1. circle each value that is a perfect square.
50 81 289 360 4 100 75 224 25
directions: find each square root.
2. $sqrt{36}$
3. $-sqrt{225}$
4. $-sqrt{64}$
5. $sqrt{324}$
6. $sqrt{121}$
7. $-sqrt{169}$
8. $sqrt{\frac{16}{9}}$
9. $sqrt{\frac{81}{400}}$
10. $-sqrt{\frac{1}{100}}$
directions: identify the two consecutive integers in which each square root lies between.
11. $sqrt{95}$
12. $sqrt{320}$
13. $-sqrt{17}$
14. $-sqrt{156}$
15. $sqrt{48}$
16. $-sqrt{249}$
directions: estimate each square root to the nearest tenth.
17. $sqrt{108}$
18. $-sqrt{372}$
19. $sqrt{61}$
0. circle each value that is a perfect cube.
27 1,000 90 1 300 72 525
ections: find each cube root.
$sqrt3{729}$
22. $sqrt3{125}$
23. $sqrt3{-1,331}$
xplain why each non - zero integer has two square roots but only one cube root.

Question

1. circle each value that is a perfect square.
50  81  289  360  4  100  75  224  25
directions: find each square root.
2. $sqrt{36}$
3. $-sqrt{225}$
4. $-sqrt{64}$
5. $sqrt{324}$
6. $sqrt{121}$
7. $-sqrt{169}$
8. $sqrt{\frac{16}{9}}$
9. $sqrt{\frac{81}{400}}$
10. $-sqrt{\frac{1}{100}}$
directions: identify the two consecutive integers in which each square root lies between.
11. $sqrt{95}$
12. $sqrt{320}$
13. $-sqrt{17}$
14. $-sqrt{156}$
15. $sqrt{48}$
16. $-sqrt{249}$
directions: estimate each square root to the nearest tenth.
17. $sqrt{108}$
18. $-sqrt{372}$
19. $sqrt{61}$
0. circle each value that is a perfect cube.
27  1,000  90  1  300  72  525
ections: find each cube root.
$sqrt3{729}$
22. $sqrt3{125}$
23. $sqrt3{-1,331}$
xplain why each non - zero integer has two square roots but only one cube root.

Accepted Answer

### Problem 2: $\boldsymbol{\sqrt{36}}$
# Explanation:
## Step1: Recall square of integers
We know that $6	imes6 = 36$ (i.e., $6^2=36$) and also $(-6)	imes(-6) = 36$, but the principal square root (the one with the $\sqrt{}$ symbol) is the non - negative root.
## Step2: Find the square root
Since $6^2 = 36$, $\sqrt{36}=6$.
# Answer:
$6$

### Problem 3: $\boldsymbol{-\sqrt{225}}$
# Explanation:
## Step1: Find the principal square root of 225
We know that $15	imes15=225$, so $\sqrt{225} = 15$.
## Step2: Apply the negative sign
We have $-\sqrt{225}=- 15$.
# Answer:
$-15$

### Problem 4: $\boldsymbol{-\sqrt{64}}$
# Explanation:
## Step1: Find the principal square root of 64
Since $8	imes8 = 64$, $\sqrt{64}=8$.
## Step2: Apply the negative sign
So, $-\sqrt{64}=-8$.
# Answer:
$-8$

### Problem 5: $\boldsymbol{\sqrt{324}}$
# Explanation:
## Step1: Recall square of integers
We know that $18	imes18=324$ (because $18^2=(20 - 2)^2=20^2-2	imes20	imes2 + 2^2=400-80 + 4 = 324$).
## Step2: Find the square root
So, $\sqrt{324}=18$.
# Answer:
$18$

### Problem 6: $\boldsymbol{\sqrt{121}}$
# Explanation:
## Step1: Recall square of integers
We know that $11	imes11 = 121$, so $\sqrt{121}=11$.
# Answer:
$11$

### Problem 7: $\boldsymbol{-\sqrt{169}}$
# Explanation:
## Step1: Find the principal square root of 169
Since $13	imes13 = 169$, $\sqrt{169}=13$.
## Step2: Apply the negative sign
So, $-\sqrt{169}=-13$.
# Answer:
$-13$

### Problem 8: $\boldsymbol{\sqrt{\frac{16}{9}}}$
# Explanation:
## Step1: Recall square of fractions
We know that for a fraction $\frac{a}{b}$, $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ (where $a\geq0,b > 0$). Here, $a = 16$ and $b = 9$.
## Step2: Find square roots of numerator and denominator
$\sqrt{16}=4$ and $\sqrt{9}=3$. So, $\sqrt{\frac{16}{9}}=\frac{4}{3}$.
# Answer:
$\frac{4}{3}$

### Problem 9: $\boldsymbol{\sqrt{\frac{81}{400}}}$
# Explanation:
## Step1: Use the property of square roots of fractions
$\sqrt{\frac{81}{400}}=\frac{\sqrt{81}}{\sqrt{400}}$.
## Step2: Find square roots of numerator and denominator
$\sqrt{81}=9$ and $\sqrt{400}=20$. So, $\sqrt{\frac{81}{400}}=\frac{9}{20}$.
# Answer:
$\frac{9}{20}$

### Problem 10: $\boldsymbol{-\sqrt{\frac{1}{100}}}$
# Explanation:
## Step1: Find the principal square root of the fraction
$\sqrt{\frac{1}{100}}=\frac{\sqrt{1}}{\sqrt{100}}=\frac{1}{10}$.
## Step2: Apply the negative sign
So, $-\sqrt{\frac{1}{100}}=-\frac{1}{10}$.
# Answer:
$-\frac{1}{10}$

### Problem 11: $\boldsymbol{\sqrt{95}}$
# Explanation:
## Step1: Find perfect squares around 95
We know that $9^2=81$ and $10^2 = 100$.
## Step2: Compare with 95
Since $81<95<100$, we have $9<\sqrt{95}<10$.
# Answer:
Between 9 and 10

### Problem 12: $\boldsymbol{\sqrt{320}}$
# Explanation:
## Step1: Find perfect squares around 320
We know that $17^2=289$ and $18^2=324$.
## Step2: Compare with 320
Since $289<320<324$, we have $17<\sqrt{320}<18$.
# Answer:
Between 17 and 18

### Problem 13: $\boldsymbol{-\sqrt{17}}$
# Explanation:
## Step1: Find perfect squares around 17
We know that $4^2 = 16$ and $5^2=25$. So, $4<\sqrt{17}<5$.
## Step2: Apply the negative sign
When we apply the negative sign, the inequality sign reverses. So, $- 5<-\sqrt{17}<-4$.
# Answer:
Between - 5 and - 4

### Problem 14: $\boldsymbol{-\sqrt{156}}$
# Explanation:
## Step1: Find perfect squares around 156
We know that $12^2=144$ and $13^2 = 169$. So, $12<\sqrt{156}<13$.
## Step2: Apply the negative sign
When we apply the negative sign, the inequality sign reverses. So, $-13<-\sqrt{156}<-12$.
# Answer:
Between - 13 and - 12

### Problem 15: $\boldsymbol{\sqrt{48}}$
# Explanation:
## Step1: Find perfect squares around 48
We know that $6^2=36$ and $7^2 = 49$.
## Step2: Compare with 48
Since $36<48<49$, we have $6<\sqrt{48}<7$.
# Answer:
Between 6 and 7

### Problem 16: $\boldsymbol{-\sqrt{249}}$
# Explanation:
## Step1: Find perfect squares around 249
We know that $15^2=225$ and $16^2=256$. So, $15<\sqrt{249}<16$.
## Step2: Apply the negative sign
When we apply the negative sign, the inequality sign reverses. So, $-16<-\sqrt{249}<-15$.
# Answer:
Between - 16 and - 15

### Problem 17: $\boldsymbol{\sqrt{108}}$ (to the nearest tenth)
# Explanation:
## Step1: Find two perfect squares around 108
We know that $10^2 = 100$ and $11^2=121$. So, $10<\sqrt{108}<11$.
## Step2: Calculate the difference
Let $x=\sqrt{108}-10$. We know that $108 - 100=8$ and $121 - 100 = 21$. So, $x\approx\frac{8}{21}\approx0.38$.
## Step3: Find the approximate value
$\sqrt{108}\approx10 + 0.38=10.38\approx10.4$ (to the nearest tenth).
# Answer:
$10.4$

### Problem 18: $\boldsymbol{-\sqrt{372}}$ (to the nearest tenth)
# Explanation:
## Step1: Find two perfect squares around 372
We know that $19^2=361$ and $20^2 = 400$. So, $19<\sqrt{372}<20$.
## Step2: Calculate the difference
Let $x=\sqrt{372}-19$. We know that $372 - 361 = 11$ and $400 - 361=39$. So, $x\approx\frac{11}{39}\approx0.28$.
## Step3: Find the approximate value and apply the negative sign
$\sqrt{372}\approx19 + 0.28 = 19.28\approx19.3$. Then $-\sqrt{372}\approx-19.3$.
# Answer:
$-19.3$

### Problem 19: $\boldsymbol{\sqrt{61}}$ (to the nearest tenth)
# Explanation:
## Step1: Find two perfect squares around 61
We know that $7^2=49$ and $8^2 = 64$. So, $7<\sqrt{61}<8$.
## Step2: Calculate the difference
Let $x=\sqrt{61}-7$. We know that $61 - 49 = 12$ and $64 - 49=15$. So, $x\approx\frac{12}{15}=0.8$.
## Step3: Find the approximate value
$\sqrt{61}\approx7 + 0.8=7.8$.
# Answer:
$7.8$

### Problem 20: Circle each value that is a perfect cube
A perfect cube is a number that can be written as $n^3$ where $n$ is an integer.
- $27 = 3^3$, so 27 is a perfect cube.
- $1000=10^3$, so 1000 is a perfect cube.
- $90$: We check if there is an integer $n$ such that $n^3=90$. $4^3 = 64$, $5^3=125$, so 90 is not a perfect cube.
- $1=1^3$, so 1 is a perfect cube.
- $300$: $6^3=216$, $7^3 = 343$, so 300 is not a perfect cube.
- $72$: $4^3=64$, $5^3 = 125$, so 72 is not a perfect cube.
- $525$: $8^3=512$, $9^3=729$, so 525 is not a perfect cube.

So we circle 27, 1000, 1.

### Problem 21: $\boldsymbol{\sqrt[3]{729}}$
# Explanation:
## Step1: Recall cube of integers
We know that $9	imes9	imes9=729$ (i.e., $9^3 = 729$).
## Step2: Find the cube root
So, $\sqrt[3]{729}=9$.
# Answer:
$9$

### Problem 22: $\boldsymbol{\sqrt[3]{125}}$
# Explanation:
## Step1: Recall cube of integers
We know that $5	imes5	imes5 = 125$ (i.e., $5^3=125$).
## Step2: Find the cube root
So, $\sqrt[3]{125}=5$.
# Answer:
$5$

### Problem 23: $\boldsymbol{\sqrt[3]{-1331}}$
# Explanation:
## Step1: Recall cube of negative integers
We know that $(-11)	imes(-11)	imes(-11)=-1331$ (i.e., $(-11)^3=-1331$).
## Step2: Find the cube root
So, $\sqrt[3]{-1331}=-11$.
# Answer:
$-11$

### Explanation for why each non - zero integer has two square roots but only one cube root:
# Brief Explanations:
For square roots: If $n^2=a$ ($a
eq0$), then $(-n)^2=a$ too. So non - zero $a$ has two square roots ($n$ and $-n$). For cube roots: If $n^3=a$, and $m^3=a$ with $m
eq n$, then $n^3 - m^3=(n - m)(n^2+nm + m^2)=0$. Since $n^2+nm + m^2
eq0$ for real $n,m$ (except $n = m = 0$), $n=m$. So only one real cube root.
# Answer:
Non - zero integers have two square roots (positive and negative) as $n^2=(-n)^2$, but only one cube root as the cube function is one - to - one (if $n^3=m^3$, then $n = m$).

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 2: $\boldsymbol{\sqrt{36}}$

Step1: Recall square of integers

Step2: Find the square root

Step1: Find the principal square root of 225

Step2: Apply the negative sign

Step1: Find the principal square root of 64

Step2: Apply the negative sign

Answer:

Problem 3: $\boldsymbol{-\sqrt{225}}$