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−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha
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17 Mar 2018, 07:26
17 Mar 2018, 07:26
Expert Reply
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Question Stats:
31% (01:56) correct
68% (02:09) wrong based on 16 sessions
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−1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. What most accurately describes the range of \(a^2\) ?
A. −16 < \(a^2\) < 11
B. −4 < \(a^2\) < 11
C. 0 < \(a^2\) < 16
D. 0 < \(a^2\) < 121
E. 16 < \(a^2\) < 121
A. −16 < \(a^2\) < 11
B. −4 < \(a^2\) < 11
C. 0 < \(a^2\) < 16
D. 0 < \(a^2\) < 121
E. 16 < \(a^2\) < 121
Drill 2
Question: 5
Page: 497
Question: 5
Page: 497
Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha
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18 Mar 2018, 07:00
18 Mar 2018, 07:00
Expert Reply
Explanation
If a range of values for a can be found, then the range of values for \(a^2\) can be found.
Start by testing the end values of b, –3 and 1.
Plug in –3 for b in the first given inequality then solve for a.
You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.
Remember to take the last step, though!
The question is looking for the range of \(a^2\), not a; \(a^2\) is always positive (i.e., \(0 < a^2\)).
Because a < 11, \(a^2 < 121\).
This means \(0 < a^2 < 121\); the answer is choice (D).
If a range of values for a can be found, then the range of values for \(a^2\) can be found.
Start by testing the end values of b, –3 and 1.
Plug in –3 for b in the first given inequality then solve for a.
You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.
Remember to take the last step, though!
The question is looking for the range of \(a^2\), not a; \(a^2\) is always positive (i.e., \(0 < a^2\)).
Because a < 11, \(a^2 < 121\).
This means \(0 < a^2 < 121\); the answer is choice (D).
Re: −1 < a − b < 10, with b an integer such that −3 ≤ b ≤ 1. Wha
[#permalink]
22 Mar 2018, 00:52
22 Mar 2018, 00:52
I don't understand how the answer can be D.
If b = -3 then we have -4 < a < 7.
If b = 1 then we have 0 < a < 11.
So - 4 < a < 11.
Squaring the inequality yields, 16 < a^{2} < 121.
a^{2} > 16 > 0.
If b = -3 then we have -4 < a < 7.
If b = 1 then we have 0 < a < 11.
So - 4 < a < 11.
Squaring the inequality yields, 16 < a^{2} < 121.
a^{2} > 16 > 0.
sandy wrote:
Explanation
If a range of values for a can be found, then the range of values for \(a^2\) can be found.
Start by testing the end values of b, –3 and 1.
Plug in –3 for b in the first given inequality then solve for a.
You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.
Remember to take the last step, though!
The question is looking for the range of \(a^2\), not a; \(a^2\) is always positive (i.e., \(0 < a^2\)).
Because a < 11, \(a^2 < 121\).
This means \(0 < a^2 < 121\); the answer is choice (D).
If a range of values for a can be found, then the range of values for \(a^2\) can be found.
Start by testing the end values of b, –3 and 1.
Plug in –3 for b in the first given inequality then solve for a.
You find that –4 < a < 7. If b = 1, 0 < a <11; b could be any integer in the range –3 ≤ b ≤ 1, this means –4 < a < 11 overall.
Remember to take the last step, though!
The question is looking for the range of \(a^2\), not a; \(a^2\) is always positive (i.e., \(0 < a^2\)).
Because a < 11, \(a^2 < 121\).
This means \(0 < a^2 < 121\); the answer is choice (D).
