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irga5000 [103]
3 years ago
9

A 70ft ladder is mounted 10ft above the ground on a fire truck. The bottom of the ladder is 40ft from the wall of a building.the

top of the ladder is touching the building. How high off the ground is the top of the ladder

Mathematics
1 answer:
Luda [366]3 years ago
8 0
Please refer to my attachments for visual guidelines.
We are going to solve your problem by using the pythagorean theorem, a^2+b^2 = c^2, where a and b are the legs of the triangle, and c is the hypotenuse (the longest side).

The length of the ladder is equal to 70ft (hypotenuse); one leg is the distance between the wall and the bottom of the ladder - 40 ft, the other leg is unknown for it is the distance between 10 ft above the ground and the top of the ladder-represented by "x". Using pythagorean theorem, a^2+b^=c^2, we have x^2+40^2 = 70^2. Solving the exponents, we have x^2 + 1600 = 4900.

Isolating the variable x, we have x^2 = 4900-1600. Futher simplying, x^2 = 3300. Thus, x = √3300 or 57.4456264654 ft.

Adding 10 ft to x, therefore, the top of the leadder is 67.4456264654 ft off the ground.

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The ratio of building heights will be the same as the ratio of shadow lengths.
  (taller building)/(shorter building) = (longer shadow)/(shorter shadow)
  (taller building)/(84 ft) = (110 ft)/(46 ft) . . . . . . fill in the given numbers
  (taller building) = (84 ft)*(110/46) ≈ 200.9 ft . . multiply by 84 ft, evaluate

The appropriate selection is
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ikadub [295]

Answer:

<em>Explicación más abajo</em>

Step-by-step explanation:

Integración Indefinida

La integral

I=\int \dfrac{x .arcsen\left(x\right)}{\sqrt{1-x^2}}

Se resuelve con el cambio de variables:

t=arcsen(x)

Una vez hechos los cambios, la integral se resuelve en función de t:

I=sen(t)-t.cos(t)+C

Hay que devolver los cambios para mostrarla en función de x.

El cambio de variables también se puede escribir:

x=sen(t)

Recordando que

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Entonces:

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Es la respuesta correcta

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