What is the irrational square root 63−−√ rewritten as the product of an integer and another irrational square root, in simplest form?(1 point)
To rewrite √63, we need to find the largest perfect square that is a factor of 63, which is 9. Rewrite √63 as √(9 × 7) = √9 × √7 = 3√7. Therefore, the product of an integer and the irrational square root √63 is 3√7.
What is the irrational cube root 81−−√3 rewritten as a product of an integer and another irrational cube root? Note: You will find the option to add a cubic root symbol in the Algebra ( × ) keyboard. (1 point)
To rewrite ∛81 as a product of an integer and another irrational cube root, we need to find the largest perfect cube that is a factor of 81, which is 27. Rewrite ∛81 as ∛(27 × 3) = ∛27 × ∛3 = 3∛3. Therefore, the product of an integer and the irrational cube root ∛81 is 3∛3.
To rewrite the irrational square root of 63 (√63) as the product of an integer and another irrational square root, we need to find a perfect square factor of 63 that we can take out of the square root.
Let's factorize 63:
63 = 9 * 7
Since 9 is a perfect square, we can simplify the square root of 63:
√63 = √(9 * 7) = √9 * √7
The square root of 9 (√9) is an integer, which is 3. Thus, we rewrite the expression as:
√63 = 3 * √7
Therefore, the irrational square root of 63 (√63) can be rewritten as the product of the integer 3 and the irrational square root of 7 (√7).