Ask question

Ask question

All categories

Artyom0805 [142]

3 years ago

15

The course for a boat race starts at point A and proceeds in the direction of S52'E for 1 hour at 8 knots to point B and then in

the direction S50°W for 1 hour at 4.5 knots to point C and finally back to A. Find the total distance of the race and find the compass bearing from point A to point to the nearest whole degree.

1 answer:

KATRIN_1 [288]3 years ago

8 0

Answer:

20.823 nautical miles

S20°E

Step-by-step explanation:

I assume you mean ° (degrees), not ' (minutes). There are 60 minutes in 1 degree.

S52°E means "south, 52° east", or 52° east of south.

S50°W means "south, 50° west", or 50° west of south.

1 knots = 1 nautical mile / hour, so the boat first travels 8 nautical miles from A to B, then 4.5 nautical miles from B to C, then finally back to A.

If we say A is at the origin, then the coordinates of B are:

(8 sin 52°, -8 cos 52°)

And the coordinates of C are:

(8 sin 52° − 4.5 sin 50°, -8 cos 52° − 4.5 cos 50°)

(2.857, -7.818)

So the distance from A to C is:

x = √(2.857² + (-7.818)²)

x ≈ 8.323

And the total distance of the race is:

d = 8 + 4.5 + 8.323

d = 20.823

The compass bearing from A to C is:

θ = atan(2.857 / 7.818)

θ ≈ S20°E

You might be interested in

What is the mean (average) of these five numbers?

IRINA_888 [86]

Answer:

C. 5

Step-by-step explanation:

Mean: Numbers over amount of numbers

(2 + 4 + 5 + 6 + 8)/5

25/5

5

4 0

2 years ago

find the centre and radius of the following Cycles 9 x square + 9 y square +27 x + 12 y + 19 equals 0

Citrus2011 [14]

Answer:

Radius: $r =\frac{\sqrt {21}}{6}$

$Center = (-\frac{3}{2}, -\frac{2}{3})$

Step-by-step explanation:

Given

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Solving (a): The radius of the circle

First, we express the equation as:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

So, we have:

$9x^2 + 9y^2 + 27x + 12y + 19 = 0$

Divide through by 9

$x^2 + y^2 + 3x + \frac{12}{9}y + \frac{19}{9} = 0$

Rewrite as:

$x^2 + 3x + y^2+ \frac{12}{9}y =- \frac{19}{9}$

Group the expression into 2

$[x^2 + 3x] + [y^2+ \frac{12}{9}y] =- \frac{19}{9}$

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

Next, we complete the square on each group.

For $[x^2 + 3x]$

1: Divide the $coefficient\ of\ x\ by\ 2$

2: Take the $square\ of\ the\ division$

3: Add this $square\ to\ both\ sides\ of\ the\ equation.$

So, we have:

$[x^2 + 3x] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}$

$[x^2 + 3x + (\frac{3}{2})^2] + [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Factorize

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y] =- \frac{19}{9}+ (\frac{3}{2})^2$

Apply the same to y

$[x + \frac{3}{2}]^2+ [y^2+ \frac{4}{3}y +(\frac{4}{6})^2 ] =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ (\frac{3}{2})^2 +(\frac{4}{6})^2$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =- \frac{19}{9}+ \frac{9}{4} +\frac{16}{36}$

Add the fractions

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{-19 * 4 + 9 * 9 + 16 * 1}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{21}{36}$

$[x + \frac{3}{2}]^2+ [y +\frac{4}{6}]^2 =\frac{7}{12}$

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

By comparison:

$r^2 =\frac{7}{12}$

Take square roots of both sides

$r =\sqrt{\frac{7}{12}}$

Split

$r =\frac{\sqrt 7}{\sqrt 12}$

Rationalize

$r =\frac{\sqrt 7*\sqrt 12}{\sqrt 12*\sqrt 12}$

$r =\frac{\sqrt {84}}{12}$

$r =\frac{\sqrt {4*21}}{12}$

$r =\frac{2\sqrt {21}}{12}$

$r =\frac{\sqrt {21}}{6}$

Solving (b): The center

Recall that:

$(x - h)^2 + (y - k)^2 = r^2$

Where

$r = radius$

$(h,k) =center$

From:

$[x + \frac{3}{2}]^2+ [y +\frac{2}{3}]^2 =\frac{7}{12}$

$-h = \frac{3}{2}$ and $-k = \frac{2}{3}$

Solve for h and k

$h = -\frac{3}{2}$ and $k = -\frac{2}{3}$

Hence, the center is:

$Center = (-\frac{3}{2}, -\frac{2}{3})$

6 0

2 years ago

Someone help me asap math 10

Ghella [55]

Answer:

x ≈ 25.4 Km

Step-by-step explanation:

Using the tangent ratio in the right triangle on the right

tan54° = $\frac{opposite}{adjacent}$ = $\frac{opp}{17}$ ( multiply both sides by 17 )

17 × tan54° = opp , thus

opp ≈ 23.4

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

Using the cosine ratio in the right triangle on the left

cos23° = $\frac{adjacent}{hypotenuse}$ = $\frac{23.4}{x}$ ( multiply both sides by x )

x × cos23° = 23.4 ( divide both sides by cos23° )

x = $\frac{23.4}{cos23}$ ≈ 25. 4 Km ( to the nearest tenth )

6 0

3 years ago

When solved for x, which inequality represents the number line?

ikadub [295]

Answer:

x<-3

The value of x is found in all the numbers that are less than - 3 or all the numbers from - 3 to negative infinity

Please note: - 3 is not included in those numbers

4 0

2 years ago

What is the area of rectangle ABCD?

lina2011 [118]

Multiply the length by the width and your answer will be there.

8 0

2 years ago

Read 2 more answers