A una granja llegaron 54 ovejas 24 vacas 30 cerdos los cuales se quieren repartir en corrales con igual numero de animales de forma que haya la mayor cantidad posible en cada corral ¿cuantos animales posibles deben ir en cada uno de los corrales?

6 animals should go in each pen.

Step-by-step explanation:

Total sheep = 54

Total cows = 24

Total pigs = 30

Highest number of animals are possible in each pen such that animals are distributed in pens with the same number = Greatest common divisior (54,24, 30)

54= 6 x 9

24= 6 x 4

30 = 6 x 5

So, Greatest common divisior (54,24, 30) = 6

Hence, 6 animals should go in each pen.

the f

last graph i believe

step-by-step explanation:

worked it out i believe

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!