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24. Find the equation of the line passing through (3 , -1) and parallel to the line
y = -2x +1

25. A $15,000 robot depreciates linearly to zero in 10 years. How much is the robot worth three years after it is purchased?

Show work and explain it to me please I am bad in Math
asked Oct 2, 2014 in PRECALCULUS by Baruchqa Pupil

2 Answers

+1 vote

(24).

The line equation is y = - 2x + 1 and the point is (3, - 1).

Compare the equation y = - 2x + 1 with slope-intercept form of line equation y = mx + b, where m = slope and b = y-intercept.

The slope of line equation y = - 2x + 1 is -2.

If the two lines are parallel then their slopes are equal.

The slope parallel line equation is - 2.

To find the line equation, substitute m = - 2 and (x1, y1) = (3, - 1) in the point-slope form of line equation y - y1 = m(x - x1), where m = slope and (x1, y1) = point.

y - (- 1) = (- 2) [x - (3)]

y + 1 = - 2(x - 3)

y + 1 = - 2x + 6

Subtract 1 from each side.

y  = - 2x + 5.

The line equation is y = - 2x + 5.

answered Oct 2, 2014 by casacop Expert
+1 vote

(25).

The cost of robot depends upon number of the years. So the number of years taken as x-coordinate and the cost of the robot taken as y-coordinate.

From the situation, the two points are (0, 15000) and (10, 0).

Find the line equation and it is represent the cost of the robot :

Two-points form of line equation : y - y1 = [(y2 - y1)/(x2 - x1)](x - x1), where m = slope and (x1, y1) and (x2, y2) are two points.

Substitute (x1, y1) = (0, 15000) and (x2, y2) = (10, 0) in the two-points form of line equation.

y - (15000) = [(0 - 15000)/(10 - 0)](x - 0)

y - 15000 = - 1500x

y = - 1500x + 15000.

To find the cost of the robot after 3 years, substitute x = 3 in the above equation.

y = - 1500(3) + 15000

y = - 4500 + 15000

y = 10500.

The cost of the robot after 3 years is $10,500.

 

answered Oct 2, 2014 by casacop Expert
edited Oct 2, 2014 by bradely

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