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.
Step-by-step explanation:
V=hpir^2
r=10
v=6283
6283=hpi10^2
6283=100hpi
divide both sides by 100
62.83=hpi
aprox pi=3.141592
divide both sides by pi
19.9994143=h
so about 20 inches
Answer:
m<B=m<K=100
m<A=m<J=133
m<L=m<C=41
m<M=m<D=86
Step-by-step explanation:
Since 
, the corresponding angles are congruent.
This implies that:
m<L=41=m<C
m<M=86=m<D
The sum of angles in a quadrilateral is 360 degrees.
m<B+133+41+86=360
m<B+260=360
m<B=360-260
m<B=m<K=100
Let’s find the unit rate of the problem
If we divide both $15 and 3 by 3 we get $5 and 1.
That means every package costs $5, so the COP or k =5
b. Using 5 as the k in y=kx, the equation is y=5x
