Consuelo is buying flavored coffee and plain coffee. The total amount of money she spends on coffee is T = 5.50p + 7f, where "p" represents the cost of a package of plain coffee and "f" represents the cost of a package of flavored coffee.
a. If Consuelo has $40 to spend on coffee, describe the constraints on the formula.
b. Solve for f.
c. If Consuelo needs to buy 3 packages of plain coffee, what is the maximum number of packages of flavored coffee she can buy?

The constraints on the formula are the costs of plain coffee and cost of flavored coffee
The equation of f is: [tex]\mathbf{f = \frac{40 -5.50p}7 }[/tex].

The function is given as:

[tex]\mathbf{T = 5.50p + 7f}[/tex]

(a) The constraints when T = 40

Substitute 40 for T in [tex]\mathbf{T = 5.50p + 7f}[/tex]

[tex]\mathbf{5.50p + 7f = 40}[/tex]

The variables in the above equation are p and f

Hence, the constraints are the cost of a package of plain coffee and the cost of a package of flavored coffee

(b) Solve for f

In (a), we have:

[tex]\mathbf{5.50p + 7f = 40}[/tex]

Subtract both sides by 5.50p

[tex]\mathbf{7f = 40 -5.50p }[/tex]

Divide both sides by 7

[tex]\mathbf{f = \frac{40 -5.50p}7 }[/tex]

Hence, the equation of f is:

[tex]\mathbf{f = \frac{40 -5.50p}7 }[/tex]

(c) The maximum number of flavored coffee when she buys 3 packages of plain coffee

In (b), we have:

[tex]\mathbf{f = \frac{40 -5.50p}7 }[/tex]

Substitute 3 for p

[tex]\mathbf{f = \frac{40 -5.50 \times 3}7 }[/tex]

[tex]\mathbf{f = \frac{40 -16.50}7 }[/tex]

[tex]\mathbf{f = \frac{23.5}7 }[/tex]

Divide

[tex]\mathbf{f = 3.36 }[/tex]

Remove the decimal part (do not approximate)

[tex]\mathbf{f = 3 }[/tex]

Hence, the maximum number of packages of flavored coffee she can buy is 3

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