1.4t−0.4(t−3.1)=5.8 SOLVE FOR Y
1 answer:
You might be interested in
Answer:
Please find attached the graph of the following function;
Step-by-step explanation:
We note that the function is linear from x = 2 to just before x = 0
The linear relationship of the function f(x) with x changes just before x = 0
At x = 0, the value of f(x) is indicated as 1
From just after x = 0, the function is a straight horizontal line y = 3
The function also changes value immediately after x = 0 to the line y = 3
The areas where the function is defined are shown in continuous lines
3:5 ratio mean there are 8 "parts" (3 + 5 = 8)
The distance between the x-coordinates is |-3 - 5| = 8.
So each "part" is 8/8 = 1 unit long in the x-direction.
You want I to be 3(1) = 3 units from D, so the x-coordinate of I is -3 + 3 = 0.
Same deal for y.
|2 - 5| = 3 is the distance between D and E
Each part is 3/8.
3(3/8) = 9/8
2 + 9/8 = 25/8
So the point I is (0, 25/8)
The distance between the x-coordinates is |-3 - 5| = 8.
So each "part" is 8/8 = 1 unit long in the x-direction.
You want I to be 3(1) = 3 units from D, so the x-coordinate of I is -3 + 3 = 0.
Same deal for y.
|2 - 5| = 3 is the distance between D and E
Each part is 3/8.
3(3/8) = 9/8
2 + 9/8 = 25/8
So the point I is (0, 25/8)