Question

for problems 13 - 16, write < or > to complete each statement.

1. $sqrt{15}-4$ ⑦ $sqrt{15}-7$
2. $2sqrt{18}$ ④ $2sqrt{21}$
3. $-sqrt3{30}$ ④ $-3$
4. $sqrt{8}+1$ ⑦ $sqrt{17}-2$

for problems 17 - 20, complete each inequality using the greatest possible integer for comparison.

1. $sqrt{42}>6$
2. $sqrt{150}>12$
3. $sqrt{245}>15$
4. $sqrt{398}>19$

for problems 21 - 24, circle the lesser of the two numbers.

1. $sqrt{30}+4$ 8
2. $sqrt{11}$ $sqrt{20}-2$
3. $sqrt{7}$ $sqrt3{40}$
4. $15 - sqrt{2}$ $sqrt{125}$

for problems 25 - 31, complete each inequality using a pair of integers.

1. _ < $sqrt{10}+4$ < _
2. _ < $sqrt{19}+8$ < _
3. _ < $sqrt{61}-5$ < _
4. _ < $sqrt{92}-11$ < _
5. _ < $sqrt3{50}-1$ < _
6. _ < $sqrt3{100}+2$ < _
7. _ < $sqrt3{17}+5$ < _
8. reason is there a limit to the number of decimals to approximate an irrational number? explain.

Explanation:

Step1: Recall square - root and cube - root approximations

We know that if \(a^2 < b < c^2\), then \(a<\sqrt{b}

Step2: Solve problem 25

Since \(3^2=9<10<16 = 4^2\), then \(3<\sqrt{10}<4\). So \(3 + 4<\sqrt{10}+4<4 + 4\), which gives \(7<\sqrt{10}+4<8\).

Step3: Solve problem 26

Since \(4^2=16<19<25 = 5^2\), then \(4<\sqrt{19}<5\). So \(4 + 8<\sqrt{19}+8<5 + 8\), which gives \(12<\sqrt{19}+8<13\).

Step4: Solve problem 27

Since \(7^2=49<61<64 = 8^2\), then \(7<\sqrt{61}<8\). So \(7-5<\sqrt{61}-5<8 - 5\), which gives \(2<\sqrt{61}-5<3\).

Step5: Solve problem 28

Since \(9^2=81<92<100 = 10^2\), then \(9<\sqrt{92}<10\). So \(9-11<\sqrt{92}-11<10 - 11\), which gives \(-2<\sqrt{92}-11<-1\).

Step6: Solve problem 29

Since \(3^3=27<50<64 = 4^3\), then \(3<\sqrt[3]{50}<4\). So \(3-1<\sqrt[3]{50}-1<4 - 1\), which gives \(2<\sqrt[3]{50}-1<3\).

Step7: Solve problem 30

Since \(4^3=64<100<125 = 5^3\), then \(4<\sqrt[3]{100}<5\). So \(4 + 2<\sqrt[3]{100}+2<5 + 2\), which gives \(6<\sqrt[3]{100}+2<7\).

Step8: Solve problem 31

Since \(2^3=8<17<27 = 3^3\), then \(2<\sqrt[3]{17}<3\). So \(2+5<\sqrt[3]{17}+5<3 + 5\), which gives \(7<\sqrt[3]{17}+5<8\).

Answer:

1. \(7<\sqrt{10}+4<8\)
2. \(12<\sqrt{19}+8<13\)
3. \(2<\sqrt{61}-5<3\)
4. \(-2<\sqrt{92}-11<-1\)
5. \(2<\sqrt[3]{50}-1<3\)
6. \(6<\sqrt[3]{100}+2<7\)
7. \(7<\sqrt[3]{17}+5<8\)