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Lapatulllka [165]
2 years ago
10

A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na

utical miles west will the ship have traveled by 6:00 PM
Mathematics
1 answer:
ira [324]2 years ago
7 0

Answer:

Approximately 58.2\; \text{nautical miles} (assuming that the bearing is {\rm S$29^{\circ}$W}.)

Step-by-step explanation:

Let v denote the speed of the ship, and let t denote the duration of the trip. The magnitude of the displacement of this ship would be v\, t.

Refer to the diagram attached. The direction {\rm S$29^{\circ}$W} means 29^{\circ} west of south. Thus, start with the south direction and turn towards west (clockwise) by 29^{\circ} to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of v\, t. The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly 29^{\circ}.

Since the hypotenuse is of length v\, t, the dashed line segment opposite to the \theta = 29^{\circ} vertex would have a length of:

\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}.

Substitute in \begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned} and t = 6\; \text{hour}:

\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}.

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Can someone help me in this please???
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The first one is 15 mornings
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1. A student took 60 minutes to answer a combination of 20 multiple-choice and extended-response questions. She took 2 minutes t
Darina [25.2K]

Answer:

1) m=15 and r =5

2) 4 and 2 ml

3) x= 0.5 and y = -1

Step-by-step explanation:

given that m multiple choice questions and r extended response questions.

Also given that a) m+r = total no of questions =20 ... i and

                        2m+6r = total time taken = 60  ... ii

b) Divide second equation by 2, m+3r = 30  ... iii

                                                  m+r  = 20   ... i

Subtract to get 2r =10 or r =5

m = 20-5 = 15

Verify: m+r = 15+5 =20

and     2m+6r = 30+30 = 60 minutes.  

Hence verified

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

2) Let a litres of 20% solution and b litres of 50% solution be mixed

a) Then total volume = 6 ml = a+b  ... i

Resulting solution = 30% of 6 ml = 1.8 = 0.2a+0.5b ... ii

b) Solve i and ii

b = 6-a: substitute in ii.

0.2a + 3-0.5a = 1.8

Or a = 4 ml and b = 2ml

Verify: Total volume = a+b =6ml and

concentration = 0.2(4)+0.5(2) = 1.8 = 30% of 6 ml.

Thus verified

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

3) 2x-3y=4 ...i

  2x-5y =6... ii

Because x term has the same coefficient in both the equations, elimination is easier.

i-ii gives 2y =-2 or y =-1

Substitute in i, 2x+3 =4 or x = 0.5

So answer is x = 0.5 and y =1


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3 years ago
A sphere with a diameter of 5 feet long is being painted. how much more paint would be needed to paint a sphere that has four ti
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well, if the diameter is 5, thus its radius must be half that, or 2.5, and therefore, the radius of the one four times as much will be (4)(2.5).

Let's simply get their difference, since that'd be how much more is needed from the smaller to larger sphere.

~\hfill \stackrel{\textit{surface area of a sphere}}{SA=4\pi r^2}\qquad \qquad r=radius~\hfill \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{\large difference of their areas}}{\stackrel{\textit{radius of (4)(2.5)}}{4\pi (4)(2.5)^2}~~ - ~~\stackrel{\textit{radius of 2.5}}{4\pi (2.5)^2}}\implies 100\pi -25\pi \implies 75\pi ~~ \approx ~~235.62~ft^2

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