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luda_lava [24]
3 years ago
6

Which set of coordinates doesn’t represent the problem?

Mathematics
1 answer:
Semmy [17]3 years ago
5 0
ThE answer is C, congrats
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What is the value of x in the equation 8+4 = 2(x-1)?<br> 5<br> 11/2<br> 13/2<br> 7
Citrus2011 [14]

Answer:

x = 7

Step-by-step explanation:

I hope this helps!

4 0
3 years ago
Which graph represents a proportional relationship?<br> A<br> B<br> C<br> D
Andre45 [30]

Answer:

D!

Step-by-step explanation:

its pretty hard to explain but let me put it like this- it crosses through (0,0) and it mirrors itself in the 1st and 3rd quadrants

6 0
3 years ago
Use properties of addition and subtraction to evaluate the expression.<br><br> −28−47−12
Brrunno [24]
The answer would be -87 
6 0
3 years ago
Find the distance between the two points (2,5) and (6,-1)
Reptile [31]

Answer:

= \sqrt{52} units

Step-by-step explanation:

x₁ = 2 ; y₁ = 5

x₂ = 6 ; y₂ = -1

D = \sqrt{(x₂ - x₁)² + (y₂ - y₁)²}

   =\sqrt{(6 - 2)² + (-1 -5)²}

   =\sqrt{4² + (-6)²}

   = \sqrt{16 + 36}

D = \sqrt{52} units

7 0
3 years ago
Read 2 more answers
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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