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17 May 2021, 17:31
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
(low)
Question Stats:
83% (01:53) correct
17%
(02:07)
wrong
based on 191
sessions
History
Date
Time
Result
Not Attempted Yet
\(\sqrt{\frac{1}{3^0+3^{-1}+3^{-2}+3^{-3}+3^{-4}}}=\)
A. \( \frac{1}{11} \)
B. \( \frac{9}{11}\)
C. \( \frac{11}{9}\)
D. 11
E. \(3^{10}\)
Screenshot 2021-05-18 082606.jpg [ 5.62 KiB | Viewed 3671 times ]
A. \( \frac{1}{11} \)
B. \( \frac{9}{11}\)
C. \( \frac{11}{9}\)
D. 11
E. \(3^{10}\)
Attachment:
Screenshot 2021-05-18 082606.jpg [ 5.62 KiB | Viewed 3671 times ]
18 May 2021, 00:35
Kudos
Bookmarks
sm1510
\(\sqrt{\frac{1}{3^0+3^{-1}+3^{-2}+3^{-3}+3^{-4}}}=\)
A. \( \frac{1}{11} \)
B. \( \frac{9}{11}\)
C. \( \frac{11}{9}\)
D. 11
E. \(3^{10}\)
A. \( \frac{1}{11} \)
B. \( \frac{9}{11}\)
C. \( \frac{11}{9}\)
D. 11
E. \(3^{10}\)
I) Logical
The denominator is surely greater than 1, as 3^0=1. Therefore, we have square root of a fraction that is less than 1.
Thus the square root will be <1.
Only A and B left.
\(\frac{1}{11} = \sqrt{\frac{1}{121}}\)
Clearly the denominator will not give us an integer value 121.
Hence A.
II) \(\sqrt{\frac{1}{3^0+3^{-1}+3^{-2}+3^{-3}+3^{-4}}}=\)
\(\sqrt{\frac{1}{3^0+\frac{1}{3^{1}}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}}}=\)
\(\sqrt{\frac{1}{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}}}=\)
\(\sqrt{\frac{1}{1+\frac{27+9+3+1}{81}}}=\)
\(\sqrt{\frac{1}{\frac{121}{81}}}=\sqrt{\frac{81}{121}}=\frac{9}{11}\)
III) \(\sqrt{\frac{1}{3^0+3^{-1}+3^{-2}+3^{-3}+3^{-4}}}=\)
To Remove fraction, multiply by 3^4
\(\sqrt{\frac{1*3^4}{3^4(3^0+3^{-1}+3^{-2}+3^{-3}+3^{-4})}}=\)
\(\sqrt{\frac{81}{3^4+3^3+3^2+3^1}}=\sqrt{\frac{81}{121}}=9/11\)
B
09 May 2023, 07:27
Kudos
Bookmarks
chetan2u
sm1510
\(\sqrt{\frac{1}{3^0+3^{-1}+3^{-2}+3^{-3}+3^{-4}}}=\)
A. \( \frac{1}{11} \)
B. \( \frac{9}{11}\)
C. \( \frac{11}{9}\)
D. 11
E. \(3^{10}\)
A. \( \frac{1}{11} \)
B. \( \frac{9}{11}\)
C. \( \frac{11}{9}\)
D. 11
E. \(3^{10}\)
I) Logical
The denominator is surely greater than 1, as 3^0=1. Therefore, we have square root of a fraction that is less than 1.
Thus the square root will be <1.
Only A and B left.
\(\frac{1}{11} = \sqrt{\frac{1}{121}}\)
Clearly the denominator will not give us an integer value 121.
Hence A.
II) \(\sqrt{\frac{1}{3^0+3^{-1}+3^{-2}+3^{-3}+3^{-4}}}=\)
\(\sqrt{\frac{1}{3^0+\frac{1}{3^{1}}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\frac{1}{3^{4}}}}=\)
\(\sqrt{\frac{1}{1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}}}=\)
\(\sqrt{\frac{1}{1+\frac{27+9+3+1}{81}}}=\)
\(\sqrt{\frac{1}{\frac{121}{81}}}=\sqrt{\frac{81}{121}}=\frac{9}{11}\)
III) \(\sqrt{\frac{1}{3^0+3^{-1}+3^{-2}+3^{-3}+3^{-4}}}=\)
To Remove fraction, multiply by 3^4
\(\sqrt{\frac{1*3^4}{3^4(3^0+3^{-1}+3^{-2}+3^{-3}+3^{-4})}}=\)
\(\sqrt{\frac{81}{3^4+3^3+3^2+3^1}}=\sqrt{\frac{81}{121}}=9/11\)
B
I think you have missed "+1" in the last step
\(\sqrt{\frac{81}{3^4+3^3+3^2+3^1+1}}\)
