The square root of 100 = 10
The square root of 225 = 15
The sum = 10 + 15 = 25
the answer is actually 29 because ask your teacher he'll ive you the work <3 Answer: PEMDAS-order of operations The square root of 100 is 10. The square root of 225 is 15. The sum of 10 + 15 is 25. The product of 6 and 25 is 250.
150
It's not. Take 49 and 16 for example. The square root of the sum is the square root of 65. The sum of the square roots is 11.
15² + 2³ = 225 + 8 = 233
18
the answer is actually 29 because ask your teacher he'll ive you the work <3 Answer: PEMDAS-order of operations The square root of 100 is 10. The square root of 225 is 15. The sum of 10 + 15 is 25. The product of 6 and 25 is 250.
150
It's not. Take 49 and 16 for example. The square root of the sum is the square root of 65. The sum of the square roots is 11.
false
To find the sum of the square roots of the digits in the number 121, we first identify its digits: 1, 2, and 1. The square roots of these digits are √1 = 1, √2 ≈ 1.41, and √1 = 1. Adding these together gives us 1 + 1.41 + 1 = 3.41. Thus, the sum of the square roots of the digits in 121 is approximately 3.41.
The sum of square roots is zero only when all the square roots involved are equal to zero. Since the square root of any non-negative number is non-negative, the only way for their sum to equal zero is if each term in the sum is zero itself. Therefore, if you have (\sqrt{a} + \sqrt{b} + \sqrt{c} = 0), it implies that (a = 0), (b = 0), and (c = 0). In any other case, the sum will be positive.
Assuming the roots are positive, then ~ 15.59.
It is: 81+100 = 181
15² + 2³ = 225 + 8 = 233
100
The sum of the first 15 positive even numbers is 240. (Simply square 15, then add 15 to the result: 15 x 15 = 225. 225 + 15 = 240).
100 + 49 + 1