The common difference if there is one is the constant difference that occurs between any term and the term before it.... in this case:
There is no common difference,
dx=18,20,16,18 the difference or velocity is not constant...
d2x=2, -4,2 the acceleration is not constant...
d3x=-6,6 the thrust is not constant
Now we might be tempted to say that:
d4x=12 and say that that is constant and we COULD make a quartic equation fit all the data points, but without further data points in the sequence there is no mathematical proof that the quartic equation would produce accurate data points outside of the range given...
And solving a system of five equations for five unknowns is tedious for such a problem...a^4+bx^3+cx^2+dx+e=y
Answer:
01/6
wd
Step-by-step explanation:
Answer:
54.6cm
Step-by-step explanation:
Arc length = radius * central angle
= 18.2*3
=54.6
Part A. Your model is as good as any. It is hard to tell if the label needs to include the descriptor (length = 12 ft, for example). Certainly your model is sufficient for Part B.
Part B. The total area is the sum of the areas of each of the 6 faces of the box. Opposite faces are the same area, so you have
A = 2(LW +WD +LD) = 2(LW +D(L +W))
Subsituting the given dimensions, you get
A = 2((12 ft)(6 ft) +(3 ft)(12 ft +6 ft))
A = 2(72 ft² +(3 ft)(18 ft))
A = 2(72 ft² +54 ft²) = 252 ft²
The least amount of paper required to cover the box is 252 ft².
