Answer:
procedure always produces 6
Step-by-step explanation:
Let 'n' be the unknown number
Add 4 to the number : n+4
multiply the sum by 3.
multiply the sum n+4 by 3
Now subtract 6, so we subtract 6 from 3n+12
finally decrease the difference by the tripe of the original number
triple of original number is 3n
so the procedure always produces 6
<span>The number of x-intercepts that appear on the graph of the function
</span>f(x)=(x-6)^2(x+2)^2 is two (2): x=6 (multiplicity 2) and x=-2 (multiplicity 2)
Solution
x-intercepts:
f(x)=0→(x-6)^2 (x+2)^2 =0
Using that: If a . b =0→a=0 or b=0; with a=(x-6)^2 and b=(x+2)^2
(x-6)^2=0
Solving for x. Square root both sides of the equation:
sqrt[ (x-6)^2] = sqrt(0)→x-6=0
Adding 6 both sides of the equation:
x-6+6=0+6→x=6 Multiplicity 2
(x+2)^2=0
Solving for x. Square root both sides of the equation:
sqrt[ (x+2)^2] = sqrt(0)→x+2=0
Subtracting 2 both sides of the equation:
x+2-2=0-2→x=-2 Multiplicity 2
Answer:
The function of f(x) is 20 for the middle problem idk the others
Step-by-step explanation:
With what? Maybe I can help
