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sineoko [7]
1 year ago
8

A boat sails on a bearing of 038°anf then 5km on a bearing of 067°.

Mathematics
1 answer:
I am Lyosha [343]1 year ago
7 0

This question is not complete

Complete Question

A boat sails 4km on a bearing of 038 degree and then 5km on a bearing of 067 degree.(a)how far is the boat from its starting point.(b) calculate the bearing of the boat from its starting point

Answer:

a)8.717km

b) 54.146°

Step-by-step explanation:

(a)how far is the boat from its starting point.

We solve this question using resultant vectors

= (Rcos θ, Rsinθ + Rcos θ, Rsinθ)

Where

Rcos θ = x

Rsinθ = y

= (4cos38,4sin38) + (5cos67,5sin67)

= (3.152, 2.4626) + (1.9536, 4.6025)

= (5.1056, 7.065)

x = 5.1056

y = 7.065

Distance = √x² + y²

= √(5.1056²+ 7.065²)

= √75.98137636

= √8.7167296826

Approximately = 8.717 km

Therefore, the boat is 8.717km its starting point.

(b)calculate the bearing of the boat from its starting point.

The bearing of the boat is calculated using

tan θ = y/x

tan θ = 7.065/5.1056

θ = arc tan (7.065/5.1056)

= 54.145828196°

θ ≈ 54.146°

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Check the picture below.

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5 0
2 years ago
Denise is constructing A square in which two of its vertices are points M and N. She has already used her straightedge and compa
yan [13]

Denise is constructing A square.

Note: A square has all sides equal.

We already given two vertices M and N of the square.

And another edge of the square is made by from N.

Because a square has all sides of equal length, the side NO should also be equal to MN side of the square.

Therefore, <em>Denise need to place the point of the compass on point N and draw an arc that intersects N O, using MN as the width for the opening of the compass. That would make the NO equals MN.</em>

Therefore, correct option is :

D) place the point of the compass on point N and draw an arc that intersects N O, using MN as the width for the opening of the compass.

6 0
2 years ago
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Yi Min is a pitcher on her softball team. This season,
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Answer:

48

Step-by-step explanation:

0.8 times 60= 48

4 0
2 years ago
Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the proporti
lesya [120]

Answer:

a) the sample size (n) = 156.25≅ 156

Step-by-step explanation:

<u>Step1 </u>:-

Given the two sample sizes are equal so n_{1} =n_{2} = n

Given the standard error (S.E) = 0.04

The standard error of the proportion of the given  sample size

S.E = \sqrt{\frac{pq}{n} }

Step 2:-

here we assume that the proportion of boys and girls are equally likely

p= 1/2 and q= 1/2

S.E = \sqrt{\frac{p(1-p)}{n} } \leq \frac{\frac{1}{2} }{\sqrt{n} }

\sqrt{n} = \frac{\frac{1}{2} }{S.E}

squaring on both sides, we get

n = \frac{1}{(2X0.04)^{2} }

on simplification, we get

n= 156.25 ≅ 156

sample size (n) = 156

<u>verification</u>:-

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4 0
1 year ago
The area of a rectangle is 25/42 and its width is 5/6. What is the length? ASAP
Serhud [2]

Answer:

<h2>length =  \frac{25}{28}</h2>

Step-by-step explanation:

Let the length of the rectangle be l

Area of a rectangle = length × width

From the question

Area = 25/42

width = 5/6

Substitute the values into the above formula and solve for the length

That's

<h3>length =  \frac{area}{width}</h3>

So we have

<h3>length =  \frac{25}{42}  \div  \frac{4}{6}  \\  =  \frac{25}{42}  \times  \frac{6}{4}  \\  =  \frac{25}{7}  \times  \frac{1}{4}</h3>

We have the final answer as

<h3>\frac{25}{28}</h3>

Hope this helps you

8 0
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