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)
Answer:
A bona fide occupational qualification defense
Explanation:
Since the store is for women clothing, the retail may prefer to employ only female to assist the customers. Under a bona fide occupational qualification defense, an employer is allowed to discriminate if a characteristic is a necessity for the performance of the job and for the business. Therefore, the store has a bona fide occupational qualification defense.
Answer:
Explanation:
R1 = 1 k; R2 = 1k; R3 = 3k; R4 = 6k; R5 = 5k; R6 = 6k and R7 = 2k
Vt = 428V
The series and parallel circuit combination is as follow:
(R6║R7 + R5) + R4 + R3║R2 + R1
(6*2/6 + 2) + 5 = 13/ k
(13/2*6/13/6 + 6) = 78/31k
78/3 + 3 = 171/3 = 57k
57k║R2 = 57k║1k = 57/58k + R1 = (57/58 + 1)k = 115/58k = 2k
It = Vt/2k = 428/2000 = 0.2A
∴ I1 = 0.2A
I1 = I2 + I3
Using current divider rules to obtain I2 and I3
∴ I2 = I1 X (I2/I2 + I3) = 0.2 X ( 1/4) = 0.05A
and I3 = I1 X (I3/I2 + I3) = 0.2 X (3/4) = 0.15A
I3 = I4 + I5, using current divider
I4 = I3 X (I4/I4 + I5) = 0.15 X (6/6 + 5) = 0.08
I5 = 0.15 - 0.08 = 0.07A
I5 = I6 + I7, using current divider
I6 = I5 X (I6/I6 + I7) = 0.07 X (6/6 + 2) = 0.05A
I7 = 0.07 - 0.05 = 0.02A
Acceleration of Car = 10 ms⁻²
Explanation:
Step 1:
The basic formula of acceleration is a = (v-u)/t ms⁻²
where, v- final velocity
u- initial velocity
t= time taken
Step 2:
Here v = 70 ms⁻¹
u = 50 ms⁻¹
t = 5 s
∴ a = ( 70 - 20)/5
a = 10 ms⁻²