K = 4pq^2
k/4p = q^2
0.5(k/p)^1/2 = q
Graph 3 is the answer. Choose a point on the line. Go down by 3. Then go to the right by 5.
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
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:
- For a constant acceleration:
, where
is the final velocity in a direction after the acceleration is applied,
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. - <em>Then for the x direction</em> it is known that the initial velocity is
5320 m/s, the acceleration (the applied by the engine) in x direction is
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: 
- In the same fashion, <em>for the y direction</em>, the initial velocity is
0 m/s, the acceleration in y direction is
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: 
Slope equals rise over the run
so if slope = 5 then the rise also equals 5 and the run is 1