A car accelerates at 3.00 m/s2. The car has a mass of 1354kg. What net force will produce the acceleration?
1 answer:
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Answer:
Step-by-step explanation:
An equation of a line is given , where m is the gradient or
The rise indicates the change in value of y (vertical change in the graph).
The run indicates the change in value of x (horizontal change in the graph).
Let point 1 of the graph be (x,y) = (0,-34) and point 2 be (x,y) = (20,0).
Substitute point 1 into the graph to get the constant c.
∴ The equation of the line is,
Answer:
Segment EF: y = -x + 8
Segment BC: y = -x + 2
Step-by-step explanation:
Given the two similar right triangles, ΔABC and ΔDEF, for which we must determine the slope-intercept form of the side of ΔDEF that is parallel to segment BC.
Upon observing the given diagram, we can infer the following corresponding sides:
We must determine the slope of segment BC from ΔABC, which corresponds to segment EF from ΔDEF.
<h2>Slope of Segment BC:</h2>In order to solve for the slope of segment BC, we can use the following slope formula:
Use the following coordinates from the given diagram:
Point B: (x₁, y₁) = (-2, 4)
Point C: (x₂, y₂) = ( 1, 1 )
Substitute these values into the slope formula:
Similar to how we determined the slope of segment BC, we will use the coordinates of points E and F from ΔDEF to find its slope:
Point E: (x₁, y₁) = (4, 4)
Point F: (x₂, y₂) = (6, 2)
Substitute these values into the slope formula:
Our calculations show that segment BC and EF have the same slope of -1. In geometry, we know that two nonvertical lines are <u>parallel</u> if and only if they have the same slope.
Since segments BC and EF have the same slope, then it means that .
The <u>y-intercept</u> is the point on the graph where it crosses the y-axis. Thus, it is the value of "y" when x = 0.
Using the slope of segment BC, m = -1, and the coordinates of point C, (1, 1), substitute these values into the <u>slope-intercept form</u> (y = mx + b) to solve for the y-intercept, <em>b. </em>
y = mx + b
1 = -1( 1 ) + b
1 = -1 + b
Add 1 to both sides to isolate b:
1 + 1 = -1 + 1 + b
2 = b
Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 2.
Therefore, the linear equation in <u>slope-intercept form of segment BC</u> is:
⇒ y = -x + 2.
<h3><u /></h3><h3><u>Segment EF:</u></h3>Using the slope of segment EF, <em>m</em> = -1, and the coordinates of point E, (4, 4), substitute these values into the <u>slope-intercept form</u> to solve for the y-intercept, <em>b. </em>
y = mx + b
4 = -1( 4 ) + b
4 = -4 + b
Add 4 to both sides to isolate b:
4 + 4 = -4 + 4 + b
8 = b
Hence, the <u><em>y-intercept</em></u> of segment BC is: <em>b</em> = 8.
Therefore, the linear equation in <u>slope-intercept form of segment EF</u> is:
⇒ y = -x + 8.