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In the xy-plane, Region R is bounded by the line segments with equations, 2x + |4y| = 20 and x = 0, whereas Region P is bounded by the line segments with equations |4x| + 2y = 20 and y = 0. If the function A(B) is defined as the area of Region B, what is the value of |A(P) – 2A(R)|?
- a)0
- b)25
- c)50
- d)75
- e)100
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
In the xy-plane, Region R is bounded by the line segments with equatio...
To find the value of |A(P) - 2A(R)|, we first need to determine the equations of the line segments that bound Regions R and P. Let's start with Region R.
Equation of the line segment 2x + |4y| = 20:
We can rewrite this equation as two separate equations:
2x + 4y = 20 (when 4y ≥ 0)
2x - 4y = 20 (when 4y < 0)
We can rewrite this equation as two separate equations:
2x + 4y = 20 (when 4y ≥ 0)
2x - 4y = 20 (when 4y < 0)
To find the x-intercept, set y = 0 in each equation:
2x = 20 -> x = 10 (x-intercept of the first equation)
2x = 20 -> x = 10 (x-intercept of the second equation)
2x = 20 -> x = 10 (x-intercept of the first equation)
2x = 20 -> x = 10 (x-intercept of the second equation)
So the line segment for 2x + |4y| = 20 is a vertical line passing through x = 10.
Equation of the line segment x = 0:
This is a vertical line passing through x = 0.
This is a vertical line passing through x = 0.
Now let's determine the equations for Region P.
Equation of the line segment |4x| + 2y = 20:
We can rewrite this equation as two separate equations:
4x + 2y = 20 (when 4x ≥ 0)
-4x + 2y = 20 (when 4x < 0)
We can rewrite this equation as two separate equations:
4x + 2y = 20 (when 4x ≥ 0)
-4x + 2y = 20 (when 4x < 0)
To find the y-intercept, set x = 0 in each equation:
2y = 20 -> y = 10 (y-intercept of the first equation)
2y = 20 -> y = 10 (y-intercept of the second equation)
2y = 20 -> y = 10 (y-intercept of the first equation)
2y = 20 -> y = 10 (y-intercept of the second equation)
So the line segment for |4x| + 2y = 20 is a horizontal line passing through y = 10.
Equation of the line segment y = 0:
This is a horizontal line passing through y = 0.
This is a horizontal line passing through y = 0.
Now we have the equations for the line segments that bound Regions R and P. We can find the area of each region by integrating between the appropriate limits. However, since we are only interested in the difference between the areas, we can use geometry to determine the areas directly.
Region R:
The line segments x = 0 and 2x + 4y = 20 form a right-angled triangle with legs of length 10 and 5. So, the area of Region R is (1/2) * 10 * 5 = 25.
The line segments x = 0 and 2x + 4y = 20 form a right-angled triangle with legs of length 10 and 5. So, the area of Region R is (1/2) * 10 * 5 = 25.
Region P:
The line segments y = 0 and |4x| + 2y = 20 form a trapezoid with bases of length 10 and 10 and height 10. So, the area of Region P is (1/2) * (10 + 10) * 10 = 100.
The line segments y = 0 and |4x| + 2y = 20 form a trapezoid with bases of length 10 and 10 and height 10. So, the area of Region P is (1/2) * (10 + 10) * 10 = 100.
Now we can calculate |A(P) - 2A(R)|:
|A(P) - 2A(R)| = |100 - 2*25| = |100 - 50| = |50| = 50.
|A(P) - 2A(R)| = |100 - 2*25| = |100 - 50| = |50| = 50.
Therefore, the value of |A(P) - 2A(R)| is 50, so the correct answer is C.
Most Upvoted Answer
In the xy-plane, Region R is bounded by the line segments with equatio...
To find the value of |A(P) - 2A(R)|, we first need to determine the equations of the line segments that bound Regions R and P. Let's start with Region R.
Equation of the line segment 2x + |4y| = 20:
We can rewrite this equation as two separate equations:
2x + 4y = 20 (when 4y ≥ 0)
2x - 4y = 20 (when 4y < 0)
We can rewrite this equation as two separate equations:
2x + 4y = 20 (when 4y ≥ 0)
2x - 4y = 20 (when 4y < 0)
To find the x-intercept, set y = 0 in each equation:
2x = 20 -> x = 10 (x-intercept of the first equation)
2x = 20 -> x = 10 (x-intercept of the second equation)
2x = 20 -> x = 10 (x-intercept of the first equation)
2x = 20 -> x = 10 (x-intercept of the second equation)
So the line segment for 2x + |4y| = 20 is a vertical line passing through x = 10.
Equation of the line segment x = 0:
This is a vertical line passing through x = 0.
This is a vertical line passing through x = 0.
Now let's determine the equations for Region P.
Equation of the line segment |4x| + 2y = 20:
We can rewrite this equation as two separate equations:
4x + 2y = 20 (when 4x ≥ 0)
-4x + 2y = 20 (when 4x < 0)
We can rewrite this equation as two separate equations:
4x + 2y = 20 (when 4x ≥ 0)
-4x + 2y = 20 (when 4x < 0)
To find the y-intercept, set x = 0 in each equation:
2y = 20 -> y = 10 (y-intercept of the first equation)
2y = 20 -> y = 10 (y-intercept of the second equation)
2y = 20 -> y = 10 (y-intercept of the first equation)
2y = 20 -> y = 10 (y-intercept of the second equation)
So the line segment for |4x| + 2y = 20 is a horizontal line passing through y = 10.
Equation of the line segment y = 0:
This is a horizontal line passing through y = 0.
This is a horizontal line passing through y = 0.
Now we have the equations for the line segments that bound Regions R and P. We can find the area of each region by integrating between the appropriate limits. However, since we are only interested in the difference between the areas, we can use geometry to determine the areas directly.
Region R:
The line segments x = 0 and 2x + 4y = 20 form a right-angled triangle with legs of length 10 and 5. So, the area of Region R is (1/2) * 10 * 5 = 25.
The line segments x = 0 and 2x + 4y = 20 form a right-angled triangle with legs of length 10 and 5. So, the area of Region R is (1/2) * 10 * 5 = 25.
Region P:
The line segments y = 0 and |4x| + 2y = 20 form a trapezoid with bases of length 10 and 10 and height 10. So, the area of Region P is (1/2) * (10 + 10) * 10 = 100.
The line segments y = 0 and |4x| + 2y = 20 form a trapezoid with bases of length 10 and 10 and height 10. So, the area of Region P is (1/2) * (10 + 10) * 10 = 100.
Now we can calculate |A(P) - 2A(R)|:
|A(P) - 2A(R)| = |100 - 2*25| = |100 - 50| = |50| = 50.
|A(P) - 2A(R)| = |100 - 2*25| = |100 - 50| = |50| = 50.
Therefore, the value of |A(P) - 2A(R)| is 50, so the correct answer is C.
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In the xy-plane, Region R is bounded by the line segments with equations, 2x + |4y| = 20 and x = 0, whereas Region P is bounded by the line segments with equations |4x| + 2y = 20 and y = 0. If the function A(B) is defined as the area of Region B, what is the value of |A(P) – 2A(R)|?a)0b)25c)50d)75e)100Correct answer is option 'C'. Can you explain this answer?
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In the xy-plane, Region R is bounded by the line segments with equations, 2x + |4y| = 20 and x = 0, whereas Region P is bounded by the line segments with equations |4x| + 2y = 20 and y = 0. If the function A(B) is defined as the area of Region B, what is the value of |A(P) – 2A(R)|?a)0b)25c)50d)75e)100Correct answer is option 'C'. Can you explain this answer? for GMAT 2025 is part of GMAT preparation. The Question and answers have been prepared according to the GMAT exam syllabus. Information about In the xy-plane, Region R is bounded by the line segments with equations, 2x + |4y| = 20 and x = 0, whereas Region P is bounded by the line segments with equations |4x| + 2y = 20 and y = 0. If the function A(B) is defined as the area of Region B, what is the value of |A(P) – 2A(R)|?a)0b)25c)50d)75e)100Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GMAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the xy-plane, Region R is bounded by the line segments with equations, 2x + |4y| = 20 and x = 0, whereas Region P is bounded by the line segments with equations |4x| + 2y = 20 and y = 0. If the function A(B) is defined as the area of Region B, what is the value of |A(P) – 2A(R)|?a)0b)25c)50d)75e)100Correct answer is option 'C'. Can you explain this answer?.
In the xy-plane, Region R is bounded by the line segments with equations, 2x + |4y| = 20 and x = 0, whereas Region P is bounded by the line segments with equations |4x| + 2y = 20 and y = 0. If the function A(B) is defined as the area of Region B, what is the value of |A(P) – 2A(R)|?a)0b)25c)50d)75e)100Correct answer is option 'C'. Can you explain this answer? for GMAT 2025 is part of GMAT preparation. The Question and answers have been prepared according to the GMAT exam syllabus. Information about In the xy-plane, Region R is bounded by the line segments with equations, 2x + |4y| = 20 and x = 0, whereas Region P is bounded by the line segments with equations |4x| + 2y = 20 and y = 0. If the function A(B) is defined as the area of Region B, what is the value of |A(P) – 2A(R)|?a)0b)25c)50d)75e)100Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for GMAT 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the xy-plane, Region R is bounded by the line segments with equations, 2x + |4y| = 20 and x = 0, whereas Region P is bounded by the line segments with equations |4x| + 2y = 20 and y = 0. If the function A(B) is defined as the area of Region B, what is the value of |A(P) – 2A(R)|?a)0b)25c)50d)75e)100Correct answer is option 'C'. Can you explain this answer?.
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ample number of questions to practice In the xy-plane, Region R is bounded by the line segments with equations, 2x + |4y| = 20 and x = 0, whereas Region P is bounded by the line segments with equations |4x| + 2y = 20 and y = 0. If the function A(B) is defined as the area of Region B, what is the value of |A(P) – 2A(R)|?a)0b)25c)50d)75e)100Correct answer is option 'C'. Can you explain this answer? tests, examples and also practice GMAT tests.
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