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16 Sep 2014, 00:13
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If \(x^2 + y^2 = 100\), all of the following could be true EXCEPT:
A. \(|x| + |y| = 10\)
B. \(|x| \gt |y|\)
C. \(|x| \gt |y| + 10\)
D. \(|x| = |y|\)
E. \(|x| - |y| = 5\)
A. \(|x| + |y| = 10\)
B. \(|x| \gt |y|\)
C. \(|x| \gt |y| + 10\)
D. \(|x| = |y|\)
E. \(|x| - |y| = 5\)
16 Sep 2014, 00:13
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Official Solution:
If \(x^2 + y^2 = 100\), all of the following could be true EXCEPT:
A. \(|x| + |y| = 10\)
B. \(|x| \gt |y|\)
C. \(|x| \gt |y| + 10\)
D. \(|x| = |y|\)
E. \(|x| - |y| = 5\)
The equation \(x^2 + y^2 = 100\) represents a circle with a radius of 10 units and centered at the origin. This means that any point (x, y) that satisfies the equation lies on the circle.
Let's analyze each option:
A. \(|x| + |y| = 10\). This is possible if one of the coordinates is 0 and the other is 10 or -10. There are four points on the circle that satisfy this condition: (0, 10), (0, -10), (10, 0), and (-10, 0).
B. \(|x| \gt |y|\). We can see that if \(x = 10\) and \(y = 0\), this option is true. It is also evident from the figure that there must exist points (x, y) on the circle such that \(|x| > |y|\) since the circle cannot consist only of points where \(|x| \leq |y|\).
C. \(|x| \gt |y| + 10\). Since \(|y|\) is more than or equal to 0, then this option implies that \(|x| > 10\), which leads to \(x^2 + y^2\) being greater than 100, contradicting the given equation. Therefore, option C is not possible.
D. \(|x| = |y|\). This option is possible since there must exist (x, y) points on the circle such that \(|x| = |y|\). For example, in the first quadrant, a point at 45 degrees has equal \(x\) and \(y\) coordinates.
E. \(|x| - |y| = 5\). To simplify, let's consider the first quadrant and check whether \(x - y = 5\) is possible. Consider a point (10, 0), which has \(x - y = 10\). As we move this point along the curve in an anticlockwise direction, both \(x\) and \(y\) coordinates change continuously. Specifically, \(x\) decreases from 10 to 0 and \(y\) increases from 0 to 10. Thus, at some point on the curve, there must exist an (x, y) point where \(x - y = 5\).
Therefore, options A, B, D, and E are possible, but option C is not possible.
Answer: C
If \(x^2 + y^2 = 100\), all of the following could be true EXCEPT:
A. \(|x| + |y| = 10\)
B. \(|x| \gt |y|\)
C. \(|x| \gt |y| + 10\)
D. \(|x| = |y|\)
E. \(|x| - |y| = 5\)
The equation \(x^2 + y^2 = 100\) represents a circle with a radius of 10 units and centered at the origin. This means that any point (x, y) that satisfies the equation lies on the circle.
Let's analyze each option:
A. \(|x| + |y| = 10\). This is possible if one of the coordinates is 0 and the other is 10 or -10. There are four points on the circle that satisfy this condition: (0, 10), (0, -10), (10, 0), and (-10, 0).
B. \(|x| \gt |y|\). We can see that if \(x = 10\) and \(y = 0\), this option is true. It is also evident from the figure that there must exist points (x, y) on the circle such that \(|x| > |y|\) since the circle cannot consist only of points where \(|x| \leq |y|\).
C. \(|x| \gt |y| + 10\). Since \(|y|\) is more than or equal to 0, then this option implies that \(|x| > 10\), which leads to \(x^2 + y^2\) being greater than 100, contradicting the given equation. Therefore, option C is not possible.
D. \(|x| = |y|\). This option is possible since there must exist (x, y) points on the circle such that \(|x| = |y|\). For example, in the first quadrant, a point at 45 degrees has equal \(x\) and \(y\) coordinates.
E. \(|x| - |y| = 5\). To simplify, let's consider the first quadrant and check whether \(x - y = 5\) is possible. Consider a point (10, 0), which has \(x - y = 10\). As we move this point along the curve in an anticlockwise direction, both \(x\) and \(y\) coordinates change continuously. Specifically, \(x\) decreases from 10 to 0 and \(y\) increases from 0 to 10. Thus, at some point on the curve, there must exist an (x, y) point where \(x - y = 5\).
Therefore, options A, B, D, and E are possible, but option C is not possible.
Answer: C
08 Dec 2014, 09:34
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plaverbach
Can this be solved visually??
1) as a circle on the xy plane
OR
2) as a tringle(pitagoras)
1) as a circle on the xy plane
OR
2) as a tringle(pitagoras)
yes it can be solved by co-ordinate geometry.
x^2 + y^2 = 100
is a circle with radius 10
A. |x| + |y|=10 are four lines which intersect the circle at (10,0),(0,10),(-10,0)&(0,-10)
B. |x|>|y| there are multiple such points I have marked one such in the attached picture
C. This is an area outside the circle. The closest to the circle is at y=0 even here x is outside the area of the circle. A suggestion that when we look at inequalities in co-ordinate geometry think in terms of area.
D. |x|=|y| essentially two lines y=x and y=-x
E. |x| - |y|=5 they are four lines again
Attaching a very crude diagram. Apologies.
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