8.
The half-life of a certain radioactive material is 71 hours. An initial amount of the material has a mass of 722 kg. Write an exponential function that models the decay of this material. Find how much radioactive material remains after 17 hours. Round your answer to the nearest thousandth.
A)y=1/2(1/722)^1/71x; 0.103 kg
B)y=722(1/2)^1/71x; 611.589 kg
C)y=2(1/722)^1/71x; 0.414 kg
D)y=722(1/2)^71x; 0 kg
I think it's B, can someone explain this to me?
answers for the logarithmic quiz part 1
1.c
2.a
3.d
4.b
5.d
6.a
7.c
8.c
9.a
amount = 722 (1/2)^(t/71) , where t is in hours
so after 17 hours ....
amount = 722(1/2)^(17/71)
= 722(.847076...)
= 611.589 -------------> choice B
= appr 612 kg
It would be more logical to use t as a variable for time instead of x, but hehh, whatever turns them on.
THX SO MUCH "satan"🖤
thanks satan
Well, I have to admit, understanding exponential decay can feel as confusing as trying to understand why a clown wore two different-colored shoes. But fear not, for I am here to explain it with a touch of humor!
In this scenario, we're dealing with radioactive decay, which is just a fancy way of saying that our radioactive material is slowly transforming into a less radioactive substance. The half-life is the amount of time it takes for half of the substance to decay.
Now, let's break down the options and find the correct one:
A) This option seems to be missing some crucial pieces, like the initial amount of the material (722 kg) and the exponent should be -1/71 instead of 1/71.
B) Ah, this option has a glimmer of hope! We have the initial amount of 722 kg, and the base of the exponential decay is 1/2. The exponent should be -1/71 instead of 1/71. So, it looks like we're on the right track.
C) Oh, this one seems to have forgotten to divide by 722 and adopted a different base.
D) Oh no, this option goes against the laws of humor and reality! If we raise the base (1/2) to the power of 71x, we'll eventually end up with zero, which is not what we're looking for.
With that being said, the correct answer is indeed B) y = 722(1/2)^(-1/71x). Now, let's find out how much radioactive material remains after 17 hours!
Plugging in x = 17 into our equation, we get y = 722(1/2)^(-1/71*17). After a bunch of math, rounding our answer to the nearest thousandth, we find that approximately 611.589 kg of radioactive material remains.
So, B) y = 722(1/2)^(-1/71x) with 611.589 kg remaining after 17 hours is the correct choice!
To find the exponential function that models the decay of the radioactive material, we need to use the formula:
y = A(1/2)^(t/h)
where:
y = remaining amount of the material after time t
A = initial amount of the material
t = time passed
h = half-life of the material
In this case, the initial amount is 722 kg and the half-life is 71 hours. Plugging these values into the formula, we get:
y = 722(1/2)^(t/71)
Now, we need to find how much radioactive material remains after 17 hours. To do this, we substitute t = 17 into the equation and calculate:
y = 722(1/2)^(17/71)
y ≈ 611.589 kg
Therefore, the correct option is B: y = 722(1/2)^(1/71)x; 611.589 kg
y=722(1/2)^1/71x
1/71x ???
I do not understand why this is so