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pochemuha
3 years ago
8

A plane decides to travel along AB towards the North with a velocity of V.

Mathematics
1 answer:
Maru [420]3 years ago
8 0

Step-by-step explanation:

The wind has a speed of w and a direction α with the vertical.  The x component of that speed is w sin α.  The y component is -w cos α.

In order to stay on the north trajectory AB, the plane must have a horizontal speed of -w sin α.  The plane's speed is v, so using Pythagorean theorem, the y component of the plane's speed is:

v² = (-w sin α)² + vᵧ²

v² = w² sin²α + vᵧ²

vᵧ = √(v² − w² sin²α)

The total vertical speed is therefore √(v² − w² sin²α) − w cos α.

If a is the length of AB, then the time is:

t = a / [√(v² − w² sin²α) − w cos α]

To rationalize the denominator, we multiply by the conjugate.

t = a / [√(v² − w² sin²α) − w cos α] × [√(v² − w² sin²α) + w cos α] / [√(v² − w² sin²α) + w cos α]

t = a [√(v² − w² sin²α) + w cos α] / (v² − w² sin²α − w² cos²α)

t = a [√(v² − w² sin²α) + w cos α] / (v² − w²)

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2 years ago
Marcie solved the following inequality and her work is shown below:. . −2(x − 5) − 12 ≤ 4 + 6(x + 3). −2x + 10 − 12 ≤ 4 + 6x + 1
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All steps that Marcie did are right, but she forgot to change sign in last step.

Consider the inequality

-2(x - 5) -12 \le 4 + 6(x + 3).

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Lucy has two star systems left to visit on her voyage, but her ship is running low on fuel. The first system, KA-77, is 12001200
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Step-by-step explanation:

Please find the attachment.

Let x represent the distance between KA-7 and KA-11.

We have been given that the first system, KA-7, is 1200 light years away while the second system, KA-11, is 1700 light years away. Lucy sees an angle of 52 degrees between KA-7 and KA-11.

We can see from our attachment that Lucy, KA-7 and KA-11 forms a triangle and we will use law of cosines to solve for x.  

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Upon substituting our given values in above formula we will get,

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