elimanation variations matrix
Answer:
The domain and the range of the function are, respectively:
![Dom\{f\} = [0\,m,5\,m]](https://tex.z-dn.net/?f=Dom%5C%7Bf%5C%7D%20%3D%20%5B0%5C%2Cm%2C5%5C%2Cm%5D)
![Ran\{f\} = [0\,m^{2}, 10\,m^{2}]](https://tex.z-dn.net/?f=Ran%5C%7Bf%5C%7D%20%3D%20%5B0%5C%2Cm%5E%7B2%7D%2C%2010%5C%2Cm%5E%7B2%7D%5D)
Step-by-step explanation:
Jina represented a function by a graphic approach, where the length, measured in meters, is the domain of the function, whereas the area, measured in square meters, is its range.
![Dom\{f\} = [0\,m,5\,m]](https://tex.z-dn.net/?f=Dom%5C%7Bf%5C%7D%20%3D%20%5B0%5C%2Cm%2C5%5C%2Cm%5D)
![Ran\{f\} = [0\,m^{2}, 10\,m^{2}]](https://tex.z-dn.net/?f=Ran%5C%7Bf%5C%7D%20%3D%20%5B0%5C%2Cm%5E%7B2%7D%2C%2010%5C%2Cm%5E%7B2%7D%5D)
C.) 27 and a half. 12 minutes times 5 equals an hour so 5 and a half pages times 5 equals 27 and a half pages per hour.
Answer:
Container B has smaller surface area.
Step-by-step explanation:
Given:
Container A
Radius = 60/2 = 30 mm
Height = 4 x 60 = 240 mm
Container B
Length = 120
Width = 120
Height = 60
Computation:
Surface area of container A (Cylinder) = 2πr[h+r]
Surface area of container A (Cylinder) = 2[22/7][60][120+60]
Surface area of container A (Cylinder) = 67,885.70 mm² (Approx)
Surface area of container B (Cuboid) = 2[lb+bh+hl]
Surface area of container B (Cuboid) = 2[(14,400)+(7,200)+(7,200)]
Surface area of container B (Cuboid) = 57,600 mm²
Container B has smaller surface area.