why dont they keep prisoners in the ocean? locate the following irrational numbers on the number line. e…

why dont they keep prisoners in the ocean? locate the following irrational numbers on the number line. e $sqrt{130}$ m $sqrt{48}$ r $sqrt{\frac{1}{2}}$ t $-sqrt{122}$ e $sqrt{60}$ s $-sqrt{6}$ i $sqrt3{130}$ c $-sqrt{56}$ r $sqrt{23}$ y $-sqrt{67}$ l $-sqrt3{65}$ a $-sqrt{\frac{2}{3}}$ v $sqrt{110}$ h $-sqrt{105}$ w $sqrt{74}$ e $-sqrt{90}$ t $-sqrt{3}$ a $sqrt{91}$ t $sqrt{2}$ c $sqrt{10}$ o $-sqrt{40}$ a $sqrt3{13}$ d $-sqrt{15}$ u $-sqrt{26}$

why dont they keep prisoners in the ocean? locate the following irrational numbers on the number line. e $sqrt{130}$ m $sqrt{48}$ r $sqrt{\frac{1}{2}}$ t $-sqrt{122}$ e $sqrt{60}$ s $-sqrt{6}$ i $sqrt3{130}$ c $-sqrt{56}$ r $sqrt{23}$ y $-sqrt{67}$ l $-sqrt3{65}$ a $-sqrt{\frac{2}{3}}$ v $sqrt{110}$ h $-sqrt{105}$ w $sqrt{74}$ e $-sqrt{90}$ t $-sqrt{3}$ a $sqrt{91}$ t $sqrt{2}$ c $sqrt{10}$ o $-sqrt{40}$ a $sqrt3{13}$ d $-sqrt{15}$ u $-sqrt{26}$

Answer

Explanation:

Step1: Find the range for each square - root number

For $\sqrt{130}$, since $11^2 = 121$ and $12^2=144$, then $11<\sqrt{130}<12$. For $\sqrt{48}$, since $6^2 = 36$ and $7^2 = 49$, then $6<\sqrt{48}<7$. For $\sqrt{\frac{1}{2}}$, since $0^2 = 0$ and $1^2 = 1$, and $\sqrt{\frac{1}{2}}\approx0.707$, then $0<\sqrt{\frac{1}{2}}<1$. For $-\sqrt{122}$, since $11^2 = 121$ and $12^2 = 144$, then $- 12<-\sqrt{122}<-11$. For $\sqrt{60}$, since $7^2 = 49$ and $8^2 = 64$, then $7<\sqrt{60}<8$. For $-\sqrt{6}$, since $2^2 = 4$ and $3^2 = 9$, then $-3<-\sqrt{6}<-2$. For $\sqrt[3]{130}$, since $5^3=125$ and $6^3 = 216$, then $5<\sqrt[3]{130}<6$. For $-\sqrt{56}$, since $7^2 = 49$ and $8^2 = 64$, then $-8<-\sqrt{56}<-7$. For $\sqrt{23}$, since $4^2 = 16$ and $5^2 = 25$, then $4<\sqrt{23}<5$. For $-\sqrt{67}$, since $8^2 = 64$ and $9^2 = 81$, then $-9<-\sqrt{67}<-8$. For $-\sqrt[3]{65}$, since $4^3 = 64$ and $5^3 = 125$, then $-5<-\sqrt[3]{65}<-4$. For $-\sqrt{\frac{2}{3}}$, since $0^2 = 0$ and $1^2 = 1$, and $-\sqrt{\frac{2}{3}}\approx - 0.816$, then $-1<-\sqrt{\frac{2}{3}}<0$. For $\sqrt{110}$, since $10^2 = 100$ and $11^2 = 121$, then $10<\sqrt{110}<11$. For $-\sqrt{105}$, since $10^2 = 100$ and $11^2 = 121$, then $-11<-\sqrt{105}<-10$. For $\sqrt{74}$, since $8^2 = 64$ and $9^2 = 81$, then $8<\sqrt{74}<9$. For $-\sqrt{90}$, since $9^2 = 81$ and $10^2 = 100$, then $-10<-\sqrt{90}<-9$. For $-\sqrt{3}$, since $1^2 = 1$ and $2^2 = 4$, then $-2<-\sqrt{3}<-1$. For $\sqrt{91}$, since $9^2 = 81$ and $10^2 = 100$, then $9<\sqrt{91}<10$. For $\sqrt{2}$, since $1^2 = 1$ and $2^2 = 4$, then $1<\sqrt{2}<2$. For $\sqrt{10}$, since $3^2 = 9$ and $4^2 = 16$, then $3<\sqrt{10}<4$. For $-\sqrt{40}$, since $6^2 = 36$ and $7^2 = 49$, then $-7<-\sqrt{40}<-6$. For $\sqrt[3]{13}$, since $2^3 = 8$ and $3^3 = 27$, then $2<\sqrt[3]{13}<3$. For $-\sqrt{15}$, since $3^2 = 9$ and $4^2 = 16$, then $-4<-\sqrt{15}<-3$. For $-\sqrt{26}$, since $5^2 = 25$ and $6^2 = 36$, then $-6<-\sqrt{26}<-5$.

Step2: Mark the numbers on the number - line

Based on the above - calculated ranges, mark each of the irrational numbers on the given number - line.

Answer:

Mark each irrational number on the number - line according to the ranges calculated in Step1. For example, mark $\sqrt{130}$ between 11 and 12, mark $\sqrt{48}$ between 6 and 7, etc.