Question 1:
For this case we have that the function [tex]f (x) = \frac {5x} {x ^ 2-25}[/tex] is undefined or discontinuous where the denominator equals 0.
[tex]x ^ 2-25 = 0\\x ^ 2 = 25\\x = \pm \sqrt {25}\\x_ {1} = + 5\\x_ {2} = - 5[/tex]
Thus, the function is undefined or discontinuous at +5 and -5.
To find the zeros of the function we match the function to zero and clear "x":
[tex]\frac {5x} {x ^ 2-25} = 0[/tex]
Factoring the denominator, taking into account that the roots are -5 and +5:
[tex]\frac {5x} {(x + 5) (x-5)} = 0[/tex]
We multiply by[tex](x + 5) (x-5)[/tex]on both sides of the equation:
[tex]5x = 0\\x = 0[/tex]
ANswer:
Discontinuity: + 5, -5
Zero: x = 0
Question 2:
For this case we propose a system of equations:
x: Be the variable that represents the yellow fish
y: Be the variable that represents the green fish
[tex]x = y-6\\x = 0.4 (x + y)[/tex]
We manipulate the second equation:
[tex]x = 0.4x + 0.4y\\x-0.4x = 0.4y\\0.6x = 0.4y\\y = \frac {0.6} {0.4} x\\y = 1.5x[/tex]
We substitute in the first equation:
[tex]x = y-6\\x = 1.5x-6\\x-1.5x = -6\\-0.5x = -6\\x = \frac {-6} {- 0.5}\\x = 12[/tex]
So, we have 12 yellow fish in the aquarium.
[tex]y = 1.5 * 12\\y = 18[/tex]
So, we have 18 green fish.
Answer:
12 yellow fish
18 green fish