You bought a car 5 years ago for $10,000. The car depreciates 12% a year. What is the car worth now?
1 answer:
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To find the rate of change, or slope, pick two points. The slope, is change in y coordinates over change in x coordinates.
.
In this case, the slope is
.
Remember that subtracting a negative number equals adding the number without the negative sign.
a-(-b) = a+b
Now for the y-intercept. The y-intercept is where the graph intersects the y-axis, where x = 0.
You have the coordinate (0, -1), where x = 0. So the y-intercept is -1.
Putting these values into the slope-intercept form y = mx+b, the equation is
y = -5x - 1
In this case, the slope is
Remember that subtracting a negative number equals adding the number without the negative sign.
a-(-b) = a+b
Now for the y-intercept. The y-intercept is where the graph intersects the y-axis, where x = 0.
You have the coordinate (0, -1), where x = 0. So the y-intercept is -1.
Putting these values into the slope-intercept form y = mx+b, the equation is
y = -5x - 1
By running in a bearing of 127°, Paul motion is in the south eastern direction.
Correct responses:
a. Please find attached the required diagram created with MS Visio
b. Paul's distance travelled east is approximately 11.98 km
- Paul's distance travelled south is approximately 0.903 km
Given:
The distance Paul runs = 1.5 km
The bearing Paul is running = 127°
a. The bearing of path Paul runs, 127°, is the angle measured in the clockwise direction relative to the northern direction
Please find attached the diagram representing the situation, created with MS Visio.
b. Paul's distance travelled east relative to the starting point is given by the component of Paul's path in the x-axis direction which is found as follows;
- Paul's distance east = 15 km × cos(37°) ≈ <u>11.98 km</u>
Pau's distance travelled in the south direction relative to the starting point
is given by the vertical distance travelled relative to the x-axis, which is
given as follows;
Distance Paul travels south = 1.5 km × sin(37°) ≈ <u>0.903 km</u>
Learn more about bearings in trigonometry here:
