we are given

we can check each options
option-A:
-1,1
we can plug x=-1 and x=1 and check whethet f(x)=0
At x=-1:


At x=1:


so, this is TRUE
option-B:
0,1
we can plug x=0 and x=1 and check whethet f(x)=0
At x=0:


At x=1:


so, this is FALSE
option-C:
-2,-1
we can plug x=-2 and x=-1 and check whethet f(x)=0
At x=-2:


At x=-1:


so, this is FALSE
option-D:
-1,0
we can plug x=-1 and x=0 and check whethet f(x)=0
At x=-1:


At x=0:


so, this is FALSE
Three added to the product of four and twelve
or
The product of four and twelve added by three
Answer:
The Answer is A, a line segment that has both endpoints on a circle.
Step-by-step explanation:
Answer: 2 meters.
Step-by-step explanation:
Let w = width of the cement path.
Dimensions of pool : Length = 15 meters , width = 9 meters
Area of pool = length x width = 15 x 9 = 135 square meters
Along width cement path, the length of region = 
width = 
Area of road with pool = 

Area of road = (Area of road with pool ) -(area of pool)
![\Rightarrow\ 112 =4w^2+48w+135- 135\\\\\Rightarrow\ 112= 4w^2+48w\\\\\Rightarrow\ 4 w^2+48w-112=0\\\\\Rightarrow\ w^2+12w-28=0\ \ \ [\text{Divide both sides by 4}]\\\\\Rightarrow\ w^2+14w-2w-28=0\\\\\Rightarrow\ w(w+14)-2(w+14)=0\\\\\Rightarrow\ (w+14)(w-2)=0\\\\\Rightarrow\ w=-14\ or \ w=2](https://tex.z-dn.net/?f=%5CRightarrow%5C%20112%20%3D4w%5E2%2B48w%2B135-%20135%5C%5C%5C%5C%5CRightarrow%5C%20112%3D%204w%5E2%2B48w%5C%5C%5C%5C%5CRightarrow%5C%204%20w%5E2%2B48w-112%3D0%5C%5C%5C%5C%5CRightarrow%5C%20w%5E2%2B12w-28%3D0%5C%20%5C%20%5C%20%5B%5Ctext%7BDivide%20both%20sides%20by%204%7D%5D%5C%5C%5C%5C%5CRightarrow%5C%20w%5E2%2B14w-2w-28%3D0%5C%5C%5C%5C%5CRightarrow%5C%20w%28w%2B14%29-2%28w%2B14%29%3D0%5C%5C%5C%5C%5CRightarrow%5C%20%28w%2B14%29%28w-2%29%3D0%5C%5C%5C%5C%5CRightarrow%5C%20%20w%3D-14%5C%20or%20%5C%20w%3D2)
width cannot be negative, so w=2 meters
Hence, the width of the road = 2 meters.
0.74=74/100=37/50 in simplest form.