Math / Algebra

QuestionExpressions with Irrational Numbers: Tutorial ? Question Drag each expression to the correct location. Estimate the value of each irrational expression to the nearest integer. $\begin{array}{lllll} -3 \sqrt{4} \cdot \sqrt{3}+2 \sqrt{9} \cdot \sqrt{3} & \frac{-\sqrt{20}}{-\sqrt{21}} & \frac{2 \sqrt{2}+3 \sqrt{2}}{5 \sqrt{2}} & -\frac{3 \pi}{3 \sqrt{10}} & 4 \cdot 3 \sqrt{2}-2 \sqrt{2} \cdot \sqrt{36} \end{array}$ $-\frac{3 \pi}{3 \sqrt{10}}$

Studdy Solution

STEP 1

What is this asking? We need to estimate the value of each of these square root and pi expressions, rounding them to the nearest integer! Watch out! Don't forget to simplify the expressions before estimating the irrational numbers, otherwise things could get tricky!

STEP 2

1. Simplify and Estimate
2. More Simplification and Estimation
3. Even More Simplification and Estimation
4. Yet More Simplification and Estimation
5. One Last Simplification and Estimation

STEP 3

Let's start with $-3 \sqrt{4} \cdot \sqrt{3}+2 \sqrt{9} \cdot \sqrt{3}$ .
First, we can simplify $\sqrt{4}$ to $2$ and $\sqrt{9}$ to $3$ because $2 \cdot 2 = 4$ and $3 \cdot 3 = 9$ .
So, we have $-3 \cdot 2 \cdot \sqrt{3} + 2 \cdot 3 \cdot \sqrt{3}$ , which becomes $-6\sqrt{3} + 6\sqrt{3}$ .
Adding those together gives us $0$ !
That was easy!

STEP 4

Next up is $\frac{-\sqrt{20}}{-\sqrt{21}}$ .
The negatives cancel out since a negative divided by a negative is a positive, leaving us with $\frac{\sqrt{20}}{\sqrt{21}}$ .
Since $20$ and $21$ are very close, this fraction is approximately $1$ .

STEP 5

Now for $\frac{2 \sqrt{2}+3 \sqrt{2}}{5 \sqrt{2}}$ .
We can add the terms in the numerator: $2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}$ .
So, the expression becomes $\frac{5\sqrt{2}}{5\sqrt{2}}$ .
Since the numerator and denominator are the same, the whole thing equals $1$ !

STEP 6

Let's tackle $-\frac{3 \pi}{3 \sqrt{10}}$ .
The threes cancel out, leaving us with $-\frac{\pi}{\sqrt{10}}$ .
We know that $\pi$ is approximately $3.14$ and $\sqrt{10}$ is a little bigger than $3$ , since $3 \cdot 3 = 9$ .
So, we have roughly $- \frac{3.14}{3}$ , which is approximately $-1$ .

STEP 7

Finally, we have $4 \cdot 3 \sqrt{2}-2 \sqrt{2} \cdot \sqrt{36}$ .
We can simplify $\sqrt{36}$ to $6$ .
This gives us $12\sqrt{2} - 2\sqrt{2} \cdot 6$ , which simplifies further to $12\sqrt{2} - 12\sqrt{2}$ .
This also equals $0$ !

STEP 8

$-3 \sqrt{4} \cdot \sqrt{3}+2 \sqrt{9} \cdot \sqrt{3} \approx 0$ $\frac{-\sqrt{20}}{-\sqrt{21}} \approx 1$ $\frac{2 \sqrt{2}+3 \sqrt{2}}{5 \sqrt{2}} = 1$ $-\frac{3 \pi}{3 \sqrt{10}} \approx -1$ $4 \cdot 3 \sqrt{2}-2 \sqrt{2} \cdot \sqrt{36} = 0$

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Studdy Solution

STEP 1

What is this asking? We need to estimate the value of each of these square root and pi expressions, rounding them to the nearest integer! Watch out! Don't forget to simplify the expressions before estimating the irrational numbers, otherwise things could get tricky!

STEP 2

1. Simplify and Estimate
2. More Simplification and Estimation
3. Even More Simplification and Estimation
4. Yet More Simplification and Estimation
5. One Last Simplification and Estimation

STEP 3

STEP 4

Next up is $\frac{-\sqrt{20}}{-\sqrt{21}}$ .
The negatives cancel out since a negative divided by a negative is a positive, leaving us with $\frac{\sqrt{20}}{\sqrt{21}}$ .
Since $20$ and $21$ are very close, this fraction is approximately $1$ .

STEP 5

Now for $\frac{2 \sqrt{2}+3 \sqrt{2}}{5 \sqrt{2}}$ .
We can add the terms in the numerator: $2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2}$ .
So, the expression becomes $\frac{5\sqrt{2}}{5\sqrt{2}}$ .
Since the numerator and denominator are the same, the whole thing equals $1$ !

STEP 6

Let's tackle $-\frac{3 \pi}{3 \sqrt{10}}$ .
The threes cancel out, leaving us with $-\frac{\pi}{\sqrt{10}}$ .
We know that $\pi$ is approximately $3.14$ and $\sqrt{10}$ is a little bigger than $3$ , since $3 \cdot 3 = 9$ .
So, we have roughly $- \frac{3.14}{3}$ , which is approximately $-1$ .

STEP 7

Finally, we have $4 \cdot 3 \sqrt{2}-2 \sqrt{2} \cdot \sqrt{36}$ .
We can simplify $\sqrt{36}$ to $6$ .
This gives us $12\sqrt{2} - 2\sqrt{2} \cdot 6$ , which simplifies further to $12\sqrt{2} - 12\sqrt{2}$ .
This also equals $0$ !

STEP 8

$-3 \sqrt{4} \cdot \sqrt{3}+2 \sqrt{9} \cdot \sqrt{3} \approx 0$ $\frac{-\sqrt{20}}{-\sqrt{21}} \approx 1$ $\frac{2 \sqrt{2}+3 \sqrt{2}}{5 \sqrt{2}} = 1$ $-\frac{3 \pi}{3 \sqrt{10}} \approx -1$ $4 \cdot 3 \sqrt{2}-2 \sqrt{2} \cdot \sqrt{36} = 0$

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Studdy Solution

STEP 1

What is this asking? We need to estimate the value of each of these square root and pi expressions, rounding them to the nearest integer! Watch out! Don't forget to simplify the expressions *before* estimating the irrational numbers, otherwise things could get tricky!

STEP 2

1. Simplify and Estimate2. More Simplification and Estimation3. Even More Simplification and Estimation4. Yet More Simplification and Estimation5. One Last Simplification and Estimation

STEP 3

STEP 4

Next up is −20−21\frac{-\sqrt{20}}{-\sqrt{21}}−21​−20​​.The negatives **cancel out** since a negative divided by a negative is a positive, leaving us with 2021\frac{\sqrt{20}}{\sqrt{21}}21​20​​.Since 202020 and 212121 are very close, this fraction is approximately 111.

STEP 5

STEP 6

STEP 7

STEP 8

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

What is this asking? We need to estimate the value of each of these square root and pi expressions, rounding them to the nearest integer! Watch out! Don't forget to simplify the expressions before estimating the irrational numbers, otherwise things could get tricky!

1. Simplify and Estimate
2. More Simplification and Estimation
3. Even More Simplification and Estimation
4. Yet More Simplification and Estimation
5. One Last Simplification and Estimation

Next up is $\frac{-\sqrt{20}}{-\sqrt{21}}$ .
The negatives cancel out since a negative divided by a negative is a positive, leaving us with $\frac{\sqrt{20}}{\sqrt{21}}$ .
Since $20$ and $21$ are very close, this fraction is approximately $1$ .