hubby signed up for this course quite a while ago, but hasnt gotten started yet (busy with work and procrastinating).
http://www.georgebrown.ca/Marketing/FTCal/caet/T902.aspx
does this sound accurate:
http://www.emcourse.com/
?
what are his chances of getting a decent paying (more...
k thanks :)
but could i write that as:
[(x+1)^2]/4 + [(y-2)^2]/16 = -1
(i don't think its standard form to have a "-" infront of the equation)
and for this one, b^2 = 16 ?
(i have to do other calculations)
~Amy
thanks. i expanded it and got:
4x^2 - y^2 + 8x + 4y + 4 = 0
so.. either something went wrong when i was expanding, or the standard form i figured out is incorrect?
~Amy
if you don't mind.. one more hyperbola question:
4x^2 - y^2 + 8x + 4y + 16 = 0
what is this in standard form?
in standard form:
(x+1)^2/4 - (y-2)^2/16 = 1
does that look accurate?
~Amy
thanks
i meant to say hyperbola :shy:
i have 8 lines of work and my typing numbers is slow.. so here's part of my calculations:
x^2/ 25 - (4-x^2)/ 9 = 1
225(x^2/25) - 225 ((4-x)^2/9) = 225(1)
9x^2 - 25(16 - 4x - 4x + x^2) = 225
9x^2 - 25(16 - 8x + x^2) = 225
9x^2 - 400 + 200x -...
the parabola equation is:
(x^2/25) - (y^2/9) = 1
the line is y = 4-x
according to my calculations, if i point y - 4-x into the equation, i get
-16x^2 + 200x - 175 = 0. is that right so far?
~Amy
"a parabolic arch has an equation x^2 + 10y - 10 = 0. the arch is on a hill with equation y = 0.1x-1. (measurements are in metres).
"find the points of intersection"
for this i substituted y =0.1x - 1 into the equation x^2 + 10y - 10 = 0:
x^2 + 10(0.1x - 1) - 10 = 0
x^2 + x - 10 - 10 = 0...
thanks chaoeverlasting, but i didnt quite understand everything you said.
here's what I've done:
if x = 1, y = 1.37 (1, 1.37)
foci (-2, 0), (2, 0)
d = *square root*(x2 - x1)^2 + (y2 - y1)2
( i do this for (1, 1.37) and (-2, 0). and then for (1, 1.37) and (2, 0))
= 3.30, and...
thank you very much!
i have another quick question:
an elliptical pool table is 5m at its longest point, and 3m wide at its widest point. the pool table has two holes at the position of the foci.
so for the equation i have figured out:
x^2/6.25 + y^2/2.25 = 1.
and the foci are at...
thanks!
i won't show all of it (half page long), but here's the main parts:
x^2 - 4x + 4y^2 + 8y = 60
(x^2 - 4X + 4) + (4(y^2 - 2y + 1) = 68
(i got the 68 by added 60 + 4 + 4(1))
~Amy
Homework Statement
an ellipse is represented by the equation:
x|^2 + 4y^2 - 4 x + 8y - 60 = 0
express the equation in standard form:
((x-2)^2 / 68) + ((y-4)^2/17) = 1
can anyone tell me if this is accurate? thanks
~Amy