When a linear equation is in the form y = mx + c, the c, or constant, is the intercept on the y axis, meaning it crosses the y axis at (0, 1).
The gradient (1/3 in this case) is how much the y increments (or decrements) per increase of 1 of the value of x.
This would mean that there would be one point at (0, 1), and another at (3, 2). Draw a line from these two points and beyond, and that is the graph sketched.
Let's define variables:
s = original speed
s + 12 = faster speed
The time for the half of the route is:
60 / s
The time for the second half of the route is:
60 / (s + 12)
The equation for the time of the trip is:
60 / s + 60 / (s + 12) + 1/6 = 120 / s
Where,
1/6: held up for 10 minutes (in hours).
Rewriting the equation we have:
6s (60) + s (s + 12) = 60 * 6 (s + 12)
360s + s ^ 2 + 12s = 360s + 4320
s ^ 2 + 12s = 4320
s ^ 2 + 12s - 4320 = 0
We factor the equation:
(s + 72) (s-60) = 0
We take the positive root so that the problem makes physical sense.
s = 60 Km / h
Answer:
The original speed of the train before it was held up is:
s = 60 Km / h
The line x = -5 is a vertical line, therefore the line perpendicular to this line is the horizontal line of the form y = a.
We know that it passes through the point (1, π) → x = 1, y = π.
Therefore we have answer: A) y = π.
Two.
b1 = - 4 + sqrt(19), b2 = - 4 - sqrt(19)
