To rewrite the irrational cube root ∛40 as a product of an integer and another irrational cube root, we first need to find a perfect cube factor of 40.
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40.
However, none of these factors are perfect cubes.
Now, let's calculate the cube root of 40: ∛40 ≈ 3.107
Since 40 is not a perfect cube, ∛40 is an irrational number. It cannot be rewritten as a product of an integer and another irrational cube root.
what is the irrational cube root 3√40 rewritten as a product of an integer and another irrational cube root
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do it right
I apologize for my previous mistake. Let's correct it.
To rewrite the irrational cube root ∛40 as a product of an integer and another irrational cube root, we need to find a perfect cube factor of 40.
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40.
The largest perfect cube factor of 40 is 8, which is equal to 2^3.
So, we can rewrite ∛40 as ∛(8×5).
This can be further simplified as 2∛5.
Therefore, the irrational cube root ∛40 can be rewritten as a product of an integer (2) and another irrational cube root (∛5).
To rewrite the irrational cube root ∛40 as a product of an integer and another irrational cube root, we need to find a perfect cube factor of 40.
The factors of 40 are: 1, 2, 4, 5, 8, 10, 20, 40.
The largest perfect cube factor of 40 is 8, which is equal to 2^3.
So, we can rewrite ∛40 as ∛(8×5).
This can be further simplified as 2∛5.
Therefore, the irrational cube root ∛40 can be rewritten as a product of an integer (2) and another irrational cube root (∛5).
1 answer