Question

wednesday -
additive or multiplicative?

thursday -
complete the table.

rides	tickets
3	9
5
	27

friday -
a package of 6 batteries costs $7.50. at this rate, how much would 4 batteries cost?

Explanation:

Wednesday: Determine Additive or Multiplicative

Step1: Analyze the graph's pattern

Check if the change is constant addition (additive) or constant multiplication (multiplicative). The graph is a straight line with a non - zero y - intercept. For additive, the rate of change (slope) is constant. Let's take two points. At \(x = 0\), \(y\approx3\); at \(x = 2\), \(y = 5\); the change in \(y\) is \(5 - 3=2\), change in \(x\) is \(2 - 0 = 2\), slope \(=\frac{2}{2}=1\). At \(x = 2\), \(y = 5\); at \(x = 4\), \(y = 7\), change in \(y = 7 - 5 = 2\), change in \(x=4 - 2=2\), slope \( = 1\). So it's additive (linear, constant rate of addition).

Step1: Find the relationship between Rides and Tickets

For \(x = 1\) (Rides), \(y = 3\) (Tickets); \(x = 3\), \(y = 9\). The ratio \(\frac{9}{3}=3\), \(\frac{3}{1}=3\). So the relationship is \(Tickets=3\times Rides\) (multiplicative relationship, \(y = 3x\)).

Step2: Find Tickets for 5 Rides

Using \(y = 3x\), when \(x = 5\), \(y=3\times5 = 15\).

Step3: Find Rides for 27 Tickets

Using \(y = 3x\), \(27=3x\), solve for \(x\): \(x=\frac{27}{3}=9\).

Step1: Find the cost per battery

A package of 6 batteries costs \(\$7.50\). So cost per battery \(=\frac{7.50}{6}\) dollars per battery. \(\frac{7.50}{6}=1.25\) dollars per battery.

Step2: Find the cost of 4 batteries

Cost of 4 batteries \(=4\times1.25\) dollars. \(4\times1.25 = 5.00\) dollars.

Answer:

Additive

Thursday: Complete the Table