We are given with
a1 = 2
r = 4
These are components of a geometric series. The first term is 2 and the common ratio is 4. To get the first six terms, we use the formula:
an = a1 r^(n-1)
a1 = 2 (4)^(1-1) = 2
a2 = 2 (4)^(2-1) = 8
a3 = 2 (4)^(3-1) = 32
a4 = 2 (4)^(4-1) = 128
a5 = 2 (4)^(5-1) = 512
a6 = 2 (4)^(6-1) = 2048
Answer: It was the “high light” of his career
Answer:
- The smallest area the field could be is 6,400 m²
- The largest area the field could be is 8,250 m²
Step-by-step explanation:
Given;
smallest possible length of the international soccer field, L₀ = 100 m
smallest possible breadth of the international soccer field, B₀ = 64 m
Largest possible length of the international soccer field, L₁ = 110 m
Largest possible breadth of the international soccer field, B₁ = 75 m
Area of a rectangle is given by;
A = L x B
The smallest area the field could be is calculated as;
A₀ = L₀ x B₀
A₀ = 100 m x 64 m
A₀ = 6,400 m²
The largest area the field could be is calculated as;
A₁ = L₁ x B₁
A₁ = 110 m x 75 m
A₁ = 8,250 m²
Answer:
6.75
Step-by-step explanation: