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Answer:
The side length of the large square is √2 times larger than the side length of the small square.
Step-by-step explanation:
Suppose we have a small square (square 1) and a large square (square 2). The area of the large square is twice that of the small square, that is,
A₂ = 2 A₁
A₂/A₁ = 2 [1]
The area of a square is equal to the length of the side (l) raised to the second power.
A = l²
l = √A
The ratio of l₂ to l₁ is:
l₂/l₁ = √A₂ / √A₁ = √(A₂/A₁)
We can replace [1] in the previous expression.
l₂/l₁ = √2
The side length of the large square is √2 times larger than the side length of the small square.
Answer:
f(x) = 500 • 1.1x
Step-by-step explanation:
To solve the question, we need to understand that:
- The population of ants is going to increase with time as they mate and have a population positive growth rate.
- The initial number of ants is 500.
- The rate at which it increases will be multiplied with x months
We will work our way out by removing the wrong answers first
Since,
The population cannot be decreased so we can neglect the last two options involving 0.9 as it will result in a decrease in the population.
Now we are left with two options:
a) f(x) = 500 • 1.1x
b) f(x) = 1.1 • 500x
Since in the option (b) , 500 is multiplied with x , it is not possible as 500 is a constant initial amount and not a factor of growth of population. The growth factor is supposed to be multiple of x months.
So, the correct answer is f(x) = 500 • 1.1x
Answer:
6^ (1/12)
Step-by-step explanation:
6 ^ 1/3
-------------
6 ^ 1/4
x^a / x^b = x^ (a-b)
6^(1/3-1/4)
Getting a common denominator)
6^(4/12-3/12)
6^1/12
