How to find area of region "bounded" (for a lack of a better word) by the following equation (in comments)?
 anonymous
How to find area of region "bounded" (for a lack of a better word) by the following equation (in comments)?
 Stacey Warren  Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
 katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
 amistre64
\[\int f(x)g(x)dx\]i believe
 anonymous
\[\left 3x18 \right+\left 2y+7 \right\le3\]
 amistre64
or triangles :)
Looking for something else?
Not the answer you are looking for? Search for more explanations.
More answers
 anonymous
That looks pretty difficult to graph...
 amistre64
looks can be deceiving ....
 amistre64
3x18 + 2y+7  3 <= 0
 amistre64
dw:1328570964268:dwbetter try this on the wolf
 amistre64
http://www.wolframalpha.com/input/?i=%7C3x18%7C+%2B+%7C2y%2B7%7C++3+%3C%3D+0
 amistre64
what methods for area can we use? vectors, determinants, simpler stuff?
 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
algebra most likely; we know when x=6 the the left  drops out
 amistre64
3x18 + 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
then 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
ohh, thtas actually what the graph of the wplf gives;
 amistre64
if i could type that would read more impressive lol
 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
wait, so how did out find the vertices (btw, great work, it's really easy to follow)
 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
i zeroed out each ... by setting its variable to something that would make it zero
 amistre64
then i solved for the remaining .... and its results are the other parts of our orered pairs
 myininaya
i think i have something easier
 amistre64
im sure there is an easier way, but thats just not how a roll :)
 anonymous
ohhh, ok, thanks amistre64 for your help!
 amistre64
youre welcome :)
 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
well we can go through all those given and check to see if the inequality holds
this way would be much much easier :)
 amistre64
check x=6, y=3 as a solution
 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
those 4 points of interest create parallelgram or some sort of shape that confinces the solution set
 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
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
well, it aint a square, but its alteast rhombused
 amistre64
dw:1328573417841:dw
 amistre64
my next idea, brilliantly exectuted lol is to find the area of just one of those triangle and the 4x it
 amistre64
height = 1.5
base = 1
A = bh/2 = 1.5/2 = .75 right?
.75*4 = total area
 amistre64
im going with 3 :)
Looking for something else?
Not the answer you are looking for? Search for more explanations.