4 square meters equals 4.784 square yards
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: 
2x+2 < x+3
2x-x < 3-2
x <1
Yes it would be 59. 59 is the correct answer. Hopefully
Kudos to the poster on the typeset math.
With cube roots we don't have to worry that things are positive.
![\sqrt[3]{16 x^4} - \sqrt[3]{128 x} = \sqrt[3]{2^4 x^4} - \sqrt[3]{2^7 x} = \sqrt[3]{2^3 x^3 2x} - \sqrt[3]{4^3 2 x} = 2x\sqrt[3]{2x} - 4\sqrt[3]{2 x}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7B16%20x%5E4%7D%20-%20%5Csqrt%5B3%5D%7B128%20x%7D%20%3D%20%20%5Csqrt%5B3%5D%7B2%5E4%20x%5E4%7D%20-%20%5Csqrt%5B3%5D%7B2%5E7%20x%7D%20%3D%20%5Csqrt%5B3%5D%7B2%5E3%20x%5E3%202x%7D%20-%20%5Csqrt%5B3%5D%7B4%5E3%202%20x%7D%20%3D%202x%5Csqrt%5B3%5D%7B2x%7D%20-%204%5Csqrt%5B3%5D%7B2%20x%7D)
![= (2x-4)\sqrt[3]{2 x}](https://tex.z-dn.net/?f=%3D%20%282x-4%29%5Csqrt%5B3%5D%7B2%20x%7D%20%20%20)
