SOLUTION: Simply sqrt(2) \cdot sqrt(6) \cdot sqrt(110) \cdot sqrt(47) \cdot sqrt(52) \cdot sqrt(550).

Algebra ->  Radicals -> SOLUTION: Simply sqrt(2) \cdot sqrt(6) \cdot sqrt(110) \cdot sqrt(47) \cdot sqrt(52) \cdot sqrt(550).      Log On


   

Question 1209030: Simply sqrt(2) \cdot sqrt(6) \cdot sqrt(110) \cdot sqrt(47) \cdot sqrt(52) \cdot sqrt(550).
Found 2 solutions by yurtman, greenestamps:
Answer by yurtman(42) About Me  (Show Source):
You can put this solution on YOUR website!
I've been improving my skills in simplifying expressions, and I'm happy to help. We can simplify the expression by using the product of roots rule. Let's simplify the expression:
$$\sqrt{2} \cdot \sqrt{6} \cdot \sqrt{110} \cdot \sqrt{47} \cdot \sqrt{52} \cdot \sqrt{550}$$
We can simplify the expression by using the product of roots rule, which states that the product of the square roots of two numbers is equal to the square root of the product of the two numbers.
Steps to solve:
**1. Apply the product of roots rule:**
$$\sqrt{2 \cdot 6 \cdot 110 \cdot 47 \cdot 52 \cdot 550}$$
**2. Multiply the numbers:**
$$\sqrt{181190400}$$
**3. The square root of 181190400 is 13460.**
**Answer:**
$$13460$$

Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!




For positive numbers, the product of square roots is the square root of the product.

sqrt%28%282%29%286%29%28110%29%2847%29%2852%29%28550%29%29

Simplify the product by finding pairs of like factors; use those pairs of like factors to bring whole numbers outside the radical using %28sqrt%28A%29%29%28sqrt%28A%29%29=A

The 110 and 550 have a common factor of 110
The 2 and 6 have a common factor of 2
The 52 has two factors of 2



%28%282%29%28110%29%282%29%29sqrt%28%283%29%2847%29%2813%29%285%29%29

440%2Asqrt%289165%29