Hi. I pretty much have the answer for these problems that I worked out just need to know if there right. Thank you all!
16�ã27 + 7�ã108 = 90�ã3
16�ã9�ã3 + 7�ã36�ã3
16 . 3�ã3 + 7 . 6�ã3
(48 +42)�ã3
90�ã3;
And
�ã600x^4 = �ã100 . �ãx^2 . �ã6 = 10x^2�ã6
To verify the correctness of the given expressions, let's break down each problem step by step.
Problem 1: 16�ã27 + 7�ã108 = 90�ã3
To simplify this expression, start by finding the square root of both numbers inside the radicals:
16�ã27 = 16�ã(3^3) = 16�ã3 = 4�ã3
7�ã108 = 7�ã(2^2 * 3^3) = 7�ã(2^2 * 3^2 * 3) = 7 * 2 * 3�ã3 = 42�ã3
Now, add the two simplified radicals together:
4�ã3 + 42�ã3 = (4 + 42)�ã3 = 46�ã3
Therefore, the expression 16�ã27 + 7�ã108 simplifies to 46�ã3.
Problem 2: 16�ã9�ã3 + 7�ã36�ã3
Similar to the previous problem, start by simplifying each radical:
16�ã9�ã3 = 16�ã(3^2)�ã3 = 16�ã3�ã3 = 16�ã3
7�ã36�ã3 = 7�ã(6^2)�ã3 = 7�ã6�ã3 = 7�ã6
Now, add the two simplified radicals together:
16�ã3 + 7�ã6 = 16�ã3 + 7�ã6
Therefore, the expression 16�ã9�ã3 + 7�ã36�ã3 does not simplify any further.
Problem 3: 16 . 3�ã3 + 7 . 6�ã3
Multiply the values outside the radicals with their respective radicals:
16 . 3�ã3 = 48�ã3
7 . 6�ã3 = 42�ã3
Now, add the two simplified radicals together:
48�ã3 + 42�ã3 = (48 + 42)�ã3 = 90�ã3
Therefore, the expression 16 . 3�ã3 + 7 . 6�ã3 simplifies to 90�ã3.
Problem 4: (48 + 42)�ã3
First, find the sum inside the parentheses:
48 + 42 = 90
Now, simplify the expression:
(48 + 42)�ã3 = 90�ã3
Therefore, the expression (48 + 42)�ã3 simplifies to 90�ã3.
Problem 5: �ã600x^4 = �ã100 . �ãx^2 . �ã6 = 10x^2�ã6
To simplify this expression, start by breaking down the radical part:
�ã100 = 10
�ãx^2 = x
�ã6 = �ã(2^2 * 3) = 2�ã3
Now, multiply the simplified radicals together:
10 * x * 2�ã3 = 20x�ã3
Therefore, the expression �ã600x^4 simplifies to 20x�ã3.
In summary:
1. 16�ã27 + 7�ã108 = 90�ã3
2. 16�ã9�ã3 + 7�ã36�ã3 does not simplify any further.
3. 16 . 3�ã3 + 7 . 6�ã3 = 90�ã3
4. (48 + 42)�ã3 = 90�ã3
5. �ã600x^4 = 20x�ã3