We have a graph that relates the number of games with the total yearly cost.
The graph only shows results for up to 6 games, so we have to find the equation of the line to extrapolate it to 10 games.
We can find the equation of the line in slope-intercept form:
![y=mx+b](https://img.qammunity.org/2023/formulas/mathematics/high-school/smsb8cbft03lwblmi49nf2l6jby2ofxzws.png)
where y is the yearly total cost and x is the number of games.
The y-intercept b is the value of y when x=0. We can see it in the graph: when x=0 (no games), the yearly cost is $90.
So b=y(0)=90.
We can find the slope in many ways now, but we will find it by replacing x and y with known values of a point of the graph, like (x,y)=(1,120), that is the point that indicates that 1 game (x=1) costs a total $120 (y=120).
![\begin{gathered} y=mx+b \\ 120=m\cdot1+90 \\ m=120-90 \\ m=30 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/h4569u02w7f7gfazwcacke63w8vhctb0b5.png)
The slope has a value of m=30.
Then, we have the equation as:
![y=30x+90](https://img.qammunity.org/2023/formulas/mathematics/college/8kfq1d0pt8n8vq2bw3aw2jgqqjw4jhyoo3.png)
We can find the cost of joining and playing 10 games by replacing x with 10 and calculating for y:
![\begin{gathered} y(10)=30(10)+90 \\ y(10)=300+90 \\ y(10)=390 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/6hvclpt9md5tqux3si84nel7odgwy4udoa.png)
Answer: $390