6х + Зу = 12 Slope-intercept

1 answer:

4 0

Answer:

x=2,y=4

Step-by-step explanation:

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Anon25 [30]

Answer:

D, 22x^9

Step-by-step explanation:

Here's a key for end behavior: Look at the leading term

x^even = x -> neg inf, f(x) -> pos inf; x -> pos inf, f(x) -> pos inf

-x^even = x -> neg inf, f(x) -> neg inf; x -> pos inf, f(x) -> neg inf

x^odd = x -> neg inf, f(x) -> neg inf; x -> pos inf, f(x) -> pos inf

-x^odd = x -> neg inf, f(x) -> pos inf; x -> pos inf, f(x) -> neg inf

6 0

2 years ago

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:

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.
Then for the x direction 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: $v_{fx}=v_{0x}+a_{x}t=5320\frac{m}{s} +1.79\frac{m}{s^{2} }*739s=6642.81\frac{m}{s}$
In the same fashion, for the y direction, 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: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$