Answer:
∠FJH ≅ ∠BJA
Step-by-step explanation:
From the image we can see that the only congruent relation ∠FJH has is with ∠BJA which is due to these angles being vertically opposite
So, ∠FJH and ∠BJA are vertically opposite and hence, congruent. We can say that ∠FJH ≅ ∠BJA
Answer:
La persona está a 5 kilómetros con respecto al punto de partida.
Step-by-step explanation:
Considérese que la dirección norte coincide con el semieje +y y que la dirección este coincide con el semieje +x. A continuación, obtenemos las formas vectoriales equivalentes de cada afirmación:
(i) Una persona camina 7 kilómetros hacia el norte:
![\vec r_{1} = 7\,\hat{j}\,[km]](https://tex.z-dn.net/?f=%5Cvec%20r_%7B1%7D%20%3D%207%5C%2C%5Chat%7Bj%7D%5C%2C%5Bkm%5D)
(ii) Después 3 kilómetros hacia el este:
![\vec r_{2} = 3\,\hat{i}\,[km]](https://tex.z-dn.net/?f=%5Cvec%20r_%7B2%7D%20%3D%203%5C%2C%5Chat%7Bi%7D%5C%2C%5Bkm%5D)
(iii) Y luego, 3 kilómetros hacia el sur:
![\vec r_{3} = -3\,\hat{j}\,[km]](https://tex.z-dn.net/?f=%5Cvec%20r_%7B3%7D%20%3D%20-3%5C%2C%5Chat%7Bj%7D%5C%2C%5Bkm%5D)
El vector resultante de desplazamiento se construye a partir de la siguiente suma de vectores:
(1)
![\vec R = 3\,\hat{i} + 4\,\hat{j}\,[km]](https://tex.z-dn.net/?f=%5Cvec%20R%20%3D%203%5C%2C%5Chat%7Bi%7D%20%2B%204%5C%2C%5Chat%7Bj%7D%5C%2C%5Bkm%5D)
Asumiendo que la distancia coincide con el desplazamiento resultante, calculamos la distancia con respecto al punto de partida mediante el Teorema de Pitágoras:


La persona está a 5 kilómetros con respecto al punto de partida.
Answer:690 580 470 360 250 140
Step-by-step explanation:
These digits are all numbers that the hundreds digit is three less than the tens digit, the ones digit is also zero
the solid is made up of 2 regular octagons, 8 sides, joined up by 8 rectangles, one on each side towards the other octagonal face.
from the figure, we can see that the apothem is 5 for the octagons, and since each side is 3 cm long, the perimeter of one octagon is 3*8 = 24.
the standing up sides are simply rectangles of 8x3.
if we can just get the area of all those ten figures, and sum them up, that'd be the area of the solid.
![\bf \textit{area of a regular polygon}\\\\ A=\cfrac{1}{2}ap~~ \begin{cases} a=apothem\\ p=perimeter\\[-0.5em] \hrulefill\\ a=5\\ p=24 \end{cases}\implies A=\cfrac{1}{2}(5)(24)\implies \stackrel{\textit{just for one octagon}}{A=60} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{two octagon's area}}{2(60)}~~+~~\stackrel{\textit{eight rectangle's area}}{8(3\cdot 8)}\implies 120+192\implies 312](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B2%7Dap~~%20%5Cbegin%7Bcases%7D%20a%3Dapothem%5C%5C%20p%3Dperimeter%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D5%5C%5C%20p%3D24%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B2%7D%285%29%2824%29%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bjust%20for%20one%20octagon%7D%7D%7BA%3D60%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btwo%20octagon%27s%20area%7D%7D%7B2%2860%29%7D~~%2B~~%5Cstackrel%7B%5Ctextit%7Beight%20rectangle%27s%20area%7D%7D%7B8%283%5Ccdot%208%29%7D%5Cimplies%20120%2B192%5Cimplies%20312)