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3 years ago

Write an equation for this graph please help me I'm begging u

Mathematics

1 answer:

bekas [8.4K]3 years ago

4 0

First, solve for the slope. This can be found by looking at the y and x intercepts. At x = 0, y = 1.5. At x = 2, y = 0.

Slope is defined as Δy/Δx, or the change in y over the change in x. This means that in order to calculate the slope, you must find the difference between the values of y and divide it by the difference in the values of x for the two points to determine the slope between them.

(0 - 1.5)/(2-0) = (-1.5)/2 = -0.75 or -3/4

Now that you have the slope, we can write the equation in slope intercept form, y = mx + b, where m is the slope we calculated and b is the y intercept, 1.5.

y = (-3/4)x + 1.5

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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:

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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: $v_{fy}=v_{0y}+a_{y}t=0\frac{m}{s} +7.18\frac{m}{s^{2} }*739s=5306.02\frac{m}{s}$

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3 years ago