Suppose a line has slope -2/3 and passes through the point (1, 8). which other point must also be on the graph?
1 answer:
<h3>Answer: Choice A. (7,4)</h3>
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Explanation:
Use the slope and given point to find the y intercept
y = mx+b
8 = (-2/3)*(1) + b
8 = -2/3 + b
8 + 2/3 = b
24/3 + 2/3 = b
26/3 = b
b = 26/3
The equation of the line is y = (-2/3)x + 26/3
To confirm this, plug in x = 1 and we should get y = 8, due to the point (1,8)
y = (-2/3)x + 26/3
y = (-2/3)*1 + 26/3
y = -2/3 + 26/3
y = (-2+26)/3
y = 24/3
y = 8
So that verifies we have the correct equation.
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Next, go through each answer choice to see if the x coordinate of the point leads to the y coordinate.
If we try x = 7, then,
y = (-2/3)x + 26/3
y = (-2/3)(7) + 26/3
y = -14/3 + 26/3
y = (-14+26)/3
y = 12/3
y = 4
This shows that (7,4) is on the line. Choice A is the answer
That rules out choice B.
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If we tried x = -5, then,
y = (-2/3)x + 26/3
y = (-2/3)(-5) + 26/3
y = 10/3 + 26/3
y = 36/3
y = 12
meaning that (-5,12) is on the line. That rules out choices C and D.
Refer to the graph below. It visually confirms that of the four answer choices, only point A is on the line. I used GeoGebra to make the graph.
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Answer: The quotient is (x-2).
Step-by-step explanation:
Since we have given that
Now, we have to find the quotient of the above expression.
So, here we go:
Now, we will divide the above simplest form with g(x):
Hence, the quotient is (x-2).