Answer:
Step-by-step explanation:
1. slope=-3/2-3 is at the top,2 is at the bottom
y intercept=4-The y intercepts or goes past at 4,
equation=y=-3/2+4-We put it all together and we get y=-3/2+4!
2. slope=-3/2-It starts at 3 again,and ends at two again.
y intercept=0-The y touches the line at 0,or the middle
equation=y=-3/2-y equals the slope and we subtract y.
Answer:
a.) Between 0.5 and 3 seconds.
Step-by-step explanation:
So I just went ahead and graphed this quadratic on Desmos so you could have an idea of what this looks like. A negative quadratic, and we're trying to find when the graph's y-values are greater than 26.
If you look at the graph, you can easily see that the quadratic crosses y = 26 at x-values 0.5 and 3. And, you can see that the quadratic's graph is actually above y = 26 between these two values, 0.5 and 3.
Because we know that the quadratic's graph models the projectile's motion, we can conclude that the projectile will also be above 26 feet between 0.5 and 3 seconds.
So, the answer is a.) between 0.5 and 3 seconds.
Answer:
y = -40x+2600
Step-by-step explanation:
Writing the values as coordinate (m, s)
If after flying for 11 seconds, he is 2160 meters from her, the required coordinate is (11, 2160)
Then at 19 seconds he is 1840 meters, the required coordinate is (19, 1840)
Get the slope;
m = 2160-1840/11-19
m = 320/-8
m = -40
Substitute m = -40 and (11, 2160) into the equations;
y-y0 = m(x-x0)
y -2160 = -40(x-11)
y - 2160 = -40x+440
y = -40x+440 + 2160
y = -40x+2600
Hence the requires equation is y = -40x+2600