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Vitek1552 [10]
3 years ago
8

A slope of -2 passing through the point (-11,-17)

Mathematics
1 answer:
Leona [35]3 years ago
5 0

Answer:

y =  - 2x - 39

Step-by-step explanation:

y = mx + b

- 17 =  - 2( - 11) + b

- 17 = 22 + b

- 39 = b

y =  - 2x - 39

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Factor the following , it is not prime
MrMuchimi
So 1/9-16z^2 this is a diffirence of two perfect squares thing
so (1/3)^2-(4z)^2
so ((1/3)-4z)((1/3)+4z)

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
The coordinates of point A on a coordinate grid are (−2, −3). Point A is reflected across the y-axis to obtain point B and acros
amid [387]

Answer:

B) B(2,-3) and C(-2,3)

Step-by-step explanation:

The given point A, has coordinates (-2,-3).

When point A(-2,-3) is reflected over the y-axis to obtain point B, then the coordinates of B is obtained by negating the x-coordinate of A.

Therefore B will have coordinates (2,-3).

 When point A(-2,-3) is reflected over the x-axis to obtain point C, then the coordinates of C is obtained by negating the y-coordinate of A.

Hence the coordinates of C are (-2,3)

3 0
3 years ago
Out of every 6 students surveyed, 1 listen to country music. At that rate, how many students in a school of 1200 listens to coun
Degger [83]

Answer: 200

Step-by-step explanation:

1200 divided by 6 = 200

6 0
3 years ago
Suppose there is a 1.6°F drop in temperature for every thousand feet that an airplane climbs into the sky. If the temperature on
pogonyaev

Answer: 55.1°F

Step-by-step explanation:

We are informed that there is a 1.6°F drop in temperature for every thousand feet that an airplane climbs into the sky, if the plane reaches an altitude of 2,000ft, the drop in temperature will be:

= 1.6° × 2

= 3.2°F

If the temperature on the ground is 58.3°F​, then the temperature when the plane reaches an altitude of 2,000 ft will be:

= 58.3°F - 3.2°F

= 55.1°F

5 0
3 years ago
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