Ask question

Ask question

All categories

3 years ago

6

Pls im going to fail pls help me

1 answer:

elixir [45]3 years ago

4 0

Answer:

what do you need help with?

You might be interested in

A survey conducted by Black Flag asked whether or not the action of a certain type of roach disk was effective in killing roache

stiv31 [10]

79% caraca bastante né bons estudos survey

6 0

3 years ago

Factorise 36x³y-225xy³

blagie [28]

<h3>hi</h3><h3>here is ur answer:</h3>

6 0

3 years ago

Read 2 more answers

Write -2i + (9 - 3i) - (6 - 10i) as a complex number in standard form.

Wittaler [7]

The answer is b for your question

4 0

3 years ago

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

3 years ago

Please help me with this homework

nignag [31]

$2.40 add all the values of the coins together

8 0

3 years ago