Given that Angelo spends the same amount everyday from the amount in
the lunch card, the function of the amount remaining is a linear function.
- The constant rate of change of the function is; <u>-5.25</u>
- The two ordered pair used to find the constant rate of change are; <u>(1, 44.75) and (2, 39.5)</u>
Reasons:
The amount Angelo's mother put on the lunch card = $50
A possible table of values to the question is presented as follows;
![\begin{tabular}{r|c|c|c|c|}Days&0&1&2&3\\Money Remaining&&44.75&39.5&\end{array}\right]](https://tex.z-dn.net/?f=%5Cbegin%7Btabular%7D%7Br%7Cc%7Cc%7Cc%7Cc%7C%7DDays%260%261%262%263%5C%5CMoney%20Remaining%26%2644.75%2639.5%26%5Cend%7Barray%7D%5Cright%5D)
Required:
The constant rate of the function that gives the amount remaining from the
amount Angelo's mother put on his lunch card.
Solution:
The two ordered pair that can be used to find the slope or constant rate of change are;
(x₁, y₁) = (1, 44.75), and (x₂, y₂) = (2, 39.5)
With the above two ordered pairs, we have the constant rate of change of the function given as follows;
![Constant \ rate \ of \ change\ (The \ slope) =\dfrac{39.5-44.75}{2-1} = \mathbf{ -5.25}](https://tex.z-dn.net/?f=Constant%20%5C%20rate%20%5C%20of%20%5C%20change%5C%20%20%28The%20%5C%20slope%29%20%3D%5Cdfrac%7B39.5-44.75%7D%7B2-1%7D%20%3D%20%5Cmathbf%7B%20-5.25%7D)
The constant rate of change for the function that gives the amount remaining in the lunch card is; <u>-5.25</u>
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