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3 years ago

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A tourist standing on the New York City coast sees the Statue of Liberty, which is 305 feet tall. She must tilt her head 9 degre

es upwards to look at its head. How far away is she from the base of the statue? Round your answer to the nearest whole foot.

1 answer:

mixer [17]3 years ago

4 0

Given:

Height of the Statue of Liberty = 305 feet

Angle of elevation = 9 degrees

To find:

The between the tourist and the Statue of Liberty.

Solution:

Using the given information, draw a figure as shown below.

In a right triangle,

$\tan \theta=\dfrac{Perpendicular}{Base}$

In the triangle ABC,

$\tan 9^\circ=\dfrac{305}{x}$

$x=\dfrac{305}{\tan 9^\circ}$

$x=1925.6942$

$x\approx 1926$

Therefore, the tourist is standing 1926 feet from the Statue of Liberty.

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GIVEN: Miranda has a bag of marbles with $5$ blue marbles, $1$ white marbles, and

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,b)A red, then a white Preview

,c) A Blue, then a Blue, then a Blue.

SOLUTION:

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(a)

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As marble is not returned to bag,

Probability of drawing red marble $=\frac{\text{total red marble}}{\text{total marbles}}$

$=\frac{1}{6}$

(b)

Probability of drawing red marble $=\frac{\text{total red marble}}{\text{total marbles}}$

$=\frac{1}{7}$

As marble is not returned to bag,

Probability of drawing white marble $=\frac{\text{total white marble}}{\text{total marbles}}$

$=\frac{1}{6}$

(c)

Probability of drawing blue marble $=\frac{\text{total blue marble}}{\text{total marbles}}$

$=\frac{5}{7}$

As marble is not returned to bag,

Probability of drawing second blue marble $=\frac{\text{total blue marble}}{\text{total marbles}}$

$=\frac{4}{6}=\frac{2}{3}$

As marble is not returned to the bag

Probability of drawing third blue marble $=\frac{\text{total blue marble}}{\text{total marbles}}$

$=\frac{3}{5}$

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3 years ago

Graph this rational equation. Identify the points of discontinuity, holes, vertical asymptotes, x-intercepts, and horizontal asy

irakobra [83]

Step-by-step explanation:

We have given,

A rational function : f(x) = $\frac{x-2}{x-4}$

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Plug y=f(x) = ∞ in f(x) to get vertical asymptote {We can us writing ∞ = $\frac{1}{0}$ }

i.e ∞ = $\frac{x-2}{x-4}$

or $\frac{1}{0}=\frac{x-2}{x-4}$

or x-4 =0

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Plug f(x)=0 , to get x intercept.

i.e 0 = $\frac{x-2}{x-4}$

or x - 2 =0

or x = 2

Hence at x=2, f(x) has an x intercept

Horizontal asymptote:

Plug x = ∞ in f(x) to get horizontal asymptote.

i.e f(x) = $\frac{x-2}{x-4}$ = $\frac{x(1-\frac{2}{x} )}{x(1-\frac{4}{x} )}$

or f(x) = $\frac{(1-\frac{2}{∞} )}{(1-\frac{4}{∞} )}$

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