write the slope-intercept form of the equation of the line passing through the point(-5,-2) and parallel to the line y=2x-2
1 answer:
You might be interested in
The solution to a system of (linear) equations is the point where the graphs intersect. Consider two parallel lines. By definition, two parallel lines never intersect each other, but all pairs of non-parallel lines will eventually intersect. That means they will also have a solution.
Let's consider what makes a line parallel to another line. It basically looks identical, having the same steepness (slope), but the graph is just shifted over. That is, a parallel line would have the same slope and a different y-intercept. For our equation
, or 
in slope-intercept form, a parallel line will be of the form 
.
That describes the form of a parallel line, which we do not want. Any other line, however, will give a solution to our system, so we merely want a line where the slope does not equal 2.
We can have any equation of the form
.
Let's consider what makes a line parallel to another line. It basically looks identical, having the same steepness (slope), but the graph is just shifted over. That is, a parallel line would have the same slope and a different y-intercept. For our equation
That describes the form of a parallel line, which we do not want. Any other line, however, will give a solution to our system, so we merely want a line where the slope does not equal 2.
We can have any equation of the form
Answer:
The other factor of the polynomial is (5x -6)
Step-by-step explanation:
Here, we want to find the other factor of this binomial expression
To find the other factor, we simply proceed to divide the polynomial by the binomial
That would be;
(10x^2 + 18x - 36)/(2x + 6)
= 5x - 6
Answer:
s = 22.5 m
Step-by-step explanation:
the equation for the speed change of a coach moving along a straight section of the road and starting braking at a speed of 20 m / s has the form v (t) = 25-5t. Using integral calculus, determine the coach's braking distance.
v (t) = 25 - 5 t
at t = 0 , v = 20 m/s
Let the distance is s.
Let at t = t, the v = 20
So,
20 = 25 - 5 t
t = 1 s
So, s = 25 x 1 - 2.5 x 1 = 22.5 m