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2 years ago

PLSS HELP ITS DUE IN HALF AN HOURRR!!

Mathematics

1 answer:

natulia [17]2 years ago

6 0

Answer

Step-by-step explanation:

$\mathrm{Convert\:mixed\:numbers\:to\:improper\:fractions}:\quad6\frac{2}{3}=\frac{20}{3}$

=8\div \frac{20}{3}

=\frac{8}{1}\div \frac{20}{3}

=\frac{8}{1}\times \frac{3}{20}

=\frac{2}{1}\times \frac{3}{5}

=\frac{2\times \:3}{1\times \:5}

=\frac{6}{1\times \:5}

=\frac{6}{5}

Convert improper fractions to mixed numbers

=1\frac{1}{5}

Brainliest plz

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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 straight 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 inverse 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 straight 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

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2 years ago

A)0.5 B)2.5 C)-0.5 D)-2.5

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Answer:

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When x = 3 the actual = 6

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2 years ago

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