|r+3| = (r+3) when r ≥ -3 so for r ≥ -3
r + 3 ≥ 7
r ≥ 4
|r+3| = -(r+3) when r < -3 so for r < -3
-(r + 3) ≥ 7
r + 3 ≤ -7
r ≤ -10
The answer to this question is r ≤ -10, r ≥ 4
Answer:
2x for the first term
The whole thing is 2x+3
Step-by-step explanation:
Think of it like normal division
What will get the closest answer to 2x^2+x?
2x*x-1=2x^2-2x
Now the expression is 3x-3 so if you but 3 than it becomes 3x-3 which would be a remainder of 0
Answer:
The answer is 97
Step-by-step explanation:
According to PEMDAS, we will solve the multiplication first. 2 × 42 is 84. Next, we will add or subtract from left to right. 9 + 84 + 6 – 2 is 97, the answer to our problem.
