# Qlt1 Task 2

Submitted By tankerd
Words 274
Pages 2
x 0 45 30

Y = (-2/3)x + 30 30 0 10 Y intercept: (0,30) Y = -2/3(x) + 30 Y = 0 + 30 Y= 0 X intercept: (45,0) 0 = -2/3 (x) + 5 0 – 30 = -2/3 (x) +30 -30 -30 = -2/3 x 2/3(x) = 30 X = 30 ● 3/2 X = 45

Height of the beam 30 ft. away from the face of building is 10 ft. Y = -2/3 (x) + 30 Y = -2/3 30 + 30 Y = -20 + 30 Y = 10 Y represents the height above the ground in feet.
35 Y intercept (0, 30) 30 25

The graph depicts a visual representation of the path of the laser beam using only quadrant I. Quadrants ll, lll, and lV are not needed to show the direction of the laser beam.

Y = (-2/3)x + 30

Y = (-2/3)x + 30 20

When the laser beam is shined at the ground at 30 ft. up it will hit the ground at 45 ft. from the face of the building using the equation Y=(-2/3)x + 30.

15 10 (30, 10) 5 X Axis 0 0 Y Axis 10 20 30

The height of the beam 30 ft. away from the face of the building is 10 ft.

X intercept (45, 0) 40 50

X represents the distance from face of bldg. in ft.

The laser beam hits the side of the building at 30 ft. which is the Y intercept (0, 30) and the laser beam hits the ground at 45 ft. from the face of the building at the X intercept (45, 0).

X intercept (45, 0).

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