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gtnhenbr [62]
3 years ago
11

Because of hiring a plumber, y, to work x hours on a repair project can be modeled using a linear function. the plumber charges

a fixed cost of $80 plus an additional cost of $45 per hour. the plumber works a maximum of 8 hours per day
Mathematics
1 answer:
seraphim [82]3 years ago
4 0

Answer:

The plumbers cost would be equal to the equation 80+45x=y with y being the total amount of money he made and x being how many hours he worked.

Step-by-step explanation:

In this equation to find the amount the plumber was paid all you need to do is plug in the amount of hours he worked for x.

If the plumber worked his maximum of 8 hours then the equation would change from 80+45x=y to 80+(45×8)=440.

So, if the plumber worked for 8 hours he would make $440.

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79+82+85+91+97+73+88+87+85+92+90+85+86+91+94+85+92=1482 divided by 17 ( the amount of numbers) = 87.17 
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Answer:

Liam can paint 18 bird houses.

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What is the equation of the line that best fits the given data? A graph has points (1, 7), (1, 6), (2, 5), (2.5, 5), (3, 3), (4,
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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

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2 years ago
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The answer it -2
:::::::::::::::::::::::::::::::::::
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3 years ago
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