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Paul [167]
3 years ago
10

List the steps used to solve the equation below.

Mathematics
1 answer:
pychu [463]3 years ago
4 0

Answer:

- 1

Step-by-step explanation:

2x + 3 = x - 4

2x - x = - 4 + 3

x = - 1

The solution to the above equation is - 1.

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Solve for q. k = 4pq²
9966 [12]
K = 4pq^2
k/4p = q^2
0.5(k/p)^1/2 = q
7 0
4 years ago
Read 2 more answers
The line represented by the equation 3x + 5y = 2 has a slope of -3/5. Which shows the graph of this equation?
Daniel [21]
Graph 3 is the answer. Choose a point on the line. Go down by 3. Then go to the right by 5.
4 0
3 years ago
What is the value of p+9 when p=16​
Tasya [4]

Answer:

25 :)

Step-by-step explanation:

well, if it's p+9 and you know that p=16 then you plug 16 in for p so the equation will be 16+9 and then you solve from there which is 25

4 0
3 years ago
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
When a linear function has a slope of 5, what is the "run" part of the slope?
alukav5142 [94]
Slope equals rise over the run
so if slope = 5 then the rise also equals 5 and the run is 1


7 0
3 years ago
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