Which graph represents a line with a slope of and a y-intercept equal to that of the line y = x – 2? HELP!!!!!!!!!!!!!!!!!!!!!!!
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30 POINTS!!!
AND BRAINLEIST!!!!
30 POINTS!!!
AND BRAINLEIST!!!!
2 answers:
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Step-by-step explanation:
Given that,
Radius of circle, r = 38 cm = 0.38 m
It rotates form 10 degrees to 100 degrees in 11 seconds i.e.
Let is the angular velocity of the particle such that,
We need to find the instantaneous velocity of the particle. The relation between the angular velocity and the linear velocity is given by :
v = 0.053 m/s
So, the instantaneous velocity of the particle is 0.053 m/s. Hence, this is the required solution.
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.