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vekshin1
3 years ago
6

An airplane pilot over the Pacific sights an atoll at an angle of depression of 5. At this time, the horizontal distance from th

e airplane to the atoll is 4629 meters. What is the height of the plane to the nearest meter?

Mathematics
2 answers:
Vlad1618 [11]3 years ago
8 0
X=4629*tan(5)=404.98
=405 meters
Lelu [443]3 years ago
4 0

Answer:

Step-by-step explanation:

Alright, lets get started.

Please refer the diagram I have attached.

Using SOH CAH TOA

Tan angle = \frac{opposite}{adjacent}

Tan 5 = \frac{x}{4629}

x = 4629*Tan5

x = 4629*0.087488

x = 404.98

To the nearest meters, the height is 405 meters.   :  Answer

Hope it will help :)

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Nutka1998 [239]

Answer:

This is ezz lol.

Step-by-step explanation:

Your answer would be 1/5

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2 years ago
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An open box with a square base is to have a volume of 18 ft^3.
kotegsom [21]

Answer:

2.29 ft of side length and 1.14 height

Step-by-step explanation:

a) Volume V = x2h, where x is side of square base and h is hite.

Then surface area S = x2 + 4xh because box is open.

b) From V = x2h = 6 we have h = 6/x2.

Substitude in formula for surface area: S = x2 + 4x·6/x2, S = x2 + 24/x.

We get S as function of one variable x. To get minimum we have to find derivative S' = 2x - 24/x2 = 0, from here 2x3 - 24 = 0, x3 = 12, x = (12)1/3 ≅ 2.29 ft.

Then h = 6/(12)2/3 = (12)1/3/2 ≅ 1.14 ft.

To prove that we have minimum let get second derivative: S'' = 2 + 48/x3, S''(121/3) = 2 + 48/12 = 6 > 0.

And because by second derivative test we have minimum: Smin = (12)2/3 + 4(12)1/3(12)1/3/2 = 3(12)2/3 ≅ 15.72 ft2

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2 years ago
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There were 976 students that attended assemblies on bullying prevention throughout the
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3 years ago
4. Andrés desea embaldosar el piso de su casa que tiene 375 cm de ancho y 435 cm de largo. Calcula la longitud del lado que tend
svetlana [45]

Answer:

Sabemos que el piso es un rectángulo de 435 cm de largo y 375 cm de ancho.

Recordar que para un rectángulo de largo L, y ancho W, el área es:

A = L*W

Entonces el área del piso, será:

A = 435cm*375cm = 163,125 cm^2

Primero, sabemos que se utilizaran baldosas (las cuales son cuadradas) y queremos saber la longitud de lado que tendrían las baldosas.

No tenemos ningún criterio para encontrar este lado, solo que (si queremos usar un número entero de baldosas) el largo L del lado de la baldosa deberá ser un divisor de tanto el ancho como el largo del suelo.

Dicho de otra forma

el largo, 435cm, tiene que ser múltiplo de L

el ancho, 375cm, tiene que ser múltiplo de L.

Por ejemplo, ambos números son múltiplos de 5, entonces podríamos tomar L = 5cm

En este caso, el área de cada baldosa es:

a = L^2 = 5cm*5cm = 25cm^2

Y el número total de baldosas que necesitaría usar esta dado por el cociente entre el área del suelo y el area de cada baldosa.

N = ( 163,125 cm^2)/(25cm^2) = 6,525 baldosas.

También sabemos que ambos números (435cm y 375cm) son múltiplos de 15cm

Entonces las baldosas podrían tener 15cm de lado.

En este caso, el área de cada baldosa es:

A = (15cm)^2 = 225cm

En este caso el número total de baldosas necesarias será:

N =  ( 163,125 cm^2)/(225cm^2) = 725 baldosas.

5 0
3 years ago
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