Pre-Calculus Pre-Calculus Math Forum

 September 5th, 2009, 09:47 AM #1 Newbie   Joined: Sep 2009 Posts: 18 Thanks: 0 Please help me on my precalc HW~ SOO CONFUSED! ty Hey, I just came back from vacation and I'm panicing because school starts in 2 days and none of this makes sense. I've spent hours and hours trying to do those pages and I've managed to get 4/8 pages done but the other 4 are very confusing. Will someone who has taken Math Analysis (Precalc) please explain to me those questions? Thanks! Any help is greatly appreciated. Since the foums cuts half of the image, here is the direct link http://img32.imageshack.us/img32/7130/hwproblems.jpg THANKS AGAIN TO ANYONE WHO OFFERS HELP.
 September 5th, 2009, 11:00 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,950 Thanks: 1141 Math Focus: Elementary mathematics and beyond Re: Please help me on my precalc HW~ SOO CONFUSED! ty I didn't see the problems at the link you gave. You can upload the file using the "Upload attachment" button below the text box.
 September 5th, 2009, 11:39 AM #3 Newbie   Joined: Sep 2009 Posts: 18 Thanks: 0 Re: Please help me on my precalc HW~ SOO CONFUSED! ty You can open the file up with winrar. I have 5 scans of my problems and my internet is too slow to upload them one by one. I'm using my neighbor's wifi since my parents for got to pay the bills for the internet.
 September 5th, 2009, 04:56 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,809 Thanks: 2150 The scans are a pain to read. Most are just text that you could type here and post easily. They seem to be intended to be straightforward questions that are based on various definitions, formulas and examples you've already been given and should read first.
 September 5th, 2009, 07:38 PM #5 Newbie   Joined: Sep 2009 Posts: 18 Thanks: 0 Re: Please help me on my precalc HW~ SOO CONFUSED! ty I've spent the entire day trying to figure these questions out. I think I got afew more done...or not. I think they're all wrong anyways... I have a picture hosted on imageshak now.
 September 6th, 2009, 07:58 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,809 Thanks: 2150 26)  The mid-point of AB = ((-4 + 6)/2, (5 - 9)/2) = (1, -2). The gradient of the perpendicular bisector of AB = (6 - (-4))/(5 - (-9)) = 5/7. Hence the equation of the perpendicular bisector of AB is 5x - 7y = 5 - 7(-2) = 19 and all points equidistant from A and B lie on this line. Putting x = n, y = 2 and solving for n gives n = (19 + 7(2))/5 = 33/5 = 6 3/5 = 6.6. (I don't know if the above is the method you've been shown, but as another question asks you to find the equation of a perpendicular bisector, I chose to show how to do that.) 27)  The line 3x - 2y = 11 has gradient 3/2, so any perpendicular to that line has gradient -2/3. The line DE has gradient (k - (-9))/(8 - 3), so (k + 9)/5 = -2/3. Hence k = -5(2/3) - 9 = -37/3 = -12 1/3. Comparing the above with what you jotted down suggests you were being careless, so take your time and make sure each step you do is correct. 2  Do it again, taking more care, and post a neat answer for us to check. 29)  M(-6, 4) = ((-3 + x)/2, (-8 + y)/2), so x = 2(-6) + 3 = -9 and y = 2(4) + 8 = 16. 31)  Correct. 32)  The equation does have the form y = -(3/7)x + c, but you got the wrong value for c. Try again. 34)  Your method was okay, but you made a careless arithmetical error; try again. 35)  Try again, doing what the question asked.
 September 6th, 2009, 01:44 PM #7 Newbie   Joined: Sep 2009 Posts: 18 Thanks: 0 Re: Please help me on my precalc HW~ SOO CONFUSED! ty Thanks a million for your help. I really hope I can understand all of this before school starts. I have no logic in math related things at all. It isn't really that much carelessness as much as cluelessness. Again, thank you.
 September 6th, 2009, 03:24 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,809 Thanks: 2150 35)  The altitude is perpendicular to NP, not to the horizontal axis. 36)  NP has equation x + 12y = -26, so the altitude through Q(6, 7) has equation 12x - y = 65. By solving these equations, the point of intersection is (26/5, -13/5). 40)  NQ has gradient 2/5. PQ has gradient -5/2. As these have product -1, NP is perpendicular to PQ, i.e., triangle NPQ has a right-angle at Q. 41)  The area of triangle NPQ is exactly 116. Rounding it to 3 s.f. makes no difference! If a triangle's vertices are $(x_1,\,y_1),\,(x_2,\,y_2)$ and $(x_3,\,y_3),$ the triangle's area is $|x_1(y_3\,-\,y_2)\,+\,x_2(y_1\,-\,y_3)\,+\,x_3(y_2\,-\,y_1|/2.$
 December 11th, 2014, 10:07 PM #9 Newbie   Joined: Dec 2014 From: England Posts: 1 Thanks: 0 M(-6, 4) = ((-3 + x)/2, (-8 + y)/2), so x = 2(-6) + 3 = -9 and y = 2(4) + 8 = 16. hehehe? Inconsistant

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