Let us denote the semi arcs as congruent angles. This means that angles FEJ and EFJ are congruent (That is, they have the same measure). Since angles FEJ and EFJ have the same measure, this implies that sides EJ and FJ are equal. Since angles EJK and FJH are supplementary angles to angle EJF, this implies that EJK and FJH have the same measure.
Using the Angle Side Angle (SAS) criteria, we determine that triangles EKJ and triangle FJH are congruent. This implies that sides EK and FH are equal and that angles EKJ and FHJ are congruent. Note that angle EKJ is the same as EKF and that FHJ is the same as FHE.
Once again, since angles EKF and FHJ are congruent, and angle EKD is supplementary to the angle EKJ when angle FHG is supplementary to angle FHJ, then we have that angles EKD and angle FHG are congruent.
Using again the SAS criteria, we determine that triangles EKD and FHG are congruent.
From this reasoning, we have proved the following facts:
Triangle DEK is congruent to triangl GFH
Angle EKF is congruent to angle FHE
Segment EK is the same as segment FH
Answer:
Step-by-step explanation:
Multiply numerator and denominator by √2
Answer: HEEEEEEYYYYYYYYYYYYYYYYYYYYYYYYYY!!!!!!!!!!!!!!!!!
Step-by-step explanation:
I order to solve this you have to find out how how much root beer there is to the total amount of candy. 12/27. Then you find out what the percentage of root beer there is by dividing 12 by 27. It’ll give you a decimal point. Percentage has a maximum of 100%. And you’ll find out what percent based on this decimal. Factor it out of 1. The percentage you get is .44. By factoring out of 1 you can find out that the percentage is 44%. So the probability of finding a root beer out of all the candy is roughly 44%
Responder:
Contenedor 24
Explicación paso a paso:
Para obtener la cantidad de contenedores que deberán llenar para completar sus respectivos paquetes al mismo tiempo; obtener el mínimo común múltiplo de la agrupación adoptada por el primer y segundo trabajador;
Múltiplos de:
8: 8, 16, 24, 32, 40, 48, 56, ...
12:12, 24, 26, 48, 60, ....
Por lo tanto, el mínimo común múltiplo de 8 y 12 es 24.
Llenarán 24 contenedores
