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dimulka [17.4K]
3 years ago
11

A number line going from negative 4 to positive 4. What is the distance between 0 and –2? Use the number line to answer the ques

tion. -2 0 1 2
Mathematics
2 answers:
vaieri [72.5K]3 years ago
5 0

Answer:

A number line going from negative 4 to positive 4.

What is the distance between 0 and –2?

Use the number line to answer the question.

-2  

0

1

2

Step-by-step explanation:

-2 is correct

batgirl1000000000000000000000
2 years ago
it is not -2 it is just 2
Nana76 [90]3 years ago
4 0

Answer:

I wish <em>i could help </em>

Step-by-step explanation:

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Answer:

The velocities after 739 s of firing of each engine would be 6642.81 m/s in the x direction and 5306.02 in the y direction

Step-by-step explanation:

  1. For a constant acceleration: v_{f}=v_{0}+at, where  v_{f} is the final velocity in a direction after the acceleration is applied, v_{0} is the initial velocity in that direction  before the acceleration is applied, a is the acceleration applied in such direction, and t is the amount of time during where that acceleration was applied.
  2. <em>Then for the x direction</em> it is known that the initial velocity is v_{0x} = 5320 m/s, the acceleration (the applied by the engine) in x direction is a_{x} 1.79 m/s2 and, the time during the acceleration was applied (the time during the engines were fired) of the  is 739 s. Then: v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}
  3. In the same fashion, <em>for the y direction</em>, the initial velocity is  v_{0y} = 0 m/s, the acceleration in y direction is a_{y} 7.18 m/s2, and the time is the same that in the x direction, 739 s, then for the final velocity in the y direction: v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}
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