<span>Ending Amt = Bgng Amt * e ^-0.03t
In this equation, the "-0.03" is the decay factor or "k"
We can now solve for half-life by this equation:
</span>t = <span>(<span>ln [y(t) ÷ a]<span>)<span> ÷ -k (we can say beginning amount = 200 and ending amount = 100
</span></span></span></span>t = <span>(<span>ln [200 ÷ 100]<span>)<span> ÷ -k
</span></span></span></span>t = <span>(<span>ln [2]<span>)<span> ÷ -k
</span></span></span></span>t = 0.69314718056<span> ÷ --.03
t =</span><span><span><span> 23.1049060187
</span>
about 23 years
</span></span>
Answer: There are no solutions.
Step-by-step explanation:
Annie's total earnings from her initial savings, $12, and from babysitting should be equal or more than 30. Annie's total earnings from babysitting may be expressed as $6n. The inequality should be,
12 + 6n <span>≥ 30
Solving for x,
6n </span><span>≥ 30 - 12
</span><span>
6n </span>≥ 18 ; n <span>≥ 3
</span><span>
Thus, the answer is the second among the choices.
</span>
Answer:
1 / 8
Step-by-step explanation:
I'm assuming that is meant to be 3/4 and 1/6...
Area = Length * Width
A = 3/4 * 1/6
A = 3 / 24
simplify the fraction to get your answer
Answer:
Please read the complete procedure below:
Step-by-step explanation:
You have the following initial value problem:
a) The algebraic equation obtain by using the Laplace transform is:
![L[y']+2L[y]=4L[t]\\\\L[y']=sY(s)-y(0)\ \ \ \ (1)\\\\L[t]=\frac{1}{s^2}\ \ \ \ \ (2)\\\\](https://tex.z-dn.net/?f=L%5By%27%5D%2B2L%5By%5D%3D4L%5Bt%5D%5C%5C%5C%5CL%5By%27%5D%3DsY%28s%29-y%280%29%5C%20%5C%20%5C%20%5C%20%281%29%5C%5C%5C%5CL%5Bt%5D%3D%5Cfrac%7B1%7D%7Bs%5E2%7D%5C%20%5C%20%5C%20%5C%20%5C%20%282%29%5C%5C%5C%5C)
next, you replace (1) and (2):
(this is the algebraic equation)
b)
![sY(s)+2Y(s)-8=\frac{4}{s^2}\\\\Y(s)[s+2]=\frac{4}{s^2}+8\\\\Y(s)=\frac{4+8s^2}{s^2(s+2)}](https://tex.z-dn.net/?f=sY%28s%29%2B2Y%28s%29-8%3D%5Cfrac%7B4%7D%7Bs%5E2%7D%5C%5C%5C%5CY%28s%29%5Bs%2B2%5D%3D%5Cfrac%7B4%7D%7Bs%5E2%7D%2B8%5C%5C%5C%5CY%28s%29%3D%5Cfrac%7B4%2B8s%5E2%7D%7Bs%5E2%28s%2B2%29%7D)
(this is the solution for Y(s))
c)
![y(t)=L^{-1}Y(s)=L^{-1}[\frac{4}{s^2(s+2)}+\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+L^{-1}[\frac{8}{s+2}]\\\\=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}](https://tex.z-dn.net/?f=y%28t%29%3DL%5E%7B-1%7DY%28s%29%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%2B%5Cfrac%7B8%7D%7Bs%2B2%7D%5D%5C%5C%5C%5C%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%5D%2BL%5E%7B-1%7D%5B%5Cfrac%7B8%7D%7Bs%2B2%7D%5D%5C%5C%5C%5C%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%5D%2B8e%5E%7B-2t%7D)
To find the inverse Laplace transform of the first term you use partial fractions:
Thus, you have:
![y(t)=L^{-1}[\frac{4}{s^2(s+2)}]+8e^{-2t}\\\\y(t)=L^{-1}[\frac{-1}{s}+\frac{2}{s^2}]+L^{-1}[\frac{1}{s+2}]+8e^{-2t}\\\\y(t)=-1+2t+e^{-2t}+8e^{-2t}=-1+2t+9e^{-2t}](https://tex.z-dn.net/?f=y%28t%29%3DL%5E%7B-1%7D%5B%5Cfrac%7B4%7D%7Bs%5E2%28s%2B2%29%7D%5D%2B8e%5E%7B-2t%7D%5C%5C%5C%5Cy%28t%29%3DL%5E%7B-1%7D%5B%5Cfrac%7B-1%7D%7Bs%7D%2B%5Cfrac%7B2%7D%7Bs%5E2%7D%5D%2BL%5E%7B-1%7D%5B%5Cfrac%7B1%7D%7Bs%2B2%7D%5D%2B8e%5E%7B-2t%7D%5C%5C%5C%5Cy%28t%29%3D-1%2B2t%2Be%5E%7B-2t%7D%2B8e%5E%7B-2t%7D%3D-1%2B2t%2B9e%5E%7B-2t%7D)
(this is the solution to the differential equation)
