Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/11/08 at 8:34 pm
so i'm trying to solve this word problem.
"a rectangular sheet of cardboard fora poster measures 5 feet by 4 feet. The margins at the top and bottom and on the sides are to be the same width. if the overall perimeter of the poster is 1 1/2 times the perimeter of the picture area, how wide are the margins?"
okay, so the perimeter of the poster is 5(2)+4(2), or 18. the perimeter of the poster is 1 1/2 times as big as the perimeter of the picture, so if the perimeter of the picture is x, we get 3x/2=18? is that right? i think this is the part i can't get.
let's see. you'd have to multiply the perimeter of the picture by 1 1/2 to yield the perimeter of the poster.
so. 3x=32 or x=10 2/3. so far so good. the proportion of the picture will be the same as the proportion of the poster so if we define y so that 5y yields the length of the tops and bottoms and 4y the length of the sides of the picture, we end up with, er... 2(5y)+2(4y)=10 2/3. or 18y=10 2/3. (22/3)/18 is 22/54, or 11/12. oo! actually, that makes sense. sorta. cuz, like, a foot has 12 inches.
so, er, the width of the tops and bottoms of the picture is 5(11/12), or 55/12, or 4 7/12 feet. the margin is going to be half the difference between the width of the poster's top or bottom and the pictures top or bottom. so (5 - 4 7/12)/2, or (5/12)/2, 5/24, or two and a half inches.
this has been bugging me all afternoon.
ha! i'm looking back over this and i fudged up like 5 times. really bad basic fudge ups. it's like karaokelgebra.
22/54=11/22? dude, that's so BAD. :-[
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Red Ant on 03/11/08 at 9:41 pm
that was a "B" difficulty question. here's a "C".
"Starting from the same point and at the same time, three men travel in the same direction around a circular track 90 yards in circumference. If the men move at the respective average rates of 9 yards, 6 yards, and 3 yards per second, where and when will they first be together again?"
and the answer is... um... fudge if i know.
Well, this one isn't too bad I don't think...I have been wrong before though :D
Man A = 9 yards/sec, therefore can complete a circuit in 10 seconds.
Man B = 6 yards/sec, therefore can complete a circuit in 15 seconds.
Man C = 3 yards/sec, therefore can complete a circuit in 30 seconds.
Note that 30 seconds is the lowest common multiple, so based on this, I believe the men will meet up at the same point at the same time at 30 seconds...but that sounds too simple so there's gotta be a trick to it!
You are correct, but only so far as the time portion, so there is a "trick". You solved for the "when", but not the "where".
Another way to look at is Man A is running 3x faster than man C, and Man B is running 2x faster than Man C. Taking your previous calculations, man A runs three laps, Man B 2 laps and Man C 1 lap, but they all meet at the start/finish line.
This is, of course, assuming they run identical paths, which breaks the laws of physics (three people cannot occupy the exact same space at the same time). Running different paths, as in a real race, presents a much more challenging equation, and switching paths mid race adds in variables that fry even my nerd brain. :D
(unless they happen to run in what looks like a 3-phase power wave (i.e., like braiding hair), in which case they conveniently wind up having travelled the same distance despite their 120* offsets. This still doesn't solve the physics portion of the equation as two people would occupy the same space at the same time, but close enough... )
http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/Semi/03269.png
Ant
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/11/08 at 9:55 pm
You are correct, but only so far as the time portion, so there is a "trick".
Another way to look at is Man A is running 3x faster than man C, and Man B is running 2x faster than Man C. Taking your previous calculations, man A runs three laps, Man B 2 laps and Man C 1 lap, but they all meet at the start/finish line.
This is, of course, assuming they run identical paths, which breaks the laws of physics (three people cannot occupy the exact same space at the same time). Running different paths, as in a real race, presents a much more challenging equation, and switching paths mid race adds in variables that fry even my nerd brain. :D
Ant
sounds like you might be ready for the wriggly worm problem i was reading about on the metro back from work today. it's the idea that a worm gets restless at night so it wriggles randomly, and its parents need to find a blanket that will cover it up from the cold, but blanket material is expensive so they're trying to find the smallest blanket possible to cover the wriggling worm.
first solution is that they assume the wriggling worm wriggles from one of the endpoints on its body, so a circle with a radius equivalent to the fully stretched length of the worm is assumed to be the blanket. this is actually a very bad solution, but it has the advantage of being simple.
then some wiseacre sez if you assume the worm is stationary in the middle then you come up with an ellipse defined by both endpoints of the wriggling worm, head and tail. yeah? i followed as long as that but then it went into this whole bit about "Cantor sets," which emerge from this repetitive operation of trisecting line segments, leaving the middle segment empty, and then trisecting the two resulting line segments on the ends and repeating the result iteratively. you end up with this line that's all endpoint and no segment, or something. it got really weird at that point and i just started staring out of the window wondering when i'd reach my metro stop. interesting book though.
http://www.amazon.com/Game-Set-Math-Conundrums-Mathematics/dp/0140132376
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Jessica on 03/11/08 at 9:56 pm
My brain just exploded.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/11/08 at 9:59 pm
My brain just exploded.
quick! someone compute the minimum area necessary to design a blanket to cover jessica's exploded brain parts! assuming that said blanket is conveyed by three paramedics, each of whom runs one-half again faster than the paramedic preceding him or her.
please show your work.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: loki 13 on 03/11/08 at 10:02 pm
You are correct, but only so far as the time portion, so there is a "trick".
Another way to look at is Man A is running 3x faster than man C, and Man B is running 2x faster than Man C. Taking your previous calculations, man A runs three laps, Man B 2 laps and Man C 1 lap, but they all meet at the start/finish line.
This is, of course, assuming they run identical paths, which breaks the laws of physics (three people cannot occupy the exact same space at the same time). Running different paths, as in a real race, presents a much more challenging equation, and switching paths mid race adds in variables that fry even my nerd brain. :D
(unless they happen to run in what looks like a 3-phase power wave (i.e., like braiding hair), in which case they conveniently wind up having travelled the same distance despite their 120* offsets. This still doesn't solve the physics portion of the equation as two people would occupy the same space at the same time, but close enough... )
http://www.eng.cam.ac.uk/DesignOffice/mdp/electric_web/Semi/03269.png
Ant
This is all well and good but either you are over thinking the problem or I'm under thinking it. I assume the men starting as the
same point....90 yards from this finish line. So this means the starting postions would be at different locations just like a regular
track meet. They start in a straight line in a dash rash, any race around the track is scattered to allow a fair race for the outer
racers. Horses, on the other hand, are SOL.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Rice_Cube on 03/11/08 at 10:08 pm
quick! someone compute the minimum area necessary to design a blanket to cover jessica's exploded brain parts! assuming that said blanket is conveyed by three paramedics, each of whom runs one-half again faster than the paramedic preceding him or her.
please show your work.
Is there enough slack in the blanket to allow for differing paramedic velocities?
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/11/08 at 10:11 pm
Is there enough slack in the blanket to allow for differing paramedic velocities?
yes but it must be absolutely held to a minimum because of the aforementioned blanket-being-expensive provision.
i cant friggin solve word problems to save my life but i think i've managed to actually come up with one that would get albert einstein himself going, um, dude, no clue here. :P
karaokelgebra! i think i can make it catch on.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Red Ant on 03/11/08 at 10:29 pm
This is all well and good but either you are over thinking the problem or I'm under thinking it. I assume the men starting as the
same point....90 yards from this finish line. So this means the starting postions would be at different locations just like a regular
track meet. They start in a straight line in a dash rash, any race around the track is scattered to allow a fair race for the outer
racers. Horses, on the other hand, are SOL.
Yes, they do start at the same point. If it were a 90 yard dash, the problem of concentricity would not be an issue: each runner runs 90 yards in a straight line (in that case it's irrevelant what lane they start in because they each travel 90 yards).
There also would be a different answer: they would never meet (until sometime after all three crossed the finish line and drank Gatorade togther: an undefined/undeterminable answer).
Seeing as how they run in a circle, the runners in the outside lanes have to run a longer distance than those on the inside, which is why real races always have runners taking the shortest path (the inside lane). Staying in the outermost lane of a typical 8-lane run would put a runner at an extreme disadvantage (auto racing, like NASCAR, is a different animal altogether, but that's an algebra problem for another day).
As a purely mathematical problem, Kin and my answers make the whole problem solved. Bring in physics, and this hypothetical race becomes a nightmare to correctly calculate.
sounds like you might be ready for the wriggly worm problem i was reading about on the metro back from work today. it's the idea that a worm gets restless at night so it wriggles randomly, and its parents need to find a blanket that will cover it up from the cold, but blanket material is expensive so they're trying to find the smallest blanket possible to cover the wriggling worm.
first solution is that they assume the wriggling worm wriggles from one of the endpoints on its body, so a circle with a radius equivalent to the fully stretched length of the worm is assumed to be the blanket. this is actually a very bad solution, but it has the advantage of being simple.
then some wiseacre sez if you assume the worm is stationary in the middle then you come up with an ellipse defined by both endpoints of the wriggling worm, head and tail. yeah? i followed as long as that but then it went into this whole bit about "Cantor sets," which emerge from this repetitive operation of trisecting line segments, leaving the middle segment empty, and then trisecting the two resulting line segments on the ends and repeating the result iteratively. you end up with this line that's all endpoint and no segment, or something. it got really weird at that point and i just started staring out of the window wondering when i'd reach my metro stop. interesting book though.
http://www.amazon.com/Game-Set-Math-Conundrums-Mathematics/dp/0140132376
I'm not familiar with "Cantor Sets" either, but assuming the blanket can hold down the worm so that it can't escape, I imagine the answer would be close to the area of two circles with a combined area of 2 x 1/2L2 x Pi, as opposed to Pi x r2 (plus a little undetermined extra for the middle point of the worm, which might not be covered by the circles)
IOW, if the worm if 2 ft long, pi r2 would be a blanket of ~12.56 sq ft (the worm moves as in the first solution you presented: length = radius).
My solution would be 2 x (1ft2 x pi) + (undetermined negligable amount for middle section), or ~6.28 sq ft.
Of course, that again is assuming the blanket can hold down the worm. As it wiggles randomly, I don't think there is a true solution to this problem.
My brain just exploded.
Lemme guess: you divided by zero? ;D
Ant
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Rice_Cube on 03/11/08 at 10:35 pm
Lemme guess: you divided by zero? ;D
Ant
Her issue appears to be undefined.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Red Ant on 03/11/08 at 10:57 pm
Her issue appears to be undefined.
TTotM? 3-7 days of variables no mathematician will ever completely solve for... ;D
*ducks sharp and heavy objects thrown at me by Jess*
Ant
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Jessica on 03/11/08 at 11:10 pm
http://i123.photobucket.com/albums/o283/Nerdprincess1980/Funny/catnip.jpg
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: La Roche on 03/11/08 at 11:19 pm
quick! someone compute the minimum area necessary to design a blanket to cover jessica's exploded brain parts! assuming that said blanket is conveyed by three paramedics, each of whom runs one-half again faster than the paramedic preceding him or her.
please show your work.
I like to consider myself more the artistic sort, but the equation is fool proof.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/12/08 at 8:13 am
I'm not familiar with "Cantor Sets" either, but assuming the blanket can hold down the worm so that it can't escape, I imagine the answer would be close to the area of two circles with a combined area of 2 x 1/2L2 x Pi, as opposed to Pi x r2 (plus a little undetermined extra for the middle point of the worm, which might not be covered by the circles)
http://en.wikipedia.org/wiki/Cantor_set
now it's clear as mud! you're welcome.
the upshot of what the book was saying, i think, is that questions asking to determine the least possible area of something are often hard to solve, i think for reasons kinda like why it's hard to prove a negative. you can establish that something is non-zero (that's sorta what the cantor set is, i think? nonzero but as close to zero as makes no odds) but beyond that it's easy to propose a solution that might be the least possible area, but hard to prove that there isn't some undiscovered alternative that could be less still.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/12/08 at 9:46 am
"Let the regions within or on the indicated rectangles in the adjoining diagram represent the subsets R, S and T of a set U. Redraw and shade the region representing the subset of U specified in each exercise."
http://i28.tinypic.com/5dut6c.gif
1.
http://i31.tinypic.com/zmdrub.gif
http://i27.tinypic.com/10cvreb.gif
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/12/08 at 9:54 am
http://i31.tinypic.com/2rpzmhl.gif
http://i31.tinypic.com/eqv3wy.gif
that's a difficulty C. seemed pretty easy to me.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Reynolds1863 on 03/12/08 at 10:14 am
What exactly is Cramer's rule?
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: wsmith4 on 03/12/08 at 10:24 am
I think, probably, the best way to answer this question would be using the following method:
http://upload.wikimedia.org/math/e/9/6/e96379eeb8832176cbe89b39245f22a4.png
http://upload.wikimedia.org/math/d/d/e/dde2390688d8f36ec284e0ab274e3ad8.png
http://upload.wikimedia.org/math/9/5/b/95ba2256785a60ad355fc33cf8c0fdb8.png
http://upload.wikimedia.org/wikipedia/en/6/60/CentralTendencyLV.jpg
http://performingarts.ufl.edu/wp-content/uploads/2007/05/parsons-dance-company-man-flying-cropped.jpg
http://earthscience.files.wordpress.com/2007/05/tornado.jpg
=4
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/12/08 at 10:25 am
that must be new math!
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Reynolds1863 on 03/12/08 at 10:34 am
I think, probably, the best way to answer this question would be using the following method:
http://upload.wikimedia.org/math/e/9/6/e96379eeb8832176cbe89b39245f22a4.png
http://upload.wikimedia.org/math/d/d/e/dde2390688d8f36ec284e0ab274e3ad8.png
http://upload.wikimedia.org/math/9/5/b/95ba2256785a60ad355fc33cf8c0fdb8.png
http://upload.wikimedia.org/wikipedia/en/6/60/CentralTendencyLV.jpg
http://performingarts.ufl.edu/wp-content/uploads/2007/05/parsons-dance-company-man-flying-cropped.jpg
http://earthscience.files.wordpress.com/2007/05/tornado.jpg
=4
It's going to take me a few hours to get it. Eventually I will.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/12/08 at 10:37 am
why does math hate babboons? :\'(
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Reynolds1863 on 03/12/08 at 10:39 am
why does math hate babboons? :\'(
Because math have a Superiority complex, anything less is nonexistant.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/12/08 at 11:14 am
Because math have a Superiority complex, anything less is nonexistant.
http://i32.tinypic.com/34e8b2r.gif
:(
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Jessica on 03/12/08 at 11:31 am
Good math.
http://static.pyzam.com/img/funnypics/misc/1442.jpg
http://static.pyzam.com/img/funnypics/misc/fun32.jpg
http://static.pyzam.com/img/funnypics/misc/evilgirls.jpg
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: wsmith4 on 03/12/08 at 12:38 pm
why does it say "leet" for my # of posts???
oh, that's weird...after i posted this it went back to normal. it was because the # was "1337" and for some reason it changed to letters and said "leet"... strange!
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/12/08 at 12:47 pm
why does it say "leet" for my # of posts???
oh, that's weird...after i posted this it went back to normal. it was because the # was "1337" and for some reason it changed to letters and said "leet"... strange!
rice should probably field this one. :P
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Rice_Cube on 03/12/08 at 1:54 pm
rice should probably field this one. :P
I am no longer allowed to de-leet to be leet :P
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Howard on 03/12/08 at 2:41 pm
http://i30.tinypic.com/6h01zr.jpg
oops. :-[ this is hard!
What the hell is this? Are we suppose to solve all that? ??? ::)
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: wsmith4 on 03/12/08 at 2:51 pm
What the hell is this? Are we suppose to solve all that? ??? ::)
this one is easier...
you + me = ?
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/12/08 at 2:59 pm
howard and bill, taking a cue from the politicians in office, decide to start paying one another for their nights of love. in the beginning they pay each other equal amounts. howard receives $200 more per year at the end of every half-year, while bill receives $400 more per year at the end of each year. By how much will howard's love dividend over ten years exceed bill's?
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Howard on 03/12/08 at 3:29 pm
this one is easier...
you + me = ?
8-P
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Howard on 03/12/08 at 3:30 pm
howard and bill, taking a cue from the politicians in office, decide to start paying one another for their nights of love. in the beginning they pay each other equal amounts. howard receives $200 more per year at the end of every half-year, while bill receives $400 more per year at the end of each year. By how much will howard's love dividend over ten years exceed bill's?
Don't answer that.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/12/08 at 3:30 pm
Don't answer that.
actually i dont know how. :-[ i bet rice or red ant can do it.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Howard on 03/12/08 at 3:33 pm
actually i dont know how. :-[ i bet rice or red ant can do it.
I bet they couldn't even if they tried.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/12/08 at 3:52 pm
A side of an equilateral triangle is 20 inches long. A second equilateral triangle is inscribed in it by joining the midpoints of the sides of the first triangle. The process is continued, as shown in the accompanying diagram. Find the perimeter of the fourth inscribed circle.
http://i32.tinypic.com/6nzb5h.gif
Since the perimeter of each triangle is 3t, where t is the length of its side, the first triangle has a perimeter of 60. The factor of r is 3/2 (I think) because the perimeter of each triangle is 3 times ½ the length of the side of the preceding triangle. Since we want to look at the perimeter of the fourth inscribed triangle n will be 4. Plug all of this into the formula
http://i29.tinypic.com/10o40p3.gif
And we get
http://i27.tinypic.com/2cg0q3a.gif
Crap, that can’t be right. The triangles are supposed to be getting smaller, not monstrously huge.
Okay, the question of perimeter vs. side is trivial anyway since the one is just three times the other. So let’s just figure out what the sides are. This means r will be ½ and a(1) becomes 20.
http://i26.tinypic.com/6gfsbc.gif
and i think the perimeter would be three times this, or 7 1/2.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Rice_Cube on 03/12/08 at 4:09 pm
What ARE you doing? :D :D
It's an equilateral triangle. That means that if you join midpoints of each side, that new line would make another equilateral triangle and so on.
Triangle 1: side = 20
Triangle 2: side = 10
Triangle 3: side = 5
Triangle 4: side = 2.5
Unless I totally screwed that up, geometry was a long time ago.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/12/08 at 4:17 pm
What ARE you doing? :D :D
It's an equilateral triangle. That means that if you join midpoints of each side, that new line would make another equilateral triangle and so on.
Triangle 1: side = 20
Triangle 2: side = 10
Triangle 3: side = 5
Triangle 4: side = 2.5
Unless I totally screwed that up, geometry was a long time ago.
i already did that one! admittedly the long way round.
forget that one. do the howard and bill love dividend one.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Henk on 03/12/08 at 4:21 pm
A side of an equilateral triangle is 20 inches long. A second equilateral triangle is inscribed in it by joining the midpoints of the sides of the first triangle. The process is continued, as shown in the accompanying diagram. Find the perimeter of the fourth inscribed circle.
http://i32.tinypic.com/6nzb5h.gif
Since the perimeter of each triangle is 3t, where t is the length of its side, the first triangle has a perimeter of 60. The factor of r is 3/2 (I think) because the perimeter of each triangle is 3 times ½ the length of the side of the preceding triangle. Since we want to look at the perimeter of the fourth inscribed triangle n will be 4. Plug all of this into the formula
http://i29.tinypic.com/10o40p3.gif
And we get
http://i27.tinypic.com/2cg0q3a.gif
Crap, that can’t be right. The triangles are supposed to be getting smaller, not monstrously huge.
Two things. First, r should be 1/2 - not 3/2. Second, involution goes before multiplication, as I've been taught (so you get (1/2)^3 times 60, not (60 x 1/2)^3).
If you alter these two factors, it also results in 7.5.
Wow...Did I actually do that? :o I'm surprised myself.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Howard on 03/13/08 at 6:28 am
i already did that one! admittedly the long way round.
forget that one. do the howard and bill love dividend one.
No.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/13/08 at 12:01 pm
No.
Okay im going to give it a shot.
howard and bill, taking a cue from the politicians in office, decide to start paying one another for their nights of love. in the beginning they pay each other equal amounts. howard receives $200 more per year at the end of every half-year, while bill receives $400 more per year at the end of each year. By how much will howard's love dividend over ten years exceed bill's?
since they begin paying each other the same amount which isn’t specified, we might as well assume it’s zero. Therefore for both bill and howard a1=0. I think you’d have to set up two equations, one for bill and another for Howard. For howard you’d have to make each increment of n to equal one half year? I think? Since you’re raising d by 200 dollars at the end of every half year, but for the whole year? And then call d 200? And for Bill n could be per year with d equaling 400. I feel like this reasoning is flawed somewhere but that’s the magic of karaokelgebra. Okay, so we plug both of these values into the series equation
http://i28.tinypic.com/riykcn.gif
So that for Howard the equation would be
http://i25.tinypic.com/20aq781.gif
and for Bill the equation would be
http://i31.tinypic.com/whjipv.gif
Okay, I think that’s wrong because that’s quite a big difference. Damn math.
you know what i think it is? you have to divide howard's number in half because it's done by half years rather than full years. this would lead to 38,000/2=19,000, which is a thousand more than bill would get. I think that might make sense because if you raise the amount of love dividend twice a year rather than once, that means that for half the year howard will be making $200 a year more than bill. and half of 200 is 100, times ten for ten years is 1000.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Howard on 03/13/08 at 3:05 pm
Could you stop with this stupid equation? ::)
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/13/08 at 3:06 pm
Could you stop with this stupid equation? ::)
i think i solved it!!!!!!!! :)
anyway, you get more money. i thought you'd be pleased.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Reynolds1863 on 03/13/08 at 3:07 pm
Could you stop with this stupid equation? ::)
This is the algebra thread Howard. Expect thing to be done in equation or line graph.
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Reynolds1863 on 03/13/08 at 3:11 pm
Okay im going to give it a shot.
howard and bill, taking a cue from the politicians in office, decide to start paying one another for their nights of love. in the beginning they pay each other equal amounts. howard receives $200 more per year at the end of every half-year, while bill receives $400 more per year at the end of each year. By how much will howard's love dividend over ten years exceed bill's?
since they begin paying each other the same amount which isn’t specified, we might as well assume it’s zero. Therefore for both bill and howard a1=0. I think you’d have to set up two equations, one for bill and another for Howard. For howard you’d have to make each increment of n to equal one half year? I think? Since you’re raising d by 200 dollars at the end of every half year, but for the whole year? And then call d 200? And for Bill n could be per year with d equaling 400. I feel like this reasoning is flawed somewhere but that’s the magic of karaokelgebra. Okay, so we plug both of these values into the series equation
http://i28.tinypic.com/riykcn.gif
So that for Howard the equation would be
http://i25.tinypic.com/20aq781.gif
and for Bill the equation would be
http://i31.tinypic.com/whjipv.gif
Okay, I think that’s wrong because that’s quite a big difference. Damn math.
you know what i think it is? you have to divide howard's number in half because it's done by half years rather than full years. this would lead to 38,000/2=19,000, which is a thousand more than bill would get. I think that might make sense because if you raise the amount of love dividend twice a year rather than once, that means that for half the year howard will be making $200 a year more than bill. and half of 200 is 100, times ten for ten years is 1000.
I'm impressed. I feel like graphing that on my calculator. :)
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Howard on 03/13/08 at 3:13 pm
i think i solved it!!!!!!!! :)
anyway, you get more money. i thought you'd be pleased.
How much money? ??? ::)
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Tia on 03/13/08 at 3:32 pm
How much money? ??? ::)
this much!
http://i28.tinypic.com/riykcn.gif
http://i25.tinypic.com/20aq781.gif
http://i31.tinypic.com/whjipv.gif
http://upload.wikimedia.org/math/9/5/b/95ba2256785a60ad355fc33cf8c0fdb8.png
http://i25.tinypic.com/w2hemf.gif
http://i25.tinypic.com/w2hemf.gif
http://i30.tinypic.com/6h01zr.jpg
Subject: Re: algebra!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1
Written By: Philip Eno on 04/04/08 at 5:42 am
this much!
http://i28.tinypic.com/riykcn.gif
http://i25.tinypic.com/20aq781.gif
http://i31.tinypic.com/whjipv.gif
http://upload.wikimedia.org/math/9/5/b/95ba2256785a60ad355fc33cf8c0fdb8.png
http://i25.tinypic.com/w2hemf.gif
http://i25.tinypic.com/w2hemf.gif
http://i30.tinypic.com/6h01zr.jpg
= 42.