cw 8.5 mixed practice continued
16) $ \sqrt{6} \cdot \sqrt{18} $
17) $ -7\sqrt{3p^5} \cdot 2\sqrt{33p} $
18) $ 2\sqrt{6y^3} \cdot 3\sqrt{30} $
19) $ \frac{\sqrt{300}}{\sqrt{6}} $
20) $ \frac{\sqrt{45n^5}}{\sqrt{3n}} $
21) $ \frac{10}{\sqrt{2}} $
22) $ \frac{10\sqrt{21}}{25\sqrt{7}} $
23) $ \frac{8\sqrt{5}}{\sqrt{2}} $
24) $ \frac{\sqrt{2n^2}}{\sqrt{11}} $
25) $ \frac{2\sqrt{a^4}}{\sqrt{6}} $
26) $ \frac{\sqrt{25x^3}}{\sqrt{3x}} $

Question

Accepted Answer

Let's solve these radical problems one by one. We'll use properties of square roots like $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$, $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$, and simplifying radicals by factoring out perfect squares.

### Problem 16: $\boldsymbol{\sqrt{6}\cdot\sqrt{18}}$
## Step 1: Use the property $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$
$\sqrt{6}\cdot\sqrt{18}=\sqrt{6\times18}$
## Step 2: Simplify the product inside the square root
$6\times18 = 108$, so we have $\sqrt{108}$
## Step 3: Factor 108 into perfect square and another factor
$108 = 36\times3$, so $\sqrt{108}=\sqrt{36\times3}$
## Step 4: Use the property $\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}$ (for $a$ a perfect square)
$\sqrt{36\times3}=\sqrt{36}\cdot\sqrt{3}=6\sqrt{3}$
# Answer: $6\sqrt{3}$

### Problem 17: $\boldsymbol{-7\sqrt{3p^{5}}\cdot2\sqrt{33p}}$
## Step 1: Multiply the coefficients and the radicals separately
$-7\times2=-14$ and $\sqrt{3p^{5}}\cdot\sqrt{33p}=\sqrt{3p^{5}\times33p}$
## Step 2: Simplify the product inside the radical
$3\times33 = 99$ and $p^{5}\times p=p^{6}$, so $\sqrt{99p^{6}}$
## Step 3: Factor 99 into perfect square and another factor
$99 = 9\times11$, so $\sqrt{99p^{6}}=\sqrt{9\times11\times p^{6}}$
## Step 4: Simplify the perfect square parts
$\sqrt{9}=3$ and $\sqrt{p^{6}}=p^{3}$, so $\sqrt{9\times11\times p^{6}} = 3p^{3}\sqrt{11}$
## Step 5: Multiply with the coefficient from Step 1
$-14\times3p^{3}\sqrt{11}=-42p^{3}\sqrt{11}$
# Answer: $-42p^{3}\sqrt{11}$

### Problem 18: $\boldsymbol{2\sqrt{6y^{3}}\cdot3\sqrt{30}}$
## Step 1: Multiply the coefficients and the radicals separately
$2\times3 = 6$ and $\sqrt{6y^{3}}\cdot\sqrt{30}=\sqrt{6y^{3}\times30}$
## Step 2: Simplify the product inside the radical
$6\times30 = 180$ and $y^{3}$ remains, so $\sqrt{180y^{3}}$
## Step 3: Factor 180 into perfect square and another factor
$180 = 36\times5$, so $\sqrt{180y^{3}}=\sqrt{36\times5\times y^{3}}$
## Step 4: Simplify the perfect square and $y^{3}$ ( $y^{3}=y^{2}\times y$)
$\sqrt{36}=6$, $\sqrt{y^{2}} = y$, so $\sqrt{36\times5\times y^{3}}=6y\sqrt{5y}$
## Step 5: Multiply with the coefficient from Step 1
$6\times6y\sqrt{5y}=36y\sqrt{5y}$
# Answer: $36y\sqrt{5y}$

### Problem 19: $\boldsymbol{\frac{\sqrt{300}}{\sqrt{6}}}$
## Step 1: Use the property $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$
$\frac{\sqrt{300}}{\sqrt{6}}=\sqrt{\frac{300}{6}}$
## Step 2: Simplify the fraction inside the square root
$\frac{300}{6}=50$, so $\sqrt{50}$
## Step 3: Factor 50 into perfect square and another factor
$50 = 25\times2$, so $\sqrt{50}=\sqrt{25\times2}$
## Step 4: Simplify the perfect square part
$\sqrt{25\times2}=\sqrt{25}\cdot\sqrt{2}=5\sqrt{2}$
# Answer: $5\sqrt{2}$

### Problem 20: $\boldsymbol{\frac{\sqrt{45n^{5}}}{\sqrt{3n}}}$
## Step 1: Use the property $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$
$\frac{\sqrt{45n^{5}}}{\sqrt{3n}}=\sqrt{\frac{45n^{5}}{3n}}$
## Step 2: Simplify the fraction inside the square root
$\frac{45}{3}=15$ and $\frac{n^{5}}{n}=n^{4}$, so $\sqrt{15n^{4}}$
## Step 3: Simplify the perfect square part ($n^{4}=(n^{2})^{2}$)
$\sqrt{15n^{4}}=\sqrt{15}\cdot\sqrt{n^{4}}=n^{2}\sqrt{15}$
# Answer: $n^{2}\sqrt{15}$

### Problem 21: $\boldsymbol{\frac{10}{\sqrt{2}}}$
## Step 1: Rationalize the denominator by multiplying numerator and denominator by $\sqrt{2}$
$\frac{10\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\frac{10\sqrt{2}}{2}$
## Step 2: Simplify the fraction
$\frac{10\sqrt{2}}{2}=5\sqrt{2}$
# Answer: $5\sqrt{2}$

### Problem 22: $\boldsymbol{\frac{10\sqrt{21}}{25\sqrt{7}}}$
## Step 1: Simplify the coefficients and use the property $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$
$\frac{10}{25}=\frac{2}{5}$ and $\frac{\sqrt{21}}{\sqrt{7}}=\sqrt{\frac{21}{7}}$
## Step 2: Simplify the fraction inside the square root
$\frac{21}{7}=3$, so $\sqrt{3}$
## Step 3: Multiply the simplified coefficient and radical
$\frac{2}{5}\times\sqrt{3}=\frac{2\sqrt{3}}{5}$
# Answer: $\frac{2\sqrt{3}}{5}$

### Problem 23: $\boldsymbol{\frac{8\sqrt{5}}{\sqrt{2}}}$
## Step 1: Rationalize the denominator by multiplying numerator and denominator by $\sqrt{2}$
$\frac{8\sqrt{5}\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}=\frac{8\sqrt{10}}{2}$
## Step 2: Simplify the fraction
$\frac{8\sqrt{10}}{2}=4\sqrt{10}$
# Answer: $4\sqrt{10}$

### Problem 24: $\boldsymbol{\frac{\sqrt{2n^{2}}}{\sqrt{11}}}$
## Step 1: Use the property $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$
$\frac{\sqrt{2n^{2}}}{\sqrt{11}}=\sqrt{\frac{2n^{2}}{11}}$
## Step 2: Simplify $\sqrt{n^{2}}$ (assuming $n\geq0$)
$\sqrt{n^{2}} = n$, so $\sqrt{\frac{2n^{2}}{11}}=n\sqrt{\frac{2}{11}}$
## Step 3: Rationalize the denominator (optional, but common)
$n\sqrt{\frac{2}{11}}=n\frac{\sqrt{22}}{11}=\frac{n\sqrt{22}}{11}$
# Answer: $\frac{n\sqrt{22}}{11}$ (or $n\sqrt{\frac{2}{11}}$)

### Problem 25: $\boldsymbol{\frac{2\sqrt{a^{4}}}{\sqrt{6}}}$
## Step 1: Simplify $\sqrt{a^{4}}$
$\sqrt{a^{4}}=a^{2}$ (since $(a^{2})^{2}=a^{4}$)
## Step 2: Rationalize the denominator by multiplying numerator and denominator by $\sqrt{6}$
$\frac{2a^{2}\times\sqrt{6}}{\sqrt{6}\times\sqrt{6}}=\frac{2a^{2}\sqrt{6}}{6}$
## Step 3: Simplify the fraction
$\frac{2a^{2}\sqrt{6}}{6}=\frac{a^{2}\sqrt{6}}{3}$
# Answer: $\frac{a^{2}\sqrt{6}}{3}$

### Problem 26: $\boldsymbol{\frac{\sqrt{25x^{3}}}{\sqrt{3x}}}$
## Step 1: Use the property $\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$
$\frac{\sqrt{25x^{3}}}{\sqrt{3x}}=\sqrt{\frac{25x^{3}}{3x}}$
## Step 2: Simplify the fraction inside the square root
$\frac{25}{3}$ and $\frac{x^{3}}{x}=x^{2}$, so $\sqrt{\frac{25x^{2}}{3}}$
## Step 3: Simplify the perfect square part ($x^{2}$)
$\sqrt{\frac{25x^{2}}{3}}=\frac{5x}{\sqrt{3}}$
## Step 4: Rationalize the denominator
$\frac{5x}{\sqrt{3}}=\frac{5x\sqrt{3}}{3}$
# Answer: $\frac{5x\sqrt{3}}{3}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 16: \(\boldsymbol{\sqrt{6}\cdot\sqrt{18}}\)

Step 1: Use the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\)

Step 2: Simplify the product inside the square root

Step 3: Factor 108 into perfect square and another factor

Step 4: Use the property \(\sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\) (for \(a\) a perfect square)

Answer:

Problem 17: \(\boldsymbol{-7\sqrt{3p^{5}}\cdot2\sqrt{33p}}\)

Step 1: Multiply the coefficients and the radicals separately

Step 2: Simplify the product inside the radical

Step 3: Factor 99 into perfect square and another factor

Step 4: Simplify the perfect square parts

Step 5: Multiply with the coefficient from Step 1