Answer:
a:45
b:8
might be wrong. hope it helps
We start with
![\dfrac{4x^3+5x^2+3x}{x}](https://tex.z-dn.net/?f=%5Cdfrac%7B4x%5E3%2B5x%5E2%2B3x%7D%7Bx%7D)
We can factor x at the numerator:
![\dfrac{x(4x^2+5x+3)}{x}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%284x%5E2%2B5x%2B3%29%7D%7Bx%7D)
So, assuming
(otherwise the expression would make no sense) we can simplify it:
![\dfrac{x(4x^2+5x+3)}{x} = 4x^2+5x+3](https://tex.z-dn.net/?f=%5Cdfrac%7Bx%284x%5E2%2B5x%2B3%29%7D%7Bx%7D%20%3D%204x%5E2%2B5x%2B3)
This equation has no roots, so we can't simplify it any further.
B wpuld be (-1,3), H would be (-4,4), and D would be (-4,-1). Hope this helps and please give Brainliest!
8x - 3 = -19
8x = - 19 + 3
8x = -16
x = -16/8
x = -2
answer: the number is negative 2
lx-1 l ≤ 13
There are 2 solutions:
x-1 ≤ 13
and
x-1 ≥ -13
Solve each one
x-1 ≤ 13
add 1 to both sides of the equation
x-1+1≤ 13+1
x ≤ 14
x-1 ≥ -13
Add 1 to both sides:
x-1+1 ≥-13+1
x ≥ -12
-12 ≤ x ≤ 14