Question

for problems 21 - 24, circle the lesser of the two numbers. 21. $sqrt{30}+4$ 8 22. $sqrt{11}$ $sqrt{20}-2$ 23. $sqrt{7}$ $sqrt3{40}$ 24. $15 - sqrt{2}$ $sqrt{125}$

Explanation:

Step1: Compare $\sqrt{30}+4$ and $8$

First, estimate $\sqrt{30}$. Since $5^2 = 25$ and $6^2=36$, $5<\sqrt{30}<6$. Then $\sqrt{30}+4$ is between $5 + 4=9$ and $6 + 4 = 10$. So $\sqrt{30}+4>8$.

Step2: Compare $\sqrt{11}$ and $\sqrt{20}-2$

Estimate $\sqrt{11}\approx3.32$ (since $3.32^2=10.9264$) and $\sqrt{20}\approx4.47$ (since $4.47^2 = 19.9809$), then $\sqrt{20}-2\approx4.47 - 2=2.47$. So $\sqrt{20}-2<\sqrt{11}$.

Step3: Compare $\sqrt{7}$ and $\sqrt[3]{40}$

Estimate $\sqrt{7}\approx2.65$ (since $2.65^2=7.0225$) and $\sqrt[3]{40}\approx3.42$ (since $3.42^3=39.959168$). So $\sqrt{7}<\sqrt[3]{40}$.

Step4: Compare $15-\sqrt{2}$ and $\sqrt{125}$

Estimate $\sqrt{2}\approx1.41$, so $15-\sqrt{2}\approx15 - 1.41 = 13.59$. And $\sqrt{125}\approx11.18$ (since $11.18^2=124.9924$). So $\sqrt{125}<15-\sqrt{2}$.

Answer:

1. 8
2. $\sqrt{20}-2$
3. $\sqrt{7}$
4. $\sqrt{125}$