A motorcycle and a car leave an intersection at the same time. the motorcycle heads north at an average speed of 20 miles per h
1 answer:
First, determine the distance of the motorcycle and the car from the start point. The distance could be determined using
d stands for distance, v stands for speed, t stands for time
The car
d = 48 × t
d = 48t
The motorcycle
d = 20 × t
d = 20t
At the end of t hours, the car is 48t miles (east) from the start point and the motorcycle is 20t miles (north) from the start point.
Second, determine the distance between 48t miles at east and 20t miles at north using pythagoras
distance =
distance =
distance =
distance = 52t
The expression for their distance apart at the end of t hours is 52t
d stands for distance, v stands for speed, t stands for time
The car
d = 48 × t
d = 48t
The motorcycle
d = 20 × t
d = 20t
At the end of t hours, the car is 48t miles (east) from the start point and the motorcycle is 20t miles (north) from the start point.
Second, determine the distance between 48t miles at east and 20t miles at north using pythagoras
distance =
distance =
distance =
distance = 52t
The expression for their distance apart at the end of t hours is 52t
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Answer:
(x, y) = (3, -6)
Step-by-step explanation:
I like a good graphing calculator for solving systems of equations by graphing.
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If you're solving these by hand, you need to graph the equations. It can be convenient to put the equations into "intercept form" so you can use the x- and y-intercepts to draw your graph.
That form is ...
x/(x-intercept) +y/(y-intercept) = 1
Dividing a standard-form equation by the constant on the right will put it in this form.
x/(-12/2) +y/(-12/3) = 1 . . . . . . divide the first equation by -12
x/(-6) +y/(-4) = 1 . . . . . . . . . . . the x-intercept is -6; the y-intercept is -4
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x/(12/10) +y/(12/3) = 1 . . . . . . divide the second equation by 12
x/1.2 +y/4 = 1 . . . . . . . . . . . . . the x-intercept is 1.2*; the y-intercept is 4
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The locations of these intercepts and the slopes of the lines tell you that the solution will be in the fourth quadrant. The lines intersect at (x, y) = (3, -6).
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* It can be difficult to draw an accurate graph using an intercept point that is not on a grid line. It may be desirable to put the second equation into slope-intercept form, so you can see the rise/run values that let you choose grid points on the line. That equation is y =-10/3x +4. A "rise" of -10 for a "run" of +3 will get you to (3, -6) starting from the y-intercept of (0, 4).
