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Ipatiy [6.2K]
2 years ago
10

A ship is stationary at sea.

Mathematics
1 answer:
Leni [432]2 years ago
8 0

The distance of the yacht from the ship is 35 km

<h3>How to determine the distance?</h3>

See attachment for the diagram that represents the given parameters.

The distance of the yacht from the ship is then calculated using the following law of cosine.

YS^2 = ST^2 + YT^2 - 2 * ST * YT * cos(T)

This gives

YS^2 = 24^2 + 12^2 - 2 * 24 * 12 * cos(155)

Evaluate the exponents and the products

YS^2 = 576 + 144 + 522

Evaluate the sum

YS^2 = 1242

Take the square root of both sides

YS = 35

Hence, the distance of the yacht from the ship is 35 km

Read more about law of cosine at:

brainly.com/question/4372174

#SPJ1

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GuDViN [60]

Answer:

Answer:

Since you don't show your triangles, I'm guessing it is this one...?

△DWP  D = (-2, 4), W = (-1, 2), P = (3, 3)

△MJS  M = (-4, -4), J = (-2, -3), S = (-3, 1)

There are 2 steps:

1. 90° counterclockwise rotation about the origin

2. Translation downward 2 units

Step-by-step explanation:

1. 90° counterclockwise rotation about the origin = (x, y) to (-y, x),

D moves from (-2, 4) to (-4, -2)

W moves from (-1, 2) to (-2, -1)

P moves from (3, 3) to (-3, 3)

2. Translation downward 2 units = (x, y) to (x, y-2),

D moves from (-4, -2) to (-4, -4) = M

W moves from (-2, -1) to (-2, -3) = J

P moves from (-3, 3) to (-3, 1) = S

Therefore △DWP IS congruent to △MJS.

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grandymaker [24]

Answer:

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or

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Step-by-step explanation:

I used my ti-84 plus ce calculator to solve

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Answer:

<h2><u><em>a²+2ab+b²-c²</em></u></h2>

Step-by-step explanation:

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(a²+ab-ac+ab+b²-bc+ac+bc-c²)=

a²+ab-ac+ab+b²-bc+ac+bc-c²=

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Answer is provided in the image attached.

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