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2 years ago

11

What is the radius of a sphere that has a surface area of 951.1 in2?

2 answers:

Archy [21]2 years ago

8 0

Answer:

The answer is D. 8.7

Step-by-step explanation:

A=4πr2

Solving fore

r=1

2A

π=1

2·951.1

π≈8.69978in

dusya [7]2 years ago

4 0

Answer:

FORMULA FOR SURFACE AREA OF SPHERE = 4 pi r^2

r = 6 " area = 4 pi X 6 X 6 = 144 pi " squared

r = 12 "

area = 4 X pi X 12 X 12 = 576 pi" squared

ANSWER Area will be 4 times the original

Step-by-step explanation:

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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 <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:

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

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

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

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