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Semmy [17]
4 years ago
15

A sailboat leaves port and sails 12 kilometers west and then 9 kilometers north. The sailboat is now kilometers from port

Mathematics
1 answer:
Luden [163]4 years ago
5 0

Answer:  The sailboat is at a distance of 15 km from the port.

Step-by-step explanation:  Given that a sail boat leaves port and sails 12 kilometers west and then 9 kilometers north.

We are to find the distance between the sailboat from the port in kilometers.

Since the directions west and north are at right-angles, we can visualize the movement of the sailboat in the form of a right-angled triangle as shown in the attached figure.

The sailboat moves leaves the port at P and reach O after sailing 12 km west. From point O, again it moves towards north 9 km and reach the point S.

PS = ?

Using the Pythagoras theorem, we have from right-angled triangle SOP,

PS^2=OS^2+OP^2=9^2+12^2=81+144=225\\\\\Rightarrow PS=15~\textup{km}.

Thus, the sailboat is at a distance of 15 km from the port.

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Interpreting the inequality, it is found that the correct option is given by F.

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  • The first equation is of the line.
  • The equal sign is present in the inequality, which means that the line is not dashed, which removes option G.

In standard form, the equation of the line is:

x + 2y = 6

2y = 6 - x

y = -0.5x + 2

Thus it is a decreasing line, which removes options J.

  • We are interested in the region on the plane below the line, that is, less than the line, which removes option H.

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  • As for the second equation, the normalized equation is:

3x^2 + 3y^2 = 12

3(x^2 + y^2) = 12

x^2 + y^2 = 4

  • Thus, a circle centered at the origin and with radius 2.
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  • First, we find the following distance:

d = \sqrt{\frac{|ax + by + c|}{a^2 + b^2}}

  • Considering the coefficients of the line and the center of the circle.

d = \sqrt{\frac{|1(0) + 2(0) - 6|}{1^2 + 2^2}} = \sqrt{\frac{6}{5}} = 1.1

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A similar problem is given at brainly.com/question/16505684

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