Question

which inequality correctly orders the numbers $-\frac{9}{3}$, $-\frac{3}{9}$, and $-1$? choose 1 answer: a $-\frac{9}{3} > -1 > -\frac{3}{9}$ b $-1 > -\frac{9}{3} > -\frac{3}{9}$ c $-1 > -\frac{3}{9} > -\frac{9}{3}$ d $-\frac{3}{9} > -1 > -\frac{9}{3}$

Explanation:

Step1: Simplify the fractions

Simplify $-\frac{9}{3}$ and $-\frac{3}{9}$. We know that $-\frac{9}{3} = - 3$ and $-\frac{3}{9}=-\frac{1}{3}$. Now we have the numbers $-3$, $-\frac{1}{3}$, and $-1$.

Step2: Compare the numbers

For negative numbers, the larger the absolute value, the smaller the number. The absolute values are $|-3| = 3$, $|-\frac{1}{3}|=\frac{1}{3}$, and $|-1| = 1$. Since $\frac{1}{3}<1 < 3$, the order of the negative numbers from largest to smallest is $-\frac{1}{3}>-1>-3$. Substituting back the original fractions, we get $-\frac{3}{9}>-1>-\frac{9}{3}$.

Answer:

D. $-\frac{3}{9} > -1 > -\frac{9}{3}$