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3 years ago

At a currency exchange, 11 U.S. dollars can be exchanged for 10 Euros. How many U.S. dollars will you receive for 1 Euro?

Mathematics

1 answer:

Slav-nsk [51]3 years ago

3 0

Answer:

$1.22096

Step-by-step exanation:

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torisob [31]

Answer:

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Step-by-step explanation:

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2 years ago

Read 2 more answers

What does it mean to compose two functions?

bazaltina [42]

In mathematics, function composition is the pointwise application of one function to the result of another to produce a third function. ... Intuitively, composing two functions is a chaining process in which the output of the inner function becomes the input of the outer function.

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3 years ago

What is 5/12 of a full rotation

wlad13 [49]

Answer:

150 degrees

Step-by-step explanation:

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3 years ago

The shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the Sun is 60°. Find the ang

Naddika [18.5K]

Answer:

$30^{\circ}$ .

Step-by-step explanation:

Let $\theta$ denote the unknown angle of elevation. Let $h$ denote the height of the tower.

Refer to the diagram attached. In this diagram, ${\sf A}$ denotes the top of the tower while ${\sf B}$ denote the base of the tower; ${\sf BC}$ and ${\sf BD}$ denote the shadows of the tower when the angle of elevation of the sun is $60^{\circ}$ and $\theta$ , respectively. The length of segment ${\sf AB}$ is $h$ ; $\angle {\sf ACB} = 60^{\circ}$ , $\angle {\sf ADB} = \theta$ , and ${\sf BD} = 3\, {\sf BC}$ ..

Note that in right triangle $\triangle {\sf ABC}$ , segment ${\sf AB}$ (the tower) is opposite to $\angle {\sf ACB}$ . At the same time, segment ${\sf BC}$ (shadow of the tower when the angle of elevation of the sun is $60^{\circ}$ ) is adjacent to $\angle {\sf ACB}$ .

Thus, the ratio between the length of these two segments could be described with the tangent of $m\angle {\sf ACB} = 60^{\circ}$ :

$\begin{aligned}\tan(\angle {\sf ACB}) &= \frac{\text{opposite}}{\text{adjacent}} = \frac{{\sf AB}}{{\sf BC}}\end{aligned}$ .

$\begin{aligned}\frac{{\sf AB}}{{\sf BC}} = \tan(60^{\circ}) = \sqrt{3}\end{aligned}$ .

The length of segment ${\sf AB}$ is $h$ . Rearrange this equation to find the length of segment ${\sf BC}$ :

$\begin{aligned} {\sf BC} &= \frac{{\sf AB}}{\tan(\angle ACB)} \\ &= \frac{h}{\tan(60^{\circ})}\\ &= \frac{h}{\sqrt{3}} \\ &\end{aligned}$ .

Therefore:

$\begin{aligned}{\sf BD} &= 3\, {\sf BC} \\ &= \frac{3\, h}{\sqrt{3}} \\ &= (\sqrt{3})\, h\end{aligned}$ .

Similarly, in right triangle ${\sf ABD}$ , segment ${\sf AB}$ (the tower) is opposite to $\angle {\sf ADB}$ . Segment ${\sf BD}$ (shadow of the tower, with $\theta$ as the angle of elevation of the sun) is adjacent to $\angle {\sf ADB}$ .

$\begin{aligned}\tan(\angle {\sf ADB}) &= \frac{\text{opposite}}{\text{adjacent}} = \frac{{\sf AD}}{{\sf BD}}\end{aligned}$ .

$\begin{aligned}\frac{{\sf AB}}{{\sf BD}} = \tan(\theta) \end{aligned}$ .

Since ${\sf AB} = h$ while ${\sf BD} = (\sqrt{3})\, h$ :

$\begin{aligned}\tan(\theta) &= \frac{{\sf AB}}{{\sf BD}} \\ &= \frac{h}{(\sqrt{3})\, h} \\ &= \frac{1}{\sqrt{3}}\end{aligned}$ .

Therefore:

$\begin{aligned}\theta &= \arctan\left(\frac{1}{\sqrt{3}}\right) \\ &= 30^{\circ}\end{aligned}$ .

In other words, the angle of elevation of the sun at the time of the longer shadow would be $30^{\circ}$ .

8 0

2 years ago

Please help!!

Inessa [10]

Answer: $\text{x}\sqrt{3\text{x}}$ which is Choice A

=========================================

Work Shown:

Let $y = \sqrt{3\text{x}}$

So,

$4\text{x}\sqrt{3\text{x}}-\text{x}\sqrt{3\text{x}}-2\text{x}\sqrt{3\text{x}}\\\\4\text{x}y-\text{x}y-2\text{x}y\\\\(4\text{x}-\text{x}-2\text{x})y\\\\\text{x}y\\\\\text{x}\sqrt{3\text{x}}\\\\$

This therefore means:

$4\text{x}\sqrt{3\text{x}}-\text{x}\sqrt{3\text{x}}-2\text{x}\sqrt{3\text{x}}=\text{x}\sqrt{3\text{x}}\\\\$

as long is x is nonnegative.

6 0

1 year ago