Tienes que hallar el mínimo común múltiplo de las 3 cantidades.
18= 2×3²
24= 2³×3
36= 2²×3²
mcm(18,24,36) = 2³×3²=8×9= 72
Eso quier decir que si partieron a la misma hora se encontraran de nuevo en el punto de partida 72 minutos después de la salida.
Las vueltas que habrán realizado será el resultado de dividir 72 entre el tiempo que tardan en dar una vuelta
<span>Mayor: </span> = 4
<span>Mediano: </span> = 3
<span>Pequeño: </span> = 2
Soluciónes:
se vuelven a encontrar a los 72 min de la salida
<span>El mayor dió 4 vueltas, el mediano 3 y el pequeño 2</span>
Answer:
Step-by-step explanation:
The question is incomplete, as the angles of rotation are not stated.
However, I will list the angles less than 360 degrees that will carry the hexagon and the nonagon onto itself
We have:
Divide 360 degrees by the number of sides in each angle, then find the multiples.
<u>Nonagon</u>
List the multiples of 40
<u>Hexagon</u>
List the multiples of 60
List out the common angles
This means that, only a rotation of 
will lift both shapes onto themselves, when applied to both shapes.
The other angles will only work on one of the shapes, but not both at the same time.
So what we do is
area that remains=total area-triangle area that was cut out
we need to find 2 things
total area
triangle area
total area=rectange=base times height
area=(3x+4) times (2x+3)
FOIL or distribute
6x^2+8x+9x+12=6x^2+17x+12
triangle area=1/2 times base times height
triangle area=1/2 times (2x+2) times (x-2)=
(x+2) times (x-2)=x^2+2x-2x-4=x^2-4
so
total area=6x^2+17x+12
triangle area=x^2-4
subtract
area that remains=total area-triangle area that was cut out
area that remains=6x^2+17x+12-(x^2-4)=
6x^2+17x+12-x^2+4=
6x^2-x^2+17x+12+4=
5x^2+17x+16
area that remains is 5x^2+17x+16
