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jok3333 [9.3K]
2 years ago
12

My phone rang exactly midway between 10:53 AM one day and 1:05 PM the next day. What time did it ring?

Mathematics
2 answers:
djverab [1.8K]2 years ago
8 0

Answer:

11:59 AM

Step-by-step explanation:

We must determined the halfway point of 10:53 AM and 1:05 PM. The total time between those to times can be determine by changiing 1:05 to 13:05. We can subtract the times:

=13:05-10:53=2:12

The total time was 2 hours and 12 minutes. We can then add half of this time to 10:53 PM.

=10:53+1:06=11:59

The time the phone rang was 11:59 AM

marishachu [46]2 years ago
4 0
The time it rang was 11:59 PM. 
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Solve for z a(t+z)=45z+67
1. Expand
at+az=45z+67

2. Subtract at from both sides
az=45z+67-at

3. Subtract 45z from both sides
az-45z=67-at

4. Factor out the common term z
z(a-45)=67-at

5. Divide both sides by a - 45
z= \frac{67-at}{ya-45}

Answer: 
z= \frac{67-at}{a-45}
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If h=3 the vertex shifts to (3,0). 

If h=-5 the vertex is shifted to (-5,0)

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Whats 1797.26 rounded to the nearest 100
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Answer:1797.3


Step-by-step explanation:


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Performance task: A parade route must start And and at the intersections shown on the map. The city requires that the total dist
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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;

l = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}

Where for the Broadway potion of the parade route, we have;

(x₁, y₁) = (12, 3)

(x₂, y₂) = (6, 0)

l_1 = \sqrt{\left (0 -3\right )^{2}+\left (6-12 \right )^{2}} = 3 \cdot \sqrt{5}

For the Central Avenue potion of the parade route, we have;

(x₁, y₁) = (6, 0)

(x₂, y₂) = (2, 4)

l_2 = \sqrt{\left (4 -0\right )^{2}+\left (2-6 \right )^{2}} = 4 \cdot \sqrt{2}

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;

l_1 = \sqrt{\left (0 -y\right )^{2}+\left (6-x \right )^{2}} = 12 - 4 \cdot \sqrt{2}

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

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Solving using a graphing calculator, gives;

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For central avenue;

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For Broadway;

Camera location = ((6 + 9.941)/2, (0 + 4.970)/2) = (7.97, 2.49).

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