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:
False ; the answer to that equation is 56
Step-by-step explanation:
We know that
Right triangles PBM and MTF are similar
because
angle PMB=angle TMF
and
angle BPM=angle FTM
and
angle B =angle F=90 degrees
so
corresponding sides are
BM and MF
PB and TF
PM and MT
(PB/TF)=BM/MF
solve for PB
PB=(TF*BM)/MF
where
TF=6ft
BM=20 ft
MF=3 ft
so
PB=(6*20)/3------> 40 ft
the answer is
<span>the height of the peak is 40 ft</span>
Answer: 15
REMEMBER: The mode is the value that appears most frequently in a data set.
300 - 30 3/10 = 269 7/10
hope this helps
