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MArishka [77]
3 years ago
11

8. Expo markers cost $4.99 per pack. There is a 20%

Mathematics
1 answer:
Alexus [3.1K]3 years ago
8 0

Answer:

34.17

Step-by-step explanation:

8 packs x 4.99/pack=39.92

39.92x.8(100-20% discount)=31.94

31.94 x 1.07% (includes 7% sales tax)=34.17

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Select all ratios equivalent to 5:3.<br> 13:9<br> 9:15<br> 20:12
torisob [31]

Answer:

20:12

brainlest if it helped

Step-by-step explanation:

7 0
2 years ago
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8.
lukranit [14]

Answer:

4

Step-by-step explanation:

The y-coordinate of the point where both lines intersect will give us the same solution for x.

At y = 4, both equations will have a corresponding x-value of 1.

Therefore, the value of y that that will make both equations have the same solution of x is 4.

7 0
3 years ago
-6x-16 &lt; 2x please help asap! and help me solve it
ICE Princess25 [194]

1. Divide both sides by 2

2. Add 16 to each side

Answer  x > -2

         

3 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
I need help pleaseeeeee​
MA_775_DIABLO [31]

Answer:

D

Step-by-step explanation:

5 0
3 years ago
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