Question

order the following values from greatest (1) to least (4). a) - 3π b) ∛64 c) 0.056 d) -√77

Explanation:

Step1: Evaluate each value

• a) $- 3\pi\approx - 3\times3.14=-9.42$.
• b) $\sqrt[3]{64} = 4$ since $4\times4\times4 = 64$.
• c) $0.056$.
• d) $-\sqrt{77}$. We know that $8^2=64$ and $9^2 = 81$, so $\sqrt{77}$ is between $8$ and $9$, and $-\sqrt{77}$ is between $- 9$ and $-8$.

Step2: Compare the values

We have $\sqrt[3]{64}=4$ is the largest. Then $0.056$. Next, since $-8>-\sqrt{77}>-9$ and $-9.42<-9$, we have $-3\pi$ is the smallest. So the order from greatest to least is $\sqrt[3]{64},0.056,-\sqrt{77},-3\pi$.

Answer:

1. b) $\sqrt[3]{64}$
2. c) $0.056$
3. d) $-\sqrt{77}$
4. a) $-3\pi$