17000.106 is how you write the number above in standard form
The answer is B, because A and C are automatically eliminated because they are subtracting, and the equation that represents the situation is an addition inequality. B is correct because the two parenthesis numbers represents the students attending the fair paying $0.50, and there are 200 students. The 2.25a represents an amount of adults, each paying $2.25.
Answer: B
Since a right triangle has special rules to it such that the hypotenuse squared equals the sum of the squares of the other 2 sides. In other words, if the hypotenuse = c, and the 2 smaller sides are a and b, then:
![{c}^{2} = {a}^{2} + {b}^{2}](https://tex.z-dn.net/?f=%20%7Bc%7D%5E%7B2%7D%20%20%3D%20%20%7Ba%7D%5E%7B2%7D%20%20%2B%20%20%7Bb%7D%5E%7B2%7D%20)
Solving for c (hypotenuse), we get:
![\sqrt{({c}^{2})} = \sqrt{({a}^{2}+{b}^{2} )} \\ c = \sqrt{({a}^{2}+{b}^{2} )}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%28%7Bc%7D%5E%7B2%7D%29%7D%20%3D%20%20%20%5Csqrt%7B%28%7Ba%7D%5E%7B2%7D%2B%7Bb%7D%5E%7B2%7D%20%29%7D%20%5C%5C%20c%20%3D%20%20%5Csqrt%7B%28%7Ba%7D%5E%7B2%7D%2B%7Bb%7D%5E%7B2%7D%20%29%7D%20)
Therefore c is the root of the square of the other sides. So by having root18, it's like saying:
![\sqrt{18} = \sqrt{({a}^{2} + {b}^{2} )} \\ 18 = {a}^{2} + {b}^{2}](https://tex.z-dn.net/?f=%20%5Csqrt%7B18%7D%20%3D%20%20%5Csqrt%7B%28%7Ba%7D%5E%7B2%7D%20%20%2B%20%20%7Bb%7D%5E%7B2%7D%20%29%7D%20%20%5C%5C%2018%20%3D%20%7Ba%7D%5E%7B2%7D%20%20%2B%20%20%7Bb%7D%5E%7B2%7D%20%20)
Getting rid of the square root, so sides a and b must have their squares total 18:
the only squares < 18 are: 16 (4×4), 9 (3×3), 4 (2×2), and 1 (1×1)
of those above added in any order of two of them, only 16+4
![18 = 17 + 1 - - > \sqrt{17} \: and \: 1 \\ 18 = 16 + 2 - - > \sqrt{16} \: and \: \sqrt{2} \\ ... \: 4 \: and \: \sqrt{2} \\ 18 = 15 + 3 - - > \sqrt{15} \: and \: \sqrt{3}](https://tex.z-dn.net/?f=18%20%3D%2017%20%2B%201%20-%20%20-%20%20%3E%20%20%5Csqrt%7B17%7D%20%20%5C%3A%20and%20%5C%3A%201%20%5C%5C%2018%20%3D%2016%20%2B%202%20-%20%20-%20%20%3E%20%20%5Csqrt%7B16%7D%20%5C%3A%20and%20%5C%3A%20%20%5Csqrt%7B2%7D%20%20%5C%5C%20...%20%5C%3A%204%20%5C%3A%20and%20%5C%3A%20%20%5Csqrt%7B2%7D%20%20%5C%5C%2018%20%3D%2015%20%2B%203%20-%20%20-%20%20%3E%20%20%5Csqrt%7B15%7D%20%5C%3A%20%20and%20%5C%3A%20%20%5Csqrt%7B3%7D%20)
![18 = 14 + 4 - - > \sqrt{14} \: and \: \sqrt{4} \\ ... \sqrt{14} \: and \: 2 \\ 18 = 13 + 5 - - > \sqrt{13} \: and \: \sqrt{5} \\ 18 = 12+ 6 - - > \sqrt{12} \: and \: \sqrt{6}](https://tex.z-dn.net/?f=18%20%3D%2014%20%2B%204%20-%20%20-%20%20%3E%20%20%5Csqrt%7B14%7D%20%5C%3A%20%20and%20%5C%3A%20%20%5Csqrt%7B4%7D%20%20%5C%5C%20...%20%5Csqrt%7B14%7D%20%5C%3A%20and%20%5C%3A%202%20%20%5C%5C%2018%20%3D%2013%20%2B%205%20-%20%20-%20%20%3E%20%20%5Csqrt%7B13%7D%20%5C%3A%20and%20%5C%3A%20%20%5Csqrt%7B5%7D%20%20%20%5C%5C%2018%20%3D%2012%2B%206%20-%20%20-%20%20%3E%20%20%5Csqrt%7B12%7D%20%5C%3A%20%20and%20%5C%3A%20%20%5Csqrt%7B6%7D%20)
![18 = 11 + 7 - - > \sqrt{11} \: and \: \sqrt{7} \\ 18 = 10 + 8 - - > \sqrt{10} \: and \: \sqrt{8} \\ ... \sqrt{10} \: and \: 2 \sqrt{2} \\ 18 = 9 + 9 - - > \sqrt{9} = 3 \: and \: 3](https://tex.z-dn.net/?f=18%20%3D%2011%20%2B%207%20-%20%20-%20%20%3E%20%20%5Csqrt%7B11%7D%20%5C%3A%20%20and%20%5C%3A%20%20%5Csqrt%7B7%7D%20%20%5C%5C%2018%20%3D%2010%20%2B%208%20-%20%20-%20%20%3E%20%20%5Csqrt%7B10%7D%20%5C%3A%20%20and%20%5C%3A%20%20%5Csqrt%7B8%7D%20%20%5C%5C%20...%20%5Csqrt%7B10%7D%20%5C%3A%20%20and%20%5C%3A%202%20%5Csqrt%7B2%7D%20%20%5C%5C%2018%20%3D%209%20%2B%209%20-%20%20-%20%20%3E%20%20%5Csqrt%7B9%7D%20%20%3D%203%20%5C%3A%20and%20%5C%3A%203%20)
I HOPE THAT IS ALONG THE LINE THAT WAS ASKED FOR!!! :-D
What am I solving? Like what is the question?
Answer:
Hala would have 10 candies, Ahmed would have 35, and Eman would have 45.
Step-by-step explanation:
if the candy is given out 2:7:9, add the numbers and devide. 90 ÷ 18 is five so multiply each number by 5 to get 10:35:45