Question

directions: estimate each square root to the nearest tenth. 17. √108 18. -√372 19. √61 20. circle each value that is a perfect cube. 1 90 300 72 525 216

Explanation:

Step1: Find perfect - square bounds for $\sqrt{108}$

We know that $10^2 = 100$ and $11^2=121$. Since $108$ is closer to $100$ than to $121$, $\sqrt{108}\approx10.4$.

Step2: Find perfect - square bounds for $-\sqrt{372}$

We know that $19^2 = 361$ and $20^2 = 400$. Since $372$ is closer to $361$ than to $400$, $\sqrt{372}\approx19.3$, so $-\sqrt{372}\approx - 19.3$.

Step3: Find perfect - square bounds for $\sqrt{61}$

We know that $7^2 = 49$ and $8^2=64$. Since $61$ is closer to $64$ than to $49$, $\sqrt{61}\approx7.8$.

Step4: Determine perfect cubes

A perfect cube is a number of the form $n^3$ where $n$ is an integer.
$1^3=1$, $6^3 = 216$. For $90$, there is no integer $n$ such that $n^3=90$. For $300$, there is no integer $n$ such that $n^3 = 300$. For $72$, there is no integer $n$ such that $n^3=72$. For $525$, there is no integer $n$ such that $n^3 = 525$.
The perfect cubes are $1$ and $216$.

Answer:

1. $10.4$
2. $-19.3$
3. $7.8$
4. Circle $1$ and $216$