pa adv.alg/trig: notes & practice
unit 4: inverses & radical functions
name:
date:
4.4.c independent practice & hw
1. simplify.
a. $\sqrt{\frac{64}{25}}$
b. $\sqrt{\frac{16}{100}}$
c. $\sqrt{\frac{72}{121}}$
d. $\sqrt{\frac{147}{144}}$
e. $\sqrt{\frac{405}{36}}$
f. $\sqrt3{\frac{64}{216}}$
g. $\sqrt3{\frac{729}{512}}$
h. $\sqrt3{\frac{108}{8}}$
i. $\sqrt3{\frac{3000}{343}}$
2. simplify.
a. $\sqrt{\frac{25r^3}{36t^2}}$
b. $\sqrt{\frac{192j^4k}{64m^8n^{12}}}$
c. $\sqrt{\frac{180x^7y^{13}}{81w^{16}z^{14}}}$
d. $\sqrt3{\frac{729m^{12}}{64n^6}}$
e. $\sqrt3{\frac{432p^{10}r^2}{125q^3s^9}}$
f. $\sqrt3{\frac{250g^{11}h^{19}}{27e^{15}f^{24}}}$

Question

Accepted Answer

### Problem 1a: Simplify $\boldsymbol{\sqrt{\frac{64}{25}}}$
# Explanation:
## Step1: Use the property $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
$\sqrt{\frac{64}{25}}=\frac{\sqrt{64}}{\sqrt{25}}$
## Step2: Simplify square roots
$\sqrt{64} = 8$, $\sqrt{25}=5$, so $\frac{8}{5}$
# Answer:
$\frac{8}{5}$

### Problem 1b: Simplify $\boldsymbol{\sqrt{\frac{16}{100}}}$
# Explanation:
## Step1: Apply $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
$\sqrt{\frac{16}{100}}=\frac{\sqrt{16}}{\sqrt{100}}$
## Step2: Simplify square roots
$\sqrt{16} = 4$, $\sqrt{100}=10$, so $\frac{4}{10}=\frac{2}{5}$
# Answer:
$\frac{2}{5}$

### Problem 1c: Simplify $\boldsymbol{\sqrt{\frac{72}{121}}}$
# Explanation:
## Step1: Use $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
$\sqrt{\frac{72}{121}}=\frac{\sqrt{72}}{\sqrt{121}}$
## Step2: Simplify $\sqrt{72}$ and $\sqrt{121}$
$\sqrt{72}=\sqrt{36	imes2}=6\sqrt{2}$, $\sqrt{121}=11$, so $\frac{6\sqrt{2}}{11}$
# Answer:
$\frac{6\sqrt{2}}{11}$

### Problem 1d: Simplify $\boldsymbol{\sqrt{\frac{147}{144}}}$
# Explanation:
## Step1: Apply $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
$\sqrt{\frac{147}{144}}=\frac{\sqrt{147}}{\sqrt{144}}$
## Step2: Simplify square roots
$\sqrt{147}=\sqrt{49	imes3}=7\sqrt{3}$, $\sqrt{144}=12$, so $\frac{7\sqrt{3}}{12}$
# Answer:
$\frac{7\sqrt{3}}{12}$

### Problem 1e: Simplify $\boldsymbol{\sqrt{\frac{405}{36}}}$
# Explanation:
## Step1: Use $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
$\sqrt{\frac{405}{36}}=\frac{\sqrt{405}}{\sqrt{36}}$
## Step2: Simplify square roots
$\sqrt{405}=\sqrt{81	imes5}=9\sqrt{5}$, $\sqrt{36}=6$, so $\frac{9\sqrt{5}}{6}=\frac{3\sqrt{5}}{2}$
# Answer:
$\frac{3\sqrt{5}}{2}$

### Problem 1f: Simplify $\boldsymbol{\sqrt[3]{\frac{64}{216}}}$
# Explanation:
## Step1: Apply $\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$
$\sqrt[3]{\frac{64}{216}}=\frac{\sqrt[3]{64}}{\sqrt[3]{216}}$
## Step2: Simplify cube roots
$\sqrt[3]{64} = 4$, $\sqrt[3]{216}=6$, so $\frac{4}{6}=\frac{2}{3}$
# Answer:
$\frac{2}{3}$

### Problem 1g: Simplify $\boldsymbol{\sqrt[3]{\frac{729}{512}}}$
# Explanation:
## Step1: Use $\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$
$\sqrt[3]{\frac{729}{512}}=\frac{\sqrt[3]{729}}{\sqrt[3]{512}}$
## Step2: Simplify cube roots
$\sqrt[3]{729} = 9$, $\sqrt[3]{512}=8$, so $\frac{9}{8}$
# Answer:
$\frac{9}{8}$

### Problem 1h: Simplify $\boldsymbol{\sqrt[3]{\frac{108}{8}}}$
# Explanation:
## Step1: Apply $\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$
$\sqrt[3]{\frac{108}{8}}=\frac{\sqrt[3]{108}}{\sqrt[3]{8}}$
## Step2: Simplify cube roots
$\sqrt[3]{108}=\sqrt[3]{27	imes4}=3\sqrt[3]{4}$, $\sqrt[3]{8}=2$, so $\frac{3\sqrt[3]{4}}{2}$
# Answer:
$\frac{3\sqrt[3]{4}}{2}$

### Problem 1i: Simplify $\boldsymbol{\sqrt[3]{\frac{3000}{343}}}$
# Explanation:
## Step1: Use $\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$
$\sqrt[3]{\frac{3000}{343}}=\frac{\sqrt[3]{3000}}{\sqrt[3]{343}}$
## Step2: Simplify cube roots
$\sqrt[3]{3000}=\sqrt[3]{1000	imes3}=10\sqrt[3]{3}$, $\sqrt[3]{343}=7$, so $\frac{10\sqrt[3]{3}}{7}$
# Answer:
$\frac{10\sqrt[3]{3}}{7}$

### Problem 2a: Simplify $\boldsymbol{\sqrt{\frac{25r^3}{36t^2}}}$
# Explanation:
## Step1: Apply $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
$\sqrt{\frac{25r^3}{36t^2}}=\frac{\sqrt{25r^3}}{\sqrt{36t^2}}$
## Step2: Simplify square roots
$\sqrt{25r^3}=\sqrt{25r^2\cdot r}=5r\sqrt{r}$, $\sqrt{36t^2}=6t$, so $\frac{5r\sqrt{r}}{6t}$
# Answer:
$\frac{5r\sqrt{r}}{6t}$

### Problem 2b: Simplify $\boldsymbol{\sqrt{\frac{192j^4k}{64m^8n^{12}}}}$
# Explanation:
## Step1: Apply $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
$\sqrt{\frac{192j^4k}{64m^8n^{12}}}=\frac{\sqrt{192j^4k}}{\sqrt{64m^8n^{12}}}$
## Step2: Simplify square roots
$\sqrt{192j^4k}=\sqrt{64	imes3j^4k}=8j^2\sqrt{3k}$, $\sqrt{64m^8n^{12}}=8m^4n^6$, so $\frac{8j^2\sqrt{3k}}{8m^4n^6}=\frac{j^2\sqrt{3k}}{m^4n^6}$
# Answer:
$\frac{j^2\sqrt{3k}}{m^4n^6}$

### Problem 2c: Simplify $\boldsymbol{\sqrt{\frac{180x^7y^{13}}{81w^{16}z^{14}}}}$
# Explanation:
## Step1: Apply $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$
$\sqrt{\frac{180x^7y^{13}}{81w^{16}z^{14}}}=\frac{\sqrt{180x^7y^{13}}}{\sqrt{81w^{16}z^{14}}}$
## Step2: Simplify square roots
$\sqrt{180x^7y^{13}}=\sqrt{36	imes5x^6\cdot x y^{12}\cdot y}=6x^3y^6\sqrt{5xy}$, $\sqrt{81w^{16}z^{14}}=9w^8z^7$, so $\frac{6x^3y^6\sqrt{5xy}}{9w^8z^7}=\frac{2x^3y^6\sqrt{5xy}}{3w^8z^7}$
# Answer:
$\frac{2x^3y^6\sqrt{5xy}}{3w^8z^7}$

### Problem 2d: Simplify $\boldsymbol{\sqrt[3]{\frac{729m^{12}}{64n^6}}}$
# Explanation:
## Step1: Apply $\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$
$\sqrt[3]{\frac{729m^{12}}{64n^6}}=\frac{\sqrt[3]{729m^{12}}}{\sqrt[3]{64n^6}}$
## Step2: Simplify cube roots
$\sqrt[3]{729m^{12}}=\sqrt[3]{729}\cdot\sqrt[3]{m^{12}}=9m^4$, $\sqrt[3]{64n^6}=\sqrt[3]{64}\cdot\sqrt[3]{n^6}=4n^2$, so $\frac{9m^4}{4n^2}$
# Answer:
$\frac{9m^4}{4n^2}$

### Problem 2e: Simplify $\boldsymbol{\sqrt[3]{\frac{432p^{10}r^2}{125q^3s^9}}}$
# Explanation:
## Step1: Apply $\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$
$\sqrt[3]{\frac{432p^{10}r^2}{125q^3s^9}}=\frac{\sqrt[3]{432p^{10}r^2}}{\sqrt[3]{125q^3s^9}}$
## Step2: Simplify cube roots
$\sqrt[3]{432p^{10}r^2}=\sqrt[3]{216	imes2p^9\cdot p r^2}=6p^3\sqrt[3]{2p r^2}$, $\sqrt[3]{125q^3s^9}=5q s^3$, so $\frac{6p^3\sqrt[3]{2p r^2}}{5q s^3}$
# Answer:
$\frac{6p^3\sqrt[3]{2p r^2}}{5q s^3}$

### Problem 2f: Simplify $\boldsymbol{\sqrt[3]{\frac{250g^{11}h^{19}}{27e^{15}f^{24}}}}$
# Explanation:
## Step1: Apply $\sqrt[3]{\frac{a}{b}}=\frac{\sqrt[3]{a}}{\sqrt[3]{b}}$
$\sqrt[3]{\frac{250g^{11}h^{19}}{27e^{15}f^{24}}}=\frac{\sqrt[3]{250g^{11}h^{19}}}{\sqrt[3]{27e^{15}f^{24}}}$
## Step2: Simplify cube roots
$\sqrt[3]{250g^{11}h^{19}}=\sqrt[3]{125	imes2g^9\cdot g^2h^{18}\cdot h}=5g^3h^6\sqrt[3]{2g^2h}$, $\sqrt[3]{27e^{15}f^{24}}=3e^5f^8$, so $\frac{5g^3h^6\sqrt[3]{2g^2h}}{3e^5f^8}$
# Answer:
$\frac{5g^3h^6\sqrt[3]{2g^2h}}{3e^5f^8}$

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QUESTION IMAGE

Question

Explanation:

Problem 1a: Simplify $\boldsymbol{\sqrt{\frac{64}{25}}}$

Step1: Use the property $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

Step2: Simplify square roots

Step1: Apply $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

Step2: Simplify square roots

Step1: Use $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$

Step2: Simplify $\sqrt{72}$ and $\sqrt{121}$

Answer:

Problem 1b: Simplify $\boldsymbol{\sqrt{\frac{16}{100}}}$