So sorry I replied so late but I'll leave the answer here just in case.
Answer:
7. Domain: [0,30] Range: [0,7]
8. f(20) = 4.67 or 4 2/3; After 20 seconds, the squirrel is 4.67 meters away from the camera.
9. x = 1.25 and x = 21.43; The quirrel is 4 meters away from the camera at 1.25 seconds and 21.43 seconds.
10. At 2.5 and 19.29 seconds
11. 10 seconds
12. The squirrel went to the back of the camera, away from its line of view.
Step-by-step explanation:
7. The line is between 0 and 30 on the x-axis and is between 0 and 7 on the y-axis.
8. The slope of the third part of the line is not a whole number, so you need to calculate it. The rise is -7 meters and run is 15 seconds, so the slope is -7/15. Plugin -7/15 into the slope-intercept form (y = mx+b) and the equation is f(x) = -7/15 x+14. Plugin x=20 into the equation:
f(20) = -7/15(20)+14
f(20) = -9.33+14
f(20) = 4.67
9. The first slope of the graph will also need to be calculated. Rise is 7-3 (4) and the run is 5. Slope = rise/run = 4/5 = 0.8. Plugin the slope into y = mx+b with the y-intercept, and the equation is f(x) = 0.8x+3. Replace f(x) with 4 and solve the equation.
4 = 0.8x+3
1 = 0.8x
x = 1.25
There is another x value that matches with f(x) = 4 in the third part of the graph, so we'll use the slope-intercept form for the previous question for this one.
4 = -7/15 x+14
-10 = -7/15 x
x = 21.43
10. The squirel is 5 meters away from the camera twice, so we'll be using the same equations for the previous question.
5 = 0.8x+3
2 = 0.8x
x = 2.5
5 = -7/15 x+14
-9 = -7/15 x
x = 19.29
11. The squirrel is 7 meters away between 5 and 15 seconds. 15-5 = 10.
12. The squirrel most likely went somehwere where the camera can't see it, and it's probably the back of it.
