Hello,
Let's z=0.1234567891011121314151617181920.....(never end)
2/5=(2/(5z)*z<3241/1000 *z
3/4=(3/(4z))*z>6075/1000*z
So here is 2836 irrational numbers:
3241*z/1000
3242*z/1000
3243*z/1000
...
6073*z/1000
6074*z/1000
6075*z/1000
Your file attachment isnt working so i cant help sorry.
Answer:
(f+g) (x) =0 for x=-2
Step-by-step explanation:
f(x) = x^2 – 2x, g(x) = 6x + 4,
(f+g) (x) =f(x) +g(x) =
(x^2 – 2x) + (6x + 4) =
x^2 – 2x+ 6x + 4=
x^2 +4x+4
(f+g)(x) =x^2 +4x+4=0 (quvadratic equation)
x^2 +4x+4=x^2 +2x*2+2^2=(x+2)^2
then
(x+2)^2 =0
x+2=0
x=-2
(f+g) (x) =0 for x=-2
