining after [tex]\( x \)[/tex] days.

1. Interpret the function:
- The function [tex]\( c(x) = -10x + 140 \)[/tex] starts with 140 pieces of candy (when [tex]\( x = 0 \)[/tex]).
- As each day passes (increasing [tex]\( x \)[/tex]), the number of remaining pieces decreases by 10 pieces per day (since the coefficient of [tex]\( x \)[/tex] is -10).

2. Find the range of [tex]\( c(x) \)[/tex]:
- When [tex]\( x = 0 \)[/tex], the function value is:
[tex]\( c(0) = -10(0) + 140 = 140 \)[/tex]
- For any positive value of [tex]\( x \)[/tex]:
[tex]\( x > 0 \)[/tex], the term [tex]\(-10x\)[/tex] will be subtracted from 140, which means the function value will decrease.

3. Determine the lowest possible value of [tex]\( c(x) \)[/tex]:
- [tex]\( c(x) \)[/tex] will continue to decrease until it reaches zero. To find the point at which [tex]\( c(x) = 0 \)[/tex]:
[tex]\[ -10x + 140 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -10x = -140 \\ x = 14 \][/tex]
- Therefore, after 14 days, the number of pieces of candy will be zero.

4. Confirm the range:
- Before day 14, [tex]\( c(x) \)[/tex] will be a value between 0 and 140.
- After 14 days, the number of pieces of candy cannot become negative, meaning the lowest value [tex]\( c(x) \)[/tex] can take is 0.

5. Conclusion:
The range of the function [tex]\( c(x) \)[/tex] is from 0 to 140, inclusive.

Therefore, the correct answer is:
[tex]\[ [0, 140] \][/tex]
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