Answer: last option/D
Step-by-step explanation:
It says to do h(t) divided by g(t) so
1- you see that h(t) is 4t +2
2- you see that g(t) is 2t-1
3- you put h(t) on top of g(t)
So, it is 4t+2/2t-1
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
6.05
Step-by-step explanation:
r=squareroot a/π= square root 115/π = 6.05
Answer:
96
Step-by-step explanation:
Basically you're multiplying how many coins you have that year by eight.
The fist year you had 4 coins. 4 multiplied by 8 equals 32. Mrs Bradford had 32 coins.
The second year you had 6 coins. 6 multiplied by 8 equals 48. Mrs Bradford had 48 coins.
The third year you had 8 coins. 8 multiplied by 8 equals 64. Mrs Bradford had 64 coins
For the fourth and fifth year you have to add 2 for each year.
So for the 4th year you had 10 coins and 10 multiplied by 8 equals 80. Mr.Bradford had 80 coins.
And Finally for year five you had 12 coins. 12 multiplied by 8 equals 96. Mr.Bradford had 96 coins in year five.