Step-by-step explanation:
f ( X) = ( 2X - I ) / 2X ^ 2 - 9X + 10
= 2X-1 ) /( 2X- 5 ) ( x - 2 )
set of zeros of denominator are :
2X- 5 . = 0
X= 5/2
or
X-2=0
X=2
set of zeros of denominator = { 5/2 ,2 }
domain = R - { 5/2,2 }
Answer:
r ≤ -5 and r ≥ 1
Step-by-step explanation:
The solution has two parts:
1) that derived from | 2 +r | ≥ 3 when (2 + r) is already positive. Then:
2 + r ≥ 3, or r ≥ 1
and
that derived from | 2 +r | ≥ 3 when (2 + r) is negative. If (2 + r) is negative, then
|(2 + r)| = -(2 + r) = -2 -r, which is ≥ 3. Therefore, -2 -r ≥ 3, or -r ≥ 5. To solve this for r, divide both sides by -1 and reverse the direction of the inequality sign: r ≤ -5
Answer:
2y -8
Step-by-step explanation:
−2y−8+4y
Combine like terms
-2y +4y -8
2y -8
Answer:
Step-by-step explanation:
- 8
- 6
- 4
- 1