Question

for problems 17 - 20, complete each inequality using the greatest possible integer for comparison. 17. √42 > 18. √150 > 19. √245 > 20. √398 > for problems 21 - 24, circle the lesser of the two numbers.

Explanation:

Step1: Find the closest perfect - square integers.

For $\sqrt{42}$, we know that $6^2 = 36$ and $7^2=49$. So $6<\sqrt{42}<7$.

Step2: For $\sqrt{150}$, we know that $12^2 = 144$ and $13^2 = 169$. So $12<\sqrt{150}<13$.

Step3: For $\sqrt{245}$, we know that $15^2=225$ and $16^2 = 256$. So $15<\sqrt{245}<16$.

Step4: For $\sqrt{398}$, we know that $19^2 = 361$ and $20^2=400$. So $19<\sqrt{398}<20$.

Step5: Compare the integers.

$6<12<15<19$

Answer:

1. $6$
2. $12$
3. $15$
4. $19$