Q4 q19.) use the model to determine how many grams of carbon-14 will be present in 5670 years.

Hey there!

As according to your carbon-14 model, your equation to find the amount of Carbon-14 left after a certain number of years (t) is
A=16e^-0.000121t

Because you are trying to find the amount of carbon left after 5670 years, you can plug 5670 in for the value of t since t is the amount of years.

Your equation should look like this now:
A=16e^-0.000121(5670)

Now, simplify your equation to solve for the value of A, the amount of carbon left after t years:
A=16e^-0.000121(5670)
A=16e^-0.68607
A=16(0.503551)
A≈8.0568...

Therefore, there will be about 8 grams of carbon present after 5670 years.

Hope this helps and have a nice day! :)

The volume of the box is L * W * H = 1680 
Let the width be W. But Height, H, = W + 7 and Length, L, = 2W - 2 
So we have (2W - 2) * W * (W + 7) = 1680 
So we have (2W^2 - 2W)(W + 7) = 1680 
2W^3 - 2W^2 + 14W^2 - 14W = 1680 
2W^3 + 12W^2 - 14W = 1680 
W^3 + 6W^2 - 7W = 840 
W^3 + 6W^2 - 7W - 840 = 0 
Let f(w) = W^3 + 6W^2 - 7W - 840  
The root of the polynomial occurs when f(w) = 0 
So by trial and error let w = 8 
Then we have 8^3 + 6(8)^2 - 7(8) - 840 = 0 
Hence w = 8 is a root of the equation.  
So we divide our polynomial by the root to obtain other roots, I.e W^3 + 6W^2
- 7W - 840 divide by w- 8. Doing this we have w = 15 and W = 14.
 Hence the dimensions of the box is 8 * 15 * 14