Answer:
Explanation:
Since there are six points, the minimum distance from all points would be the centroid of polygon formed by A,B,C,D,E,F
To find the coordinates of centroid of a polygon we use the following formula. Let A be area of the polygon.
where i=1 to N-1 and N=6
A area of the polygon can be found by the following formula
where i=1 to N-1
![A=\frac{1}{2}[ (x_{1} y_{2} -x_{2} y_{1})+ (x_{2} y_{3} -x_{3} y_{2})+(x_{3} y_{4} -x_{4} y_{3})+(x_{4} y_{5} -x_{5} y_{4})+(x_{5} y_{6} -x_{6} y_{5})]](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%5B%20%28x_%7B1%7D%20%20y_%7B2%7D%20-x_%7B2%7D%20%20y_%7B1%7D%29%2B%20%28x_%7B2%7D%20%20y_%7B3%7D%20-x_%7B3%7D%20%20y_%7B2%7D%29%2B%28x_%7B3%7D%20%20y_%7B4%7D%20-x_%7B4%7D%20%20y_%7B3%7D%29%2B%28x_%7B4%7D%20%20y_%7B5%7D%20-x_%7B5%7D%20%20y_%7B4%7D%29%2B%28x_%7B5%7D%20%20y_%7B6%7D%20-x_%7B6%7D%20%20y_%7B5%7D%29%5D)
A=0.5[(20×25 -25×15) +(25×32 -13×25)+(13×21 -4×32)+(4×8 -18×21)+(18×14 -25×8)
A=225.5 miles²
Now putting the value of area in Cx and Cy
![C_{x} =\frac{1}{6A}[ [(x_{1}+x_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(x_{2}+x_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(x_{3}+x_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(x_{4}+x_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(x_{5}+x_{6})(x_{5} y_{6} -x_{6} y_{5})]]](https://tex.z-dn.net/?f=C_%7Bx%7D%20%3D%5Cfrac%7B1%7D%7B6A%7D%5B%20%5B%28x_%7B1%7D%2Bx_%7B2%7D%29%28x_%7B1%7D%20%20y_%7B2%7D%20-x_%7B2%7D%20%20y_%7B1%7D%29%5D%2B%20%5B%28x_%7B2%7D%2Bx_%7B3%7D%29%28x_%7B2%7D%20%20y_%7B3%7D%20-x_%7B3%7D%20%20y_%7B2%7D%29%5D%2B%5B%28x_%7B3%7D%2Bx_%7B4%7D%29%28x_%7B3%7D%20%20y_%7B4%7D%20-x_%7B4%7D%20%20y_%7B3%7D%29%5D%2B%5B%28x_%7B4%7D%2Bx_%7B5%7D%29%28x_%7B4%7D%20%20y_%7B5%7D%20-x_%7B5%7D%20%20y_%7B4%7D%29%5D%2B%5B%28x_%7B5%7D%2Bx_%7B6%7D%29%28x_%7B5%7D%20%20y_%7B6%7D%20-x_%7B6%7D%20%20y_%7B5%7D%29%5D%5D)
putting the values of x's and y's you will get
For Cy
![C_{y} =\frac{1}{6A}[ [(y_{1}+y_{2})(x_{1} y_{2} -x_{2} y_{1})]+ [(y_{2}+y_{3})(x_{2} y_{3} -x_{3} y_{2})]+[(y_{3}+y_{4})(x_{3} y_{4} -x_{4} y_{3})]+[(y_{4}+y_{5})(x_{4} y_{5} -x_{5} y_{4})]+[(y_{5}+y_{6})(x_{5} y_{6} -x_{6} y_{5})]]](https://tex.z-dn.net/?f=C_%7By%7D%20%3D%5Cfrac%7B1%7D%7B6A%7D%5B%20%5B%28y_%7B1%7D%2By_%7B2%7D%29%28x_%7B1%7D%20%20y_%7B2%7D%20-x_%7B2%7D%20%20y_%7B1%7D%29%5D%2B%20%5B%28y_%7B2%7D%2By_%7B3%7D%29%28x_%7B2%7D%20%20y_%7B3%7D%20-x_%7B3%7D%20%20y_%7B2%7D%29%5D%2B%5B%28y_%7B3%7D%2By_%7B4%7D%29%28x_%7B3%7D%20%20y_%7B4%7D%20-x_%7B4%7D%20%20y_%7B3%7D%29%5D%2B%5B%28y_%7B4%7D%2By_%7B5%7D%29%28x_%7B4%7D%20%20y_%7B5%7D%20-x_%7B5%7D%20%20y_%7B4%7D%29%5D%2B%5B%28y_%7B5%7D%2By_%7B6%7D%29%28x_%7B5%7D%20%20y_%7B6%7D%20-x_%7B6%7D%20%20y_%7B5%7D%29%5D%5D)
putting the values of x's and y's you will get
So coordinates for the fire station should be (15.36,22.55)
When a slender member is subjected to an axial compressive load, it may fail by a ... Consider a column of length, L, cross-sectional Moment of Inertia, I, having Young's Modulus, E. Both ends are pinned, meaning they can freely rotate ... p2EI L2 ... scr, is the Euler Buckling Load divided by the columns cross-sectional area
Answer:
Obviously you shouldn't rely just on the meter for your safety. You'd disconnect wall fuses or kill main switches before you start, using the meter just gives you some extra protection: with the meter you might notice for example that you've disconnected the wrong fuse and the unit is still live.
Explanation:
Hope it helps! :)
Answer:
Indicators for ineffective system engineering are as follows
1.Requirement trends
2.System definition change backlog trends
3.interface trends
4.Requirement validation trends
5.Requirement verification trends
6.Work product approval trends
7.Review action closure trends
8.Risk exposure trends
9.Risk handling trends
10.Technology maturity trends
11.Technical measurement trends
12.System engineering skills trends
13.Process compliance trends
Answer:
a) it is periodic
N = (20/3)k = 20 { for K =3}
b) it is Non-Periodic.
N = ∞
c) x(n) is periodic
N = LCM ( 5, 20 )
Explanation:
We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.
then the period of the signal is given as
N = ( 2π/w₀)K
k is least integer for which N is also integer
Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2
now
a) cos(2π(0.15)n)
w₀ = 2π(0.15)
Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3
so, it is periodic
N = (20/3)k = 20 { for K =3}
b) cos(2n);
w₀ = 2
Now, 2π/w₀ = 2π/2) = π
so, it is Non-Periodic.
N = ∞
c) cos(π0.3n) + cos(π0.4n)
x(n) = x1(n) + x2(n)
x1(n) = cos(π0.3n)
x2(n) = cos(π0.4n)
so
w₀ = π0.3
2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3
∴ N1 = 20
AND
w₀ = π0.4
2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5
∴ N² = 5
so, x(n) is periodic
N = LCM ( 5, 20 )
