It is a bit tedious to write 6 equations, but it is a straightforward process to substitute the given point values into the form provided.
For segment ab. (x1, y1) = (1, 1); (x2, y2) = (3, 4).
... x = 1 + t(3-1)
... y = 1 + t(4-1)
ab = {x=1+2t, y=1+3t}
For segment bc. (x1, y1) = (3, 4); (x2, y2) = (1, 7).
... x = 3 + t(1-3)
... y = 4 + t(7-4)
bc = {x=3-2t, y=4+3t}
For segment ca. (x1, y1) = (1, 7); (x2, y2) = (1, 1).
... x = 1 + t(1-1)
... y = 7 + t(1-7)
ca = {x=1, y=7-6t}
Answer:
24 mph for both 60 miles and 40 miles.
Step-by-step explanation:
We need to find how long it takes to get to college and from it.
Speed x time (x) = Distance (60)
30x = 60
x=2
It takes two hours to get to college
20x = 60
x = 3
It takes 3 hours to get back. It's a total of 5 hours, 2 hours at 30mph and 3 at 20. Now we need to find the average
30+30+20+20+20=120
120/5= 24
The average speed for the round trip is 24.
To do the second part, we follow the same process, but replace 60 with 40 for distance.
30x=40
x= 1.33
20x=40
x=2
We can add the total distance and divide by total time to find the average.
The total distance is 80. Time is 3.33
80/3.33=24
The average speed is still 24.
Answer:
ADE=59º
Step-by-step explanation:
4x+15+13x+7=90
17x+22=90
17x=68
x=4
angle B=angle ADE so we plug in x=4 into 13x+7
13(4)+7=59º
You would fill in 9 for x so your equation would be 2(9)+3 and you would then get 21. Therefore the answer is f(x)=21
Any second part or anything else to look at?
