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3 years ago

231 km 811 m + 485 km 829 m

Mathematics

2 answers:

asambeis [7]3 years ago

7 0

Add them all together to get ur answer a
then just try to figure out the units say

Eva8 [605]3 years ago

6 0

231+811+485+829=2356km

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Please help!! This is super important!!

MAVERICK [17]

Answer:

tell me the choices so i can help you............

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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:

Bearing: The angle (in degrees) measured clockwise from north.

Trigonometric ratios

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

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)

Cosine rule

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

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.

Question 1

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.

Question 2

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.

Question 3

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.

Question 4

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

Read 2 more answers

I NEED HELP!!! PLEASE!!!!!!!!!!!!!!!!!

ale4655 [162]

The awnser to ur question is A because take 2.3 time 3 and get 6.9 and then add the 108 and -3 and get 105

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3 years ago

A rectangular swimming pool is 40 feet long and 20 feet wide. The pool is bordered by square tiles that are x feet in length sur

olga2289 [7]

Answer:

(40 + 2x)feet

Step-by-step explanation:

see attachment for the figure.

Let 'L' represents the length of the rectangular swimming pool i.e 40ft wide.

and pool is bordered by square tiles that are x feet in length surrounding the pool

therefore,

In order to find the longest side of the pool and the tile border, take the sum of length of the swimming pool i.e 40 adding two times the wide of the square tiles

(40+2x)feet.

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4 years ago

In a bag of marbles, 1/2 are red, 1/4 are blue, 1/6 are green and 1/12 are yellow. You pick a marble without looking. What color

igomit [66]

Answer:

Its red becasue 1/2 is greater than all of them

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3 years ago