What is the irrational cupe root root(40, 3) rewritten as a product of an integer and another irrational cube root? (1 point)
O 2 * root(5, 3)
O - 2 * root(5, 3)
O 8 * root(5, 3)
O 2 * root(10, 3)
The correct answer is:
O 2 * root(10, 3)
To answer this question, we need to simplify the expression "cube root root(40, 3)" as a product of an integer and another irrational cube root.
Let's break it down step-by-step:
Step 1: Simplify the expression "root(40, 3)".
To simplify, we can rewrite it as "40^(1/3)".
Step 2: Simplify "40^(1/3)".
We can take the cube root of 40, which is 2. So "40^(1/3)" equals 2.
Step 3: Rewrite "2" as a product of an integer and another cube root.
The answer is "2 * root(1, 3)".
Therefore, the correct answer is "O 2 * root(1, 3)".
To rewrite the expression "cube root(40, 3)" as a product of an integer and another irrational cube root, we need to simplify the cube root.
First, let's find the prime factors of 40:
40 = 2 * 2 * 2 * 5.
Now, let's group the prime factors in sets of 3:
(2 * 2 * 2) * (5) = 8 * 5 = 40.
So, the expression "cube root(40, 3)" can be rewritten as "cube root(8 * 5, 3)".
Now, let's split the cube root into two separate cube roots:
cube root(8 * 5, 3) = cube root(8, 3) * cube root(5, 3).
The cube root of 8 is 2 because 2 * 2 * 2 = 8.
Therefore, the answer is:
2 * cube root(5, 3).
So, the correct option is:
O 2 * root(5, 3).