Two pathways meet at 30° to each other. One pathway has lighting and the other does not. The distance between successive lights on the lighted pathway is 5 metres. Each light has a
range of effective illumination of 6 metres.
What length of the pathway without lights is illuminated by the pathway with lights?

12 meters

Step-by-step explanation:

Draw a picture of the two pathways 30° apart.  Add a circle representing the illumination of the light on one of the pathways.  Draw a line from the center of the circle to the last point where it intersects the pathway without lights.

This forms a triangle.  One leg is x, the length of the pathway without lights.  Another leg is 5n, where n is an integer.  This represents how far the light is from where the pathways meet.  The third and final leg is 6, the radius of the illumination.

Use law of cosine to solve:

6² = x² + (5n)² − 2x(5n) cos 30°

36 = x² + 25n² − 5√3 xn

0 = x² − 5√3 xn + 25n² − 36

In order to have a solution, the discriminant must be greater than or equal to 0.

b² − 4ac ≥ 0

(-5√3 n)² − 4(1)(25n² − 36) ≥ 0

75n² − 100n² + 144 ≥ 0

144 − 25n² ≥ 0

144 ≥ 25n²

144/25 ≥ n²

12/5 ≥ n

So n must be an integer less than 12/5, or 2.4.  Therefore, the largest value of n is 2.  Substituting:

0 = x² − 5√3 x(2) + 25(2)² − 36

0 = x² − 10√3 x + 64

Solve with quadratic formula:

x = [ 10√3 ± √(300 − 4(1)(64)) ] / 2(1)

x = (10√3 ± √44) / 2

x = 5√3 ± √11

x ≈ 5.34 or 11.98

We want the larger value of x.  So approximately 12 meters of the pathway without lights is illuminated.

think about it deeply and passionately about this property

i am sorry

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.