I think But I'm not 100% sure yet it's a guess
Y=3+x
Root: (-3, 0)
Intercept:(0, 3)
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.
Parallel slope = 2
This is because the original line has slope 2. Parallel lines have equal slopes.
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Perpendicular slope = -1/2
The original slope 2 or 2/1 turns into -1/2 after flipping the fraction and the sign. Note the original slope and perpendicular slope multiply to -1. This is true for any pair of perpendicular lines where neither line is vertical.
20% of 3km is 0.6 Kilometer
The answer is C. 8. Then both can have a denominator that looks like this..
6/8 and 5/8. Hope it helps!
