1.Find the polynomial f(x) of degree three that has zeroes at 1, 2, and 4 such that f(0) = -16.
2.For the following equation, find the interval(s) where f(x) %26lt; 0.
f(x) = 1 Divided by x^2-2x-8
3.The polynomial f(x) divided x - 3 results in a quotient of x^2+3x-5 with a remainder of 2. Find f(3).
4.Express the following statement as a formula with the value of the constant of proportionality determined with the given conditions: w varies directly as x and inversely as the square of y. If x = 15 and y = 5, then w = 36.
5.Find the quotient and remainder of f(x) = x^4 - 2 divided by p(x) = x - 1.
Find the polynomial f(x) of degree three that has zeroes at 1, 2, and 4 such that f(0) = -16.?
Too many questions. Here's the answer for the one in the title.
Zeroes at 1, 2, 4 means (x-1), (x-2), and (x-4) are factors. So f(x) = c(x-1)(x-2)(x-4) for some constant c. Multiplying the terms out gives c(x^3 -7x^2 + 14x - 8). Then f(0) = -8c, so c = 2 gives f(0) = -16. That means f(x) = 2x^3 - 14x^2 + 28x - 16.
Thursday, November 19, 2009
Finding "Hard-To-Find" Torrents?
I'm pretty good at finding things I want to download, but certain torrents are seemingly impossible to find......
I usually can find anything w/ isohunt and mybittorrents, but not the following:
28 Days (the Movie w/ Sandra Bullock)
First Wives Club
Yesterday's Children (Jane Seymour)
Wife Swap
Honey, We're Killing The Kids
How Clean Is Your House?
Rich Girls
Amish In The City
Mystery Diagnosis
What Not To Wear
Can you help????
Finding "Hard-To-Find" Torrents?
there could be alot of reasons ill give u a mini guide:::::::
get utorrent 1.6.1 not the latest release and not 1.6
http://www.download3000.com/download-uto...
then go to
www.ilovetorrents.com
www.iptorrents.com
www.leecherslair.com
www.filemp3.org
www.mixfiend.com
reason being these sites make the users upload as much as they download so they keep their upload speeds blazing and that results in higher download speed for you.
if you plan on using non private torrents then
www.torrentspy.com
www.mininova.org
www.thepiratebay.org
www.isohunt.com
but for good speeds on these sights look for lots of seeds with low leechers
Reply:Just use one site. One site to search them all. xD
http://www.plentyoftorrents.com
I usually can find anything w/ isohunt and mybittorrents, but not the following:
28 Days (the Movie w/ Sandra Bullock)
First Wives Club
Yesterday's Children (Jane Seymour)
Wife Swap
Honey, We're Killing The Kids
How Clean Is Your House?
Rich Girls
Amish In The City
Mystery Diagnosis
What Not To Wear
Can you help????
Finding "Hard-To-Find" Torrents?
there could be alot of reasons ill give u a mini guide:::::::
get utorrent 1.6.1 not the latest release and not 1.6
http://www.download3000.com/download-uto...
then go to
www.ilovetorrents.com
www.iptorrents.com
www.leecherslair.com
www.filemp3.org
www.mixfiend.com
reason being these sites make the users upload as much as they download so they keep their upload speeds blazing and that results in higher download speed for you.
if you plan on using non private torrents then
www.torrentspy.com
www.mininova.org
www.thepiratebay.org
www.isohunt.com
but for good speeds on these sights look for lots of seeds with low leechers
Reply:Just use one site. One site to search them all. xD
http://www.plentyoftorrents.com
Find the equation of the line which passes through A(3, -2) and B(-2, 4).?
i need the formula for the following 4 questions:
find the equation of the line which passes through A(3, -2) and B(-2, 4).
Find the slope of the line with equation 3x + 5y = -2
Find what "a" is on a right angle triangle with the coordinates being A(3, 2) B(-1,-3) C(7,a). With A being the point of the right angle.
Find the intersection of 2x+y=5 and 3x-2y=11.
thankyou.
Find the equation of the line which passes through A(3, -2) and B(-2, 4).?
slope = (4+2) /(-2-3) = 6 / -5
- 6/5 = (y + 2 ) / (x -3)
- 6 (x -3) = 5 (y + 2 )
18 - 6x = 5 y + 10
6x + 5y -8 = 0
3x + 5y = -2
5y = - 3x -2
y = - 3/5 x - 2/5
slope = - 3/5
Reply:line through (a, b) and (c, d) is
y -- b = [(d -- b)/(c - a)](x -- a)
slope of line ax + by = c is
m = --b/a
Formula to find a in your question is
(slope of AC)(slope of BC) = --1 and slope you found in question 1
Intersection of ax + by + c = 0 and
dx + ey + f = 0 is given by
[(bf --ce)/(ae -- bd), (cd -- af)/(ae -- bd)]
Reply:A(3,-2) and B(-2,4)
m = (4 - (-2))/(-2 - 3)
m = (4 + 2)/(-2 + (-3))
m = -6/5
A(3,-2), m = (-6/5)
-2 = (-6/5)(3) + b
-2 = (-18/5) + b
-10 = -18 + 5b
5b = 8
b = (8/5)
ANS : y = (-6/5)x + (8/5)
--------------------------------------...
3x + 5y = -2
5y = -3x - 2
y = (-3/5)x - (2/5)
ANS : m = (-3/5)
--------------------------------------...
A(3,2), B(-1,-3), C(7,a)
AB = sqrt((-1 - 3)^2 + (-3 - 2)^2)
AB = sqrt((-1 + (-3))^2 + (-3 + (-2))^2
AB = sqrt((-4)^2 + (-5)^2)
AB = sqrt(16 + 25)
AB = sqrt(41)
AC = sqrt((7 - 3)^2 + (a - 2)^2)
AC = sqrt(4^2 + (a - 2)^2)
AC = sqrt(16 + (a - 2)^2)
AC = sqrt(16 + ((a - 2)(a - 2)))
AC = sqrt(16 + (a^2 - 2a - 2a + 4))
AC = sqrt(16 + (a^2 - 4a + 4))
AC = sqrt(16 + a^2 - 4a + 4)
AC = sqrt(a^2 - 4a + 20)
BC = sqrt((7 - (-1))^2 + (a - (-3))^2)
BC = sqrt((7 + 1)^2 + (a + 3)^2)
BC = sqrt(8^2 + ((a + 3)(a + 3)))
BC = sqrt(64 + (a^2 + 6a + 9))
BC = sqrt(a^2 + 6a + 73)
(AC)^2 + (BC)^2 = (AC)^2
(sqrt(a^2-4a+20))^2 + (sqrt(a^2+6a+73))^2 = (sqrt(41))^2
a^2 - 4a + 20 + a^2 + 6a + 73 = 41
2a^2 + 2a + 93 = 41
2a^2 + 2a + 52 = 0
2(a^2 + a + 26) = 0
a^2 + a + 26 = 0
Sorry, i can't find a solution for this one.
--------------------------------------...
2x + y = 5
3x - 2y = 11
Multiply top by 2
4x + 2y = 10
3x - 2y = 11
7x = 21
x = 3
2(3) + y = 5
6 + y = 5
y = -1
ANS : (3,-1)
Reply:equation is ax+by+c = 0
3a-2b + c =0
-2a+4b+c = 0
subtract
5a - 6b = 0
so a:b = 6:5
allow a =6,b=5, then see what c has to be
(because we only need the ratio of a:b:c)
6x+5y+c = 0
18-10+c=0 c = -8
6x+5y+c=0
-12+20+c = 0
6x+5y-8 = 0
Reply:I'll do the last one for you.
For intersection questions you just have to make one of the equations in terms of either x or y (that means it says x = something or y=something) and then sub that into the other one.
2x+y=5 .... (1)
3x-2y=11 .... (2)
so with 1 you get y = 5-2x
sub that into 2
3x - 2(5-2x) = 11 (see how they're all x's now)
3x - 10 +4x = 11
7x = 21
x = 21/7
x=3
then if you sub that into either 1 or 2 you will get an answer for y
and that will give you a point (3,-1) that is their intesection
p.s you'd be better off asking each question seperate, more chance of getting each one answered
Reply:To find the equation of a line given two points (x1,y1), (x2,y2):
1. Find the slope m = (y1-y2)/(x1-x2)
In our case: m = (-2-4)/(3-(-2)) = -6/5
2. Point-slope form: Y-y1 = m(X-x1)
In our case: Y-(-2) = -6/5(X-3)
3. Solve for Y (slope-intercept form)
In our case:
Y = -6/5X + 18/5 -2
Y = -6/5X + 8/5
------------------------------
To find the slope, solve for y and then the coefficient of x will be the slope:
3x+5y=-2
5y = -3x - 2
y = -3/5 x - 2
Slope is -3/5
--------------------------------------...
First, note that angle BAC is right, as stated. Therefore, AB is perpendicular to AC.
The slope of AB = (2-(-3))/(3-(-1)) = 5/4.
The slope of AC = -4/5 (negative reciprocal because they are perpendicular)
We also know that the slope of AC is: (2-a)/(3-7). Thus:
(2-a)/(3-7) = -4/5
(2-a)/(-4) = -4/5
2-a = 16/5
a = 2-16/5
a = -6/5
----------------------------------
To find the intersection of two lines, you have to manipulate the two in such a way to eliminate a variable to solve for the other:
3x-2y=11
(2x+y=5) * 2 = 4x+2y = 10
Add 3x-2y=11 to 4x+2y=10:
7x = 21
x = 3
Now, plug in x=3 to either equation to solve for y:
3(3)-2y=11
9 - 2y = 11
-2y = 2
y = -1
Thus, the intersection is (x,y) = (3,-1)
Reply:Slope of line can be detemined with 2 points (x1,y1) %26amp; (x2, y2)
For A (3,-2) B(-2, 4)
slope m = (y2 - y1)/(x2-x1)
= (-2 - 4)/(3-(-2))
= -6/5
substitute into y = m x + b one of the points, say A
-2 = -6/5 * 3 + b
or -2 = -18/5 + b
b = -2 + 18/5 = 8/5
so equation of line is
y = -6/5 x + 8/5
Slope of 3x + 5y = -2
rearrange to give 5y = -3x -2
or y = -3/5 x - 2/5
Slope is -3/5
Line 1: 2x + y = 5
Line 2: 3x -2y = 11
Rearrange line 1 to terms of y
to give y = -2x + 5
Substitute into line 2 equation giving
3x - 2 ( -2x + 5) = 11
3x - (-4x +10) = 11
3x + 4x - 10 = 11
or 7x = 21 =%26gt; x = 3
substitute into y = -2 x + 5 gives y = -2 *3 + 5 = -6 + 5 = -1
so intersections is (3, -1)
teeth grinding
find the equation of the line which passes through A(3, -2) and B(-2, 4).
Find the slope of the line with equation 3x + 5y = -2
Find what "a" is on a right angle triangle with the coordinates being A(3, 2) B(-1,-3) C(7,a). With A being the point of the right angle.
Find the intersection of 2x+y=5 and 3x-2y=11.
thankyou.
Find the equation of the line which passes through A(3, -2) and B(-2, 4).?
slope = (4+2) /(-2-3) = 6 / -5
- 6/5 = (y + 2 ) / (x -3)
- 6 (x -3) = 5 (y + 2 )
18 - 6x = 5 y + 10
6x + 5y -8 = 0
3x + 5y = -2
5y = - 3x -2
y = - 3/5 x - 2/5
slope = - 3/5
Reply:line through (a, b) and (c, d) is
y -- b = [(d -- b)/(c - a)](x -- a)
slope of line ax + by = c is
m = --b/a
Formula to find a in your question is
(slope of AC)(slope of BC) = --1 and slope you found in question 1
Intersection of ax + by + c = 0 and
dx + ey + f = 0 is given by
[(bf --ce)/(ae -- bd), (cd -- af)/(ae -- bd)]
Reply:A(3,-2) and B(-2,4)
m = (4 - (-2))/(-2 - 3)
m = (4 + 2)/(-2 + (-3))
m = -6/5
A(3,-2), m = (-6/5)
-2 = (-6/5)(3) + b
-2 = (-18/5) + b
-10 = -18 + 5b
5b = 8
b = (8/5)
ANS : y = (-6/5)x + (8/5)
--------------------------------------...
3x + 5y = -2
5y = -3x - 2
y = (-3/5)x - (2/5)
ANS : m = (-3/5)
--------------------------------------...
A(3,2), B(-1,-3), C(7,a)
AB = sqrt((-1 - 3)^2 + (-3 - 2)^2)
AB = sqrt((-1 + (-3))^2 + (-3 + (-2))^2
AB = sqrt((-4)^2 + (-5)^2)
AB = sqrt(16 + 25)
AB = sqrt(41)
AC = sqrt((7 - 3)^2 + (a - 2)^2)
AC = sqrt(4^2 + (a - 2)^2)
AC = sqrt(16 + (a - 2)^2)
AC = sqrt(16 + ((a - 2)(a - 2)))
AC = sqrt(16 + (a^2 - 2a - 2a + 4))
AC = sqrt(16 + (a^2 - 4a + 4))
AC = sqrt(16 + a^2 - 4a + 4)
AC = sqrt(a^2 - 4a + 20)
BC = sqrt((7 - (-1))^2 + (a - (-3))^2)
BC = sqrt((7 + 1)^2 + (a + 3)^2)
BC = sqrt(8^2 + ((a + 3)(a + 3)))
BC = sqrt(64 + (a^2 + 6a + 9))
BC = sqrt(a^2 + 6a + 73)
(AC)^2 + (BC)^2 = (AC)^2
(sqrt(a^2-4a+20))^2 + (sqrt(a^2+6a+73))^2 = (sqrt(41))^2
a^2 - 4a + 20 + a^2 + 6a + 73 = 41
2a^2 + 2a + 93 = 41
2a^2 + 2a + 52 = 0
2(a^2 + a + 26) = 0
a^2 + a + 26 = 0
Sorry, i can't find a solution for this one.
--------------------------------------...
2x + y = 5
3x - 2y = 11
Multiply top by 2
4x + 2y = 10
3x - 2y = 11
7x = 21
x = 3
2(3) + y = 5
6 + y = 5
y = -1
ANS : (3,-1)
Reply:equation is ax+by+c = 0
3a-2b + c =0
-2a+4b+c = 0
subtract
5a - 6b = 0
so a:b = 6:5
allow a =6,b=5, then see what c has to be
(because we only need the ratio of a:b:c)
6x+5y+c = 0
18-10+c=0 c = -8
6x+5y+c=0
-12+20+c = 0
6x+5y-8 = 0
Reply:I'll do the last one for you.
For intersection questions you just have to make one of the equations in terms of either x or y (that means it says x = something or y=something) and then sub that into the other one.
2x+y=5 .... (1)
3x-2y=11 .... (2)
so with 1 you get y = 5-2x
sub that into 2
3x - 2(5-2x) = 11 (see how they're all x's now)
3x - 10 +4x = 11
7x = 21
x = 21/7
x=3
then if you sub that into either 1 or 2 you will get an answer for y
and that will give you a point (3,-1) that is their intesection
p.s you'd be better off asking each question seperate, more chance of getting each one answered
Reply:To find the equation of a line given two points (x1,y1), (x2,y2):
1. Find the slope m = (y1-y2)/(x1-x2)
In our case: m = (-2-4)/(3-(-2)) = -6/5
2. Point-slope form: Y-y1 = m(X-x1)
In our case: Y-(-2) = -6/5(X-3)
3. Solve for Y (slope-intercept form)
In our case:
Y = -6/5X + 18/5 -2
Y = -6/5X + 8/5
------------------------------
To find the slope, solve for y and then the coefficient of x will be the slope:
3x+5y=-2
5y = -3x - 2
y = -3/5 x - 2
Slope is -3/5
--------------------------------------...
First, note that angle BAC is right, as stated. Therefore, AB is perpendicular to AC.
The slope of AB = (2-(-3))/(3-(-1)) = 5/4.
The slope of AC = -4/5 (negative reciprocal because they are perpendicular)
We also know that the slope of AC is: (2-a)/(3-7). Thus:
(2-a)/(3-7) = -4/5
(2-a)/(-4) = -4/5
2-a = 16/5
a = 2-16/5
a = -6/5
----------------------------------
To find the intersection of two lines, you have to manipulate the two in such a way to eliminate a variable to solve for the other:
3x-2y=11
(2x+y=5) * 2 = 4x+2y = 10
Add 3x-2y=11 to 4x+2y=10:
7x = 21
x = 3
Now, plug in x=3 to either equation to solve for y:
3(3)-2y=11
9 - 2y = 11
-2y = 2
y = -1
Thus, the intersection is (x,y) = (3,-1)
Reply:Slope of line can be detemined with 2 points (x1,y1) %26amp; (x2, y2)
For A (3,-2) B(-2, 4)
slope m = (y2 - y1)/(x2-x1)
= (-2 - 4)/(3-(-2))
= -6/5
substitute into y = m x + b one of the points, say A
-2 = -6/5 * 3 + b
or -2 = -18/5 + b
b = -2 + 18/5 = 8/5
so equation of line is
y = -6/5 x + 8/5
Slope of 3x + 5y = -2
rearrange to give 5y = -3x -2
or y = -3/5 x - 2/5
Slope is -3/5
Line 1: 2x + y = 5
Line 2: 3x -2y = 11
Rearrange line 1 to terms of y
to give y = -2x + 5
Substitute into line 2 equation giving
3x - 2 ( -2x + 5) = 11
3x - (-4x +10) = 11
3x + 4x - 10 = 11
or 7x = 21 =%26gt; x = 3
substitute into y = -2 x + 5 gives y = -2 *3 + 5 = -6 + 5 = -1
so intersections is (3, -1)
teeth grinding
Where can I find the full version of this video?
I need help to find the full video version of the youtube video with the following id "aYVjhJ3dOCQ" or "qXtN4aJafcY". Here is what happened:
I work with blogs, so I'm used to find every kind of information I want. Two weeks ago a friend of mine told he couldn't find the full version of that video ("aYVjhJ3dOCQ"), he told me he was searching it for a month and couldn't find it. So I suggested:
" I bet US$60 that I can find it in two weeks ",
And he accepted. Nine days have passed and I still cant find it. It's not a big deal but I would like to win that bet so i anyone can help me I still have five days.
If someone wants to know my blog adress just send me an e-mail.
Where can I find the full version of this video?
You are looking for an Asian stripping on a webcam? That is pretty lame.
Reply:try limewire
I work with blogs, so I'm used to find every kind of information I want. Two weeks ago a friend of mine told he couldn't find the full version of that video ("aYVjhJ3dOCQ"), he told me he was searching it for a month and couldn't find it. So I suggested:
" I bet US$60 that I can find it in two weeks ",
And he accepted. Nine days have passed and I still cant find it. It's not a big deal but I would like to win that bet so i anyone can help me I still have five days.
If someone wants to know my blog adress just send me an e-mail.
Where can I find the full version of this video?
You are looking for an Asian stripping on a webcam? That is pretty lame.
Reply:try limewire
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
Ok. Where can I find a site where they give you a real scenery and they put a picture of you in it so it looks like your actually there? I really want to try that out. If you know site please post it here. Thanks!
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
blingee .com aadil
Reply:Download the program GIMP. It's free. You can use the layer feature to put in a background scene of your choice. Then you can get a picture of yourself, erase the background and put yourself into any scene you want. It will take time to erase the background unless you take a photo of yourself in front of a plain color.
Reply:i didn't answer your questions all the other times you asked it i don't know.
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
Ok. Where can I find a site where they give you a real scenery and they put a picture of you in it so it looks like your actually there? I really want to try that out. If you know site please post it here. Thanks!
Where can I find a site where I put in my picture and they place me in different scenes like red carpet, etc.?
blingee .com aadil
Reply:Download the program GIMP. It's free. You can use the layer feature to put in a background scene of your choice. Then you can get a picture of yourself, erase the background and put yourself into any scene you want. It will take time to erase the background unless you take a photo of yourself in front of a plain color.
Reply:i didn't answer your questions all the other times you asked it i don't know.
Where to find restored emails?
I have just restored some emails that I had deleted by accident. They went into the recycle bin. I clicked on 'restore all' but now can't find them. I checked through previous answers on here and find that the majority of people replied by saying 'back where they were' - well they aren't! They WERE in my inbox and sent items folders but those remain empty! The only answer that I found to be helpful was the one in the link, which just explains it all (and I understand) but the bottom line is how can I find them - I am prepared to 'go to a lot of trouble'!!!
This is the question that I found the helpful answer in:
http://uk.answers.yahoo.com/question/ind...
Where to find restored emails?
I think that the recycle bin would be unable to restore your deleted e-mails back to outlook but has probably restored them to a folder on your c drive.
I can only suggest that you do a full search of your c drive - all files and folders - word or phrase in file (your e-mail address).(it will be on both your in-box e-mails and your sent items)
Hope this helps you find them.
Reply:Hi there,
I am assuming that you are using Outlook as your email client since you never mentioned anything.
If you delete any emails from within Outlook they arent placed in the normal Recycle Bin (on your desktop) because they are treated differently from normal file (like what the link says).
They are instead kept in the "Deleted Item" INSIDE your Outlook. Try to look for your emails there (i hope when you delete your emails, you didnt hold down the SHIFT button).
This is the question that I found the helpful answer in:
http://uk.answers.yahoo.com/question/ind...
Where to find restored emails?
I think that the recycle bin would be unable to restore your deleted e-mails back to outlook but has probably restored them to a folder on your c drive.
I can only suggest that you do a full search of your c drive - all files and folders - word or phrase in file (your e-mail address).(it will be on both your in-box e-mails and your sent items)
Hope this helps you find them.
Reply:Hi there,
I am assuming that you are using Outlook as your email client since you never mentioned anything.
If you delete any emails from within Outlook they arent placed in the normal Recycle Bin (on your desktop) because they are treated differently from normal file (like what the link says).
They are instead kept in the "Deleted Item" INSIDE your Outlook. Try to look for your emails there (i hope when you delete your emails, you didnt hold down the SHIFT button).
Where can I find 'boyfriend' material guys???
I am a 20yo female.
I'm finding it hard to find a boyfriend! All the men I meet are after one thing and that turns me off.
I have no idea where to look to find a good, decent guy.
Im not un-attractive, I get checked out/ hit on basically on a regular basis, I just cant seem to find that one guy to be with.
Its getting very frustrating as I really want someone in my life to love and etc. Where do I find the right guy?
Ive been told that love happens unexpectedly...I honestly cant wait that long!
Any suggestions on what I should do???
Keep in my mind I live in Australia :)
Apparently im a great catch, then why the hell isnt anyone fishing for me??? Is it something that im doing or is it that good, decent men are hard to find???
Where should I be looking exactly?
Any tips on what I should be doing?
I have an age limit of 18-30 (not really into the younger fellas or the older fellas but age is just a number)
HELP!!!
Where can I find 'boyfriend' material guys???
Err Joe that was a line.....not an answer LOL This is funny!
As for you Matie, I believe the same thing is true for all of us.. we all want a 'man'. I think the answer is in your 'activities' and what you are interested in.. find your passion and start 'joining clubs' that center around your passion. I love birds, so I'd hit the bird clubs up ;) (as an example for myself)
Reply:Contact me joeonline_2005@yahoo.com
Joe
Gimmie a Smile I will give you my life
joeonline_2005@yahoo.com
teeth bleaching
I'm finding it hard to find a boyfriend! All the men I meet are after one thing and that turns me off.
I have no idea where to look to find a good, decent guy.
Im not un-attractive, I get checked out/ hit on basically on a regular basis, I just cant seem to find that one guy to be with.
Its getting very frustrating as I really want someone in my life to love and etc. Where do I find the right guy?
Ive been told that love happens unexpectedly...I honestly cant wait that long!
Any suggestions on what I should do???
Keep in my mind I live in Australia :)
Apparently im a great catch, then why the hell isnt anyone fishing for me??? Is it something that im doing or is it that good, decent men are hard to find???
Where should I be looking exactly?
Any tips on what I should be doing?
I have an age limit of 18-30 (not really into the younger fellas or the older fellas but age is just a number)
HELP!!!
Where can I find 'boyfriend' material guys???
Err Joe that was a line.....not an answer LOL This is funny!
As for you Matie, I believe the same thing is true for all of us.. we all want a 'man'. I think the answer is in your 'activities' and what you are interested in.. find your passion and start 'joining clubs' that center around your passion. I love birds, so I'd hit the bird clubs up ;) (as an example for myself)
Reply:Contact me joeonline_2005@yahoo.com
Joe
Gimmie a Smile I will give you my life
joeonline_2005@yahoo.com
teeth bleaching
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