What is the irrational square root sqrt(28) simplified? (1 point)
2sqrt(14)
2sqrt(7)
sqrt(4) * sqrt(7)
4sqrt(7)
7 answers
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)
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.
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)
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.
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)
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.
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.
1 answer