√48 can be simplified as √(16*3) which is equal to √16 * √3.
√16 is equal to 4 since 4*4=16, so the square root of 16 is an integer.
Therefore, √48 can be rewritten as 4√3.
Rewrite the irrational square root 48−−√ as the product of an integer and another irrational square root.(1 point)
10 answers
What is the irrational square root 28−−√ simplified?
√28 can be simplified as √(4*7), which is equal to √4 * √7.
√4 is equal to 2 since 2*2=4, so the square root of 4 is an integer.
Therefore, √28 can be simplified as 2√7.
√4 is equal to 2 since 2*2=4, so the square root of 4 is an integer.
Therefore, √28 can be simplified as 2√7.
Rewrite the irrational cube root 48−−√3 as a product of an integer and another irrational cube root.
To rewrite ∛48 as a product of an integer and another irrational cube root, we can find a perfect cube that divides 48.
48 does not have a perfect cube factor, so we can simplify it by factoring out the largest perfect cube, which is 2³.
∛48 = ∛(2³ * 6)
Since ∛(2³) simplifies to 2, and the irrational cube root of 6 remains the same, we can rewrite it as:
2∛6
48 does not have a perfect cube factor, so we can simplify it by factoring out the largest perfect cube, which is 2³.
∛48 = ∛(2³ * 6)
Since ∛(2³) simplifies to 2, and the irrational cube root of 6 remains the same, we can rewrite it as:
2∛6
What is the irrational cube root −24−−−−√3 rewritten as a product of an integer and another irrational cube root?
To rewrite −24−−−−√3 as a product of an integer and another irrational cube root, we can find a perfect cube that divides -24.
-24 does not have a perfect cube factor, so we can simplify it by factoring out the largest perfect cube, which is -2³.
−24−−−−√3 = -2³∛3
Therefore, the irrational cube root −24−−−−√3 can be written as -2∛3.
-24 does not have a perfect cube factor, so we can simplify it by factoring out the largest perfect cube, which is -2³.
−24−−−−√3 = -2³∛3
Therefore, the irrational cube root −24−−−−√3 can be written as -2∛3.
What is the irrational cube root 40−−√3 rewritten as a product of an integer and another irrational cube root?
To rewrite ∛40 as a product of an integer and another irrational cube root, we can find a perfect cube that divides 40.
40 does not have a perfect cube factor, so we need to represent it as a product of perfect cubes.
We can write 40 as (8 * 5). Notice that 8 is a perfect cube because it can be expressed as 2^3.
∛40 = ∛(8 * 5)
Using the property of radicals, we can separate the product of radicals and simplify:
∛(8 * 5) = ∛8 * ∛5
Since √8 simplifies to 2 and the irrational cube root of 5 remains the same, we can rewrite it as:
2∛5
Therefore, the irrational cube root ∛40 can be written as 2∛5.
40 does not have a perfect cube factor, so we need to represent it as a product of perfect cubes.
We can write 40 as (8 * 5). Notice that 8 is a perfect cube because it can be expressed as 2^3.
∛40 = ∛(8 * 5)
Using the property of radicals, we can separate the product of radicals and simplify:
∛(8 * 5) = ∛8 * ∛5
Since √8 simplifies to 2 and the irrational cube root of 5 remains the same, we can rewrite it as:
2∛5
Therefore, the irrational cube root ∛40 can be written as 2∛5.
hooraY! its all RIGHTT
1 answer