educating someone in geometry, I have a friend who likes the game of pool but her geometry is poor.  How would I go about doing this
201232
The whole thing in pool is that the angle of inflection is equal to the angle of reflection.  I wonder, I've never tried this, if using a laser pointer to show the line of sight and the place on the ball where the cue ball should hit.  You could also use a mirror and a laser to find the position on a the bumper to shoot the cue ball for a bank shot.  If you move the mirror along the bank with a laser behind the cue ball pointing at the mirror, the only time it will shine on the ball intended to be struck will be at the position you would like to aim the cue ball in.  \n\nI just made all this up, so I don't know if it would work.  But, it would be a good visual way to describe what you are trying to explain to your friend.
489268
good- geometry is hard for alot of kids-just try to bring that one grade up to a C and I'd say your golden
573646
Ive noticed you were asking quite alot of these geometry questions. I think you should pay more attention in class and you wouldnt need to ask home work questions, and plus, just knowing the answer wont help, youll have to know how to do it, especially if this is for a final exam...
413056
you can probably searth that like search Homework Help/ Tutoring websites for Geom/honors    and tht will probably work.
420555
There are many "Diamond systems" out there. You may want to check out Robert Byrne's books. Also for general pool physics, I would suggest the site below. Tons of articles on everything from basic aiming and tangent lines, to more advances stuff, including some videos.
636773
buddy..go to www.cbse.nic.in\nand then to:sample papers...u wil get loads of them!\nall the best!!waisae,co ordinate geometry,is the easiest!contact me for more help!\nall the best...
201089
Aha, geometry!\n\nDraw a diagram and follow the steps below.\n\nThe trick is to show that the triangles DBE and \ntriangle DAF are similar.\n\nProof That triangles DBE and triangle DAF are similar.\n\nLet <ABC> donote angle ABC.\nWe need to show any two corresponding angles of these traingles are equal.\n\nFirst angle,\n<FDA> = <BDE>  because D,B,F collinear and D,E,A collinear.\n\nSecond angle,\n<DAF> = <DAC> because A,F,C collinear\n      = (pi) - <ACD> - <ADC> angles in triangle ACD\n      = (pi) - <ACD> - <ADC>  ...(1)\n\nNow, separately  manipulate,<ACD> and <ADC>.\n\n<ACD> = <ACB> + <BCD> splitting <ACD> with line CB\n      = <ACB> + <CAB>, tangent property <BCD> = <CAB> \n      = (pi) - <CBA>  angles in triangle ABC\n      = <ABE>  because C,B,E are all colinear\n\nNow, similarly\n\n<ADC> = <ADB> + <BDC> splitting <ADC> with line DB\n      = <ADB> + <BAD>, tangent property <BDC> = <BAD> \n      = (pi) - <ABD>  angles in triangle ABD\n         \nSubstituiting these two info into (1), we get,  \n\n<DAF> = (pi) - <ACD> - <ADC>\n      = (pi) - <ABE> - [ (pi) - <ABD> ]\n      = <ABD> - <ABE>\n      = <BDE> because BE splits <ABD> \n\nDone. In showing the triangle DAF and triangle DBE are similar.\n\nHence, by ratio for similar triangles,\n\n        DA/DF = DB/DE\n\nOr Rewriting we get\n \n      DA * DE = DB * DF\n\nProved.\n\nNot so difficult, was it. hehehe.\n\nRegards
144142
Here is a way to find more problems than you probably want to do - go to Google, and search for:\n\n"algebra I practice questions"   \n\n(grade 9)\n\n"geometry practice questions"\n\n(grade 10)\n\nBe sure to use the quotes.  I just tried and hundreds of nice sites woth good problems appeared.  Good for you for wanting to learn on your own!
377442
Simply draw a chart that has each of the 6 variables at right angles to each other, draw lines between the linked variables such coke-greenbike (#3), then use process of elimination.\n\nYeah, right... draw a 6-dimensional hypercube....\n\n Jeez, I've spent about 3 hours on this, and so far it looks like Carla, who was on the yellow bike which was 1st in line, was Bigbird and brought rootbeer and jelly beans.  Either Emily, who was dressed as either dracula or the clown, or Daniel dressed as the other, brought the candycorn.  But at the moment I have a contradiction, which means that somewhere I made a logical error.  I'll try to get back to this sometime later - good luck
478692
draw FBDs of the 2 blocks independently\nbe careful\nfind the accelrations of the  2 blocks using CONSTRAINT relations\nand comes ur answer\n\ndon't wait for the solution\nit lessens ur thinking power n creativity\n\nif it's still problematic, consult ur lecturer\n\nconcentrate on ur studies \nur still a college guy
441422
Why not leave your house and actually play real pool with real people for once.\n\nIt's not just a computer game you know.
101248
http://www.schoolingkids.com/analytical-co-ordinate-cartesian-geometry/study-notes/
461956
Hopefully you'll find some help here: http://www.coursework4you.co.uk/sprtlitra11.htm
663781
Go get one of those French for beginers books to have him practice. Or make a game out of it yourself. My teacher plays Hexagons with us. Cut a bunch of hexagons out and put French words on some sides and English words on the others. It's like a puzzle. The French words have to match up with the English words. \nOr play Attaque. Make a gameboard. Some squares say "Je" "Tu" "Il" "Elle" "Nous" "Vous" "Ils" et "Elles". He will have to then draw a card when he lands on one of those spots. The cards have verbs on them. He will have to write the correct form of the verb. For example, say he lands on "Tu" and picks up the card "Manger". He will then have to write down "Tu manges". If he gets it right, he gets 3 points. Then other squares say Attaque. Then he'll have to pick up a verb card and write all the forms of the verb. EX: "manger" Je mange, tu manges, il/elle/on mange, nous mangeons, vous mangez, et ils, elles mangent. He either gets 5 points or can steal someone else's points if he wins.The other squares say "Quel est ceci?" or "What is this?". If he lands on these squares, he picks up a card and says what the verb is in english. EX: "manger" = "to eat" He gets 1 point for this.
445079
Well hey, you need help, ask the teacher! I know how to do this, with help from my nerdy Calculus BC friend, but it's really not that hard once you think about it...
469917
It is always understandable when a student feels overwhelmed by "word problems" because they are often presented in an intimidating manner.  If only they could be FUN again ... if YOU wrote the problem, I believe you might find it fun!  :-)\n\nBut let's allow ourselves some joy once we uncover all the unknowns in this problem.  Perhaps we'll get results from converting "written words" into "mathematical expressions."\n\nWhat's most important is that you understand the shape involved (think "shoe box") and how the "openings" are placed (think "windows" on three faces of the shoe box).  All the openings are understood to be identical.\n\nUsing only the letter X -- which stands for the length of an opening -- we can indeed come up with expressions for every other dimension of the container, as well as the openings.\n\nX = Opening LENGTH (given)\nX-2 = Opening HEIGHT ("length is 2 more than height")\n\n2X+8 = Container LENGTH ("8 more than twice X")\n3 = Container HEIGHT (given)\nX+2 = Container WIDTH ("2 more than length of opening")\n\nThe surface area of the container's sides (minus the three openings) is given as 300 square meters.  The equation is also given as 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings), so let's just set these equal:\n\n300 = 2 (sides) + 2 (front/back) + 2 (top/bottom) - 3 (openings)\n\nAll we need to do is to write, and then substitute, expressions for the areas in parentheses.\n\nSide area = length * height = 3 (2X+8) = 6X + 24\nFront or back area = width * height = 3 * (X+2) = 3X + 6\nTop or bottom area = length * width = (2X+8) * (X+2)\n____________________________ = 2X^2 + 12X + 16\nOpening area = length * height = X (X-2) = X^2 - 2X\n\nTime to substitute:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n\nNow that this (long) equation has been written, it is necessary to solve for X, since we need that value in order to calculate what the problem asks for:  The area of each element.\n\nCollect like terms:\n\n300 = 2 (6X+24) + 2 (3X+6) + 2 (2X^2+12X+16) - 3 (X^2-2X)\n300 = 12X + 48 + 6X + 12 + 4X^2 + 24X + 32 - 3X^2 + 6X\n300 = X^2 + 48X + 92\nX^2 + 48X - 208 = 0\n\nWhat we hope is that the trinomial on the right-hand side of this equation is able to be factored easily!  Does it?\nMy approach is to factor '-208' and see whether I can 'build' the linear term's coefficient ('48') from the factors.\n\n-208 = -2 * 104 = -2 * 2 * 52 = -2 * 2 * 2 * 26 = -2 * 2 * 2 * 2 * 13\n\nCan we group these factors into (one positive, one negative) numbers that produce 48 when added together?  Yes!\n\n(-2*2) + (2*2*13) = -4 + 52 = 48\n\nWe got lucky ... the '-4' and '52' provide the perfect pair of factors needed to create a 48 (through addition), so we can now factor.\n\n0 = (X-4)(X+52)\n\nThings are getting better still, because only one of the zeros (some call them 'roots') from the right-hand side will be positive.  The zeros are 4 and -52, and we will only choose the 4, since it represents a physical measurement of length (of an opening).\n\nWe can go back to the expressions above and substitute X=4.\n\nSide area = 6X + 24 = 6(4)+24 = 48 sq. m\nFront/back area = 3X + 6 = 3(4)+6 = 18 sq. m\nTop/bottom area = 2X^2+12X+16 = 2(4^2)+12(4)+16 = 96 sq. m\nOpening area = X^2 - 2X = 4^2-2(4) = 8 sq. m\n\nI hope this helps!
491645
Check the image I attached...\n\nHere I've made the polygon into a complete rectangle.  The figure in red is the added triangle.  It has sides 6 and 8 and is a right triangle.  If you remember your Pythagorean ratios, you should know 3-4-5 and 6-8-10.  Or you can prove it using:\n\n6² + 8² = x²\n36 + 64 = 100\nx² = 100\nx = 10\n\nTo get the perimeter, just add up the sides:\nPerimeter = 1 + 10 + 9 + 4 + 10\nPerimeter = 34 units\n\nTo get the area, take the big rectangle (9 x 10) and remove the area of the triangle (1/2 b * h = 6 x 8 / 2)\nArea = 9 x 10 - (6 x 8)/2\nArea = 90 - 24\nArea = 66 sq. units