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In the xy plane, find the area of the region enclosed by the graph of
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05 Feb 2018, 10:09
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In the xy plane, find the area (in sq. units) of the region enclosed by the graph of x1 + y+2=3. A) 4.5 B) 9 C) 18 D) 36 E) 42
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In the xy plane, find the area of the region enclosed by the graph of
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05 Feb 2018, 18:36
souvonik2k wrote: In the xy plane, find the area (in sq. units) of the region enclosed by the graph of x1 + y+2=3. A) 4.5 B) 9 C) 18 D) 36 E) 42 See attached figure..the point of intersection is (1,2) When x1 is 0, \(y+2=3.....y=32=1\) and \(y=23=5\) So this becomes base \(1(5)=6\) Now let y+2=0, \(x1=3.....x=4\) and \(x=13=2\) Two triangles with base 6..Area of the one with altitude 4 = \(\frac{1}{2} *4*6=12\) Area of other =\(\frac{1}{2} *6*2=6\).. Total =\(12+6=18\) C
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In the xy plane, find the area of the region enclosed by the graph of
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05 Feb 2018, 20:26
chetan2u wrote: souvonik2k wrote: In the xy plane, find the area (in sq. units) of the region enclosed by the graph of x1 + y+2=3. A) 4.5 B) 9 C) 18 D) 36 E) 42 See attached figure When x is 0, \(y+2=3.....y=32=1\) and \(y=23=5\) So this becomes base \(1(5)=6\) Now let y=0, \(x1=3.....x=4\) and \(x=13=2\) Two triangles with base 6..Area of the one with altitude 4 = \(\frac{1}{2} *4*6=12\) Area of other =\(\frac{1}{2} *6*2=6\).. Total =\(12+6=18\) C Hi chetan2u, Please correct me if I am wrong in my understanding. Here, we have to take \(x  1 = 0\) (Instead of \(x = 0\)) & \(y + 2 = 0\) (instead of \(y = 0\)). When, \(x  1 = 0, y+2 = 3, y = 1,5\) \(y + 2 = 0, x1= 3, x = 4,2.\) However, in this case the origin should be transformed to \((1,2)\)
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In the xy plane, find the area of the region enclosed by the graph of
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05 Feb 2018, 20:35
souvonik2k wrote: In the xy plane, find the area (in sq. units) of the region enclosed by the graph of x1 + y+2=3. A) 4.5 B) 9 C) 18 D) 36 E) 42 Good Question souvonik2k.. +1 to you..!! Let \(X = x  1, Y = y + 2\) We have, \(X + Y = 3.\) Now, take \(X = 0, Y = 3\). Take \(Y = 0, X = 3\). Now, Draw a line from \((0,3) & (3,0)\). (Alternatively, You can also think of X intercept & Y intercept in Ist Quadrant). So, area of triangle in first Quadrant will be \(A = \frac{1}{2}*3*3.\) As this is mod, total area will be A = \(4*\frac{1}{2}*3*3 = 18\)
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Re: In the xy plane, find the area of the region enclosed by the graph of
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05 Feb 2018, 20:43
rahul16singh28 wrote: chetan2u wrote: souvonik2k wrote: In the xy plane, find the area (in sq. units) of the region enclosed by the graph of x1 + y+2=3. A) 4.5 B) 9 C) 18 D) 36 E) 42 See attached figure When x is 0, \(y+2=3.....y=32=1\) and \(y=23=5\) So this becomes base \(1(5)=6\) Now let y=0, \(x1=3.....x=4\) and \(x=13=2\) Two triangles with base 6..Area of the one with altitude 4 = \(\frac{1}{2} *4*6=12\) Area of other =\(\frac{1}{2} *6*2=6\).. Total =\(12+6=18\) C Hi chetan2u, Please correct me if I am wrong in my understanding. Here, we have to take \(x  1 = 0\) (Instead of \(x = 0\)) & \(y + 2 = 0\) (instead of \(y = 0\)). When, \(x  1 = 0, y+2 = 3, y = 1,5\) \(y + 2 = 0, x1= 3, x = 4,2.\) However, in this case the origin should be transformed to \((1,2)\) Hi.. Rahul, you are absolutely correct, the origin will shift to (1,2).. Although answer will remain the same as there are two perpendicular diagonals of length 6, so area = 1/2*6*6=18.. Kudos for your observation.
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1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
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Re: In the xy plane, find the area of the region enclosed by the graph of
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05 Feb 2018, 22:54
souvonik2k wrote: In the xy plane, find the area (in sq. units) of the region enclosed by the graph of x1 + y+2=3. A) 4.5 B) 9 C) 18 D) 36 E) 42 x1 + y+2=3 gives us 4 lines. x > 1 and y > 2 (x  1) + (y + 2) = 3 y = x + 2 x < 1 and y > 2 (x  1) + (y + 2) = 3 y = x x > 1 and y < 2 (x  1)  (y + 2) = 3 y = x 6 x < 1 and y < 2 (x  1)  (y + 2) = 3 y = x  4 This is the figure the 4 lines enclose: Attachment:
IMG_8406.jpg [ 1.38 MiB  Viewed 1448 times ]
They form a square with each side as \(3\sqrt{2}\) so area of the figure will be \((3\sqrt{2})^2 = 18\) Answer (C)
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Re: In the xy plane, find the area of the region enclosed by the graph of
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25 Feb 2018, 02:47
VeritasPrepKarishma wrote: souvonik2k wrote: In the xy plane, find the area (in sq. units) of the region enclosed by the graph of x1 + y+2=3. A) 4.5 B) 9 C) 18 D) 36 E) 42 x1 + y+2=3 gives us 4 lines. x > 1 and y > 2 (x  1) + (y + 2) = 3 y = x + 2 x < 1 and y > 2 (x  1) + (y + 2) = 3 y = x x > 1 and y < 2 (x  1)  (y + 2) = 3 y = x 6 x < 1 and y < 2 (x  1)  (y + 2) = 3 y = x  4 This is the figure the 4 lines enclose: Attachment: IMG_8406.jpg They form a square with each side as \(3\sqrt{2}\) so area of the figure will be \((3\sqrt{2})^2 = 18\) Answer (C) Hi Karishma, Thank you for the explanation. How did you determine the side of the square???



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Re: In the xy plane, find the area of the region enclosed by the graph of
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26 Feb 2018, 03:43
Kezia9 wrote: VeritasPrepKarishma wrote: souvonik2k wrote: In the xy plane, find the area (in sq. units) of the region enclosed by the graph of x1 + y+2=3. A) 4.5 B) 9 C) 18 D) 36 E) 42 x1 + y+2=3 gives us 4 lines. x > 1 and y > 2 (x  1) + (y + 2) = 3 y = x + 2 x < 1 and y > 2 (x  1) + (y + 2) = 3 y = x x > 1 and y < 2 (x  1)  (y + 2) = 3 y = x 6 x < 1 and y < 2 (x  1)  (y + 2) = 3 y = x  4 This is the figure the 4 lines enclose: Attachment: IMG_8406.jpg They form a square with each side as \(3\sqrt{2}\) so area of the figure will be \((3\sqrt{2})^2 = 18\) Answer (C) Hi Karishma, Thank you for the explanation. How did you determine the side of the square??? The length of the diagonals is 6 (from 2,2 to 4, 2) which means half the diagonal is 3. Considering each side of the square to be the hypotenuse of a right triangle, since the two legs are 3 each, the hypotenuse will be \(3\sqrt{2}\).
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Re: In the xy plane, find the area of the region enclosed by the graph of
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26 Feb 2018, 22:30
Hi Chetan, I am confused as to how you did this problem. How did you manage to figure out the altitude is 4? Can you provide more color on how you did this? I tried to draw the figure but it takes a long time.
Best,
Jimmy




Re: In the xy plane, find the area of the region enclosed by the graph of &nbs
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