QUESTION IMAGE
Question
select all the correct locations on the graph.
oliver is riding his bike to visit some friends. every two minutes, he bikes 1,760 feet. identify the pair of points you can use to create the line representing how far he has traveled in terms of time spent biking.
Step1: Determine the rate
Oliver bikes 1,760 feet every 2 minutes. So the rate is $\frac{1760}{2} = 880$ feet per minute.
Step2: Analyze the points
- The origin (0, 0) represents 0 time and 0 distance.
- For time $t = 2$ minutes, distance $d = 1760$ feet, so the point (2, 1760) should be on the line.
- For time $t = 4$ minutes, distance $d = 1760\times2 = 3520$ feet? Wait, no, wait. Wait, every 2 minutes 1760 feet. So at $t = 2$, $d = 1760$; at $t = 4$, $d = 1760\times2 = 3520$? Wait, no, looking at the graph, the y - axis has 0, 880, 1760, 2640, 3520, 4400, 5280. Wait, maybe my initial rate calculation was wrong. Wait, if at $t = 2$, $d = 1760$, then the slope (rate) is $\frac{1760}{2}=880$ feet per minute. So at $t = 0$, $d = 0$ (point (0,0)). At $t = 2$, $d = 1760$ (point (2,1760)). At $t = 4$, $d = 880\times4 = 3520$? But in the graph, there are points. Wait, the red points: one at (0,0), then at x = 1, y = 1760? No, wait the graph: x - axis is time (minutes), y - axis is distance (feet). Wait, the points: (0,0), (2,1760), (4,3520)? Wait, but in the graph, there are red dots. Let's re - examine. The problem says "every two minutes, he bikes 1,760 feet". So the relationship is linear, with slope (rate) $m=\frac{1760}{2}=880$ feet per minute. So the equation is $d = 880t$, where $d$ is distance and $t$ is time. So when $t = 0$, $d = 0$ (point (0,0)). When $t = 2$, $d = 880\times2 = 1760$ (point (2,1760)). When $t = 4$, $d = 880\times4 = 3520$? Wait, but in the graph, there are points: (0,0), then at x = 1, y = 1760? No, maybe the x - axis is in 2 - minute intervals? Wait, no, the x - axis is labeled 0,1,2,3,4,5,6,7. Wait, maybe the rate is 1760 feet every 2 minutes, so per minute it's 880 feet. So the correct points should satisfy $d = 880t$. So (0,0) and (2,1760) are on the line, or (0,0) and (4,3520), but looking at the graph, the points: (0,0), (2,1760), (4,3520) – but in the graph, there are red dots. Wait, the user's graph: y - axis: 0, 880, 1760, 2640, 3520, 4400, 5280. x - axis: 0,1,2,3,4,5,6,7. The red dots: one at (0,0), then at x = 1, y = 1760? No, that can't be. Wait, maybe the time is in 2 - minute increments? No, the x - axis is labeled 0,1,2,3,4,5,6,7, so minutes. Wait, maybe I made a mistake. Let's think again. The problem says "every two minutes, he bikes 1,760 feet". So the change in time $\Delta t = 2$ minutes, change in distance $\Delta d=1760$ feet. So the slope $m=\frac{\Delta d}{\Delta t}=\frac{1760}{2} = 880$ feet per minute. So the equation is $d = 880t$. So when $t = 0$, $d = 0$ (point (0,0)). When $t = 2$, $d = 880\times2=1760$ (point (2,1760)). When $t = 4$, $d = 880\times4 = 3520$ (point (4,3520)). But in the graph, there are red dots: (0,0), (2,1760), (4,3520) – but looking at the graph, the red dots: one at (0,0), then at x = 2, y = 1760; x = 4, y = 3520? Wait, but in the user's graph, there are red dots: (0,0), then at x = 1, y = 1760? No, maybe the x - axis is in 2 - minute units? No, the labels are 0,1,2,3,4,5,6,7. Wait, maybe the correct points are (0,0) and (2,1760), because every 2 minutes 1760 feet. So the pair of points is (0,0) and (2,1760), or (2,1760) and (4,3520), but (0,0) is a clear point (the origin) and (2,1760) is a point with y = 1760 and x = 2.
Step3: Identify the correct points
From the graph, the origin (0,0) and the point (2,1760) should be on the line, as well as (4,3520) etc. But based on the graph's red dots, the points that satisfy $d = 880t$ are (0,0) (since at t = 0, d = 0) and (2,1760) (since at t = 2, d = 1760), and also (4,3520) but let's check the graph. The y -…
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The correct points are (0, 0) and (2, 1760) (and also (4, 3520) etc., but the key pair based on the 2 - minute interval and 1760 - foot distance is (0, 0) and (2, 1760)). So on the graph, select the point at (0, 0) (the origin) and the point at (2, 1760).