QUESTION IMAGE
Question
patterns in proportional relationships
lesson 3: comparing proportional and non - proportional relationships
- cheese pizzas cost $7.00 each, and there is a one - time charge of $5.00 to have your
pizzas delivered. answer the following questions about this situation.
a. complete the table and graph for this pizza and delivery charge relationship.
b. find an algebraic rule that represents
the relationship between the number
of pizzas, x, and the total cost, y.
- reinforce use your work in question 3 to answer the following questions.
d. how much would it cost to buy 10 pizzas and have the pizzas delivered?
e. if you have exactly $75 to spend in this situation, how many pizzas can you order?
f. what if you buy 0 pizzas? how much would that cost? does your answer make sense?
why or why not?
Step1: Complete the table
- For \(x = 1\) (number of pizzas):
The cost of pizzas is \(7\times1=\$7\), and with a delivery - charge of \(\$5\), the total cost \(y=7\times1 + 5=\$12\).
- For \(x = 2\) (number of pizzas):
The cost of pizzas is \(7\times2=\$14\), and with a delivery - charge of \(\$5\), the total cost \(y=7\times2+5=\$19\).
- For \(x = 3\) (number of pizzas):
The cost of pizzas is \(7\times3=\$21\), and with a delivery - charge of \(\$5\), the total cost \(y=7\times3 + 5=\$26\).
- For \(x = 4\) (number of pizzas):
The cost of pizzas is \(7\times4=\$28\), and with a delivery - charge of \(\$5\), the total cost \(y=7\times4+5=\$33\).
Step2: Find the algebraic rule
The cost of \(x\) pizzas is \(7x\) (since each pizza costs \(\$7\)), and there is a fixed delivery charge of \(\$5\). So the linear equation representing the relationship between the number of pizzas \(x\) and the total cost \(y\) is \(y = 7x+5\).
Step3: Cost for 10 pizzas
Substitute \(x = 10\) into the equation \(y=7x + 5\).
\(y=7\times10+5=70 + 5=\$75\).
Step4: Number of pizzas for \(y = 75\)
Set \(y = 75\) in the equation \(y=7x+5\).
\(75=7x+5\).
Subtract 5 from both sides: \(75 - 5=7x\), so \(70=7x\).
Divide both sides by 7: \(x=\frac{70}{7}=10\).
Step5: Cost for \(x = 0\)
Substitute \(x = 0\) into the equation \(y=7x+5\).
\(y=7\times0+5=\$5\).
This does not make sense in the context of buying pizzas, because if you buy \(0\) pizzas, you should not be charged for the pizzas. The delivery - charge is a one - time charge for having pizzas delivered, but if there are no pizzas, the delivery - charge conceptually should not apply in the context of pizza - buying.
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- Table:
| Number of pizzas | Process | Cost in dollars |
|---|---|---|
| 2 | \(7\times2 + 5\) | 19 |
| 3 | \(7\times3+5\) | 26 |
| 4 | \(7\times4+5\) | 33 |
- Algebraic rule: \(y = 7x+5\)
- Cost for 10 pizzas: \(\$75\)
- Number of pizzas for \(\$75\): \(10\)
- Cost for 0 pizzas: \(\$5\), and it does not make sense (explanation as above).