tag:blogger.com,1999:blog-6651514617266100245.post7016822152447210206..comments2024-01-03T04:58:04.221-05:00Comments on I Hope This Old Train Breaks Down...: Some Juicy Problems from Exeter Math 3Unknownnoreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6651514617266100245.post-74202948613089465202011-08-31T00:14:42.158-04:002011-08-31T00:14:42.158-04:00I didn't guess and check, but I'd have to ...I didn't guess and check, but I'd have to sit down and re-think about the problem to figure out what I did. Did you consider the fact that circles tangent to both axes must have the form<br /><br />(X - k)^2 + (Y - k)^2 = k^2?? I think if you plug (9, 2) in there you can solve for k, and that's probably what I did.untilnextstophttps://www.blogger.com/profile/15285583728476473117noreply@blogger.comtag:blogger.com,1999:blog-6651514617266100245.post-35354348255404431462011-08-30T21:26:44.320-04:002011-08-30T21:26:44.320-04:00Thank you!
Hey on the very first one (two circles...Thank you!<br /><br />Hey on the very first one (two circles through (9, 2) - the only way I can figure it out is by guess and check. Is there a snazzier way? I know you're way busy so if you don't have time to respond don't sweat it.Katehttps://www.blogger.com/profile/14229054922453438248noreply@blogger.comtag:blogger.com,1999:blog-6651514617266100245.post-16200990754468869142011-08-21T07:15:28.970-04:002011-08-21T07:15:28.970-04:00Kate: Here you go. Let me know if I skipped a prob...Kate: Here you go. Let me know if I skipped a problem accidentally or if I left out a decimal point when I copied and pasted.<br /><br />(5, 5), 5; (17, 17), 17<br />----<br />14371 sq ft<br />----<br />center (1,5); mag = 3<br /><br />y = 5; 4x+3y = 19; 53.1 degrees<br />----<br />4/9 = 44.4%<br />----<br />5 in by 10 in<br />----<br />(3.301, 2.934, 4.401)<br />----<br />(44/13, 66/13, 0)<br />----<br />circle of radius sqrt(65) centered at the origin; it has the given points at opposite ends of a diameter<br />----<br />25pi.<br /><br />the shaded areas are both pi*h^2<br /><br />the cone volume is one third of the<br />cylinder volume; the hemisphere volume is therefore two thirds of the cylinder volume<br />----<br />sin(A + B) = sin(A)cos(B) + sin(B)cos(A)<br />cos(A + B) = cos(A)cos(B) - sin(A)sin(B)<br />----<br />90.7%<br />----<br />the two rows that contain 0 must be deleted<br /><br />40; 33.3; 7.75<br /><br />answers will vary the (the roll most likely to end the game is the 22nd ) <br />----<br />Ln = (1 / sqrt(2))^(n-1) ; L20 = 0.00138<br /><br />An = (1/2)^(n-1) is also geometric<br />----<br />G = 392.6 cps; 262*2^(n/12)<br /><br />the ratios are all 2^(7/12)=1.498<br />----<br />area = 15pi<br /><br />pi*ab<br />----<br />a.) 460 in Exeter; 540 in Hampton <br />b.) 538 in Exeter; 462 in Hampton<br /><br />a.) how to calculate the expected Tuesday am distribution from the Monday am distribution <br /><br />a.) expected distribution on Wed morning <br />b.) distribution a week from Thursday<br />c.) upper right corner is the probability that an Exeter car finishes its day in Hampton <br />d.) no cars enter or leave the system<br /><br />x = 4000/7 = 571<br />----<br />a.) P is equidistant from F and N<br />b.) except for P, all points on the parabola are closer to F than they are to N <br />c.) triangle FPN is isosceles <br />d.) angle between λ and μ equals angle between FP and λ<br />----<br />1717 ft<br />----<br />$3207.14 <br /><br />$3025.60, $2854.34, $2692.77, … $ 1000(1.06)^k<br /><br />$39992.73 <br /><br />the sum of a geometric series is (first – last*multiplier)/(1 – multiplier)untilnextstophttps://www.blogger.com/profile/15285583728476473117noreply@blogger.comtag:blogger.com,1999:blog-6651514617266100245.post-90377365899883742852011-08-20T11:54:18.837-04:002011-08-20T11:54:18.837-04:00Mimi you said you have a solution guide right? Is ...Mimi you said you have a solution guide right? Is that online somewhere? Or can you post answers to these? I don't really need whole solutions, I just want to make sure I'm not making dumb arithmetic errors.Katehttps://www.blogger.com/profile/14229054922453438248noreply@blogger.comtag:blogger.com,1999:blog-6651514617266100245.post-53746462395927013022011-08-19T01:03:45.727-04:002011-08-19T01:03:45.727-04:00On Facebook Sam was reminiscing how in h.s. he rea...On Facebook Sam was reminiscing how in h.s. he really went to town when he assumed a circular fence!untilnextstophttps://www.blogger.com/profile/15285583728476473117noreply@blogger.comtag:blogger.com,1999:blog-6651514617266100245.post-33924529342596072482011-08-18T18:22:43.925-04:002011-08-18T18:22:43.925-04:00Yikes, some of those sound hard! I played with a f...Yikes, some of those sound hard! I played with a few, didn't finish. I love the goat problems. I first saw them in The Wohascum County Problem Book. But I never saw one with that fence in it. Nice twist.Sue VanHattumhttps://www.blogger.com/profile/10237941346154683902noreply@blogger.com