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Question

Accepted Answer

### Part a)
# Explanation:
## Step 1: Simplify $\sqrt{28}$
We know that $28 = 4\times7$, so we can use the property $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ (where $a = 4$ and $b=7$) and $\sqrt{4}=2$.
$\sqrt{28}=\sqrt{4\times7}=\sqrt{4}\times\sqrt{7}=2\sqrt{7}$
# Answer:
$2$

### Part b)
Like radicals have the same radical part. The radical part of $7\sqrt{2}$ is $\sqrt{2}$, so any expression with $\sqrt{2}$ as the radical part will be a like radical. For example, $3\sqrt{2}$ (we can choose any coefficient, here we choose $3$ for simplicity).
# Brief Explanations:
Like radicals have the same radical (the part under the square root). The given radical is $7\sqrt{2}$, so a radical with $\sqrt{2}$ as the radical part is a like radical. We choose $3\sqrt{2}$ (any coefficient with $\sqrt{2}$ works).
# Answer:
$3\sqrt{2}$ (any expression of the form $k\sqrt{2}$ where $k$ is a real number is correct)

### Part c)
First, simplify $- 4\sqrt{2m}$. Like radicals have the same radical part $\sqrt{2m}$. So any expression with $\sqrt{2m}$ as the radical part will be a like radical. For example, $5\sqrt{2m}$ (we can choose any coefficient, here we choose $5$ for simplicity).
# Brief Explanations:
Like radicals have the same radical (the part under the square root). The given radical is $-4\sqrt{2m}$, so a radical with $\sqrt{2m}$ as the radical part is a like radical. We choose $5\sqrt{2m}$ (any coefficient with $\sqrt{2m}$ works).
# Answer:
$5\sqrt{2m}$ (any expression of the form $k\sqrt{2m}$ where $k$ is a real number is correct)

### Part d)
First, simplify $\sqrt[3]{32a^{5}}$. We can factor $32a^{5}$ as $8a^{3}\times4a^{2}$, and use the property $\sqrt[3]{ab}=\sqrt[3]{a}\times\sqrt[3]{b}$ (where $a = 8a^{3}$ and $b = 4a^{2}$) and $\sqrt[3]{8a^{3}}=2a$.
$\sqrt[3]{32a^{5}}=\sqrt[3]{8a^{3}\times4a^{2}}=\sqrt[3]{8a^{3}}\times\sqrt[3]{4a^{2}} = 2a\sqrt[3]{4a^{2}}$
Like radicals have the same radical part $\sqrt[3]{4a^{2}}$. So any expression with $\sqrt[3]{4a^{2}}$ as the radical part will be a like radical. For example, $3a\sqrt[3]{4a^{2}}$ (we can choose any coefficient, here we choose $3a$ for simplicity).
# Explanation:
## Step 1: Simplify $\sqrt[3]{32a^{5}}$
Factor $32a^{5}$ as $8a^{3}\times4a^{2}$.
$\sqrt[3]{32a^{5}}=\sqrt[3]{8a^{3}\times4a^{2}}=\sqrt[3]{8a^{3}}\times\sqrt[3]{4a^{2}}$
Since $\sqrt[3]{8a^{3}} = 2a$, we have $\sqrt[3]{32a^{5}}=2a\sqrt[3]{4a^{2}}$
## Step 2: Choose a like radical
A like radical will have the same radical part $\sqrt[3]{4a^{2}}$. We can choose $3a\sqrt[3]{4a^{2}}$ (any coefficient - radical part combination with $\sqrt[3]{4a^{2}}$ works).
# Answer:
$3a\sqrt[3]{4a^{2}}$ (any expression of the form $k\sqrt[3]{4a^{2}}$ where $k$ is a real - valued expression in $a$ is correct)

### Part e)
First, simplify $-8x\sqrt{18}$. We know that $18 = 9\times2$, so $\sqrt{18}=\sqrt{9\times2}=\sqrt{9}\times\sqrt{2}=3\sqrt{2}$. Then $-8x\sqrt{18}=-8x\times3\sqrt{2}=-24x\sqrt{2}$. Like radicals have the same radical part $\sqrt{2}$. So any expression with $\sqrt{2}$ as the radical part will be a like radical. For example, $5x\sqrt{2}$ (we can choose any coefficient, here we choose $5x$ for simplicity).
# Explanation:
## Step 1: Simplify $-8x\sqrt{18}$
Factor $18$ as $9\times2$.
$\sqrt{18}=\sqrt{9\times2}=\sqrt{9}\times\sqrt{2}=3\sqrt{2}$
Then $-8x\sqrt{18}=-8x\times3\sqrt{2}=-24x\sqrt{2}$
## Step 2: Choose a like radical
A like radical will have the same radical part $\sqrt{2}$. We can choose $5x\sqrt{2}$ (any coefficient - radical part combination with $\sqrt{2}$ works).
# Answer:
$5x\sqrt{2}$ (any expression of the form $k\sqrt{2}$ where $k$ is a real - valued expression in $x$ is correct)

### Part f)
First, simplify $\sqrt{20xy^{3}}$. We know that $20 = 4\times5$ and $y^{3}=y^{2}\times y$. So $\sqrt{20xy^{3}}=\sqrt{4\times5\times x\times y^{2}\times y}=\sqrt{4}\times\sqrt{y^{2}}\times\sqrt{5xy}=2y\sqrt{5xy}$. Like radicals have the same radical part $\sqrt{5xy}$. So any expression with $\sqrt{5xy}$ as the radical part will be a like radical. For example, $3y\sqrt{5xy}$ (we can choose any coefficient, here we choose $3y$ for simplicity).
# Explanation:
## Step 1: Simplify $\sqrt{20xy^{3}}$
Factor $20$ as $4\times5$ and $y^{3}$ as $y^{2}\times y$.
$\sqrt{20xy^{3}}=\sqrt{4\times5\times x\times y^{2}\times y}$
Using the property $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$, we get $\sqrt{4}\times\sqrt{y^{2}}\times\sqrt{5xy}$
Since $\sqrt{4} = 2$ and $\sqrt{y^{2}}=y$ (assuming $y\geq0$), we have $\sqrt{20xy^{3}}=2y\sqrt{5xy}$
## Step 2: Choose a like radical
A like radical will have the same radical part $\sqrt{5xy}$. We can choose $3y\sqrt{5xy}$ (any coefficient - radical part combination with $\sqrt{5xy}$ works).
# Answer:
$3y\sqrt{5xy}$ (any expression of the form $k\sqrt{5xy}$ where $k$ is a real - valued expression in $x$ and $y$ is correct)

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QUESTION IMAGE

Question

Explanation:

Part a)

Step 1: Simplify \(\sqrt{28}\)

Answer:

Part b)