write the equaion of a line in slope-intercept form if the slope is 1/5 and the y-intercept is -9 explain please
1 answer:
Answer:
Step-by-step explanation:
The slope-intercept form is y = mx + b, where:
m = slope
b = y-intercept.
The slope (m) tells you the steepness of the line. It is the average rate of change which measures how the y-value changes for each one-unit change in the x-value. Hence, slope . So the given slope of 1/5 means that for every 1 unit change in the y-value, the x-value changes by 5 units (you go up 1 unit, and "run" 5 units to the right).
Next, the y-intercept is the point on the graph where it crosses the y-axis, and has the coordinates, (0, b). It is also the value of y when x = 0. Since you're given the y-intercept of -9, then that means that it is the y-coordinate of (0, <em>b</em>). So, it becomes (0, -9).
Now that we have our slope (<em>m </em>) = 1/5, and the y-intercept (<em>b </em>) = -9, we can write the equation of the line as:
(I'm also including a screenshot of the line where it shows the y-intercept of (0, -9) on the graph).
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The distance of a train from the city depends on the speed, the time of
travel, and direction of motion of the train.
Correct responses:
- Distance Train A is from the city: <u>750 - 75·t</u>
- Distance Train B is from the city: 50·t
- After <u>six</u> hours, the two trains are the same distance from the city
- At that time both trains will be <u>300</u> miles away
1) From the time of departure of Train B to just before 6 hours after Train B's departure; 0 ≤ t < 6 hours
2) At time period, t > 6 hours
3) 250 miles
<h3>Method used to find the above response</h3>Given:
The distance of train A from the city = 750 miles
Speed of train A = 75 mph
Time at which Train B leaves the city = When Train A is 750 miles from the city
Speed of Train B = 50 mph
Solution:
The equations are;
- Distance of Train A from the city is; <u>x₁ = 750 - 75·t</u>
- Distance of Train B from the city is <u>x₂ = 50·t</u>
When the train are the same distance from the city, we have;
750 - 75·t = 50·t
750 = 50·t + 75·t = 125·t
Therefore;
The time it takes the two trains to be the same distance from the city is 6 hours.
Which gives;
- After <u>6</u> hours, the two trains are the same distance from the city.
The distance the trains will be after 6 hours is therefore;
750 - 75 × 6 = 300
The distance of the trains from the city after 6 hours = 300 miles
Therefore, we have;
- At that time both trains will be <u>300</u> miles away
1) The time period Train A is further from the city is before the first 6 hours elapse; <u>0 hours ≤ t < 6 hours</u>
2) The Train B is further from the city after 6 hours of its departure from the city; <u>t > 6 hours</u>
3) The distance of Train A from the city after 4 hours is given as follows;
x₁ = 750 - 75 × 4 = 450
x₁ = 450 miles
The distance of Train B from the city after 4 hours is; x₂ = 50 × 4 = 200
x₂ = 200 miles
Therefore;
- The distance between the trains after 4 hours = 450 miles - 200 miles = <u>250 miles</u>
Learn more about distance and time relationship equation here:
