find each value.
1. $\sqrt{81}$
2. $\sqrt3{125}$

determine the two positive integers that each square root falls between.
3. $\sqrt{11}$
4. $\sqrt{34}$

5. estimate $\sqrt{65}$ to the nearest tenth.

Question

find each value.
1. $\sqrt{81}$
2. $\sqrt3{125}$

determine the two positive integers that each square root falls between.
3. $\sqrt{11}$
4. $\sqrt{34}$

5. estimate $\sqrt{65}$ to the nearest tenth.

Accepted Answer

### Problem 1: $\boldsymbol{\sqrt{81}}$
# Explanation:
## Step1: Recall square of integers
We know that $9	imes9 = 81$, so by the definition of square root, $\sqrt{81}$ is the number whose square is 81.
$\sqrt{81}=9$
# Answer:
9

### Problem 2: $\boldsymbol{\sqrt[3]{125}}$
# Explanation:
## Step1: Recall cube of integers
We know that $5	imes5	imes5=125$, so by the definition of cube root, $\sqrt[3]{125}$ is the number whose cube is 125.
$\sqrt[3]{125} = 5$
# Answer:
5

### Problem 3: Determine two positive integers between which $\boldsymbol{\sqrt{11}}$ lies
# Explanation:
## Step1: Find perfect squares near 11
We know that $3^2=9$ and $4^2 = 16$.
## Step2: Compare with $\sqrt{11}$
Since $9<11<16$, taking square roots (since square root is an increasing function for non - negative numbers), we get $\sqrt{9}<\sqrt{11}<\sqrt{16}$, which simplifies to $3 < \sqrt{11}<4$.
# Answer:
3 and 4

### Problem 4: Determine two positive integers between which $\boldsymbol{\sqrt{34}}$ lies
# Explanation:
## Step1: Find perfect squares near 34
We know that $5^2 = 25$ and $6^2=36$.
## Step2: Compare with $\sqrt{34}$
Since $25 < 34<36$, taking square roots (since square root is an increasing function for non - negative numbers), we get $\sqrt{25}<\sqrt{34}<\sqrt{36}$, which simplifies to $5<\sqrt{34}<6$.
# Answer:
5 and 6

### Problem 5: Estimate $\boldsymbol{\sqrt{65}}$ to the nearest tenth
# Explanation:
## Step1: Find perfect squares near 65
We know that $8^2=64$ and $9^2 = 81$. So $\sqrt{64}=8$ and $\sqrt{81}=9$, and $64<65<81$, so $8<\sqrt{65}<9$.
## Step2: Calculate the decimal approximation
Let's calculate $\sqrt{65}$. We know that $8.0^2=64$, $8.1^2=8.1	imes8.1 = 65.61$. Since $64<65<65.61$, let's find the difference between 65 and 64: $65 - 64=1$. The difference between $8.1^2$ and $8.0^2$ is $65.61 - 64 = 1.61$.
We can use linear approximation. Let $f(x)=\sqrt{x}$, $f^\prime(x)=\frac{1}{2\sqrt{x}}$. At $x = 64$, $f(64)=8$, $f^\prime(64)=\frac{1}{16}$. We want to find $\Delta x$ such that $f(64+\Delta x)\approx f(64)+f^\prime(64)\Delta x$. We know that $f(64+\Delta x)=\sqrt{65}$, and $64+\Delta x = 65$, so $\Delta x = 1$. Then $f(65)\approx8+\frac{1}{16}	imes1=8+\ 0.0625 = 8.0625$.
Since $8.0625$ is closer to $8.1$? Wait, no. Wait, $8.0^2 = 64$, $8.1^2=65.61$. The value of $\sqrt{65}\approx8.06$, and to the nearest tenth, we look at the hundredth place. The digit in the hundredth place is 6, which is greater than 5. So we round up the tenth place. So $\sqrt{65}\approx8.1$.
# Answer:
8.1

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QUESTION IMAGE

Question

Explanation:

Problem 1: $\boldsymbol{\sqrt{81}}$

Step1: Recall square of integers

Step1: Recall cube of integers

Step1: Find perfect squares near 11

Step2: Compare with $\sqrt{11}$

Answer:

Problem 2: $\boldsymbol{\sqrt[3]{125}}$