Consider a prism that has n faces. Write an expression that represents the number of lateral faces.

a plane flies 5.0km due north from a point O and then 6.0km on a bearing of 060° the pilot then changes course on a bearing of 1

Virty [35]

The plane flies a distance of approximately 10.536 kilometers in <em>straight</em> line and with a bearing of approximately 035°.

A plane that travels a distance $r$ , in kilometers, with a bearing of $\theta$ sexagesimal degrees can be represented in standard position by means of the following expression:

$\vec r = r\cdot (\sin\theta, \cos \theta)$ (1)

We can obtain the resulting vector ( $\vec R$ ) by the principle of superposition:

$\vec R = \Sigma_{i=1}^{n} [r_{i}\cdot (\sin \theta_{i}, \cos \theta_{i})]$ (2)

If we know that $r_{1} = 5\,km$ , $\theta_{1} = 0^{\circ}$ , $r_{2} = 6\,km$ , $\theta_{2} = 60^{\circ}$ , $r_{3} = 4\,km$ and $\theta_{3} = 120^{\circ}$ , then the resulting vector is:

$\vec R = 5\cdot (\sin 0^{\circ}, \cos 0^{\circ}) + 6\cdot (\sin 60^{\circ}, \cos 60^{\circ}) + 4\cdot (\sin 120^{\circ}, \cos 120^{\circ})$

$\vec R = (5\sqrt{3}, 6) \,[km]$

The magnitude of the resultant is found by Pythagorean theorem:

$\|\vec R\| = \sqrt{R_{x}^{2}+R_{y}^{2}}$

And the bearing is determined by the following <em>inverse</em> trigonometric relationship:

$\theta_{R} = \tan^{-1} \left(\frac{R_{y}}{R_{x}}\right)$ (3)

If we know that $R_{x} = 5\sqrt{3}\,km$ and $R_{y} = 6\,km$ , then the magnitude and the bearing of the resultant is:

$\|\vec R\| = \sqrt{(5\sqrt{3})^{2}+6^{2}}$

$\|\vec R\| \approx 10.536\,km$

$\theta_{R} = \tan^{-1} \left(\frac{6}{5\sqrt{3}} \right)$

$\theta_{R} \approx 34.715^{\circ}$

The plane flies a distance of approximately 10.536 kilometers in <em>straight</em> line and with a bearing of approximately 035°.

To learn more on vectors, we kindly invite to check this verified question: brainly.com/question/21925479

6 0

2 years ago

Aaron answered 7 out of 12 questions correctly on his last English quiz. Find his grade

givi [52]

Divide 7 by 12
which would give you 0.58
so that’s 58%

7 0

2 years ago

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Two cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is a King and the se

34kurt

Answer:

1 / 51

Step-by-step explanation:

Given that :

Number of cards in a deck = 52

Number of heart suit = 13

Number of kings = 4

Recall:

Probability = (required outcome / Total possible outcomes)

P(first card is king) = 4/52 = 1/13

P(second card is heart) = 13/51

Hence,

P(first card is king) * P(second card is heart)

(1/ 13) * (13/51) = 13 / 663 = 1/ 51

5 0

3 years ago

Which of the following statements is true concerning ABC below?

zhuklara [117]

Angle B is the largest angle. , Option D is the correct answer.

<h3>What is an Isosceles Triangle ?</h3>

A triangle that has two sides equal is called an Isosceles Triangle.

It is given in the question a triangle ABC

With BA =BC

therefore angle A is equal to angle C

In a triangle the angle opposite to the longest side is the biggest angle

Therefore angle B is the largest angle.

Option D is the correct answer.

To know more about Isosceles Triangle

brainly.com/question/2456591

#SPJ1

8 0

2 years ago

What value of x satisfies

LenaWriter [7]

Answer:

See explanation

Step-by-step explanation:

The given trigonometric equation is $cos(90-x)=-\frac{\sqrt{3} }{3}$ .

We take the inverse cosine of both sides to get:

$90-x=cos^{-1}(-\frac{\sqrt{3} }{3})$

$90-x=125$

$x=125-90$

$x=-35\degree=325\degree$ to the nearest degree

None of the options satisfies the given equation.

But if the question is actually;

$cos(90-x)=-\frac{\sqrt{3} }{2}$

Then;

$90-x=\cos^{-1}(-\frac{\sqrt{3} }{2})$ .

$90-x=210$ .

$x=-120\degree$ .

Or

$x=360-120=240\degree$ .

In this case the answer will be B

5 0

3 years ago

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