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Vlad1618 [11]
3 years ago
8

The maximum height reached by the barnacle is _______ m.

Mathematics
2 answers:
katrin2010 [14]3 years ago
8 0

Answer:

1 answer is 1

2 answer is -1

3rd answer is 2pi

Oksanka [162]3 years ago
5 0

Answer:

1

Step-by-step explanation:

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BRAINLIEST + POINTS?!<br><br> Can someone help me please explain!
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Answer:bhwuwhn enshwh nehe

Step-by-step explanation:

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3 years ago
Solve for 2:<br> 2x + 62°<br> 52 + 38°<br> 2 =
Paladinen [302]

Answer:

The answer is 8°

Step-by-step explanation:

2x + 62° = 5x + 38°

2x – 2x + 62° – 38° = 5x – 2x + 38° – 38°

24° = 3x

3x = 24°

3x/3 = 24°/3

x = 8°

Thus, The value of x is 8°

<u>-TheUnknownScientist 72</u>

8 0
2 years ago
David has a piece of wood that is 7.28 meters long. He needs 4 pieces of wood that are each 1.75 meters long. If he cuts the woo
Tema [17]

9514 1404 393

Answer:

  yes

Step-by-step explanation:

You can go at this a couple of ways:

1. The total length David needs is (1.75 m) × 4 = 7.00 m. His 7.28 m piece is long enough to provide the pieces he needs.

__

2. The length of each piece when the wood is cut into quarters is ...

  (7.28 m)/4 = 1.82 m

This is longer than the 1.75 m needed, so each piece is long enough.

5 0
3 years ago
Encuentra la velocidad promedio de un móvil que durante su recorrido hacia el Norte tuvo las siguientes velocidades.
yan [13]

The mean or average speed of a mobile moving north is: <u>Vprom= 20.575 m/sec</u>

The average speed of a series of given speed values ​​is calculated by <u>summing</u> each of the <u>speeds</u> and then <u>dividing</u> by the number of speeds , as follows:

Speeds:

V1=18.5m/s

V2=22m/s

V3=20.3m/s

V4=21.5m/s

 V avg =?

Average V = ( V1+V2+V3 +V4)/4

 Vavg = (18.5m/s + 22m/s +20.3m/s + 21.5m/s)/4

Vavg = 20.575nm/sec

To consult visit: brainly.com/question/11100327

5 0
2 years ago
What is the value of mc013-1.jpg?<br> –4<br> –2<br> 2<br> 4<br><br> the pic is the .jpg
Black_prince [1.1K]
The answer is 2 if your dividing if not comment me and ill answer it
4 0
3 years ago
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