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## Learning objectives

• Use the product and quotient properties of radicals.
• Rationalize the denominator.
• Solve real-world problems using square root functions.

## Introduction

A radical reverses the operation of raising a number to a power. For example, to find the square of 4 we write $4^2=4 \cdot 4=16$. The reverse process is called finding the square root. The symbol for a square root is $\sqrt{}$. This symbol is also called the radical sign. When we take the square root of a number, the result is a number which when squared gives the number under the square root sign. For example,

$\sqrt{9}=3 \qquad \text{since} \qquad 3^2=3 \cdot 3=9$

Radicals often have an index in the top left corner. The index indicates which root of the number we are seeking. Square roots have an index of 2 but many times this index is not written.

$\sqrt[2]{36}=6 \qquad \text{since} \qquad 6^2=36$

The cube root of a number gives a number which when raised to the third power gives the number under the radical sign.

$\sqrt[3]{64}=4 \qquad \text{since } \qquad 4^3=4 \cdot 4 \cdot 4=64$

The fourth root of number gives a number which when raised to the power four gives the number under the radical sign.

$\sqrt[4]{81}=3 \qquad \text{since } \qquad 3^4=3 \cdot 3 \cdot 3 \cdot 3=81$

Even and odd roots

Radical expressions that have even indices are called even roots and radical expressions that have odd indices are called odd roots. There is a very important difference between even and odd roots in that they give drastically different results when the number inside the radical sign is negative.

Any real number raised to an even power results in a positive answer. Therefore, when the index of a radical is even, the number inside the radical sign must be non-negative in order to get a real answer.

On the other hand, a positive number raised to an odd power is positive and a negative number raised to an odd power is negative. Thus, a negative number inside the radical sign is not a problem. It just results in a negative answer.

Example 1

a) $\sqrt{121}$

b) $\sqrt[3]{125}$

c) $\sqrt[4]{-625}$

d) $\sqrt[5]{-32}$

Solution

a) $\sqrt{121}=11$

b) $\sqrt[3]{125}=5$

c) $\sqrt[4]{-625}$ is not a real number

d) $\sqrt[5]{-32}=-2$

## Use the Product and Quotient Properties of Radicals

Radicals can be rewritten as exponent with rational powers. The radical $y=\sqrt[m]{a^n}$ is defined as $a^{\frac{n}{m}}$.

Example 2

Write each expression as an exponent with a rational value for the exponent.

a) $\sqrt{5}$

b) $\sqrt[4]{a}$

c) $\sqrt[3]{4xy}$

d) $\sqrt[6]{x^5}$

Solution

a) $\sqrt{5}=5^{\frac{1}{2}}$

b) $\sqrt[4]{a}=a^{\frac{1}{4}}$

c) $\sqrt[3]{4xy}=(4xy)^{\frac{1}{3}}$

d) $\sqrt[6]{x^5}=x^{\frac{5}{6}}$

As a result of this property, for any non-negative number $\sqrt[n]{a^n}=a^{\frac{n}{n}}=a.$

Since roots of numbers can be treated as powers, we can use exponent rules to simplify and evaluate radical expressions. Let’s review the product and quotient rule of exponents.

$\text{Raising a product to a power}&& (x \cdot y)^n &=x^n \cdot y^n \\\text{Raising a quotient to a power }&&\left( \frac{x}{y}\right )^n &=\frac{x^n}{y^n}$

In radical notation, these properties are written as

$\text{Raising a product to a power}&&\sqrt[m]{x \cdot y}&=\sqrt[m]{x}\cdot \sqrt[m]{y}\\\text{Raising a quotient to a power }&&\sqrt[m]{\frac{x}{y}} &=\frac{\sqrt[m]{x}}{\sqrt[m]{y}}$

A very important application of these rules is reducing a radical expression to its simplest form. This means that we apply the root on all the factors of the number that are perfect roots and leave all factors that are not perfect roots inside the radical sign.

For example, in the expression $\sqrt{16}$, the number is a perfect square because $16=4^2$. This means that we can simplify.

$\sqrt{16}=\sqrt{4^2}=4$

Thus, the square root disappears completely.

On the other hand, in the expression, the number $\sqrt{32}$ is not a perfect square so we cannot remove the square root. However, we notice that $32=16 \cdot 2$, so we can write 32 as the product of a perfect square and another number.

$\sqrt{32}=\sqrt{16 \cdot 2}=\sqrt{16 \cdot 2}$

If we apply the “raising a product to a power” rule we obtain

$\sqrt{32}=\sqrt{16 \cdot 2}=\sqrt{16}\cdot \sqrt{2}$

Since $\sqrt{16}=4$, we get $\sqrt{32}=4 \cdot \sqrt{2}=4 \sqrt{2}$.

Example 3

Write the following expression in the simplest radical form.

a) $\sqrt{8}$

b) $\sqrt{50}$

c) $\sqrt{\frac{125}{72}}$

Solution

The strategy is to write the number under the square root as the product of a perfect square and another number. The goal is to find the highest perfect square possible, however, if we don’t we can repeat the procedure until we cannot simplify any longer.

a) We can write $8=4 \cdot 2$ so $\sqrt{8}=\sqrt{4 \cdot 2}$

Use the rule for raising a product to a power $\sqrt{4 \cdot 2}= \sqrt{4}\cdot \sqrt{2}$

Finally we have, $\sqrt{8}=2 \sqrt{2}$.

b) We can write $50=25 \cdot 2$ so $\sqrt{50}=\sqrt{25 \cdot 2}$

Use the rule for raising a product to a power $=\sqrt{25}\cdot \sqrt{2}=\underline{\underline{5 \sqrt{2}}}$

c) Use the rule for raising a product to a power to separate the fraction.

$\sqrt {\frac{125}{72}}=\frac{\sqrt{125}}{\sqrt{72}}$

Rewrite each radical as a product of a perfect square and another number.

$= \frac{\sqrt{25 \cdot 5}}{9 \cdot 6}=\frac{5 \sqrt{5}}{3 \sqrt{6}}$

The same method can be applied to reduce radicals of different indices to their simplest form.

Example 4

Write the following expression in the simplest radical form.

a) $\sqrt{40}$

b) $\sqrt[4]{\frac{162}{80}}$

c) $\sqrt[3]{135}$

Solution

In these cases we look for the highest possible perfect cube, fourth power, etc. as indicated by the index of the radical.

a) Here we are looking for the product of the highest perfect cube and another number. We write

$\sqrt[3]{40}=\sqrt[3]{8 \cdot 5}=2 \sqrt[3]{5}$

b) Here we are looking for the product of the highest perfect fourth power and another number.

$\text{Rewrite as the quotient of two radicals }&&\sqrt[4]{\frac{162}{80}}&=\frac{\sqrt[4]{162}}{\sqrt[4]{80}}\\\text{Simplify each radical separately }&&& =\frac{\sqrt[4]{81 \cdot 2}}{\sqrt[4]{16 \cdot 5}}=\frac{\sqrt[4]{81}\cdot \sqrt[4]{2}}{\sqrt[4]{16}\cdot \sqrt[4]{5}}=\frac{3 \sqrt[4]{2}}{2 \sqrt[4]{5}}\\\text{Recombine the fraction under one radical sign }&&&=\frac{3}{2}\sqrt[4]{\frac{2}{5}}$

c) Here we are looking for the product of the highest perfect cube root and another number.

Often it is not very easy to identify the perfect root in the expression under the radical sign.

In this case, we can factor the number under the radical sign completely by using a factor tree.

We see that $135=3 \cdot 3 \cdot 3 \cdot 5=3^3 \cdot 5$

Therefore $\sqrt[3]{135}=\sqrt[3]{3^3 \cdot 5}= \sqrt[3]{3^3}\cdot \sqrt[3]{5}=3 \sqrt[3]{5}$

Here are some examples involving variables.

Example 5

Write the following expression in the simplest radical form.

a) $\sqrt{12x^3y^5}$

b) $\sqrt[4]{\frac{1250x^7}{405y^9}}$

Solution

Treat constants and each variable separately and write each expression as the products of a perfect power as indicated by the index of the radical and another number.

a)

$\text{Rewrite as a product of radicals.}&&\sqrt{12x^3y^5}&=\sqrt{12}\cdot \sqrt{x^3}\cdot \sqrt{y^5}\\\text{Simplify each radical separately.}&& \left(\sqrt{4 \cdot 3}\right) \cdot \left(\sqrt{x^2 \cdot x}\right) \cdot \left(y^4 \cdot y\right) &= \left(2 \sqrt{3}\right) \cdot \left(x \sqrt{x}\right) \cdot \left(y^2 \sqrt{y}\right) \\\text{Combine all terms outside and inside the radical sign}&&&=2xy^2 \sqrt{3xy}$

b)

$\text{Rewrite as a quotient of radicals}&&\sqrt[4]{\frac{1250x^7}{405y^9}}&=\frac{\sqrt[4]{1250x^7}}{\sqrt[4]{405y^9}}\\\text{Simplify each radical separately}&&& =\frac{\sqrt[4]{625 \cdot 2}\cdot \sqrt[4]{x^4 \cdot x^3}}{\sqrt[4]{81 \cdot 5}\cdot \sqrt[4]{y^4 \cdot y^4 \cdot y}}=\frac{5 \sqrt[4]{2}\cdot x \cdot \sqrt[4]{x^3}}{3 \sqrt[4]{5}\cdot y \cdot \sqrt[4]{y}}=\frac{5x\sqrt[4]{2x^3}}{3y^2 \sqrt[4]{5y}}\\\text{Recombine fraction under one radical sign}&&& =\frac{5x}{3y^2}\sqrt[4]{\frac{2x^3}{5y}}$

When we add and subtract radical expressions, we can combine radical terms only when they have the same expression under the radical sign. This is a similar procedure to combining like terms in variable expressions. For example,

$4\sqrt{2}+ 5 \sqrt{2}=9 \sqrt{2}$ or $2 \sqrt{3}- \sqrt{2}+ 5 \sqrt{3}+ 10 \sqrt{2}=7 \sqrt{3}+ 9 \sqrt{2}$

It is important to simplify all radicals to their simplest form in order to make sure that we are combining all possible like terms in the expression. For example, the expression $\sqrt{8}- 2 \sqrt{50}$ looks like it cannot be simplified any more because it has no like terms. However, when we write each radical in its simplest form we have

$2\sqrt{2}- 10 \sqrt{2}$

This can be combined to obtain

$-8 \sqrt{2}$

Example 6

Simplify the following expressions as much as possible.

a) $4 \sqrt{3}+ 2\sqrt{12}$

b) $10\sqrt{24}- \sqrt{28}$

Solution

a)

$\text{Simplify}\ \sqrt{12}\ \text{to its simplest form.}&= 4\sqrt{3}+2\sqrt{4 \cdot 3}=4\sqrt{3}+ 2 \cdot 2\sqrt{3}=4 \sqrt{3}+ 4 \sqrt{3}\\\text{Combine like terms.}&= 8 \sqrt{3}$

b)

$\text{Simplify}\ \sqrt{24}\ \text{and}\ \sqrt{28}\ \text{to their simplest form.}=10 \sqrt{6 \cdot 4}- \sqrt{7 \cdot 4}= 20 \sqrt{6}- 2 \sqrt{7}$

There are no like terms.

Example 7

Simplify the following expressions as much as possible.

a) $4 \sqrt[3]{128}- 3 \sqrt[3]{250}$

b) $3\sqrt{x^3}- 4x\sqrt{9x}$

Solution

a) $\text{Rewrite radicals in simplest terms.} &=4 \sqrt[3]{2.64}- \sqrt[3]{2.125}=16 \sqrt[3]{2}- 5 \sqrt[3]{2}\\\text{Combine like terms.} &=11 \sqrt[3]{2}$

b) $\text{Rewrite radicals in simplest terms.} &=3 \sqrt{x^2 \cdot x}-4x \sqrt{9x}=3x\sqrt{x}- 12x\sqrt{x}\\\text{Combine like terms.} &=-9x \sqrt{x}$

When we multiply radical expressions, we use the “raising a product to a power” rule $\sqrt[m]{x \cdot y}= \sqrt[m]{x}\cdot \sqrt[n]{y}$.

In this case we apply this rule in reverse. For example

$\sqrt{6}\cdot \sqrt{8}=\sqrt{6 \cdot 8}=\sqrt{48}$

$\sqrt{48}=\sqrt{16 \cdot 3}=4 \sqrt{3}$

We will also make use of the fact that

$\sqrt{a}\cdot \sqrt{a}=\sqrt{a^2}=a.$

When we multiply expressions that have numbers on both the outside and inside the radical sign, we treat the numbers outside the radical sign and the numbers inside the radical sign separately.

For example

$a \sqrt{b}\cdot c \sqrt{d}=ac \sqrt{bd}.$

Example 8

Multiply the following expressions.

a) $\sqrt{2} \left(\sqrt{3}+ \sqrt{5}\right)$

b) $\sqrt{5} \left(5 \sqrt{3}+ 2 \sqrt{5}\right)$

c) $2 \sqrt{x} \left(3 \sqrt{y}+ \sqrt{x}\right)$

Solution

In each case we use distribution to eliminate the parenthesis.

a)

$\text{Distribute }\sqrt{2}\text{ inside the parenthesis.}&&\sqrt{2} \left(\sqrt{3}+ \sqrt{5}\right) &=\sqrt{2}\cdot \sqrt{3}+ \sqrt{2}\cdot \sqrt{5}\\ \text{Use the â€œraising a product to a powerâ€ rule. }&&&=\sqrt{2}\cdot \sqrt{3}+ \sqrt{2}\cdot \sqrt{5}\\ \text{Simplify. }&&&=\sqrt{6}+\sqrt{10}$

b)

$\text{Distribute }\sqrt{5}\text{ inside the parenthesis.}&&&=5\sqrt{5}\cdot \sqrt{3}- 2 \sqrt{5}\cdot \sqrt{5}\\ \text{Use the â€œraising a product to a powerâ€ rule.}&&5\sqrt{5\cdot 3}-2\sqrt{5\cdot5}&=5\sqrt{15}-2\sqrt{25}\\ \text{Simplify.}&&5\sqrt{15}-2\cdot5&=5\sqrt{15}-10$

c)

$\text{Distribute }2\sqrt{x}\text{ inside the parenthesis.} &=(2\cdot3)\left(\sqrt{x}\cdot\sqrt{y}\right)-2\cdot \left(\sqrt{x}\cdot \sqrt{x}\right)\\ \text{Multiply.} &=6\sqrt{xy}-2\sqrt{x^2}\\ \text{Simplify.} &=6\sqrt{xy}-2x$

Example 9

Multiply the following expressions.

a) $\left(2+\sqrt{5}\right)\left(2- \sqrt{6}\right)$

b) $\left(2\sqrt{x}-1\right)\left(5-\sqrt{x}\right)$

Solution

In each case we use distribution to eliminate the parenthesis.

a) $& \text{Distribute the parenthesis.}&&\left(2+\sqrt{5}\right)\left(2-\sqrt{6}\right)=(2.2) - \left(2.\sqrt{6}\right)+\left(2.\sqrt{5}\right)-\left(\sqrt{5}. \sqrt{6}-\sqrt{30}\right) \\& \text{Simplify.}&&4-2\sqrt{6}+2\sqrt{5}- 30$

b) $& \text{Distribute.}&& \left(2\sqrt{x}-1\right)\left(5-\sqrt{x}\right)=10 \sqrt{x}-2x-5+\sqrt{x}\\& \text{Simplify.}&&11\sqrt{x}-2x-5$

## Rationalize the Denominator

Often when we work with radicals, we end up with a radical expression in the denominator of a fraction. We can simplify such expressions even further by eliminating the radical expression from the denominator of the expression. This process is called rationalizing the denominator.

There are two cases we will examine.

Case 1 There is a single radical expression in the denominator $\frac{2}{\sqrt{3}}$.

In this case, we multiply the numerator and denominator by a radical expression that makes the expression inside the radical into a perfect power. In the example above, we multiply by the $\sqrt{3}$.

$\frac{2}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{2 \sqrt{3}}{3}$

Next, let’s examine $\frac{7}{\sqrt[3]{5}}$.

In this case, we need to make the number inside the cube root a perfect cube. We need to multiply the numerator and the denominator by $\sqrt[3]{5^2}$.

$\frac{7}{\sqrt[3]{5}}\cdot \frac{\sqrt[3]{5^2}}{\sqrt[3]{5^2}}=\frac{7^3 \sqrt{25}}{\sqrt[3]{5^3}}=\frac{7^3 \sqrt{25}}{5}$

Case 2 The expression in the denominator is a radical expression that contains more than one term.

Consider the expression $\frac{2}{2+\sqrt{3}}$

In order to eliminate the radical from the denominator, we multiply it by $\left(2 - \sqrt{3}\right)$. This is a good choice because the product $\left(2 + \sqrt{3}\right)\left(2 - \sqrt{3}\right)$ is a product of a sum and a difference which multiplies as follows.

$\left(2 + \sqrt{3}\right)\left(2 - \sqrt{3}\right)=2^2 - \left(\sqrt{3}\right)^2=4-3=1$

We multiply the numerator and denominator by $\left(2 - \sqrt{3}\right)$ and get

$\frac{2}{2+ \sqrt{3}}\cdot \frac{2 - \sqrt{3}}{2- \sqrt{3}}= \frac{2\left(2 - \sqrt{3}\right)}{4-3}= \frac{4 - 2 \sqrt{3}}{1}$

Now consider the expression $\frac{\sqrt{x}-1}{\sqrt{x}- 2 \sqrt{y}}$.

In order to eliminate the radical expressions in the denominator, we must multiply by $\sqrt{x}+ 2 \sqrt{y}$.

We obtain

$\frac{\sqrt{x}-1}{\sqrt{x}- 2 \sqrt{y}}\cdot \frac{\sqrt{x}+ 2 \sqrt{y}}{\sqrt{x}+ 2 \sqrt{y}}&=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+ 2 \sqrt{y}\right)}{\left(\sqrt{x}- 2 \sqrt{y}\right)\left(\sqrt{x}+ 2 \sqrt{y}\right)}\\&=\frac{x+ \sqrt{x}- 2\sqrt{xy}- 2\sqrt{y}}{x-4y}$

## Solve Real-World Problems Using Radical Expressions

Radicals often arise in problems involving areas and volumes of geometrical figures.

Example 10

A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square-feet. Find the dimensions of the pool and the area of the pool.

Solution

1. Make a sketch.
2. Let $x =$ the width of the pool.
3. Write an equation.

$\text{Area}=\text{length} \cdot \text{width}$

Combined length of pool and walkway $= 2x + 2$

Combined width of pool and walkway $= x + 2$

$\text{Area}=(2x + 2)(x + 2)$

Since the combined area of pool and walkway is $400 \ ft^2$ we can write the equation.

$(2x + 2)(x + 2)=400$

4. Solve the equation: $(2x + 2)(x + 2)=400$

Multiply in order to eliminate the parentheses.

$2x^2 + 4x +2x+ 4 =400$

Collect like terms.

$2x^2 + 6x + 4=400$

Move all terms to one side of the equation.

$2x^2 + 6x - 396 =0$

Divide all terms by 2.

$x^2 + 3x - 198 =0$

$x &=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\&=\frac{-3 \pm \sqrt{3^2 - 4(1)(-198)}}{2(1)}\\&=\frac{-3 \pm \sqrt{801}}{2}\approx \frac{-3 \pm 28.3}{2}\\x & \approx 12.65 \ \text{or}-15.65\ feet$

5. We can disregard the negative solution since it does not make sense for this context. Thus, we can check our answer of 12.65 by substituting the result in the area formula.

$\text{Area}=(2(12 \cdot 65)+2)+(12.65+2)=27.3 \cdot 14.65 \approx 400\ ft^2.$

Example 11

The volume of a soda can is $355 \ cm^3$. The height of the can is four times the radius of the base. Find the radius of the base of the cylinder.

Solution

1. Make a sketch.
2. Let $x =$ the radius of the cylinder base
3. Write an equation.

The volume of a cylinder is given by

$V=\pi r^2 \cdot h$

4. Solve the equation.

$355 &= \pi x^2 (4x) \\355 &= 4 \pi x^3 \\ x^3 &= \frac{355}{4\pi}\\ x &= \sqrt[3]{\frac{355}{4 \pi}}=3.046\ cm$

5. Check by substituting the result back into the formula.

$V=\pi r^2 \cdot h=\pi (3.046)^2 \cdot (4 \cdot 3 \cdot 046)=355\ cm^3$

So the volume is $355 \ cm^3$.

## Review Questions

1. $\sqrt{169}$
2. $\sqrt[4]{81}$
3. $\sqrt[3]{-125}$
4. $\sqrt[5]{1024}$

Write each expression as a rational exponent.

1. $\sqrt[3]{14}$
2. $\sqrt[4]{zw}$
3. $\sqrt{a}$
4. $\sqrt[9]{y^3}$

Write the following expressions in simplest radical form.

1. $\sqrt{24}$
2. $\sqrt{300}$
3. $\sqrt[5]{96}$
4. $\sqrt{\frac{240}{567}}$
5. $\sqrt[3]{500}$
6. $\sqrt[6]{64x^8}$
7. $\sqrt[3]{48a^3b^7}$
8. $\sqrt[3]{\frac{16x^5}{135y^4}}$

Simplify the following expressions as much as possible.

1. $3 \sqrt{8}- 6 \sqrt{32}$
2. $\sqrt{180}+ 6 \sqrt{405}$
3. $\sqrt{6}- \sqrt{27}+ 2 \sqrt{54}+ 3 \sqrt{48}$
4. $\sqrt{8x^3}- 4x\sqrt{98x}$
5. $\sqrt{48a}+ \sqrt{27a}$
6. $\sqrt[3]{4x^3}+x\sqrt[3]{256}$

Multiply the following expressions.

1. $\sqrt{6}\left(\sqrt{10}+\sqrt{8}\right)$
2. $\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)$
3. $\left(2 \sqrt{x}+5\right)\left(2 \sqrt{x}+5\right)$

Rationalize the denominator.

1. $\frac{7}{\sqrt 15}$
2. $\frac{9}{\sqrt 10}$
3. $\frac{2x}{\sqrt 5x}$
4. $\frac{\sqrt 5}{\sqrt 3y}$
5. $\frac{12}{2 - \sqrt 5}$
6. $\frac{6 - \sqrt 3}{4 - \sqrt 3}$
7. $\frac{x}{\sqrt 2+\sqrt x}$
8. $\frac{5y}{2\sqrt y-5}$
9. The volume of a spherical balloon is 950 cm-cubed. Find the radius of the balloon. (Volume of a sphere $=\frac{4}{3} \pi R^{3}$).
10. A rectangular picture is 9-inches wide and 12-inches long. The picture has a frame of uniform width. If the combined area of picture and frame is 180 in-squared, what is the width of the frame?

1. 13
2. not a real solution
3. -5
4. 4
5. $14^{\frac{1}{3}}$
6. $z^{\frac{1}{4}}w^{\frac{1}{4}}$
7. $a^{\frac{1}{2}}$
8. $y^{\frac{1}{3}}$
9. $2 \sqrt{6}$
10. $10 \sqrt{3}$
11. $2 \sqrt[5]{3}$
12. $\frac{4}{9}\sqrt{\frac{15}{7}}$
13. $5 \sqrt[3]{4}$
14. $2x. \sqrt[6]{x^2}$
15. $2ab^2 \sqrt[3]{\sqrt 6b}$
16. $\frac{2x}{3y}\sqrt{\frac{x^2}{5y}}$
17. $-18 \sqrt{2}$
18. $15 \sqrt{5}$
19. $7 \sqrt{6}+9 \sqrt{3}$
20. $-26x \sqrt{2x}$
21. $7 \sqrt{3a}$
22. $5x \sqrt[3]{4}$
23. $2\sqrt{15}+4\sqrt{3}$
24. $a - b$
25. $4x+20 \sqrt{x}+ 25$
26. $\frac{7 \sqrt{5}}{5}$
27. $\frac{9 \sqrt{10}}{10}$
28. $\frac{2 \sqrt{5x}}{5}$
29. $\frac{\sqrt{15y}}{3y}$
30. $-24 - 12 \sqrt{5}$
31. $\frac{27+10 \sqrt{3}}{13}$
32. $\frac{\sqrt{2x - x}}{2 - x}$
33. $\frac{10y \sqrt{y}+25y}{4y-25}$
34. $R=6.1 \ cm$
35. 1.5 inches

Feb 22, 2012

Jun 13, 2014

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