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Fiesta28 [93]
3 years ago
13

5. A yacht is cruising at 20 knots at bearing of 185° when a 15 knot wind starts to blow at a bearing of 290°. Find the directio

n and speed of the boat.
Mathematics
1 answer:
kicyunya [14]3 years ago
3 0

Answer:

<em>Direction of the boat is 43.05° and the speed is 21.66 knot</em>

<em></em>

Step-by-step explanation:

yacht speed = 20 knots

yacht bearing = 185° = 85°  below the negative horizontal x-axis

wind speed = 15 knots

wind bearing = 290° = 20° above the negative horizontal x-axis

we find the x and y components of the boat velocities

for yacht,

x component = -20 cos 85° = -1.74 knots

y component = -20 sin 85° = -19.92 knots

for the wind,

x component = -15 cos 20° = -14.09 knots

y component = 15 sin 20° = 5.13 knots

total x component  Vx = -1.74 + (-14.09) = -15.83 knots

total y component Vy =  -19.92 + 5.13 = -14.79 knots

Resultant speed of the boat = \sqrt{Vx^{2} + Vy^{2}  }

==> \sqrt{15.83^{2} + 14.79^{2}  } = <em>21.66 knot</em>

<em></em>

direction of boat = tan^{-1} \frac{Vy}{Vx}

==> tan^{-1} \frac{14.79}{15.83} = <em>43.05°</em>

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<h2>Hello!</h2>

The answer is:

Center: (-4,-4)

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<h2>Why?</h2>

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The given equation is:

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So, solving we have:

x^{2}+y^{2}+8x+8y=-28

(x^2+8x+(\frac{8}{2})^{2})+(y^2+8y+(\frac{8}{2})^{2})=-28+((\frac{8}{2})^{2})++(\frac{8}{2})^{2})\\\\(x^2+8x+16 )+(y^2+8y+16)=-28+16+16\\\\(x^2+4)+(y^2+4)=4

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Have a nice day!

Note: I have attached a picture for better understanding.

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jen and sarah go to lunch at the Green Grill.Their meals total $28.00 .The tax is 6%.What is the total cost including tax?
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