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sergiy2304 [10]
2 years ago
7

A boat travels with velocity vector (25, 25 StartRoot 3 EndRoot). What is the directional bearing of the boat?

Mathematics
2 answers:
GenaCL600 [577]2 years ago
7 0

Answer:

A on Egde

Step-by-step explanation:

i got it right

iogann1982 [59]2 years ago
6 0

Answer:

The directional earing is N 30° E

Step-by-step explanation:

The given velocity vector is (25, 25·√3), therefor, we have;

R = 25·i + 25·√3·j

The angle of inclination of the direction of the boat with the horizontal = Arctan (y/x)

Where;

(x, y) = The coordinate of the components of the velocity of the boat

Therefore;

Arctan (y/x)  =  Arctan (25·√3/25) = Arctan (√3) = 60° North of East

The directional bearing which is the bearing relative to the northern direction is given as follows;

The directional earing = 90° - 60° = N 30° E.

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If a cyclist increases his speed by 3 miles per hour, he can reduce the time required for a 6 mile trip by 10 minutes. What is t
raketka [301]

Answer:

Slower speed = 9 mph

Step-by-step explanation:

Let the initial speed (slower speed) = s mph

Speed 's' = \frac{\text{Distance}}{\text{Time}}

            s = \frac{6}{t} -------(1)

If the cyclist increases the speed by 3 mph then the final speed = (s + 3) mph

He can reduce the time 't' by 10 minutes (\frac{1}{6} hours)

Now, (s + 3) = \frac{6}{t-\frac{1}{6}}

s + 3 = \frac{36}{6t-1} --------(2)

Substitute the value of 's' from equation (1) to the equation (2).

\frac{6}{t}+3=\frac{36}{6t-1}

\frac{2}{t}+1=\frac{12}{6t-1}

\frac{2}{t}-\frac{12}{6t-1}=-1

\frac{2(6t-1)-12t}{t(6t-1)}=-1

\frac{12t-2-12t}{t(6t-1)}=-1

-\frac{2}{t(6t-1)}=-1

6t² - t = 2

6t² - t - 2 = 0

6t² - 4t + 3t - 2 = 0

2t(3t - 2) + 1(3t - 2) = 0

(2t + 1)(3t - 2) = 0

t = -\frac{1}{2},\frac{2}{3} hours

But the time can't be negative.

Therefore, t = \frac{2}{3} hours is the answer.

From equation (1),

s = \frac{6}{\frac{2}{3}}

s = 9 miles per hour.

Slower speed = 9 mph is the answer.

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Step-by-step explanation:

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