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.
Answer:
512
Step-by-step explanation:
8*8*8
64*8
512
Answer:
I am not sure but I think it is 3
Step-by-step explanation:
2x+29=5x+20
5x-2x+29=20
3x=-9
1x=3
Answer:
4
Step-by-step explanation:
- If you take the lowest half of the data (from the lower extreme to the median) and find the median of these numbers, that is called the lower quartile or quartile 1.
- If you take the upper half of the data (from the upper extreme to the median) and find the median of these numbers, that is called the upper quartile or quartile 3.
- The interquartile range is the difference between quartile 1 and quartile 3.
- 4, 5, 6, 8, 9, 10, 11, 13
- Median=8.5
- Q1=6.5
- Q3=10.5
- 10.5-6.5=4
I also added my PowerPoint if you didn't get what I was saying.
Answer:
D. (4,3)
Step-by-step explanation:
I calculated it logically
