1) Let's write out both expressions subtracting 4m²+2mn+8n² from 2m²+6mn+2n²
2) Note that when we subtract 4m^2 + 2mn + 8n^2 from 2m^2 + 6mn + 2n^2 we need to swap the sign by placing -1 outside the parentheses and then combine like terms adding those terms algebraically.
The least number of quadratic system can have is Two real solutions or no real solutions.
Answer:
I think the answers is B
Step-by-step explanation:
But im so sorry if its not right...
<em>H</em><em>a</em><em>v</em><em>e</em><em> </em><em>a</em><em> </em><em>n</em><em>i</em><em>c</em><em>e</em><em> </em><em>d</em><em>a</em><em>y</em>
Based on the perimeter solved, the length and width of the rectangle will be 11.34 feet and 4.67 feet respectively.
<h3>How to solve the perimeter</h3>
Width = w
Length = 2w + 2
Perimeter = 2(Length + Width)
Perimeter = 2(Length + Width)
32 = 2(2w + 2 + w)
32 = 4w + 4 + 2w
32 - 4 = 6w
28 = 6w
w = 28/6 = 4.67
Length = 2w + 2 = (2 × 4.67) + 2 = 11.34
Learn more about perimeter on:
brainly.com/question/19819849
If the question is to answer m2, than m2≈33.34 roughly
