Let
![L](https://tex.z-dn.net/?f=L)
be the length of the rectangle and
![W](https://tex.z-dn.net/?f=W)
be the width. In the problem it is given that
![L=2W-3](https://tex.z-dn.net/?f=L%3D2W-3)
. It is also given that the area
![LW=27](https://tex.z-dn.net/?f=LW%3D27)
. Substituting in the length in terms of width, we have
![W(2W-3)=27 \\ 2W^2-3W-27=0 \\ (2W-9)(W+3)=0](https://tex.z-dn.net/?f=W%282W-3%29%3D27%20%5C%5C%202W%5E2-3W-27%3D0%20%5C%5C%20%282W-9%29%28W%2B3%29%3D0)
. Using the zero product property,
![2W-9=0 \text{ or } W+3=0](https://tex.z-dn.net/?f=2W-9%3D0%20%5Ctext%7B%20or%20%7D%20W%2B3%3D0)
. Solving these we get the width
![W=4.5 \text{ or } -3](https://tex.z-dn.net/?f=W%3D4.5%20%5Ctext%7B%20or%20%7D%20-3)
. However, it doesn't make sense for the width to be negative, so the width must be
![\boxed{4.5 \text{ m}}](https://tex.z-dn.net/?f=%5Cboxed%7B4.5%20%5Ctext%7B%20m%7D%7D)
. From that we can tell the length
![L=2(4.5)-3=\boxed{6 \text{ m}}](https://tex.z-dn.net/?f=L%3D2%284.5%29-3%3D%5Cboxed%7B6%20%5Ctext%7B%20m%7D%7D)
.
A point that bisects a segment would be its midpoint. This is a case where the vocabulary (bisect as opposed to midpoint) makes it harder.
To find the midpoint, we use the midpoint formula. The midpoint formula is:
midpoint = (x₁ + x₂/2, y₁ + y/2). To find it, you add the x coordinates and then divide them by 2. Repeat for the y coordinates.
x: (-2 + 6)/2 = 4/2 = 2
y: (5 +1)/2 = 6/2 = 3
Thus the point B bisecting AC is at (2, 3).
Answer:
yes
Step-by-step explanation:
8456590
Yes..........................
Answer:
x =-1 y=4
Step-by-step explanation:
-1 ×4 is -4. 2×4 is 8
8 +-4 =4