Respuesta :
Answer:
Following are the solution to the given question:
Step-by-step explanation:
In question 1:
[tex]\to \frac{x^2+5x+6}{x+2}\\\\\to \frac{x^2+(3+2)x+6}{x+2}\\\\\to \frac{x^2+3x+2x+6}{x+2}\\\\\to \frac{x(x+3)+2(x+3)}{x+2}\\\\\to \frac{(x+3) (x+2)}{x+2}\\\\\to (x+3)[/tex]
In question 2:
[tex]\to \frac{2x+8}{x^2+5x+6}\\\\[/tex]
[tex]\to x^2+5x+6=0\\\\\to (x+3)(x+2)=0\\\\\to x+3 =0 \ \ \ \ \ \ x+2 =0\\\\\to x = -3 \ \ \ \ \ \ x=-2 \\\\[/tex]
In question 3:
[tex]\to f(x) = \frac{x-3}{x+5}\\\\\to f(x) \neq 0\\\\when\\\to x+5=0\\\\\to x=-4 \ excluded \ value \\\\\to x \neq 5[/tex]
In question 4:
[tex]\to f(x) = \frac{x^2+6x+8}{(x+4)}\\\\\to f(x) = \frac{x^2+(4+2)x+8}{(x+4)}[/tex]
[tex]\to f(x) = \frac{x^2+4x+2x+8}{(x+4)}\\\\\to f(x) = \frac{x(x+4)+2(x+4)}{(x+4)}\\\\\to f(x) = \frac{(x+4)(x+2)}{(x+4)}\\\\\to f(x) = x= -4 and -2[/tex]
In question 5:
[tex]\to f(x) =\frac{3x}{x^2-9} \\\\ \to f(x) =\frac{3x}{x^2-3^2} \\\\ \to f(x) =\frac{3x}{(x+3)(x-3)}[/tex]
In question 6:
total number of fish = x
yellow fish [tex]= \frac{40}{100}=0.4X[/tex]
green fish [tex]= \frac{60}{100}) =0.6X[/tex]
It is now the case that the yellow fish number is 6 less than the green fish number.
[tex]\therefore \\\\\to 0.6x-0.4x = 6 \\\\\to 0.2x = 6\\\\\to x= \frac{6}{0.2} \\\\\to x=30[/tex]
green fish [tex]= 0.6 \times 30=18[/tex]
