Given:
loan amount: 25,250
original interest rate: 3.4%
new interest rate: 6.8%
term: 10 years.
Assuming that simple interest formula is used.
I = P * r * t
I = interest
P = principal
r = interest rate
t = term/time
I = 25,250 * 3.4% * 10 years
I = 8,585
I = 25,250 * 6.8% * 10 years
I = 17,170
17,170 - 8,585 = 8,585 Additional interest paid using the new interest rate.
Using an online loan repayment calculator: Here are the following data:
Loan Balance:$25,250.00
Adjusted Loan Balance:$25,250.00Loan
Interest Rate:6.80%
Loan Fees:0.00%
Loan Term:10 years
Minimum Payment:$0.00
Monthly Loan Payment:$290.58
Number of Payments:120
Cumulative Payments:$34,869.23
Total Interest Paid:$9,619.23
<span><span>Loan Balance:$25,250.00
</span><span>Adjusted Loan Balance:$25,250.00
</span><span>Loan Interest Rate:3.40%
</span><span>Loan Fees:0.00%
</span><span>Loan Term:10 years
</span><span>Minimum Payment:$0.00</span>
<span>Monthly Loan Payment:$248.51
</span><span>Number of Payments:120</span>
<span>Cumulative Payments:$29,820.59
</span><span>Total Interest Paid:<span>$4,570.59</span></span></span>
There is 1 brother as they would all have the same brother.
3/7
6 divided by 2 equals 3
14 divided by 2 equals 7
Positive 28 because to negative will equal a positive number
At the start, the tank contains
(0.02 g/L) * (1000 L) = 20 g
of chlorine. Let <em>c</em> (<em>t</em> ) denote the amount of chlorine (in grams) in the tank at time <em>t </em>.
Pure water is pumped into the tank, so no chlorine is flowing into it, but is flowing out at a rate of
(<em>c</em> (<em>t</em> )/(1000 + (10 - 25)<em>t</em> ) g/L) * (25 L/s) = 5<em>c</em> (<em>t</em> ) /(200 - 3<em>t</em> ) g/s
In case it's unclear why this is the case:
The amount of liquid in the tank at the start is 1000 L. If water is pumped in at a rate of 10 L/s, then after <em>t</em> s there will be (1000 + 10<em>t</em> ) L of liquid in the tank. But we're also removing 25 L from the tank per second, so there is a net "gain" of 10 - 25 = -15 L of liquid each second. So the volume of liquid in the tank at time <em>t</em> is (1000 - 15<em>t </em>) L. Then the concentration of chlorine per unit volume is <em>c</em> (<em>t</em> ) divided by this volume.
So the amount of chlorine in the tank changes according to

which is a linear equation. Move the non-derivative term to the left, then multiply both sides by the integrating factor 1/(200 - 5<em>t</em> )^(5/3), then integrate both sides to solve for <em>c</em> (<em>t</em> ):


![\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac{c(t)}{(200-3t)^{5/3}}\right]=0](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dt%7D%5Cleft%5B%5Cdfrac%7Bc%28t%29%7D%7B%28200-3t%29%5E%7B5%2F3%7D%7D%5Cright%5D%3D0)


There are 20 g of chlorine at the start, so <em>c</em> (0) = 20. Use this to solve for <em>C</em> :

![\implies\boxed{c(t)=\dfrac1{200}\sqrt[3]{\dfrac{(200-3t)^5}5}}](https://tex.z-dn.net/?f=%5Cimplies%5Cboxed%7Bc%28t%29%3D%5Cdfrac1%7B200%7D%5Csqrt%5B3%5D%7B%5Cdfrac%7B%28200-3t%29%5E5%7D5%7D%7D)