Subtract 4 and 1
4-1=3
distance or length of this line segment is 3
Let the width be W, then the length is 4W (since the width is 1/4 the length)
The area of the original deck is
![W*4W=4W^{2}](https://tex.z-dn.net/?f=W%2A4W%3D4W%5E%7B2%7D%20)
The dimensions of the new deck are :
length = 4W+6
width=W+2
so the area of the new deck is :
![(4W+6)(W+2)= 4W^{2}+8W+6W+12= 4W^{2}+14W+12](https://tex.z-dn.net/?f=%284W%2B6%29%28W%2B2%29%3D%204W%5E%7B2%7D%2B8W%2B6W%2B12%3D%204W%5E%7B2%7D%2B14W%2B12)
"<span>the area of the new rectangular deck is 68 ft2 larger than the area of the original deck</span>" means that we write the equation:
![4W^{2}+14W+12=68+4W^{2}](https://tex.z-dn.net/?f=4W%5E%7B2%7D%2B14W%2B12%3D68%2B4W%5E%7B2%7D)
![14W+12=68](https://tex.z-dn.net/?f=14W%2B12%3D68)
![14W=68-12=56](https://tex.z-dn.net/?f=14W%3D68-12%3D56)
![W= \frac{56}{14}= 4](https://tex.z-dn.net/?f=W%3D%20%5Cfrac%7B56%7D%7B14%7D%3D%204%20%20)
the length is
![4W=4*4=16](https://tex.z-dn.net/?f=4W%3D4%2A4%3D16)
ft
Answer: width: 4, length: 16
D. Because it corresponds to the angle to the left
The answer is 1/(x+4)
Explanation:
You would factor out the denominator
So,
(X-4)(x+4)=x^2-16
So, x-4/(x+4)(x-4)
Then x-4 cancels each other out from the numerator and denominator
Leaving 1/x+4
1. . D . Identity , when you odd both properties it will always to be true on both sides