MayaLove wrote:
Yes, squaring the numbers is a valid approach to determine which one is the greatest. Since all the numbers are negative, the one with the smallest absolute value will be the greatest. Let’s square each of the numbers and compare the results.
Given numbers:
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.\(-9\sqrt{\frac{1}{3}}\)
Let's square each of these numbers:
1.\( \left(-3\sqrt{\frac{7}{3}}\right)^2 = 9 \cdot \frac{7}{3} =21\)
2.\( \left(-7\sqrt{\frac{9}{7}}\right)^2 = 49 \cdot \frac{9}{7} =63\)
3.\( \left(-\frac{3}{7}\sqrt{21}\right)^2 = \left(\frac{3}{7}\right)^2 \cdot 21 = \frac{9}{49} \cdot 21 = \frac{189}{49} = \frac{27}{7} \approx4\)
4.\( \left(-\frac{1}{3}\sqrt{63}\right)^2 = \left(\frac{1}{3}\right)^2 \cdot 63 = \frac{1}{9} \cdot 63 =7\)
5.\( \left(-9\sqrt{\frac{1}{3}}\right)^2 = 81 \cdot \frac{1}{3} =27\)
Since the squared value[m] \frac{27}{7}
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Statistics : Posted by Bunuel • on 06 Mar 2024, 02:03 • Replies 3 • Views 232
