Answer:
work is shown and pictured
If the length of a rectangle is 3m longer than its width, then:
L=W+3
Is the area really 154^2? Or is it 154m^2? If yes, then:
A=LW
154=(W+3)(W)
154=(W^2+3W)
0=W^2+3W-154
0=(W-11)(W+14)
This means either (W-11) or (W+14) is equal to zero so:
W=11 and W=-14
To find out let's substitute the numbers:
154=(11+3)(11)
154=154
Therefore, the width of the rectangle is 11m
Answer:
![\boxed{\boxed{\sf 23}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7B%5Csf%2023%7D%7D)
Step-by-step explanation:
![\boxed{\sf Hi\: there!}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csf%20Hi%5C%3A%20there%21%7D)
![\sf (w-x)^2 + 26x](https://tex.z-dn.net/?f=%5Csf%20%28w-x%29%5E2%20%2B%2026x)
![\sf w=6](https://tex.z-dn.net/?f=%5Csf%20w%3D6)
![\sf x=-1](https://tex.z-dn.net/?f=%5Csf%20x%3D-1)
_____________________
→ ![\sf \left(6-\left(-1\right)\right)^2+26\left(-1\right)](https://tex.z-dn.net/?f=%5Csf%20%5Cleft%286-%5Cleft%28-1%5Cright%29%5Cright%29%5E2%2B26%5Cleft%28-1%5Cright%29)
→ ![\sf 23](https://tex.z-dn.net/?f=%5Csf%2023)
»»————- ➴ ————-««
Answer:
2
Step-by-step explanation:
hope this helps