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mash [69]
3 years ago
15

Suppose you and some friends go to the movies and buy some snacks. The snack bar charges $2 for a box of candy and $6 for the “c

ombo.” The combo is a medium drink and popcorn. The only spending restriction you have is you must bring home some change from the money you have been given. After buying the tickets, you have $12 left to spend for snacks.
Mathematics
1 answer:
aleksley [76]3 years ago
3 0
If what you are trying to do is buy the most you can and still having some change left over then the answer would be 1 combo and 2 boxes of candy
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A S A P ! ! ! ! <br><br> I nEEd help again!!
atroni [7]

Answer:

3.00 feet

Step-by-step explanation:

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6 0
3 years ago
A team is selling discount cards as a fundraiser. The table shows the number of cards sold and the remaining amount of money nee
kondor19780726 [428]

Answer: negative.


Step-by-step explanation:

We know that the slope of the line \frac{\text{change in y}}{\text{change in x}}

Therefore, the slope for the given data=\frac{y_2-y_1}{x_2-x_1}

\Rightarrow\ Slope=\frac{795-875}{20-10}\\=\frac{-80}{10}\\=-8...,\text{which is negative.}

Hence, the word which describes the slope of the line representing the data in the table = negative.

8 0
3 years ago
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Please help need it due now!!!
ahrayia [7]

Answer:

1) 0.9, 0.99, 0.09, 0.0009

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3) 110

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4 0
2 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
What’s the distance between the points (-7,-4) and (-5,-4)?
Radda [10]

Answer:

2 units

Step-by-step explanation:

Distance

= \sqrt{(-4 - (-4))^2 + (-7 - (-5))^2}

= \sqrt{0^2 + (-2)^2}

= \sqrt{(-2)^2}

= 2

3 0
3 years ago
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