Denise is constructing A square in which two of its vertices are points M and N. She has already used her straightedge and compa
ss to construct the lines and arcs shown.
What should Denise do for her next step?
A) use a straightedge to draw a horizontal line that contains point O.
B) use a straightedge to draw MO.
C) place the point of the compass on point M and draw any arc above it.
D) place the point of the compass on point N and draw an arc that intersects N O, using MN as the width for the opening of the compass.
What should Denise do for her next step?
A) use a straightedge to draw a horizontal line that contains point O.
B) use a straightedge to draw MO.
C) place the point of the compass on point M and draw any arc above it.
D) place the point of the compass on point N and draw an arc that intersects N O, using MN as the width for the opening of the compass.
2 answers:
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Answer:
Part A: The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles
Part B: For the total distance is as close to 3 miles as possible, the start point of the parade should be at the point on Broadway with coordinates (9.941, 4.970)
Part C: The coordinates of the cameras stationed half way down each road are;
For central avenue; (4, 2)
For Broadway; (7.97, 2.49)
Step-by-step explanation:
Part A: The length of the given route can be found using the equation for the distance, l, between coordinate points as follows;
Where for the Broadway potion of the parade route, we have;
(x₁, y₁) = (12, 3)
(x₂, y₂) = (6, 0)
For the Central Avenue potion of the parade route, we have;
(x₁, y₁) = (6, 0)
(x₂, y₂) = (2, 4)
Therefore, the total length of the parade route =-3·√5 + 4·√2 = 12.265 unit
The scale of the drawing is 1 unit = 0.25 miles
Therefore;
The actual length of the initial parade =0.25×12.265 unit = 3.09 miles
The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles
Part B:
For an actual length of 3 miles, the length on the scale drawing should be given as follows;
1 unit = 0.25 miles
0.25 miles = 1 unit
1 mile = 1 unit/(0.25) = 4 units
3 miles = 3 × 4 units = 12 units
With the same end point and route, we have;
y² + (6 - x)² = 176 - 96·√2
y² = 176 - 96·√2 - (6 - x)²............(1)
Also, the gradient of l₁ = (3 - 0)/(12 - 6) = 1/2
Which gives;
y/x = 1/2
y = x/2 ..............................(2)
Equating equation (1) to (2) gives;
176 - 96·√2 - (6 - x)² = (x/2)²
176 - 96·√2 - (6 - x)² - (x/2)²= 0
176 - 96·√2 - (1.25·x²- 12·x+36) = 0
Solving using a graphing calculator, gives;
(x - 9.941)(x + 0.341) = 0
Therefore;
x ≈ 9.941 or x = -0.341
Since l₁ is required to be 12 - 4·√2, we have and positive, we have;
x ≈ 9.941 and y = x/2 ≈ 9.941/2 = 4.97
Therefore, the start point of the parade should be the point (9.941, 4.970) on Broadway so that the total distance is as close to 3 miles as possible
Part C: The coordinates of the cameras stationed half way down each road are;
For central avenue;
Camera location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)
For Broadway;
Camera location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49).