Factor using the AC method...
The answer is:
(x+6) (x+8)
Answer: 4 years
Step-by-step explanation:
A(0) has to be amount at start. Assume that's 5mg
Then A(t) = 5×(0.5)^(0.25t) = 5×2^(-t/4),
(also known as 5 exp(-λ t) with λ = ln(2)/4, incidentally).
We need to such that A(t) = 2.5mg, or 2^(-t/4) is 1/2, which happens when -t/4 is -1, or t is 4.
Answer:
0.0032
Step-by-step explanation:
We need to compute
by the help of third-degree Taylor polynomial that is expanded around at x = 0.
Given :
< e < 3
Therefore, the Taylor's Error Bound formula is given by :
, where ![$M=|F^{N+1}(x)|$](https://tex.z-dn.net/?f=%24M%3D%7CF%5E%7BN%2B1%7D%28x%29%7C%24)
![$\leq \frac{3}{(3+1)!} |-0.4|^4$](https://tex.z-dn.net/?f=%24%5Cleq%20%5Cfrac%7B3%7D%7B%283%2B1%29%21%7D%20%7C-0.4%7C%5E4%24)
![$\leq \frac{3}{24} \times (0.4)^4$](https://tex.z-dn.net/?f=%24%5Cleq%20%5Cfrac%7B3%7D%7B24%7D%20%5Ctimes%20%280.4%29%5E4%24)
![$\leq 0.0032$](https://tex.z-dn.net/?f=%24%5Cleq%200.0032%24)
Therefore, |Error| ≤ 0.0032
I did that ima go fine the papers
Answer:
D. 10r7
Explaination:
9 goes into 97 10 times, and has a remainder of 7