I got you, my answer is x<4
I believe the answer would be B. 11 because you can see a pattern. Its just skipping.
Answer:
m = 5, n = 2
Step-by-step explanation:
![5m - 3 n = 19\\m - 6n = - 7 \ => m = -7 + 6n\\\\Substitute \ m \ in \ 5m - 3n = 19\\\\5(-7 + 6n) - 3n = 19\\-35 + 30n -3n = 19\\27n = 19 + 35 \\27n = 54 \\n = 2\\\\Substitute \ n = 2 \ in \ m = -7 + 6n\\\\m = -7 + (6 \times 2) = -7 + 12 = 5](https://tex.z-dn.net/?f=5m%20-%203%20n%20%3D%2019%5C%5Cm%20-%206n%20%3D%20-%207%20%5C%20%3D%3E%20m%20%3D%20-7%20%2B%206n%5C%5C%5C%5CSubstitute%20%5C%20m%20%20%5C%20in%20%5C%20%205m%20-%203n%20%3D%2019%5C%5C%5C%5C5%28-7%20%2B%206n%29%20-%203n%20%3D%2019%5C%5C-35%20%2B%2030n%20-3n%20%3D%2019%5C%5C27n%20%3D%2019%20%2B%2035%20%5C%5C27n%20%3D%2054%20%5C%5Cn%20%3D%202%5C%5C%5C%5CSubstitute%20%5C%20n%20%3D%202%20%5C%20%20in%20%5C%20%20m%20%3D%20-7%20%2B%206n%5C%5C%5C%5Cm%20%3D%20-7%20%2B%20%286%20%5Ctimes%202%29%20%3D%20-7%20%2B%2012%20%3D%205)
Since the area of a square is equal to the square of one of its side's length, then the area should be equivalent to
![x^{2}](https://tex.z-dn.net/?f=%20x%5E%7B2%7D)
.
![A = x^{2}](https://tex.z-dn.net/?f=%20A%20%3D%20x%5E%7B2%7D)
---> equation (1)
By using pythagoras rule which states that the
![x^{2} = hyp^2 - opposite^2](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%3D%20hyp%5E2%20-%20opposite%5E2)
---> equation (2)
where the opposite side's length is 8 and the hypotenuse side's length is 10
by substituting by the values in equation (2) therefore,
![x^{2} = 10^{2} - 8^{2}](https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%3D%2010%5E%7B2%7D%20-%208%5E%7B2%7D%20)
substitute this value in equation (1) then
![A = x^{2} = 10^{2} -8^{2}](https://tex.z-dn.net/?f=%20%20A%20%3D%20x%5E%7B2%7D%20%3D%2010%5E%7B2%7D%20-8%5E%7B2%7D%20)
where A is the area of the square whose side is x
Answer and work down below. Let me know if you have any questions