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06 Mar 2024, 02:03
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Which of the numbers \(-3\sqrt{\frac{7}{3}}\), \(-7\sqrt{\frac{9}{7}}\), \(-\frac{3}{7}\sqrt{21}\), \(-\frac{1}{3}\sqrt{63}\), and \(-9\sqrt{\frac{1}{3}}\) is the greatest?
A. \(-3\sqrt{\frac{7}{3}}\)
B. \(-7\sqrt{\frac{9}{7}}\)
C. \(-\frac{3}{7}\sqrt{21}\)
D. \(-\frac{1}{3}\sqrt{63}\)
E. \(-9\sqrt{\frac{1}{3}}\)
A. \(-3\sqrt{\frac{7}{3}}\)
B. \(-7\sqrt{\frac{9}{7}}\)
C. \(-\frac{3}{7}\sqrt{21}\)
D. \(-\frac{1}{3}\sqrt{63}\)
E. \(-9\sqrt{\frac{1}{3}}\)
06 Mar 2024, 02:03
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Official Solution:
Which of the numbers \(-3\sqrt{\frac{7}{3}}\), \(-7\sqrt{\frac{9}{7}}\), \(-\frac{3}{7}\sqrt{21}\), \(-\frac{1}{3}\sqrt{63}\), and \(-9\sqrt{\frac{1}{3}}\) is the greatest?
A. \(-3\sqrt{\frac{7}{3}}\)
B. \(-7\sqrt{\frac{9}{7}}\)
C. \(-\frac{3}{7}\sqrt{21}\)
D. \(-\frac{1}{3}\sqrt{63}\)
E. \(-9\sqrt{\frac{1}{3}}\)
To compare the numbers easily, let's put all of them under square roots. Because of the minus sign, the one with the smallest absolute value will be the greatest.
\(-3\sqrt{\frac{7}{3}} = -\sqrt{\frac{3^2*7}{3}} = -\sqrt{21} \)
\(-7\sqrt{\frac{9}{7}} =-\sqrt{\frac{7^2*9}{7}}=-\sqrt{63}\)
\(-\frac{3}{7}\sqrt{21} = -\sqrt{\frac{3^2}{7^2}*21} = -\sqrt{\frac{27}{7}} \approx -\sqrt{4} \)
\(-\frac{1}{3}\sqrt{63} = - \sqrt{\frac{1}{3^2}*63}= -\sqrt{7}\)
\(-9\sqrt{\frac{1}{3}} = -\sqrt{\frac{9^2}{3}} =-\sqrt{27}\)
Therefore, the third number, \(-\frac{3}{7}\sqrt{21}\) is the greatest.
Answer: C
Which of the numbers \(-3\sqrt{\frac{7}{3}}\), \(-7\sqrt{\frac{9}{7}}\), \(-\frac{3}{7}\sqrt{21}\), \(-\frac{1}{3}\sqrt{63}\), and \(-9\sqrt{\frac{1}{3}}\) is the greatest?
A. \(-3\sqrt{\frac{7}{3}}\)
B. \(-7\sqrt{\frac{9}{7}}\)
C. \(-\frac{3}{7}\sqrt{21}\)
D. \(-\frac{1}{3}\sqrt{63}\)
E. \(-9\sqrt{\frac{1}{3}}\)
To compare the numbers easily, let's put all of them under square roots. Because of the minus sign, the one with the smallest absolute value will be the greatest.
\(-3\sqrt{\frac{7}{3}} = -\sqrt{\frac{3^2*7}{3}} = -\sqrt{21} \)
\(-7\sqrt{\frac{9}{7}} =-\sqrt{\frac{7^2*9}{7}}=-\sqrt{63}\)
\(-\frac{3}{7}\sqrt{21} = -\sqrt{\frac{3^2}{7^2}*21} = -\sqrt{\frac{27}{7}} \approx -\sqrt{4} \)
\(-\frac{1}{3}\sqrt{63} = - \sqrt{\frac{1}{3^2}*63}= -\sqrt{7}\)
\(-9\sqrt{\frac{1}{3}} = -\sqrt{\frac{9^2}{3}} =-\sqrt{27}\)
Therefore, the third number, \(-\frac{3}{7}\sqrt{21}\) is the greatest.
Answer: C
MayaLove
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27 May 2024, 10:59
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Hi Bunuel! Can we square the numbers to find out which one's the greatest?
