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2 years ago

Hey can you please give me the answer to this?

Mathematics

1 answer:

Olegator [25]2 years ago

5 0

The height of the tower is 452.66 ft

<h3>What is angle of elevation and depression?</h3>

The angle formed by the line of sight and the horizontal plane for an object above the horizontal.

The angle of depression is the angle between the horizontal line and the observation of the object from the horizontal line.

Given:

we have two angles one of which is angle of elevation and another angle depression.

Using trigonometric ratio,

tan ∅ = P / B

tan 32° = h / 425

0.624= h /425

h= 265.569 ft

again,

tan 29° = h(bottom of the tower) / 425

h(bottom of the tower) = 425 × tan 29°

h(bottom of the tower) = 235.581 ft

Thus, Height of tower = h(top of the tower) + h(bottom of the tower)

Height of tower = 265.569 + 235.581

Height of tower = 501.15 ft

Learn more about this concept here:

brainly.com/question/20890484

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Lady_Fox [76]

Answer:

3x^2−4x−4

Step-by-step explanation:

3x^2−8x−2+4x−2

=3x^2+−8x+−2+4x+−2

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=(3x^2)+(−8x+4x)+(−2+−2)

=3x^2+−4x+−4

7 0

2 years ago

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Lelechka [254]

Answer:

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Step-by-step explanation:

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3 years ago

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D -e/3d + e= e, for d

navik [9.2K]

Answer:

$\large\boxed{d=\dfrac{e^2+e}{1-3e}}$

Step-by-step explanation:

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3 years ago

In a free-fall experiment, an object is dropped from a height of h = 400 feet. A camera on the ground 500 ft from the point of i

PilotLPTM [1.2K]

Hmmm the object, is at rest, when dropped, so it has a velocity of 0 ft/s

the only force acting on the object, is gravity, using feet will then be -32ft/s²,

was wondering myself on -32 or 32.. but anyhow... we'll settle for the negative value, since it seems to be just a bit of convention issues

so, we'll do the integral to get v(t) then

$\bf \displaystyle \int -32\cdot dt\implies -32t+C
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\textit{object moves from \underline{rest}, so velocity is 0 at 0secs}
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-32(0)+C=0\implies C=0\implies \boxed{v(t)=-32t}
\\\\\\
\textit{now to get the positional s(t)}
\\\\\\
\displaystyle \int -32t\cdot dt\implies -16t^2+C
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\textit{the initial \underline{position} was 400ft away at 0secs}
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-16(0)^2+C=400\implies C=400\implies \boxed{s(t)=-16t^2+400}$

when will it reach the ground level? let's set s(t) = 0

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2 years ago

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Degger [83]

Answer:

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Step-by-step explanation:

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2 years ago