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vodka [1.7K]
3 years ago
11

Completely factor the given expression over the integers. (Correct gets brainliest)

Mathematics
1 answer:
elena-14-01-66 [18.8K]3 years ago
7 0

Answer:

50

Step-by-step explanation:

would it be 50 because you the q to the power of 2 is 2q. Then you would subtract to the other side of the equation so it would be -100=-2q. After you would divided the -2 in -2q on both sides and -100/-2 is 50. So your answer is q=50

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Answer:

A. On a coordinate plane, a curved line with a minimum value of (negative 1, negative 2), crosses the x-axis at (negative 2.5, 0) and (0.5, 0), and the y-axis at (0, negative 1).

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A. On a coordinate plane, a curved line with a minimum value of (negative 1, negative 2), crosses the x-axis at (negative 2.5, 0) and (0.5, 0), and the y-axis at (0, negative 1).

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C.  On a coordinate plane, a curved line with a maximum value of (0, 2) and a minimum value of (1.5, negative 2.75), crosses the x-axis at (negative 0.6, 0) and (0.6, 0).

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Help pls fast and give me the steps<br> -15a-19-35=-84
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Linda goes water-skiing one sunny afternoon. After skiing for 15 min, she signals to the driver of the boat to take her back to
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60 = a * (-30)^2
a = 1/15
So y = (1/15)x^2


abc)
The derivative of this function is 2x/15. This is the slope of a tangent at that point.
If Linda lets go at some point along the parabola with coordinates (t, t^2 / 15), then she will travel along a line that was TANGENT to the parabola at that point.
Since that line has slope 2t/15, we can determine equation of line using point-slope formula:
y = m(x-x0) + y0
y = 2t/15 * (x - t) + (1/15)t^2
Plug in the x-coordinate "t" that was given for any point.


d)
We are looking for some x-coordinate "t" of a point on the parabola that holds the tangent line that passes through the dock at point (30, 30).
So, use our equation for a general tangent picked at point (t, t^2 / 15):
y = 2t/15 * (x - t) + (1/15)t^2
And plug in the condition that it must satisfy x=30, y=30.
30 = 2t/15 * (30 - t) + (1/15)t^2
t = 30 ± 2√15 = 8.79 or 51.21
The larger solution does in fact work for a tangent that passes through the dock, but it's not important for us because she would have to travel in reverse to get to the dock from that point.
So the only solution is she needs to let go x = 8.79 m east and y = 5.15 m north of the vertex.
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