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learned a bit more about this stuff today i did....
x^2 = 4ay where a is the distance from the vertex..... and your vertex is at 0,0
167/4 is a so the focus is at..... 0+ 167/4
right. i have that its the first part that im having trouble with
ok.....we know that it 100 feet across, so that is 50 feet in both directions, when y = 50, what is x^2 = ?
is it headed in the x direction or the y?
the parabola opens up
then I would turn it into the normal way of viewing this: y =(1/167)x^2. when x = 50, whats y=?
so thats how deep it is?
the numbers are the same, we just repositioned them into the "normal" way that this equation would look.
the vertex is at (0,0) thats half way, half of the diameter is 50...plug it in :)
14.97 is what I get
k I'll try that. thanks!
youre welcome :)
have you ever worked with quadratic equations?
cause thats all a parabola is...
yes. im in pre calc at a jr college, this stuff is being thrown at me quick but im doing my best.
phcc sound familiar to you?
no whats that?
its the community college I go to right now :) just wondering if you were in my trig/pre calc class ....
na im at tcc
well, good luck to ya :)