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mariarad [96]
3 years ago
13

Find an angle 0 coterminal to -301°, where 0°

Mathematics
2 answers:
Zielflug [23.3K]3 years ago
6 0

Answer:

what do you mean?!??!!!!

Tems11 [23]3 years ago
5 0
I don’t understand...What do you mean
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QT 1.
Delvig [45]
Median is the middle to find it you have to organize the list.

 2, 4, 4, 5, 6, 7, 8, 8, 8.

cross them out one by one from the ends and you get 6.

your answer will be B. $6

hope this helped!

:)
6 0
3 years ago
The slope of a line is –2 and its y-intercept is (0, 3). What is the equation of the line that is parallel to the first line and
notsponge [240]

Answer:

D. y=-2x-6

Step-by-step explanation:

<u><em>First start with what we know....</em></u>

y = -2x + 3 (Slope Intercept Form)

<u><em>Because of this we can eliminate B.  </em></u>

<u><em>Parallel means that the lines wouldn't be touching which means they should have the same slope and the only one with the same slope is D. </em></u>

7 0
3 years ago
Read 2 more answers
Generate and solve another division problem with the same quotient and remainder as the two problems below.explain your strategy
Andrew [12]

Answer:

ITS AN ASSESMENT THIS VIOLATES THE HONOR CODE

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Help ASAP ILL GIVE 10 POITNS TO ANYONE THAT WILL HELP
DIA [1.3K]
End answer is five hundred and 6

3 0
3 years ago
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A spacecraft is traveling with a velocity of v0x = 5320 m/s along the +x direction. Two engines are turned on for a time of 739
OlgaM077 [116]

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}
8 0
3 years ago
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