Question

Put these in order starting with the smallest.
9 3/2
27 1/3
125 2/3​

Answer 1

Answer:

second, third,first

Step-by-step explanation:

first

3/2=1.5

9^1.5=27

second

1/3=0.333

27^0.333=3

third

2/3=0.667

125^0.667=25

Answer 2

The order with the smallest is $27^{\frac{1}{3} }<125^{\frac{2}{3} }<9^{\frac{3}{2} }$

Simplify using exponents and power rule $(a^{m} )^{n}=a^{mn}$

We have

$9^{\frac{3}{2} } \\=(3^{2} )^{\frac{3}{2} } \\=3^{3} \\=27$  

 And

$27^{\frac{1}{3} } \\=(3^{3} )^{\frac{1}{3} } \\=3$

And  

$125^{\frac{2}{3} } \\=(5^{3} )^{\frac{2}{3} } \\=5^{2} \\=25$

So, 27>25>3

Therefore order starting with the smallest is $27^{\frac{1}{3} }<125^{\frac{2}{3} }<9^{\frac{3}{2} }$

Learn more:https://brainly.com/question/13277609