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Neko [114]
3 years ago
10

Changes in the Depth of a River

Mathematics
2 answers:
devlian [24]3 years ago
6 0

It e) it rose 5 inches hope it correct


sdas [7]3 years ago
6 0

Answer:

the answer is A, this was the correct answer on usatestprep

Step-by-step explanation:


You might be interested in
If you divide -16 into 5 equal parts, how much is each part equal
zmey [24]

Answer:

-3.2

Step-by-step explanation:

-16/5=-3.2

6 0
3 years ago
A ship sails 250km due North qnd then 150km on a bearing of 075°.1)How far North is the ship now? 2)How far East is the ship now
olga_2 [115]

Answer:

1)  288.8 km due North

2)  144.9 km due East

3)  323.1 km

4)  207°

Step-by-step explanation:

<u>Bearing</u>: The angle (in degrees) measured clockwise from north.

<u>Trigonometric ratios</u>

\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}

where:

  • \theta is the angle
  • O is the side opposite the angle
  • A is the side adjacent the angle
  • H is the hypotenuse (the side opposite the right angle)

<u>Cosine rule</u>

c^2=a^2+b^2-2ab \cos C

where a, b and c are the sides and C is the angle opposite side c

-----------------------------------------------------------------------------------------------

Draw a diagram using the given information (see attached).

Create a right triangle (blue on attached diagram).

This right triangle can be used to calculate the additional vertical and horizontal distance the ship sailed after sailing north for 250 km.

<u>Question 1</u>

To find how far North the ship is now, find the measure of the short leg of the right triangle (labelled y on the attached diagram):

\implies \sf \cos(75^{\circ})=\dfrac{y}{150}

\implies \sf y=150\cos(75^{\circ})

\implies \sf y=38.92285677

Then add it to the first portion of the journey:

⇒ 250 + 38.92285677... = 288.8 km

Therefore, the ship is now 288.8 km due North.

<u>Question 2</u>

To find how far East the ship is now, find the measure of the long leg of the right triangle (labelled x on the attached diagram):

\implies \sf \sin(75^{\circ})=\dfrac{x}{150}

\implies \sf x=150\sin(75^{\circ})

\implies \sf x=144.8888739

Therefore, the ship is now 144.9 km due East.

<u>Question 3</u>

To find how far the ship is from its starting point (labelled in red as d on the attached diagram), use the cosine rule:

\sf \implies d^2=250^2+150^2-2(250)(150) \cos (180-75)

\implies \sf d=\sqrt{250^2+150^2-2(250)(150) \cos (180-75)}

\implies \sf d=323.1275729

Therefore, the ship is 323.1 km from its starting point.

<u>Question 4</u>

To find the bearing that the ship is now from its original position, find the angle labelled green on the attached diagram.

Use the answers from part 1 and 2 to find the angle that needs to be added to 180°:

\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{Total\:Eastern\:distance}{Total\:Northern\:distance}\right)

\implies \sf Bearing=180^{\circ}+\tan^{-1}\left(\dfrac{150\sin(75^{\circ})}{250+150\cos(75^{\circ})}\right)

\implies \sf Bearing=180^{\circ}+26.64077...^{\circ}

\implies \sf Bearing=207^{\circ}

Therefore, as bearings are usually given as a three-figure bearings, the bearing of the ship from its original position is 207°

8 0
2 years ago
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a boy kicks a 0.445kg soccer ball with the force of 75 newtons what is the acceleration of the soccer ball
SashulF [63]

Answer:

168.5 m/s^2

Step-by-step explanation:

6 0
3 years ago
This circle is centered at the point (4, 5), and the length of its radius is 3. What is the equation of the circle
inysia [295]
( x - 4 )^2 + ( y - 5 )^2 = 9

i am a mathematics teacher. if anything to ask please pm me
6 0
4 years ago
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Help Me for number 8
Simora [160]
One point is (-6,7) and the other one is (3,-4)
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3 years ago
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