Answer:
39.6x+26.4
Step-by-step explanation:
Step 1
Given;
![\begin{gathered} \text{Principal(p)= \$375000} \\ \text{First rate = 8\%=}\frac{8}{100}=0.08 \\ \text{Second rate = 20\%= }\frac{20}{100}=0.2 \\ \text{Time}=\frac{90}{365}=\frac{18}{73} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Ctext%7BPrincipal%28p%29%3D%20%5C%24375000%7D%20%5C%5C%20%5Ctext%7BFirst%20rate%20%3D%208%5C%25%3D%7D%5Cfrac%7B8%7D%7B100%7D%3D0.08%20%5C%5C%20%5Ctext%7BSecond%20rate%20%3D%2020%5C%25%3D%20%7D%5Cfrac%7B20%7D%7B100%7D%3D0.2%20%5C%5C%20%5Ctext%7BTime%7D%3D%5Cfrac%7B90%7D%7B365%7D%3D%5Cfrac%7B18%7D%7B73%7D%20%5Cend%7Bgathered%7D)
Required; To find the difference in interest between the two periods.
Step 2
State the formula for simple interest
![A=P(1+rt)](https://tex.z-dn.net/?f=A%3DP%281%2Brt%29)
Step 3
Find the interest when the rate is 8%
![\begin{gathered} A=375000(1+(0.08\times\frac{18}{73}) \\ A=375000(1+\frac{36}{1825}) \\ A=\text{\$}382397.26 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3D375000%281%2B%280.08%5Ctimes%5Cfrac%7B18%7D%7B73%7D%29%20%5C%5C%20A%3D375000%281%2B%5Cfrac%7B36%7D%7B1825%7D%29%20%5C%5C%20A%3D%5Ctext%7B%5C%24%7D382397.26%20%5Cend%7Bgathered%7D)
Therefore the interest is given as;
![A-P=382397.26-375000=\text{\$}7397.26](https://tex.z-dn.net/?f=A-P%3D382397.26-375000%3D%5Ctext%7B%5C%24%7D7397.26)
Step 4
Find the interest in 1980 with a 20% rate
![\begin{gathered} A=375000(1+(0.2\times\frac{18}{73}) \\ A=\text{\$}393493.15 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3D375000%281%2B%280.2%5Ctimes%5Cfrac%7B18%7D%7B73%7D%29%20%5C%5C%20A%3D%5Ctext%7B%5C%24%7D393493.15%20%5Cend%7Bgathered%7D)
The interest is given as;
![A-p=393493.15-375000=\text{\$}18493.15\text{ }](https://tex.z-dn.net/?f=A-p%3D393493.15-375000%3D%5Ctext%7B%5C%24%7D18493.15%5Ctext%7B%20%7D)
Step 5
Find the difference in interest between the two rates.
![\text{\$}18493.15-\text{\$}7397.26=\text{\$}11095.89](https://tex.z-dn.net/?f=%5Ctext%7B%5C%24%7D18493.15-%5Ctext%7B%5C%24%7D7397.26%3D%5Ctext%7B%5C%24%7D11095.89)
Hence, the difference in interest between the two rates = $11095.89
She will need 8 tables. Happy Thanksgiving! :D
The correct answer is: AB = 3.11
Explanation:
Since
![cos(\theta) = \frac{base}{hypotenuse}](https://tex.z-dn.net/?f=cos%28%5Ctheta%29%20%3D%20%20%5Cfrac%7Bbase%7D%7Bhypotenuse%7D%20)
--- (1)
![\theta](https://tex.z-dn.net/?f=%5Ctheta)
= 50°
base = 2
And hypotenuse = AB
Plug in the values in (1):
(1) => cos(50°) = 2/AB
=> AB = 2/0.643
=> AB = 3.11
If
is the amount of strontium-90 present in the area in year
, and it decays at a rate of 2.5% per year, then
![S(t+1)=(1-0.025)S(t)=0.975S(t)](https://tex.z-dn.net/?f=S%28t%2B1%29%3D%281-0.025%29S%28t%29%3D0.975S%28t%29)
Let
be the starting amount immediately after the nuclear reactor explodes. Then
![S(t+1)=0.975S(t)=0.975^2S(t-1)=0.975^3S(t-2)=\cdots=0.975^{t+1}S(0)](https://tex.z-dn.net/?f=S%28t%2B1%29%3D0.975S%28t%29%3D0.975%5E2S%28t-1%29%3D0.975%5E3S%28t-2%29%3D%5Ccdots%3D0.975%5E%7Bt%2B1%7DS%280%29)
or simply
![S(t)=0.975^ts](https://tex.z-dn.net/?f=S%28t%29%3D0.975%5Ets)
So that after 50 years, the amount of strontium-90 that remains is approximately
![S(50)=0.975^{50}s\approx0.282s](https://tex.z-dn.net/?f=S%2850%29%3D0.975%5E%7B50%7Ds%5Capprox0.282s)
or about 28% of the original amount.
We can confirm this another way; recall the exponential decay formula,
![S(t)=se^{kt}](https://tex.z-dn.net/?f=S%28t%29%3Dse%5E%7Bkt%7D)
where
is measured in years. We're told that 2.5% of the starting amount
decays after 1 year, so that
![0.975s=se^k\implies k=\ln0.975](https://tex.z-dn.net/?f=0.975s%3Dse%5Ek%5Cimplies%20k%3D%5Cln0.975)
Then after 50 years, we have
![S(50)=se^{50k}\approx0.282s](https://tex.z-dn.net/?f=S%2850%29%3Dse%5E%7B50k%7D%5Capprox0.282s)