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:
Step-by-step explanation:
To solve for the given variable 
, we need to get on one side of the equation by itself. To do so, we need to remove the -16 that is on the same side of the variable.
To get rid of the -16, we can add a 16 to that side of the equation since 
, which will get by itself; however, if we add -16 to one side of the equation, we must add it to both sides of the equation:
3. Commuinitive proterty 4. 0 5. The second one
Answer:
The formula for the volume of a sphere is V = 4/3 πr³.
Step-by-step explanation:
Surface area of sphere = 4πr2
Volume of a sphere = 4/3 πr3
You only need to know the radius to figure both the volume and surface area of a sphere.
Answers for surface area problems should always be in square units.
Answers for volume problems should always be in cubic units.
