Working attached below. Hope this helps! Any questions let me know :)
Answer:
180 cm^2
Step-by-step explanation:
For similar figures the ratio of areas is the square of the ratio of linear dimensions.
We take "tripled in size" to mean that the linear dimensions of the larger rectangle are 3 times those of the smaller rectangle. That means the area of the larger rectangle will be 3² = 9 times the area of the smaller one.
Area = (3²)(20 cm²) = 180 cm² . . . . area after tripled in size
Accelerating is not an option in executing a decision.
This is false.
Accelerating or driving in high speed does not specify any execution of any sort of action or decision. Rather it can depend on the course of driving like changing lanes or something related to emergency.
Answer:
% Remaining![= [1-(1/2)^{\frac{t}{2.6}}]x100](https://tex.z-dn.net/?f=%20%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7Bt%7D%7B2.6%7D%7D%5Dx100%20)
And replacing the value t =5.5 hours we got:
% Remaining![= [1-(1/2)^{\frac{5.5}{2.6}}]x100 =76.922\%](https://tex.z-dn.net/?f=%20%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7B5.5%7D%7B2.6%7D%7D%5Dx100%20%3D76.922%5C%25)
Step-by-step explanation:
Previous concepts
The half-life is defined "as the amount of time it takes a given quantity to decrease to half of its initial value. The term is most commonly used in relation to atoms undergoing radioactive decay, but can be used to describe other types of decay, whether exponential or not".
Solution to the problem
The half life model is given by the following expression:

Where A(t) represent the amount after t hours.
represent the initial amount
t the number of hours
h=2.6 hours the half life
And we want to estimate the % after 5.5 hours. On this case we can begin finding the amount after 5.5 hours like this:

Now in order to find the percentage relative to the initial amount w can use the definition of relative change like this:
% Remaining = 
We can take common factor
and we got:
% Remaining![= [1-(1/2)^{\frac{t}{2.6}}]x100](https://tex.z-dn.net/?f=%20%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7Bt%7D%7B2.6%7D%7D%5Dx100%20)
And replacing the value t =5.5 hours we got:
% Remaining ![= [1-(1/2)^{\frac{5.5}{2.6}}]x100 =76.922\%](https://tex.z-dn.net/?f=%3D%20%5B1-%281%2F2%29%5E%7B%5Cfrac%7B5.5%7D%7B2.6%7D%7D%5Dx100%20%3D76.922%5C%25)