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If 9 <= x <= 15
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16 May 2021, 02:35
16 May 2021, 02:35
Expert Reply
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Question Stats:
72% (00:52) correct
27% (00:53) wrong based on 18 sessions
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If \(9 \leq x \leq 15\) and \(3 \leq y \leq 5\) , what is the greatest possible value of \(\frac{7}{x-y}\) ?
Show: :: OA
\(\frac{7}{4}\)
Re: If 9 <= x <= 15
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16 May 2021, 02:49
16 May 2021, 02:49
1
Solution:
The number will be greatest when the denominator is the least. And the denominator will be least when x takes the smallest value and y take the largest one.
\(\frac{7}{9-5}\)=\(\frac{7}{4}\)
Answer: \(\frac{7}{4}\)
Hope this helps!
The number will be greatest when the denominator is the least. And the denominator will be least when x takes the smallest value and y take the largest one.
\(\frac{7}{9-5}\)=\(\frac{7}{4}\)
Answer: \(\frac{7}{4}\)
Hope this helps!
If 9 <= x <= 15
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16 May 2021, 06:41
16 May 2021, 06:41
1
Here we will get multiple values like
\(\frac{7}{9-3}\)
\(\frac{7}{9-5}\)
\(\frac{7}{15-3}\)
\(\frac{7}{15-5}\)
Among all these \(\frac{7}{4}\) is the smallest.
\(\frac{7}{9-3}\)
\(\frac{7}{9-5}\)
\(\frac{7}{15-3}\)
\(\frac{7}{15-5}\)
Among all these \(\frac{7}{4}\) is the smallest.
