What is the irrational square root sqrt(28) simplified? (1 point)
2sqrt(14)
2sqrt(7)
sqrt(4) * sqrt(7)
4sqrt(7)
2sqrt(7)
To simplify the square root of 28, sqrt(28), you can break it down into smaller factors.
1. Start by finding the largest perfect square that divides 28. In this case, it is 4 (2 * 2 = 4).
2. Rewrite the square root of 28 as the square root of the perfect square multiplied by the remaining factor. So, sqrt(28) = sqrt(4 * 7).
3. Take the square root of the perfect square, which is 2.
4. Rewrite the remaining factor outside the square root. So, sqrt(28) simplifies to 2sqrt(7).
Therefore, the simplified square root of 28 is 2sqrt(7).
To simplify the square root of 28, we can break it down into its prime factors: 2 * 2 * 7.
We can then rewrite the square root of 28 as the product of the square root of each prime factor: sqrt(2 * 2 * 7).
Since the square root of 2 * 2 is equal to 2, we can simplify the expression to: 2 * sqrt(7).
Therefore, the simplified form of the square root of 28 is 2sqrt(7).
Rewrite the irrational cube root root(48, 3) as a product of an integer and another
irrational cube root (1 point)
- 2 * root(6, 3)
2 * root(6, 3)
4 * root(3, 3)
4 * root(12, 2)
To rewrite the cube root of 48, ∛48, as a product of an integer and another irrational cube root, we need to break down 48 into its prime factors.
48 can be factored as 2 * 2 * 2 * 2 * 3.
Now, we can rewrite the cube root of 48 as the cube root of each prime factor multiplied together: ∛(2 * 2 * 2 * 2 * 3).
The cube root of 2 * 2 * 2 * 2 simplifies to 2, so the expression becomes: 2 * ∛(3).
Therefore, the simplified form of the cube root of 48 is 2∛3.
Therefore, the correct option is: 2 * ∛3.
What is the irrational cube root root(- 24, 3) rewritten as a product of an integer and
another irrational cube root? (1 point)
O - 2 * root(4, 3)
O - 2 * root(3, 3)
O 2 * root(3, 3)
O 2 * root(6, 2)
To rewrite the cube root of -24, ∛(-24), as a product of an integer and another irrational cube root, we first need to determine the prime factorization of -24.
The prime factorization of -24 is -1 * 2 * 2 * 2 * 3.
Since we are looking for the cube root, we want to group the factors in sets of 3. In this case, we can group the three 2's together.
So we can rewrite the cube root of -24 as the cube root of (-1 * 2 * 2 * 2) * 3.
The cube root of -1 * 2 * 2 * 2 simplifies to -2, so the expression becomes: -2 * ∛3.
Therefore, the irrational cube root ∛(-24) can be rewritten as -2∛3.
Therefore, the correct option is: O - 2 * ∛3.
What is the irrational cube root root(40, 3) rewritten as a product of an integer and another irrational cube root? (1 point)
O 2 * root(10, 3)
O 2 * root(5, 2)
O 8 * root(5, 2)
O - 2 * root(5, 3)
To rewrite the cube root of 40, ∛40, as a product of an integer and another irrational cube root, we first need to determine the prime factorization of 40.
The prime factorization of 40 is 2 * 2 * 2 * 5.
Since we are looking for the cube root, we want to group the factors in sets of 3. In this case, we can group two 2's together.
So we can rewrite the cube root of 40 as the cube root of (2 * 2 * 2) * 5.
The cube root of 2 * 2 * 2 simplifies to 2, so the expression becomes: 2 * ∛5.
Therefore, the irrational cube root ∛40 can be rewritten as 2∛5.
Therefore, the correct option is: O 2 * ∛5.