Since 17 isnt a factor of 42 the only factor is 1
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.
The three numbers are 12, 18 and 24
Arithmetic progression
Let the 3 number in arithmetic progression be:
a-d, d, a+d ...
If their sum is 3, then;
a-d+d+a+d = 3
2a + d = 3 ........... 1
If the sum of their squares is 11, then;
(a-d)² + d² + (a+d)² = 11
a²-2ad+d²+d²+a²+2ad+d² 11
2a²+3d² = 11 ....... 2
Solving the equations simultaneously, d = 6 and a = 12
First-term = 12
second term = 18
Thirs term = 24
Hence the three numbers are 12, 18 and 24
Hope this helps you!!!!!! :D
1. algebra x
2. _______ a -----------b _______c _________d
3. right angle
4. quadrilateral
5. algebra project a
opposites
lines
angles
right angles
defines
Answer:
yes Wyatt is very correct
