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anonymous
 4 years ago
How to find area of region "bounded" (for a lack of a better word) by the following equation (in comments)?
anonymous
 4 years ago
How to find area of region "bounded" (for a lack of a better word) by the following equation (in comments)?

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amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3\[\int f(x)g(x)dx\]i believe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\left 3x18 \right+\left 2y+7 \right\le3\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That looks pretty difficult to graph...

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3looks can be deceiving ....

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.33x18 + 2y+7  3 <= 0

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1328570964268:dwbetter try this on the wolf

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3http://www.wolframalpha.com/input/?i=%7C3x18%7C+%2B+%7C2y%2B7%7C++3+%3C%3D+0

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3what methods for area can we use? vectors, determinants, simpler stuff?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, 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
 4 years ago
Best ResponseYou've already chosen the best response.3algebra most likely; we know when x=6 the the left  drops out

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.33x18 + 2y+7 <= 3 ; x=6 2y+7 <= 3 2y+7 <=3 or 2y7 <= 3 2y <= 4 2y +7 >= 3 y <= 2 2y >= 10 y>= 5 ^    2    5  v y that might be useful .....

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3then we do the same when we drop the y side 3x18 <= 3 3x18 <= 3 or 3x18 >= 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
 4 years ago
Best ResponseYou've already chosen the best response.3ohh, thtas actually what the graph of the wplf gives;

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3if i could type that would read more impressive lol

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3x=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
 4 years ago
Best ResponseYou've already chosen the best response.0wait, so how did out find the vertices (btw, great work, it's really easy to follow)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3once 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
 4 years ago
Best ResponseYou've already chosen the best response.3i zeroed out each ... by setting its variable to something that would make it zero

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3then i solved for the remaining .... and its results are the other parts of our orered pairs

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1i think i have something easier

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3im sure there is an easier way, but thats just not how a roll :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ohhh, ok, thanks amistre64 for your help!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well, 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
 4 years ago
Best ResponseYou've already chosen the best response.1well we can go through all those given and check to see if the inequality holds this way would be much much easier :)

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3check x=6, y=3 as a solution

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3there 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
 4 years ago
Best ResponseYou've already chosen the best response.3those 4 points of interest create parallelgram or some sort of shape that confinces the solution set

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3(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
 4 years ago
Best ResponseYou've already chosen the best response.3if 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
 4 years ago
Best ResponseYou've already chosen the best response.3well, it aint a square, but its alteast rhombused

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3dw:1328573417841:dw

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3my next idea, brilliantly exectuted lol is to find the area of just one of those triangle and the 4x it

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.3height = 1.5 base = 1 A = bh/2 = 1.5/2 = .75 right? .75*4 = total area
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