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.
First, we know L and W are the same number. The in the end the Length is 6 more, therefore we add 6. The equations begins as so
2x+6=27
-6
2x=21
/2
x=10.5
+6
Therefore the Width =10.5 M
The Length = 16.5 M
The answer is 124 deg. because all the equation is 3x+40=7x-72 and when you solve for x, you get 28. Then all you have to do is plug 28 into the equation 3x+40 and you get 124..:)):
Answer:
Step-by-step explanation:
3n+2)! / (3n-3)! = (3n+2)*(3n+1)*(3n)*(3n-1)*(3n-2)*(3n-3)! divided by (3n-3)! which becomes:
23! / 18! = (23 * 22 * 21 * 20 * 19 * 18! divided by 18! which becomes:
(23 * 22 * 21 * 20 * 19)
the result is 4037880
i used my ti-84 to take 23! and divide it by 18! and i got 4037880.
this confirms the solution is correct.
Answer:
ASA theorum
Step-by-step explanation:
we are given 1 side and 1 angle, other angle is vo of it corresponding equal angle
