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amm1812
3 years ago
14

When flying at an altitude of 5 miles, the lines of sight to the horizon looking north and south make about a 173.7 degree angle

. how much of the longitude line directly under the plane is visible from 5 miles high?
Mathematics
1 answer:
oksian1 [2.3K]3 years ago
7 0
5 miles high is one of the sides of a triangle depending on accuracy level
h^2=x^2+y^2
we don't have 2 distances
Tan A=O/a
O=a tan A
We solve for O because the angle is at the top of the line going up and we want the opposite angle that is along the ground
O=5×tan(173.7/2)=90.854033512
The distance he can see is:
90.85*2~181.7 miles

Now we need to find the distance between lines:
The north south distance between each line is 69 miles
thus the number of degrees he will see will be:
181.7/69
=2 19/30
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tatuchka [14]
Convert 20% into a decimal like this >> 20/100 =0.2 now 0.2 x 60.00 =12 the answer hope it works.
7 0
2 years ago
12 x (3 + 2 to the second power) divied by 2 - 10
alexira [117]

Answer:

<h2><u><em>32</em></u></h2>

Step-by-step explanation:

Assuming that the question is:

12 * (3 + 2^2) : 2 - 10 =       remember PEMDAS

12 * (3 + 4) : 2 - 10 =

12 * 7 : 2 - 10 =

84 : 2 - 10 =

42 - 10 =

<u><em>32</em></u>

3 0
2 years ago
Read 2 more answers
I need help with this Algebra question
Ainat [17]

Step 1 = Distributive Property ( multiplying the 2 by each term in parenthesis)

Step 2 = Addition Property ( adding like terms)

Step 3 = Addition Property (adding 34x to each side)

Step 4 = Division Property. ( dividing both sides by 17)


3 0
3 years ago
What are the coordinates of the center and length of the radius of the circle whose equation is X^2+6x+4y=23?
Virty [35]

Answer:

The following equation is not a circle, but a hyperbola, because you have 4y, but no y^2

Step-by-step explanation:

Please mark for Brainliest!! :D Thanks!!

For more questions or more information, please comment below!

3 0
3 years ago
Mathematically verify the outlier(s) in the data set using the 1.5 rule.
statuscvo [17]

Given:

The data values are:

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

To find:

The outliers of the given data set.

Solution:

We have,

7, 8, 11, 13, 14, 14, 14, 15, 16, 16, 18, 19

Divide the data set in two equal parts.

(7, 8, 11, 13, 14, 14), (14, 15, 16, 16, 18, 19)

Divide each parenthesis in two equal parts.

(7, 8, 11), (13, 14, 14), (14, 15, 16), (16, 18, 19)

Now,

Q_1=\dfrac{11+13}{2}

Q_1=\dfrac{24}{2}

Q_1=12

And

Q_3=\dfrac{16+16}{2}

Q_3=\dfrac{32}{2}

Q_3=16

The interquartile range is:

IQR=Q_3-Q_1

IQR=16-12

IQR=4

The data values lies outside the interval [Q_1-1.5IQR,Q_3+1.5IQR] are known as outliers.

[Q_1-1.5IQR,Q_3+1.5IQR]=[12-1.5(4),16+1.5(4)]

[Q_1-1.5IQR,Q_3+1.5IQR]=[12-6,16+6]

[Q_1-1.5IQR,Q_3+1.5IQR]=[6,22]

All the data values lie in the interval [6,22]. So, there are no outliers.

Hence, the correct option is 4.

3 0
2 years ago
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