What is the solution, if any, to the inequality 3-14-n1 > 1? no solution
O all real numbers
On>2 or n < 6
O 2< n < 6

n > 2 or n < 6

Step-by-step explanation:

3 - | 4 - n | < 1

Let's move 3 to the other side of the equation:

- | 4 - n | < 1 - 3

- | 4 - n | < -2

Both sides of the equation are negative, so we can simplify by multiplying them with -1:

(- | 4 - n | < -2*-1

| 4 - n | < 2

We cannot get rid of the module because it's not for a single number and we dont know what n is equal to but there is a rule that we can use to remove the module:

- if | x | < a and a > 0 then -a < x < a. So if we use it we get the following:

- 2 < 4 - n < 2

These are two equations, separate and solve them separately

4 - n < 2

- n < 2 - 4

- n < -2

n > 2 (< turns into > because we changed sign)

and

- 2 < 4 - n

4 - n > -2

-n > -2 - 4

n < 6 (we change the sign of the numbers so the equation changes too)

So after solving the two equation we get the answers

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!