Answer:

Step-by-step explanation:

hope it helps..
have a great day!!
so the points are, from P1 to P2, namely P1P2, and from P2 to P3, namely P2P3, and from P3 back to P1, namely P3P1.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4})\qquad P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ P1P2=\sqrt{[8-5]^2+[-3-(-4)]^2}\implies P1P2=\sqrt{(8-5)^2+(-3+4)^2} \\\\\\ P1P2=\sqrt{3^2+1^2}\implies \boxed{P1P2=\sqrt{10}}\\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P1%28%5Cstackrel%7Bx_1%7D%7B5%7D~%2C~%5Cstackrel%7By_1%7D%7B-4%7D%29%5Cqquad%20%20P2%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%7D%29%5Cqquad%20%5Cqquad%20%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P1P2%3D%5Csqrt%7B%5B8-5%5D%5E2%2B%5B-3-%28-4%29%5D%5E2%7D%5Cimplies%20P1P2%3D%5Csqrt%7B%288-5%29%5E2%2B%28-3%2B4%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P1P2%3D%5Csqrt%7B3%5E2%2B1%5E2%7D%5Cimplies%20%5Cboxed%7BP1P2%3D%5Csqrt%7B10%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%20)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10}) \\\\\\ P2P3=\sqrt{[7-8]^2+[-10-(-3)]^2}\implies P2P3=\sqrt{(7-8)^2+(-10+3)^2} \\\\\\ P2P3=\sqrt{(-1)^2+(-7)^2}\implies P2P3=\sqrt{50}\implies \boxed{P2P3=5\sqrt{2}}\\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P2%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%7D%29%5Cqquad%20%20P3%28%5Cstackrel%7Bx_2%7D%7B7%7D~%2C~%5Cstackrel%7By_2%7D%7B-10%7D%29%20%5C%5C%5C%5C%5C%5C%20P2P3%3D%5Csqrt%7B%5B7-8%5D%5E2%2B%5B-10-%28-3%29%5D%5E2%7D%5Cimplies%20P2P3%3D%5Csqrt%7B%287-8%29%5E2%2B%28-10%2B3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P2P3%3D%5Csqrt%7B%28-1%29%5E2%2B%28-7%29%5E2%7D%5Cimplies%20P2P3%3D%5Csqrt%7B50%7D%5Cimplies%20%5Cboxed%7BP2P3%3D5%5Csqrt%7B2%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%20)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10})\qquad P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4}) \\\\\\ P3P1=\sqrt{[5-7]^2+[-4-(-10)]^2}\implies P3P1=\sqrt{(5-7)^2+(-4+10)^2} \\\\\\ P3P1=\sqrt{(-2)^2+6^2}\implies P3P1=\sqrt{40}\implies \boxed{P3P1=2\sqrt{10}}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P3%28%5Cstackrel%7Bx_2%7D%7B7%7D~%2C~%5Cstackrel%7By_2%7D%7B-10%7D%29%5Cqquad%20%20P1%28%5Cstackrel%7Bx_1%7D%7B5%7D~%2C~%5Cstackrel%7By_1%7D%7B-4%7D%29%20%5C%5C%5C%5C%5C%5C%20P3P1%3D%5Csqrt%7B%5B5-7%5D%5E2%2B%5B-4-%28-10%29%5D%5E2%7D%5Cimplies%20P3P1%3D%5Csqrt%7B%285-7%29%5E2%2B%28-4%2B10%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P3P1%3D%5Csqrt%7B%28-2%29%5E2%2B6%5E2%7D%5Cimplies%20P3P1%3D%5Csqrt%7B40%7D%5Cimplies%20%5Cboxed%7BP3P1%3D2%5Csqrt%7B10%7D%7D%20)
Answer:
second one
Step-by-step explanation:
A) The signs of the first derivative (g') tell you the graph increases as you go left from x=4 and as you go right from x=-2. Since g(4) < g(-2), one absolute extreme is (4, g(4)) = (4, 1).
The sign of the first derivative changes at x=0, at which point the slope is undefined (the curve is vertical). The curve approaches +∞ at x=0 both from the left and from the right, so the other absolute extreme is (0, +∞).
b) The second derivative (g'') changes sign at x=2, so there is a point of inflection there.
c) There is a vertical asymptote at x=0 and a flat spot at x=2. The curve goes through the points (-2, 5) and (4, 1), is increasing to the left of x=0 and non-increasing to the right of x=0. The curve is concave upward on [-2, 0) and (0, 2) and concave downward on (2, 4]. A possible graph is shown, along with the first and second derivatives.
Answer:
Step-by-step explanation:
Suppose the dimensions of the playground are x and y.
The total amount of the fence used is given and it is 780 ft. In terms of x and y this would be 3x+2y=780 (we add 3x because we want it to be cut in the middle). Therefore, y= 780/2-3/2x. Now, the total area (A )to be fenced is
A=x*y= x*(390-3/2x)=-3/2 x^2+390x
Calculating the derivative of A and setting it equals to 0 to find the maximum
A'= -3x+390=0
This yields x=130.
Therefore y=780/2-3/2*130=195
Thus, the maximum area is 130*195=25,350ft^2