the perimeter will then just be the sum of the distances of A, B and C, namely AB + BC + CA.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points}\\\\A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\qquadB(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad \qquadd = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}\\\\\\AB=\sqrt{[0-(-2)]^2+[5-(-2)]^2}\implies AB=\sqrt{(0+2)^2+(5+2)^2}\\\\\\AB=\sqrt{4+49}\implies \boxed{AB=\sqrt{53}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\B(\stackrel{x_2}{0}~,~\stackrel{y_2}{5})\qquad C(\stackrel{x_1}{3}~,~\stackrel{y_1}{1})\\\\\\BC=\sqrt{(3-0)^2+(1-5)^2}\implies BC=\sqrt{3^2+(-4)^2}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%5C%5CA%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5CqquadB%28%5Cstackrel%7Bx_2%7D%7B0%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20%5Cqquadd%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%5C%5C%5C%5C%5C%5CAB%3D%5Csqrt%7B%5B0-%28-2%29%5D%5E2%2B%5B5-%28-2%29%5D%5E2%7D%5Cimplies%20AB%3D%5Csqrt%7B%280%2B2%29%5E2%2B%285%2B2%29%5E2%7D%5C%5C%5C%5C%5C%5CAB%3D%5Csqrt%7B4%2B49%7D%5Cimplies%20%5Cboxed%7BAB%3D%5Csqrt%7B53%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5CB%28%5Cstackrel%7Bx_2%7D%7B0%7D~%2C~%5Cstackrel%7By_2%7D%7B5%7D%29%5Cqquad%20C%28%5Cstackrel%7Bx_1%7D%7B3%7D~%2C~%5Cstackrel%7By_1%7D%7B1%7D%29%5C%5C%5C%5C%5C%5CBC%3D%5Csqrt%7B%283-0%29%5E2%2B%281-5%29%5E2%7D%5Cimplies%20BC%3D%5Csqrt%7B3%5E2%2B%28-4%29%5E2%7D)
![\bf BC=\sqrt{9+16}\implies \boxed{BC=5}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\C(\stackrel{x_2}{3}~,~\stackrel{y_2}{1})\qquad A(\stackrel{x_1}{-2}~,~\stackrel{y_1}{-2})\\\\\\CA=\sqrt{(-2-3)^2+(-2-1)^2}\implies CA=\sqrt{(-5)^2+(-3)^2}\\\\\\CA=\sqrt{25+9}\implies \boxed{CA=\sqrt{34}}\\\\[-0.35em]\rule{34em}{0.25pt}\\\\~\hfill \stackrel{AB+BC+CA}{\approx 18.11}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20BC%3D%5Csqrt%7B9%2B16%7D%5Cimplies%20%5Cboxed%7BBC%3D5%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5CC%28%5Cstackrel%7Bx_2%7D%7B3%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%5Cqquad%20A%28%5Cstackrel%7Bx_1%7D%7B-2%7D~%2C~%5Cstackrel%7By_1%7D%7B-2%7D%29%5C%5C%5C%5C%5C%5CCA%3D%5Csqrt%7B%28-2-3%29%5E2%2B%28-2-1%29%5E2%7D%5Cimplies%20CA%3D%5Csqrt%7B%28-5%29%5E2%2B%28-3%29%5E2%7D%5C%5C%5C%5C%5C%5CCA%3D%5Csqrt%7B25%2B9%7D%5Cimplies%20%5Cboxed%7BCA%3D%5Csqrt%7B34%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C~%5Chfill%20%5Cstackrel%7BAB%2BBC%2BCA%7D%7B%5Capprox%2018.11%7D~%5Chfill)
Answer:

Step-by-step explanation:
Since we know we only have 640 feet of fence available, we know that L + W + L = 640,
2L + W = 640. This allows us to represent the width, W, in terms of L: W = 640 – 2L
Remember, the area of a rectangle is equal to the product of its width and length, therefore,
Notice that, quadratic has been vertically reflected, since the coefficient on the squared term is negative, so the graph will open downwards, and the vertex will be a maximum value for the area.
recall,
Since our function is A(L)=640L-2L², we get
plug in the value of a and b into the first formula:


hence,the dimensions of the pen that will maximize the area are<u> </u><u>L=</u><u>1</u><u>6</u><u>0</u><u>m</u><u> </u>and<u> </u><u>W=</u><u>7</u><u>6</u><u>8</u><u>0</u><u>0</u><u>/</u><u>1</u><u>6</u><u>0</u><u>=</u><u>4</u><u>8</u><u>0</u><u>m</u>
and we're done!
Answer:
Distance = 13
Step-by-step explanation:
Using the distance formula, plug in the two points:
d = sqrt(9-(-3))^2+(-3-2)^2
d = sqrt(12)^2+(-5)^2
d = sqrt144+25
d = sqrt169
d = 13
there is no B i see A and -5A unless its for -5A??
Answer:
a number divided by 4
Step-by-step explanation:
a number = n (it's a variable used in equations.)
so n over 4, or n divided by 4