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Murrr4er [49]
3 years ago
12

If 5 miles = 8km. how many km in 250 miles?

Mathematics
2 answers:
rjkz [21]3 years ago
7 0
The anwser is 400 km!!!
vfiekz [6]3 years ago
5 0
5 miles / 8 km=250 miles / x

x=(250 miles * 8 km) / 5 miles=400 Km.

solution: 400 Km
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which of these pairs of points defines a line whith a slope of 2/3 A. (2, 6) and (3, 0) B. (6, 2) and 3, 0) C. (6, 2) and (0, 0)
JulsSmile [24]
The answer for the question is B
8 0
3 years ago
If a = 17.0079, b = 2.05, and c = 3.1415926, then what is a/b + c?
podryga [215]
A / b + c

17.0079 / 2.05 + 3.1415926 = 
8.29653658537 + 3.1415926 =
11.4381291854 <==
4 0
3 years ago
A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th
Leno4ka [110]

The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

  • The bearing of the buoy from the is approximately <u>307.35°</u>

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}

Therefore;

\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}

Which gives;

\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}

Similarly, we can get;

\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}

The angle subtended by the distance of the buoy from the port, <em>C</em> is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ <u>307.35°</u>

Learn more about bearing in mathematics here:

brainly.com/question/23427938

5 0
2 years ago
Hey guys i really need help pls dont waste the points i actually want help to understand how to solve equations like:
tamaranim1 [39]

Answer:

your question is unclear

Step-by-step explanation:

x+2y=4

-x -x

2y=4-x

divide by 2 to get y by it's self

y=2-(1/2x)

x+2y=4

-2y -2y

x=4-2y

i don't know if u want me to help u with the second one but i did it the first one

3 0
2 years ago
Jennifer and carmen have the same amount of money to spend on carnival tickets Jennifer buys 4 tickets and has 8.40 left carmen
Sliva [168]
Do the following:

$8.40-$7.15=$1.25
Then divide 1.25 by the difference of tickets which is 3
1.25/3=0.42 (that's rounded to nearest cent)

3 0
3 years ago
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