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Alekssandra [29.7K]
3 years ago
8

What is the distance to the earth's horizon from

Mathematics
1 answer:
shepuryov [24]3 years ago
4 0

Answer:   216.2 miles

Step-by-step explanation:

Alright, lets get started.

As the diagram shows, the line from point P is tangent to earth's curvature.

It means it is making right angle.

So the triangle formed will be  right triangle.

Hence we can use Pythagorean theorem in it.

So from point P to center of earth, it will be hypotenuse of triangle.

c^2=a^2+b^2

(3959+5.9)^2=3959^2+x^2

3964.9^2=3959^2+x^2

15720432.01=15673681+x^2

x^2=46751.01

taking square root, the x will be :

x = 216.2

hence the value of x is 216.2 miles.   :   Answer

Hope it will help :)

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Answer:

The cost of 15 liters is $31.50

Step-by-step explanation:

1. Find the price for 1 liter of petrol.

Divide the cost by the number of liters:

$14.70 / 7 = $2.10 per liter.

2. Multiply the price for 1 liter by the new number of liters, 15.

$2.10 per liter x 15 liters = $31.50

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The ratio of boys and teacher in a school is 2:3 the ratio of teachers to the girls is 4:5.Find the ratio of all the three in a
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Answer:

Step-by-step explanation:

2:3 boy:teacher 8:12

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In each of Problems 5 through 10, verify that each given function is a solution of the differential equation.
WARRIOR [948]

Answer:

For First Solution: y_1(t)=e^t

y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:y_2(t)=cosht

y_2(t)=cosht  is the solution of equation y''-y=0.

Step-by-step explanation:

For First Solution: y_1(t)=e^t

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_1(t)=e^t

First order derivative:

y'_1(t)=e^t

2nd order Derivative:

y''_1(t)=e^t

Put Them in equation y''-y=0

e^t-e^t=0

0=0

Hence y_1(t)=e^t is the solution of equation y''-y=0.

For 2nd Solution:

y_2(t)=cosht

In order to prove whether it is a solution or not we have to put it into the equation and check. For this we have to take derivatives.

y_2(t)=cosht

First order derivative:

y'_2(t)=sinht

2nd order Derivative:

y''_2(t)=cosht

Put Them in equation y''-y=0

cosht-cosht=0

0=0

Hence y_2(t)=cosht  is the solution of equation y''-y=0.

3 0
3 years ago
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