Product Property of Square Roots: Number of terms

In order to simplify the given expression, apply two laws of exponents:

a. The definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for exponents applied to terms in parentheses (in reverse order):

$x^n\cdot y^n =(x\cdot y)^n$

Begin by converting the square roots to exponents using the law of exponents mentioned in a':

$\sqrt{1}\cdot\sqrt{2}\cdot\sqrt{3} \\ \downarrow\\ 1^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}=$

Due to the fact that there is a multiplication operation between three terms with identical exponents, we are able to apply the law of exponents mentioned in b' (which also applies to multiplication of several terms in parentheses) Combine them together in a multiplication operation within parentheses that are also raised to the same exponent:

$1^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}= \\ (1\cdot2\cdot3)^{\frac{1}{2}}=\\ 6^{\frac{1}{2}}=\\ \boxed{\sqrt{6}}$

In the final steps, we performed the multiplication within the parentheses and once again used the definition of root as an exponent mentioned in a' (in reverse order) to return to root notation.

Therefore, the correct answer is answer d.

Notice that in the given problem, a multiplication is performed between three terms, one of which is:

$\sqrt{0}$and let's remember that the root (of any order) of the number 0 is 0, meaning that:

$\sqrt{0}=0$and since multiplying any number by 0 will always yield the result 0,

therefore the result of the multiplication in the problem is 0, meaning:

$\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{0}=\\ \sqrt{2}\cdot\sqrt{2}\cdot0=\\ \boxed{0}$and thus the correct answer is answer C.

In order to simplify the given expression, apply two laws of exponents:

a. The definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for an exponent applied to a product in parentheses (in reverse direction):

$x^n\cdot y^n =(x\cdot y)^n$

Begin by converting the square roots to exponents using the law of exponents mentioned in a':

$\sqrt{4}\cdot\sqrt{2}\cdot\sqrt{2} \\ \downarrow\\ 4^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}=$

Due to the fact that we have a multiplication of three terms with identical exponents, we are able to apply the law of exponents mentioned in b' (which also applies to multiplying several terms in parentheses) Combine them together in a multiplication operation within parentheses that are also raised to the same exponent:

$4^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= \\ (4\cdot2\cdot2)^{\frac{1}{2}}=\\ 16^{\frac{1}{2}}=\\ \sqrt{16}=\\ \boxed{4}$

In the final steps, we first performed the multiplication within the parentheses, we then once again used the definition of root as an exponent mentioned in a' (in reverse direction) to return to root notation, and in the final stage, we calculated the known square root of 16.

Therefore, we can identify that the correct answer is answer c.

In order to simplify the given expression, apply two laws of exponents:

a. The definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for an exponent applied to a product in parentheses (in reverse direction):

$x^n\cdot y^n =(x\cdot y)^n$

Begin by converting the square roots to exponents using the law of exponents mentioned in a:

$\sqrt{10}\cdot\sqrt{2}\cdot\sqrt{5} \\ \downarrow\\ 10^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}=$

Due to the fact that there is a multiplication operation between three terms with identical exponents we are able to apply the law of exponents mentioned in b (which also applies to multiplying several terms in parentheses) Combine them together in a multiplication operation within parentheses, which are also raised to the same exponent:

$10^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}= \\ (10\cdot2\cdot5)^{\frac{1}{2}}=\\ 100^{\frac{1}{2}}=\\ \sqrt{100}=\\ \boxed{10}$

In the final steps, we first performed the multiplication within the parentheses, we then once again used the definition of root as an exponent mentioned in a (in reverse direction) to return to root notation, and in the final stage we calculated the known square root of 100.

Therefore, we can identify that the correct answer (most appropriate) is answer d.

In order to simplify the given expression, apply two laws of exponents:

a. Root definition as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for exponents applied to multiplication of terms in parentheses (in reverse order):

$x^n\cdot y^n =(x\cdot y)^n$

Begin by converting the square roots to exponents using the law of exponents mentioned in a:

$\sqrt{5}\cdot\sqrt{10}\cdot\sqrt{2}\cdot\sqrt{4}= \\ \downarrow\\ 5^{\frac{1}{2}}\cdot10^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot4^{\frac{1}{2}}=$

Due to the fact that there is a multiplication operation between four terms with identical exponents we are able to apply the law of exponents mentioned in b (which also applies to multiplication of several terms in parentheses) Combine them together in a multiplication operation inside of parentheses that are also raised to the same exponent:

$5^{\frac{1}{2}}\cdot10^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot4^{\frac{1}{2}}=\\ (5\cdot10\cdot2\cdot4)^{\frac{1}{2}}=\\ 400^{\frac{1}{2}}=\\ \sqrt{400}=\\ \boxed{20}$

In the final stages, we first performed the multiplication within the parentheses, then we once again used the root definition as an exponent mentioned earlier in a (in reverse order) to return to root notation, and in the final stage we calculated the known square root of 400.

Therefore, we can identify that the correct answer is answer c.

In order to simplify the given expression we use two laws of exponents:

A. Defining the root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$B. The law of exponents for a product of numbers with the same base (in the opposite direction):

$x^n\cdot y^n =(x\cdot y)^n$

Let's start by definging the roots as exponents using the law of exponents shown in A:

$\sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}\cdot\sqrt{2}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}=$Since we are multiplying between four numbers with the same exponents we can use the law of exponents shown in B (which also applies to a product of numbers with the same base) and combine them together in a product wit the same base which is raised to the same exponent:

$2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= \\ (2\cdot5\cdot2\cdot2)^{\frac{1}{2}}=\\ 40^{\frac{1}{2}}=\\ \boxed{\sqrt{40}}$In the last step we performed the product which is in the base, then we used again the definition of the root as an exponent shown earlier in A (in the opposite direction) to return to writing the root.

Therefore, note that the correct answer is answer C.

In order to simplify the given expression, apply two laws of exponents:

a. The definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for an exponent applied to a product in parentheses (in reverse direction):

$x^n\cdot y^n =(x\cdot y)^n$

Begin by converting the square roots to exponents using the law of exponents mentioned in a':

$\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{2}\cdot\sqrt{1}\cdot\sqrt{1}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot1^{\frac{1}{2}}\cdot1^{\frac{1}{2}}=$

Due to the fact that there is a multiplication between five terms with identical exponents we can apply the law of exponents mentioned in b' (which of course also applies to multiplying several terms in parentheses) Proceed to combine them together in a multiplication operation inside of parentheses ,which are also raised to the same exponent:

$2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot1^{\frac{1}{2}}\cdot1^{\frac{1}{2}}= \\ (2\cdot2\cdot2\cdot1\cdot1)^{\frac{1}{2}}=\\ 8^{\frac{1}{2}}=\\ \boxed{\sqrt{8}}$

In the final steps, we first performed the multiplication within the parentheses, then we once again used the definition of root as an exponent mentioned earlier in a' (in reverse direction) to return to root notation.

Therefore, we can identify that the correct answer is answer a'.

In order to simplify the given expression, apply two laws of exponents:

a. Root definition as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for exponents applied to multiplication of terms in parentheses (in reverse direction):

$x^n\cdot y^n =(x\cdot y)^n$

Begin by converting the square roots to exponents using the law of exponents mentioned in a:

$\sqrt{5}\cdot\sqrt{2}\cdot\sqrt{5}\cdot\sqrt{2}= \\ \downarrow\\ 5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}=$

Due to the fact that there is a multiplication operation between four terms with identical exponents, we are able to apply the law of exponents mentioned in b (which also applies to multiplication of multiple terms in parentheses) Combine them together in a multiplication operation within parentheses that are raised to the same exponent:

$5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= \\ (5\cdot2\cdot5\cdot2)^{\frac{1}{2}}=\\ 100^{\frac{1}{2}}=\\ \sqrt{100}=\\ \boxed{10}$

In the final steps, we first performed the multiplication within the parentheses, we then once again used the root definition as an exponent mentioned in a (in reverse direction) to return to root notation, and in the final stage, we calculated the known square root of 100.

Therefore, we can identify that the correct answer is answer d.

In order to simplify the given expression, apply two laws of exponents:

a. The definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for an exponent applied to a product in parentheses (in reverse direction):

$x^n\cdot y^n =(x\cdot y)^n$

Begin by converting the square roots to exponents using the law of exponents mentioned in a:

$\sqrt{6}\cdot\sqrt{2}\cdot\sqrt{3}\cdot\sqrt{1}= \\ \downarrow\\ 6^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}\cdot1^{\frac{1}{2}}=$

Due to the fact that we have a multiplication operation of four terms with identical exponents, we are able to apply the law of exponents mentioned in b (which also applies to multiplying several terms in parentheses) Combine them together in a multiplication operation within parentheses that are also raised to the same exponent:

$6^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}\cdot1^{\frac{1}{2}}= \\ (6\cdot2\cdot3\cdot1)^{\frac{1}{2}}=\\ 36^{\frac{1}{2}}=\\ \sqrt{36}=\\ \boxed{6}$

In the final steps, we first performed the multiplication within the parentheses, we then once again used the definition of root as an exponent mentioned in a (in reverse direction) to return to root notation, and in the final stage, we calculated the known square root of 36.

Therefore, we can identify that the correct answer is answer d.

In order to simplify the given expression, apply two laws of exponents:

a. The definition of root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for an exponent applied to a product in parentheses (in reverse direction):

$x^n\cdot y^n =(x\cdot y)^n$

Begin by converting the square roots to exponents using the law of exponents mentioned in a:

$\sqrt{10}\cdot\sqrt{2}\cdot\sqrt{5}= \\ \downarrow\\ 10^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}=$

Due to the fact that we have a multiplication of three terms with identical exponents, we are able to apply the law of exponents mentioned in b (which also applies to multiplying several terms in parentheses) Combine them together in a multiplication operation inside of parentheses that are also raised to the same exponent:

$10^{\frac{1}{2}}\cdot2^{\frac{1}{2}}\cdot5^{\frac{1}{2}}= \\ (10\cdot2\cdot5)^{\frac{1}{2}}=\\ 100^{\frac{1}{2}}=\\ \sqrt{100}=\\ \boxed{10}$

In the final steps, we first performed the multiplication within the parentheses, then we once again used the definition of root as an exponent mentioned in a (in reverse direction) to return to root notation, and in the final stage, we calculated the known square root of the number 100.

Therefore, we can identify that the correct answer is answer a.

In order to simplify the given expression, apply two laws of exponents:

a. Root definition as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

b. The law of exponents for an exponent applied to a product in parentheses (in reverse direction):

$x^n\cdot y^n =(x\cdot y)^n$

Begin by converting the square roots to exponents using the law of exponents mentioned in a':

$\sqrt{2}\cdot\sqrt{3}\cdot\sqrt{1}\cdot\sqrt{4}\cdot\sqrt{2}= \\ \downarrow\\ 2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}\cdot1^{\frac{1}{2}}\cdot4^{\frac{1}{2}}\cdot2^{\frac{1}{2}}=$

Due to the fact that there is a multiplication operation between five terms with identical exponents we are able to apply the law of exponents mentioned in b' (which of course also applies to multiplying several terms in parentheses) Combine them together in a multiplication operation inside of parentheses which are also raised to the same exponent:

$2^{\frac{1}{2}}\cdot3^{\frac{1}{2}}\cdot1^{\frac{1}{2}}\cdot4^{\frac{1}{2}}\cdot2^{\frac{1}{2}}= \\ (2\cdot3\cdot1\cdot4\cdot2)^{\frac{1}{2}}=\\ 48^{\frac{1}{2}}=\\ \boxed{\sqrt{48}}$

In the final steps, we first performed the multiplication inside of the parentheses, then we once again used the root definition as an exponent mentioned earlier in a' (in reverse direction) to return to root notation.

Therefore, we can identify that the correct answer is answer b.

To solve this problem, we'll follow these steps:

• Step 1: Combine all the square root terms into a single square root using the product property of square roots.
• Step 2: Simplify the product inside the square root.
• Step 3: Simplify the square root expression obtained after combining.

Now, let's work through each step:
Step 1: We start by combining all the terms under one square root using the identity $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$. Thus:

$\sqrt{2} \cdot \sqrt{3} \cdot \sqrt{1} \cdot \sqrt{4} \cdot \sqrt{5} \cdot \sqrt{6} = \sqrt{2 \cdot 3 \cdot 1 \cdot 4 \cdot 5 \cdot 6}$

Step 2: Calculate the product within the square root:
$2 \cdot 3 \cdot 1 \cdot 4 \cdot 5 \cdot 6 = 720$

Step 3: Now, simplify $\sqrt{720}$.
First, we find the prime factorization of 720: $720 = 2^4 \cdot 3^2 \cdot 5^1$.
Using the property that $\sqrt{a^2} = a$, we can write:

$\sqrt{720} = \sqrt{(2^2 \cdot 3)^2 \cdot 2 \cdot 5} = 2^2 \cdot 3 \cdot \sqrt{2 \cdot 5} = 4 \cdot 3 \cdot \sqrt{10}$

After simplification, the final answer is:

$4\sqrt{3}$.