anonymous
  • anonymous
How to find area of region "bounded" (for a lack of a better word) by the following equation (in comments)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
\[\int f(x)-g(x)dx\]i believe
anonymous
  • anonymous
\[\left| 3x-18 \right|+\left| 2y+7 \right|\le3\]
amistre64
  • amistre64
or triangles :)

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anonymous
  • anonymous
That looks pretty difficult to graph...
amistre64
  • amistre64
looks can be deceiving ....
amistre64
  • amistre64
|3x-18| + |2y+7| - 3 <= 0
amistre64
  • amistre64
|dw:1328570964268:dw|better try this on the wolf
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=%7C3x-18%7C+%2B+%7C2y%2B7%7C+-+3+%3C%3D+0
amistre64
  • amistre64
what methods for area can we use? vectors, determinants, simpler stuff?
anonymous
  • anonymous
Well, this is for the AMC12, where the highest level math needed is Trig/Precal and there are no calculators allowed, so is there anyway I can draw this by hand? Or find the area algebraically?
amistre64
  • amistre64
algebra most likely; we know when x=6 the the left || drops out
amistre64
  • amistre64
|3x-18| + |2y+7| <= 3 ; x=6 |2y+7| <= 3 2y+7 <=3 or -2y-7 <= 3 2y <= -4 2y +7 >= -3 y <= -2 2y >= -10 y>= -5 ^ | | | -2 | | | -5 | v y that might be useful .....
amistre64
  • amistre64
then we do the same when we drop the y side |3x-18| <= 3 3x-18 <= 3 or 3x-18 >= -3 3x <= 21 3x >= 15 x <=7 x >= 5 ^ <------|---------------> x | 5 7 | -2 | | | -5 | v y just an idea, still dont know if its gonna be helpful tho
amistre64
  • amistre64
ohh, thtas actually what the graph of the wplf gives;
amistre64
  • amistre64
if i could type that would read more impressive lol
amistre64
  • amistre64
x=6 is a vertex; and y= -7/2 is a vertex (6,-2), (6,-5), (5,-7/2),(7,-7/2) are the points of that tilted square thing
anonymous
  • anonymous
wait, so how did out find the vertices (btw, great work, it's really easy to follow)
amistre64
  • amistre64
once you know the vertices, the points, you can do tests to see if its right angled and square or if its more trapazoidal and such
amistre64
  • amistre64
i zeroed out each |...| by setting its variable to something that would make it zero
amistre64
  • amistre64
then i solved for the remaining |....| and its results are the other parts of our orered pairs
myininaya
  • myininaya
i think i have something easier
amistre64
  • amistre64
im sure there is an easier way, but thats just not how a roll :)
anonymous
  • anonymous
ohhh, ok, thanks amistre64 for your help!
amistre64
  • amistre64
youre welcome :)
anonymous
  • anonymous
well, I have the solution pamplter and there is a solution (and it was multiple choice Choices: A) 3 b) 7/2 c)4 d) 9/2 e) 5 )
myininaya
  • myininaya
well we can go through all those given and check to see if the inequality holds this way would be much much easier :)
amistre64
  • amistre64
check x=6, y=-3 as a solution
amistre64
  • amistre64
there is an area of possible created by zeroing the x part and gettting y values; then zeroing the y part and getting x values
amistre64
  • amistre64
those 4 points of interest create parallelgram or some sort of shape that confinces the solution set
amistre64
  • amistre64
(6,-2), (6,-5), (5,-7/2), (7,-7/2) -6+2 -6+2 -6+2 -6+2 --------------------------- (0,0) (0,-3) (-1,-3/2) (1,-3/2) are what we look like when we set one vertice to the origion
amistre64
  • amistre64
if I were to raise it another 3/2 we could have it centered at the origin (0,0) (0,-3) (-1,-3/2) (1,-3/2) +3/2 +3/2 +3/2 +3/2 ----------------------------- (0,1.5) (0,-1.5) (-1,0) (1,0)
amistre64
  • amistre64
well, it aint a square, but its alteast rhombused
amistre64
  • amistre64
|dw:1328573417841:dw|
amistre64
  • amistre64
my next idea, brilliantly exectuted lol is to find the area of just one of those triangle and the 4x it
amistre64
  • amistre64
height = 1.5 base = 1 A = bh/2 = 1.5/2 = .75 right? .75*4 = total area
amistre64
  • amistre64
im going with 3 :)

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