Bro just go on Symbolab and use the calculator
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:
See explanation and hopefully it answers your question.
Basically because the expression has a hole at x=3.
Step-by-step explanation:
Let h(x)=( x^2-k ) / ( hx-15 )
This function, h, has a hole in the curve at hx-15=0 if it also makes the numerator 0 for the same x value.
Solving for x in that equation:
Adding 15 on both sides:
hx=15
Dividing both sides by h:
x=15/h
For it be a hole, you also must have the numerator is zero at x=15/h.
x^2-k=0 at x=15/h gives:
(15/h)^2-k=0
225/h^2-k=0
k=225/h^2
So if we wanted to evaluate the following limit:
Lim x->15/h ( x^2-k ) / ( hx-15 )
Or
Lim x->15/h ( x^2-(225/h^2) ) / ( hx-15 ) you couldn't use direct substitution because of the hole at x=15/h.
We were ask to evaluate
Lim x->3 ( x^2-k ) / ( hx-15 )
Comparing the two limits h=5 and k=225/h^2=225/25=9.
