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09 Nov 2022, 01:59
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
83% (01:49) correct
17%
(02:32)
wrong
based on 89
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x and y are integers. 4 ≤ x < 7< y ≤ 12, what`s the range of \(( x − y)^2\) ?
A. 101
B. 102
C. 80
D. 75
E. 60
A. 101
B. 102
C. 80
D. 75
E. 60
09 Nov 2022, 03:23
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Given
Re-wording this question, We intend to find the range of the square of the minimum distance and the square of the maximum distance between x and y.
Let's see at what positions the two values can lie on a number line -
--------0--------4----------------7-----------------12--------
x can occupy the green portion of the number line and y can occupy the red portion.
The minimum distance between x and y occurs when the numbers are as close as possible with respect to each other.
x = 6 & y = 8 ; \((x-y)^2\) = \((6-8)^2\) = 4
The maximum distance between x and y occurs when x and y are as far away as possible from each other.
x = 4 & y = 12 ; \((4-12)^2\) = \((8)^2\) = 64
Range = 64 - 4 = 60
Option E
- 4 ≤ x < 7
- 7< y ≤ 12
Re-wording this question, We intend to find the range of the square of the minimum distance and the square of the maximum distance between x and y.
Let's see at what positions the two values can lie on a number line -
--------0--------4----------------7-----------------12--------
x can occupy the green portion of the number line and y can occupy the red portion.
The minimum distance between x and y occurs when the numbers are as close as possible with respect to each other.
x = 6 & y = 8 ; \((x-y)^2\) = \((6-8)^2\) = 4
The maximum distance between x and y occurs when x and y are as far away as possible from each other.
x = 4 & y = 12 ; \((4-12)^2\) = \((8)^2\) = 64
Range = 64 - 4 = 60
Option E
09 Nov 2022, 05:26
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Afroditee
Range means we have to look for largest and smallest values of \((x-y)^2\).
Largest: \((x_{min}-y_{max})^2=(4-12)^2=8^2=64\)
Smallest: \((x_{max}-y_{min})^2=(6-8)^2=2^2=4\)
Range: 64-4 = 60
E
