Check the picture below, now the distance from 2,0 to 4,0 there's no need to do much calculation since that's just 2 units, as you see there.
![~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{2}~,~\stackrel{y_1}{0})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{4}) ~\hfill d1=\sqrt{[ 1- 2]^2 + [ 4- 0]^2} \\\\\\ d1=\sqrt{(-1)^2+4^2}\implies \boxed{d1=\sqrt{17}} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=~%5Chfill%20%5Cstackrel%7B%5Ctextit%7B%5Clarge%20distance%20between%202%20points%7D%7D%7Bd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%7D~%5Chfill~%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B2%7D~%2C~%5Cstackrel%7By_1%7D%7B0%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B1%7D~%2C~%5Cstackrel%7By_2%7D%7B4%7D%29%20~%5Chfill%20d1%3D%5Csqrt%7B%5B%201-%202%5D%5E2%20%2B%20%5B%204-%200%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d1%3D%5Csqrt%7B%28-1%29%5E2%2B4%5E2%7D%5Cimplies%20%5Cboxed%7Bd1%3D%5Csqrt%7B17%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![(\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{-1}~,~\stackrel{y_2}{-2}) ~\hfill d2=\sqrt{[ -1- 1]^2 + [ -2- 4]^2} \\\\\\ d2=\sqrt{(-2)^2+(-6)^2}\implies \boxed{d2=\sqrt{40}} \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{-1}~,~\stackrel{y_1}{-2})\qquad (\stackrel{x_2}{4}~,~\stackrel{y_2}{0}) ~\hfill d3=\sqrt{[ 4- (-1)]^2 + [ 0- (-2)]^2} \\\\\\ d3=\sqrt{(4+1)^2+(0+2)^2}\implies \boxed{d3=\sqrt{29}}](https://tex.z-dn.net/?f=%28%5Cstackrel%7Bx_1%7D%7B1%7D~%2C~%5Cstackrel%7By_1%7D%7B4%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B-1%7D~%2C~%5Cstackrel%7By_2%7D%7B-2%7D%29%20~%5Chfill%20d2%3D%5Csqrt%7B%5B%20-1-%201%5D%5E2%20%2B%20%5B%20-2-%204%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d2%3D%5Csqrt%7B%28-2%29%5E2%2B%28-6%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd2%3D%5Csqrt%7B40%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%28%5Cstackrel%7Bx_1%7D%7B-1%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B4%7D~%2C~%5Cstackrel%7By_2%7D%7B0%7D%29%20~%5Chfill%20d3%3D%5Csqrt%7B%5B%204-%20%28-1%29%5D%5E2%20%2B%20%5B%200-%20%28-2%29%5D%5E2%7D%20%5C%5C%5C%5C%5C%5C%20d3%3D%5Csqrt%7B%284%2B1%29%5E2%2B%280%2B2%29%5E2%7D%5Cimplies%20%5Cboxed%7Bd3%3D%5Csqrt%7B29%7D%7D)
Answer:
The fourth graph is the graph of 
Step-by-step explanation:
As this inequality has a 
, we know two things:
The graph must have a solid line as that signifies that it includes the value.
The shaded area must be less than 2.
The fourth graph is the only one that meets our criteria
Regular heptagonSolve for <span>area
</span>A= 7/4 x a^2 x cot(180 degrees/7)
https://www.google.com/search?sourceid=chrome-psyapi2&rlz=1C1ZQQI_enUS697US697&ion=1&espv=2&ie=UTF-8&q=area%20of%20a%20heptagon%20formula&oq=area%20of%20a%20heptagon&aqs=chrome.1.69i57j0l5.4065j0j7
https://www.google.com/search?sourceid=chrome-psyapi2&rlz=1C1ZQQI_enUS697US697&ion=1&espv=2&ie=UTF-8<span>&q=area%20of%20a%20heptagon%20formula&oq=area%20of%20a%20heptagon&aqs=chrome.1.69i57j0l5.4065j0j7</span>
Step-by-step explanation:
multiply 5x+2y=19 by 3 youll get 15x+6y=57
eliminate y
7x-6y=9
<u>15x+6y=57</u><u> </u> +
22x=66
x=3
substitute x to 5x+2y=19
15+2y=19
2y=4
y=2
Your answer is #3 and #4
because both of them are bigger then 1924
hope my answer helped , feel free to ask more.
