Performance task: A parade route must start And and at the intersections shown on the map. The city requires that the total distance of the route cannot exceed 3 miles. A propos route is shown.
Part A: Why does the proposed route not meet the requirement?
Part B: Assuming that the roads used for the
route are the same and the end point is the same,
at what intersection could the parade start so the
total distance is as close to 3 miles as possible?
Part C: The city wants to station video cameras halfway down each road in the parade. Using your answer to Part B, what are the coordinates of locations for the cameras?
Answers
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;
[tex]l = \sqrt{\left (y_{2}-y_{1} \right )^{2}+\left (x_{2}-x_{1} \right )^{2}}[/tex]
Where for the Broadway potion of the parade route, we have;
(x₁, y₁) = (12, 3)
(x₂, y₂) = (6, 0)
[tex]l_1 = \sqrt{\left (0 -3\right )^{2}+\left (6-12 \right )^{2}} = 3 \cdot \sqrt{5}[/tex]
For the Central Avenue potion of the parade route, we have;
(x₁, y₁) = (6, 0)
(x₂, y₂) = (2, 4)
[tex]l_2 = \sqrt{\left (4 -0\right )^{2}+\left (2-6 \right )^{2}} = 4 \cdot \sqrt{2}[/tex]
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;
[tex]l_1 = \sqrt{\left (0 -y\right )^{2}+\left (6-x \right )^{2}} = 12 - 4 \cdot \sqrt{2}[/tex]
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).
This exercise is necessary to use the given map information and then make the distance between points, in this way we find that:
A)The proposed route does not meet requirement because it is longer than the maximum required length of 3 miles.
B) Broadway and Cedar Street
C) [tex]Central \ Avenue(4, 2)Broadway (7.97, 2.49)[/tex]
So from the distance between points and the informed map we have:
A) The equation for the distance, as follows:
[tex]l=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}[/tex]
Where for the Broadway potion of the parade route, we have:
[tex](x_1,y_1)=(12,3)\\(x_2,y_2)=(6,0)[/tex]
Applying the values in the formula given above, we have:
[tex]l_1=\sqrt{(0-3)^2+(6-12)^2}=3\sqrt{5}[/tex]
For the Central Avenue potion of the parade route, we have:
[tex](x_1, y_1) = (6, 0)(x_2, y_2) = (2, 4)[/tex]
Applying the values in the formula given above, we have:
[tex]l_2=\sqrt{(4-0)^2+(2-6)^2}=4\sqrt{2}[/tex]
Therefore, the total length of the parade route:
[tex]3\sqrt{5} + 4\sqrt{2} = 12.265 \\0.25*12.265 unit = 3.09\ miles[/tex]
B) The two streets that are the same distance apart are the streets: Broadway and Cedar Street.
C) The coordinates of the cameras stationed half way down each road are:
For Central Avenue:
[tex]Camera \ location = ((6 + 2)/2, (4 + 0)/2) = (4, 2)[/tex]
For Broadway:
[tex]Camera\ location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49)[/tex]
See more about distance at brainly.com/question/989117
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Step-by-step explanation:
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2.4
()
14 15
0
L: 2
4
А) . 1
( - ) 90
| —
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D. (2,3.5)
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Step-by-step explanation:
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Answer:
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Answers
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Answer:
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Answer:
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Step-by-step explanation:
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???
Step-by-step explanation:
???
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Answers
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Answer:
The correct answer is 21.
Step-by-step explanation:
To solve this problem, we simply substitute 3 for a in the expression.
Given the expression,
a^2 + 12
we should replace "a" with 3, as denoted by the problem, giving us:
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Enter your answer in the answer box.
Answers
Answer:
P = 12
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Answers
Answer:
The common multiplier is 24
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What is the same value as the expression 7/5+2 1/4+1.6
Answers
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Write your answer as a decimal or whole number.
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Answers
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equation
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Answers
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Step-by-step explanation:
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Answers
Answer:
Step-by-step explanation:
11). 6x + 7 = 8x - 17 [Since vertical angles are equal in measure]
8x - 6x = 17 + 7
2x = 24
x = 12
12). (11x - 15) + (5x - 13) = 180°
16x - 28 = 180°
16x = 180 + 28
16x = 208
x = 13
13). Since DB⊥AC,
m∠CBD = 90°
m∠CBE + m∠DBE = 90°
(5x - 42) + (2x - 1) = 90°
7x - 43 = 90
7x = 133
x = 19
14). Since QS bisects angle PQR,
m∠PQS = m∠RQS
10x + 1 = [tex]\frac{82}{2}[/tex]
10x + 1 = 41
10x = 40
x = 4
15). 10x - 61 = 18y + 5 [Vertical angles]
10x - 18y = 61 + 5
10x - 18y = 66
5x - 9y = 33 ------(1)
(18y + 5) + (x + 10) = 180 [linear pair of angles are supplementary]
18y + x + 15 = 180
x + 18y = 165 ------(2)
By adding equation (1)×2 and equation (2)
(10x - 18y) + (x + 18y) = 66 + 165
11x = 231
x = 21
16). (5x - 17) + (3x - 11) = 180 [[linear pair of angles are supplementary]
8x - 28 = 180
8x = 180 + 28
8x = 208
x = 26
(3x - 11)° = 78 - 11
= 67°
67° + 90° + (2y + 5)°= 180° [Sum of angles on a line]
162 + 2y = 180
2y = 180 - 162
2y = 18
y = 9
17). NP bisects ∠MNQ.
Therefore, m∠MNQ = 2(m∠PNQ)
8x + 12 = 2×78
8x + 12 = 156
8x = 156 - 12
8x = 144
x = 18
m∠MNQ = (8x + 12)° = 156°
m∠RNM = m∠ONQ [Vertical angles]
(3y - 9)° = 180° - m∠MNQ
3y - 9 = 180 - 156
3y - 9 = 24
3y = 33
y = 11
Using the knowledge of the angle relationships to create an equation, the value of x and y in the image given are:
11. x = 12
12. x = 13
13. x = 19
14. x = 4
15. x = 21; y = 8
16. x = 26; y = 9
17. x = 18; y = 11
Applying the knowledge of angle relationships, we can solve each given problem by first, creating a an equation from the relationship between angles, then solve for the variables.
11. [tex]6x + 7 $ and $ 8x - 17[/tex] are vertical angles.
Since vertical angles are congruent, therefore:[tex]6x + 7 = 8x - 17[/tex]
Solve for x[tex]6x - 8x = -7 - 17\\\\-2x = -24\\\\x = 12[/tex]
12. [tex]11x - 15 $ and $ 5x - 13[/tex] are linear pair angles.
Since, the sum of linear pair angles equals 180 degrees, therefore:[tex](11x - 15) + (5x - 13) = 180[/tex]
Solve for x[tex]11x - 15 + 5x - 13 = 180\\\\16x - 28 = 180\\\\16x - 28 + 28 = 180 + 28\\\\16x = 208\\\\x = 13[/tex]
13. <DBE and <CBE are complementary angles.
Since the some of complementary angles equals 90 degrees, therefore,[tex]m \angle DBE + m \angle CBE = 90[/tex]
Substitute[tex](2x - 1) + (5x - 42) = 90[/tex]
Solve for x[tex]2x - 1 + 5x - 42 = 90\\\\7x - 43 = 90\\\\7x = 90 + 43\\\\7x = 133\\\\x = 19[/tex]
14. Since QS bisects <PQR, therefore,
m<PQS = 1/2(PQR)
Substitute[tex]10x + 1 = \frac{1}{2} (82)[/tex]
Solve for x[tex]10x + 1 = 41\\\\10x = 41 - 1\\\\10x = 40\\\\x = 4[/tex]
15. [tex](x + 10) $ and $ (10x - 16)[/tex] are linear pair angles.
Since the sum of linear pair angles equals 180 degrees, therefore,[tex](x + 10) + (10x - 16) = 180\\[/tex]
Solve for x[tex]x + 10 + 10x - 61 = 180\\\\11x - 51 = 180\\\\11x = 180 + 51\\\\11x = 231\\\\x = 21[/tex]
Find y:Since [tex](10x - 61) $ and $ (10y + 5)[/tex] are vertical angles, they are congruent. Therefore,
[tex](10x - 61) = (10y + 5)[/tex]
Plug in the value of x and solve for y[tex]10(21) - 61 = 18y + 5\\\\149 = 18y + 5\\\\149 - 5 = 10y\\\\144 = 18y\\\\8 = y\\\\\mathbf{y = 8}[/tex]
16. Since [tex](5x - 17) $ and $ (3x - 11)[/tex] are linear pair, their sum equals 180 degrees. Therefore,
[tex](5x - 17) + (3x - 11) = 180[/tex]
Solve for x[tex]5x - 17 + 3x - 11 = 180\\\\8x - 28 = 180\\\\8x = 180 + 28\\\\8x = 208\\\\x = 26[/tex]
Find y:[tex](2y + 5) + 90 + (3x - 11) = 180[/tex] (angles on a straight line = 180 degrees)
Plug in the value of x and solve for y[tex]2y + 5 + 90 + 3(26) - 11 = 180\\\\2y + 5 + 90 + 78 - 11 = 180\\\\2y + 162 = 180\\\\2y = 180 - 162\\\\2y =18\\\\y = 9[/tex]
17. Since NP bisects <MNQ, therefore,
[tex]m \angle PNQ = \frac{1}{2}(m \angle MNQ)[/tex]
Substitute[tex]78 = \frac{1}{2}(8x + 12)[/tex]
Solve for x[tex]78 \times 2 = 8x + 12\\\\156 = 8x + 12\\\\156 - 12 = 8x \\\\144 = 8x\\\\18 = x\\\\x = 18[/tex]
[tex]m \angle RNM + m \angle MNQ = 180^{\circ}[/tex] (linear pair)
Substitute[tex](3y - 9) + (8x + 12) = 180[/tex]
Plug in the value of x and solve for y[tex]3y - 9 + 8(18) + 12 = 180\\\\3y - 9 + 144 + 12 = 180\\\\3y + 147 = 180\\\\3y = 180 - 147\\\\3y = 33\\\\y = 11[/tex]
Therefore, using the knowledge of the angle relationships to create an equation, the value of x and y in the image given are:
11. x = 12
12. x = 13
13. x = 19
14. x = 4
15. x = 21; y = 8
16. x = 26; y = 9
17. x = 18; y = 11
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What kind of transformation is illustrated in this figure?
O translation
.
O reflection
dilation
Orotation
Answers
Answer:
translation
Step-by-step explanation:
shape, size, measurement all stayed the same. they just moved locations :)