rewrite the irrational square root √48 as a product of an integer and another irrational square root
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The irrational square root √48 can be rewritten as √(16 * 3). We can simplify this by taking the square root of 16, which is 4, and leaving the square root of 3 intact. Therefore, √48 can be expressed as 4√3.
what is the irrational square root √28 simplified
To simplify the irrational square root √28, we need to find the largest perfect square that divides 28. In this case, it is 4 since 4 * 7 = 28.
Now we can rewrite the square root of 28 as follows:
√28 = √(4 * 7)
Next, we can break down the square root of 4 as follows:
√(4 * 7) = √4 * √7
The square root of 4 is 2, so we have:
√4 * √7 = 2√7
Therefore, the irrational square root √28 can be simplified as 2√7.
Now we can rewrite the square root of 28 as follows:
√28 = √(4 * 7)
Next, we can break down the square root of 4 as follows:
√(4 * 7) = √4 * √7
The square root of 4 is 2, so we have:
√4 * √7 = 2√7
Therefore, the irrational square root √28 can be simplified as 2√7.
rewrite the irrational cube root ^3√48 as a product of an integer and another irrational cube root
To rewrite the irrational cube root ^3√48 as a product of an integer and another irrational cube root, we need to find a perfect cube that can be factored out of 48. This will allow us to express ^3√48 as the product of the cube root of that perfect cube and another irrational cube root.
First, let's factor 48 into its prime factorization:
48 = 2^4 * 3^1
Now, let's identify the perfect cube factors. In this case, the perfect cube is 8 since 8 = 2^3.
We can rewrite 48 as:
48 = 8 * 6
Taking the cube root of both sides, we get:
^3√48 = ^3√(8 * 6)
Using the property of cube roots, the cube root of a product is the product of the cube roots of the factors:
^3√48 = ^3√8 * ^3√6
Simplifying, we have:
^3√48 = 2√6
Therefore, the irrational cube root ^3√48 can be expressed as the product of an integer (2) and another irrational cube root √6.
First, let's factor 48 into its prime factorization:
48 = 2^4 * 3^1
Now, let's identify the perfect cube factors. In this case, the perfect cube is 8 since 8 = 2^3.
We can rewrite 48 as:
48 = 8 * 6
Taking the cube root of both sides, we get:
^3√48 = ^3√(8 * 6)
Using the property of cube roots, the cube root of a product is the product of the cube roots of the factors:
^3√48 = ^3√8 * ^3√6
Simplifying, we have:
^3√48 = 2√6
Therefore, the irrational cube root ^3√48 can be expressed as the product of an integer (2) and another irrational cube root √6.
rewrite the irrational cube root ^3√48 as a product of an integer and another irrational cube root
A. 4^3√3
B.2^3√6
C.-2^3√6
D.4^3√12
A. 4^3√3
B.2^3√6
C.-2^3√6
D.4^3√12
To rewrite the irrational cube root ^3√48 as a product of an integer and another irrational cube root, we start by factoring 48.
48 = 2^4 * 3^1
Now, let's find the largest perfect cube that can be factored out of 48. The largest perfect cube that divides 48 is 8 since 8 = 2^3.
We can rewrite 48 as:
48 = 8 * 6
Taking the cube root of both sides, we get:
^3√48 = ^3√(8 * 6)
Using the property of cube roots, the cube root of a product is the product of the cube roots of the factors:
^3√48 = ^3√8 * ^3√6
Simplifying, we have:
^3√48 = 2√6
Therefore, the irrational cube root ^3√48 can be expressed as the product of an integer (2) and another irrational cube root √6. Therefore, the correct answer is option B: 2^3√6.
48 = 2^4 * 3^1
Now, let's find the largest perfect cube that can be factored out of 48. The largest perfect cube that divides 48 is 8 since 8 = 2^3.
We can rewrite 48 as:
48 = 8 * 6
Taking the cube root of both sides, we get:
^3√48 = ^3√(8 * 6)
Using the property of cube roots, the cube root of a product is the product of the cube roots of the factors:
^3√48 = ^3√8 * ^3√6
Simplifying, we have:
^3√48 = 2√6
Therefore, the irrational cube root ^3√48 can be expressed as the product of an integer (2) and another irrational cube root √6. Therefore, the correct answer is option B: 2^3√6.
rewrite the irrational cube root ^√-24 rewritten as a product of an integer and another irrational cube root
The expression ^√-24 is not well-defined as it involves taking the cube root of a negative number. In the real number system, cube roots of negative numbers are not real-valued. Therefore, the irrational cube root ^√-24 cannot be rewritten as a product of an integer and another irrational cube root.
sorry, i misstype it
rewrite the irrational cube root ^3√-24 rewritten as a product of an integer and another irrational cube root
rewrite the irrational cube root ^3√-24 rewritten as a product of an integer and another irrational cube root
To rewrite the irrational cube root ^3√-24 as a product of an integer and another irrational cube root, we start by factoring -24.
-24 can be factored as -1 * 2^3 * 3^1.
Now, let's rewrite ^3√-24 using this factorization:
^3√-24 = ^3√(-1 * 2^3 * 3^1)
A cube root can be taken individually for each factor, so we have:
^3√-24 = ^3√(-1) * ^3√(2^3) * ^3√(3^1)
The cube root of -1 is -1, and the cube root of 2^3 is 2. However, the cube root of 3^1 cannot be simplified further as it is an irrational number.
Therefore, ^3√-24 can be rewritten as -1 * 2 * ^3√3.
Hence, the irrational cube root ^3√-24 can be expressed as a product of an integer (-2) and another irrational cube root, which is ^3√3.
-24 can be factored as -1 * 2^3 * 3^1.
Now, let's rewrite ^3√-24 using this factorization:
^3√-24 = ^3√(-1 * 2^3 * 3^1)
A cube root can be taken individually for each factor, so we have:
^3√-24 = ^3√(-1) * ^3√(2^3) * ^3√(3^1)
The cube root of -1 is -1, and the cube root of 2^3 is 2. However, the cube root of 3^1 cannot be simplified further as it is an irrational number.
Therefore, ^3√-24 can be rewritten as -1 * 2 * ^3√3.
Hence, the irrational cube root ^3√-24 can be expressed as a product of an integer (-2) and another irrational cube root, which is ^3√3.
rewrite the irrational cube root ^3√-24 rewritten as a product of an integer and another irrational cube root
A. 4^3√6
B.-2^3√3
C.2^3√3
D.-2^3√4
A. 4^3√6
B.-2^3√3
C.2^3√3
D.-2^3√4
To rewrite the irrational cube root ^3√-24 as a product of an integer and another irrational cube root, we start by factoring -24.
-24 can be factored as -1 * 2^3 * 3^1.
Now, let's rewrite ^3√-24 using this factorization:
^3√-24 = ^3√(-1 * 2^3 * 3^1)
A cube root can be taken individually for each factor, so we have:
^3√-24 = ^3√(-1) * ^3√(2^3) * ^3√(3^1)
The cube root of -1 is -1, and the cube root of 2^3 is 2. However, the cube root of 3^1 cannot be simplified further as it is an irrational number.
Therefore, ^3√-24 can be rewritten as -1 * 2 * ^3√3.
Hence, the irrational cube root ^3√-24 can be expressed as a product of an integer (-2) and another irrational cube root, which is ^3√3.
Therefore, the correct answer is option B: -2^3√3.
-24 can be factored as -1 * 2^3 * 3^1.
Now, let's rewrite ^3√-24 using this factorization:
^3√-24 = ^3√(-1 * 2^3 * 3^1)
A cube root can be taken individually for each factor, so we have:
^3√-24 = ^3√(-1) * ^3√(2^3) * ^3√(3^1)
The cube root of -1 is -1, and the cube root of 2^3 is 2. However, the cube root of 3^1 cannot be simplified further as it is an irrational number.
Therefore, ^3√-24 can be rewritten as -1 * 2 * ^3√3.
Hence, the irrational cube root ^3√-24 can be expressed as a product of an integer (-2) and another irrational cube root, which is ^3√3.
Therefore, the correct answer is option B: -2^3√3.
rewrite the irrational cube root ^3√40 rewritten as a product of an integer and another irrational cube root
To rewrite the irrational cube root ^3√40 as a product of an integer and another irrational cube root, we need to find the largest perfect cube that divides 40. In this case, it is 8 since 8 * 5 = 40.
Now we can rewrite the cube root of 40 as follows:
^3√40 = ^3√(8 * 5)
Next, we can break down the cube root of 8 as follows:
^3√(8 * 5) = ^3√8 * ^3√5
The cube root of 8 is 2, so we have:
^3√8 * ^3√5 = 2 * ^3√5
Therefore, the irrational cube root ^3√40 can be rewritten as the product of an integer (2) and another irrational cube root (√5):
^3√40 = 2 * ^3√5.
Now we can rewrite the cube root of 40 as follows:
^3√40 = ^3√(8 * 5)
Next, we can break down the cube root of 8 as follows:
^3√(8 * 5) = ^3√8 * ^3√5
The cube root of 8 is 2, so we have:
^3√8 * ^3√5 = 2 * ^3√5
Therefore, the irrational cube root ^3√40 can be rewritten as the product of an integer (2) and another irrational cube root (√5):
^3√40 = 2 * ^3√5.
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