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\[\int f(x)-g(x)dx\]i believe
\[\left| 3x-18 \right|+\left| 2y+7 \right|\le3\]
or triangles :)
That looks pretty difficult to graph...
looks can be deceiving ....
|3x-18| + |2y+7| - 3 <= 0
|dw:1328570964268:dw|better try this on the wolf
what methods for area can we use? vectors, determinants, simpler stuff?
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?
algebra most likely; we know when x=6 the the left || drops out
|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 .....
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
ohh, thtas actually what the graph of the wplf gives;
if i could type that would read more impressive lol
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
wait, so how did out find the vertices (btw, great work, it's really easy to follow)
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
i zeroed out each |...| by setting its variable to something that would make it zero
then i solved for the remaining |....| and its results are the other parts of our orered pairs
i think i have something easier
im sure there is an easier way, but thats just not how a roll :)
ohhh, ok, thanks amistre64 for your help!
youre welcome :)
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 )
well we can go through all those given and check to see if the inequality holds this way would be much much easier :)
check x=6, y=-3 as a solution
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
those 4 points of interest create parallelgram or some sort of shape that confinces the solution set
(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
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)
well, it aint a square, but its alteast rhombused
my next idea, brilliantly exectuted lol is to find the area of just one of those triangle and the 4x it
height = 1.5 base = 1 A = bh/2 = 1.5/2 = .75 right? .75*4 = total area
im going with 3 :)