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I need help on 26 and for 37, i forget how to find the length.
Thanks
And another problem. x^2-4x-12=0, it says solve by completing the square and don't factor.
Math help
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il0seonpurpose
Korea (South)5638 Posts
I need help on 26 and for 37, i forget how to find the length. Thanks And another problem. x^2-4x-12=0, it says solve by completing the square and don't factor. | ||
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Meta
United States6228 Posts
midpoint is half the length in the direction of the other point. edit: kau figured out the midpoint, i misread one of the points >< I totally have no idea how to do 26 lol, been like 4 years since i did that algebra | ||
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il0seonpurpose
Korea (South)5638 Posts
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thedeadhaji
39489 Posts
http://mathforum.org/library/drmath/view/53160.html lol totally forgot about this method. | ||
Kau
Canada3502 Posts
length = sqrt(64+36) length = 10 midpoint_x = (-6+2))/2 = -2 midpoint_y = (-2+4)/2 = -1 midpoint = (-2,-1) (3-sqrt(2)) / (2sqrt(3)+5) (3-sqrt(2)) / (2sqrt(3)+5) * [(2sqrt(3)-5) / (2sqrt(3)-5)] (3-sqrt(2))*(2sqrt{3}-5) / (4*3-25) (6sqrt(3)-15-2sqrt(6)+5sqrt(2)) / (4*3-25) -(5sqrt(2)+6sqrt(3)-2sqrt(6)-15) / 13 x^2-4x-12=0 x^2-4x+4-4-12=0 (x-2)^2-16=0 (x-2)^2=16 x-2=4 x=6 | ||
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raiame
United States421 Posts
For 26, you multiply the top and the bottom by the conjugate of the bottom, so multiply top and bottom by 5-2rt(3). Bottom turns out to be 13, top becomes something else (too lazy). As for completing the square, you just find the square that has x^2-4x in it, which is (x-2)^2=x^2-4x+4. Make the left side into that by adding 16 to both sides, then take the square root and account for both positive and negative. | ||
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il0seonpurpose
Korea (South)5638 Posts
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il0seonpurpose
Korea (South)5638 Posts
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goldrush
Canada709 Posts
You take the zeroes, that's where y = 0, therefore: (x+2)(x)(x-3)(x-4) = y is the formula, because if x = -2, 0, 3 or 4, y = 0. | ||
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Jathin
United States3505 Posts
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il0seonpurpose
Korea (South)5638 Posts
y=(x+1)^2-3 y=lnx y=1/(x-2) | ||
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thoraxe
United States1449 Posts
On August 12 2008 07:10 goldrush wrote: 39: You take the zeroes, that's where y = 0, therefore: (x+2)(x)(x-3)(x-4) = y is the formula, because if x = -2, 0, 3 or 4, y = 0. oh wow, totally forgot how to do that one, thanks for the refresh. *All that you have to do now is multiply all those variables.* | ||
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Roffles
Pitcairn19291 Posts
On August 12 2008 07:25 il0seonpurpose wrote: Real quick, what are the domain and ranges of the following? y=(x+1)^2-3 y=lnx y=1/(x-2) Domain: X values which exist. Range: Y values which exist. Plug some numbers in each equation, find it out. Or graph it and find it yourself. You should be spanked for not trying these yourself. | ||
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il0seonpurpose
Korea (South)5638 Posts
On the y=1/(x-2), all numbers work expect 2 but I don't know how to put that in a proper way. Do I just say all real numbers but 2? | ||
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Kau
Canada3502 Posts
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blabber
United States4448 Posts
Pretty sure. Sorry it's been awhile haha (like two years ago!) | ||
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ZpuX
Sweden1230 Posts
the range for y = ln(x) should be any positive number bigger than 0. | ||
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Pressure
7326 Posts
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Kau
Canada3502 Posts
D: (-infty, infty) R: [-2, infty) y=lnx D: (0,infty) R: (-infty, infty) y=1/(x-2) D: x =/= 2 R: y =/= 0 | ||
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ZpuX
Sweden1230 Posts
On August 12 2008 07:57 Kau wrote: y=lnx D: (0,infty) R: (-infty, infty) should be D: (>0) ? | ||
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Kentor
United States5784 Posts
[0, infinity) | ||
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Polemarch
Canada1564 Posts
On August 12 2008 06:33 Kau wrote: x^2-4x-12=0 x^2-4x+4-4-12=0 (x-2)^2-16=0 (x-2)^2=16 x-2=4 x=6 Don't forget the negative root x-2 = 4 OR x-2 = -4 | ||
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