Hello!
The discriminant of quadratic functions is: b² - 4ac. Since the equation is in standard form, which is Ax² + Bx + C = 0 , we can substitute those values into our discriminant and simplify.
The value of the discriminant will tell us how many solutions there are to the given quadratic equation.
A positive discriminant will have two real solutions.
A discriminant of zero will have one real solution.
A negative discriminant will no real solutions.
1. Substitute, a = 16, b = 8, c = 1.
8² - 4(16)(1)
64 - 4(16)(1)
64 - 64(1)
64 - 64
0
Since the discriminant is zero, the answer is choice A, double root, because since it is raised to the power of 2, it must has two roots, but in this case, both of the roots the same x-values.
Answer:
D) - 11
Step-by-step explanation:

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)