Product Property of Square Roots: Number of terms

In order to simplify the given expression, we will use 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):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a':

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

We'll continue, since there is multiplication between three terms with identical exponents, we can use the law of exponents mentioned in b' (which also applies to multiplication of several terms in parentheses) and combine them together in multiplication under parentheses 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 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, we will use 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):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a':

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

We'll continue, since we have a multiplication of three terms with identical exponents, we can use the law of exponents mentioned in b' (which also applies to multiplying several terms in parentheses) and combine them together in a multiplication under parentheses that are 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, then we used again 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, we will use 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):

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

Let's start with converting the square roots to exponents using the law of exponents mentioned in a:

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

We'll continue, since there is a multiplication between three terms with identical exponents we can use the law of exponents mentioned in b (which also applies to multiplying several terms in parentheses) and combine them together in a multiplication under parentheses which are 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 used again 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, we will use 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):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a':

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

We'll continue, since there is a multiplication between five terms with identical exponents we can use the law of exponents mentioned in b' (which of course also applies to multiplying several terms in parentheses) and combine them together in a multiplication under parentheses which are 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 used again 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 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, we will use 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):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a:

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

We'll continue, since there is multiplication between four terms with identical exponents we can use the law of exponents mentioned in b (which also applies to multiplication of several terms in parentheses) and combine them together in multiplication under parentheses 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 used again 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 will use 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):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a:

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

We'll continue, since there is multiplication between four terms with identical exponents, we can use the law of exponents mentioned in b (which also applies to multiplication of multiple terms in parentheses) and combine them together in multiplication under 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, then we used again 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, we will use 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):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a:

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

We'll continue, since we have a multiplication of three terms with identical exponents, we can use the law of exponents mentioned in b (which also applies to multiplying several terms in parentheses) and combine them together in a multiplication under parentheses that are 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 used again 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, we will use 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):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a':

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

We'll continue, since there is a multiplication between five terms with identical exponents we can use the law of exponents mentioned in b' (which of course also applies to multiplying several terms in parentheses) and combine them together in a multiplication under parentheses which are 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 in parentheses, then we used again 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.

In order to simplify the given expression, we will use 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):

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

Let's start by converting the square roots to exponents using the law of exponents mentioned in a:

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

We'll continue, since we have a multiplication of four terms with identical exponents, we can use the law of exponents mentioned in b (which also applies to multiplying several terms in parentheses) and combine them together in a multiplication under parentheses that are 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, then we used again 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.