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Question 1072915: A bug is moving along the line 3x + 4y = 12 with constant speed 5 units per second. The bug crosses the x-axis when t = 0 seconds. It crosses the y-axis later. When? Where is the bug when t = 2? when t = −1? when t = 1.5? What does a negative t-value mean?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! 3x + 4y = 12
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if y = 0, then 3x = 12 and x = 4, therefore
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at t = 0, the bug is located at the point (4, 0)
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if x = 0, then 4y = 12 and y = 3, the bug is located at (0, 3)
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the distance(d) between 2 points = square root( (0-4)^2 + (3-0)^2) ) = 5
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the distance between the x and y intercepts is 5 units and the bug travels 5 units in one second
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the bug crosses the y axis at t = 1 second
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at t = 2 the bug has traveled 10 units from point (4, 0)
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10^2 = (x - 4)^2 + (y - 0)^2
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1) 100 = x^2 -8x +16 + y^2
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the point slope form of the line is
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2) y = -3x/4 +3
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use the y value from equation 2 and substitute in equation one
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100 = x^2 -8x +16 +(-3x/4 +3)^2
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100 = x^2 -8x +16 + 9x^2/16 -18x/4 +9
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100 = 25x^2/16 -50x/4 +25
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4 = x^2/16 -2x/4 +1
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64 = x^2 -8x + 16
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x^2 -8x -48 = 0
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(x-12) * (x+4) = 0
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x = 12, x = -4
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The bug is moving up the line to the left from point (4,0) since from equation 2 we know that the line has negative slope, therefore we reject the positive value for x
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substitute x = -4 in equation 2 to find the y value
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y = -3(-4)/4 +3 = 6
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at t = 2 the bug is at point (-4, 6)
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at t = -1 the bug is located at -5 units from (4, 0)
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3) 25 = x^2 -8x +16 + y^2
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25 = x^2 -8x +16 +9x^2/16 -18x/4 +9
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25 = 25x^2/16 -50x/4 +25
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1 = x^2/16 -2x/4 +1
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16 = x^2 -8x +16
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x(x -8) = 0
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x = 0 or x = 8
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a negative value for t has the bug moving down and to the right from (4, 0), therefore we reject x = 0
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y = -3(8)/4 +3 = -3
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at time t = -1 the bug is at point (8, -3)
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at time t = 1.5, the bug has moved 7.5 units
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4) 56.25 = x^2 -8x +16 + y^2
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56.25 = x^2 -8x +16 +9x^2/16 -18x/4 +9
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56.25 = 25x^2/16 -50x/4 +25
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900 = 25x^2 -200x +400
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25x^2 -200x -500 = 0
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x^2 -8x -20 = 0
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(x-10) * (x+2) = 0
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x = 10 or x = -2
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a positive value for t means that the bug is moving up and to the left from point (4, 0), therefore we reject x = 10
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y = -3(-2)/4 +3 = 4.5
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at time t = 1.5, the bug is at point (-2, 4.5)
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here is a graph of the line that the bug is traveling on
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