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Klio2033 [76]
3 years ago
13

Darnell is comparing three loan options to determine which one is best for his situation. Match each loan option with the correc

t description.
Loan Option Principal Amount Monthly Payment Loan Term
option R $14,000 $203.67 72 months
option S $15,000 $256.24 60 months
option T $16,000 $334.43 48 months
option R
option S
option T
This loan option has the lowest finance charge.
arrowRight
This loan option has the highest finance charge.
arrowRight
This loan option has neither the highest finance
charge nor the lowest monthly payment.
arrowRight
Mathematics
1 answer:
shusha [124]3 years ago
4 0

Answer:

Option T - Lowest

Option R - Highest

Option S - Neither

Explanation:

Option T $334.43 x 48 = $16,052.64 -- adding $52.64 in interest

Option R $203.67 x 72 = $14,664.24 -- adding $664.24 in interest

Option S $256.24 x 60 = $15,374.4 -- adding $374.4 in interest.

T = $52.64 (Lowest)

R = $664.24 (Highest)

S = $374.4 (Middle)

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What is the reactance of a 25 mF capacitor when the applied frequency is 400 Hz?
LuckyWell [14K]

The Reactance of a Capacitor is t<u>he resistance it has when alternating current (A/C) with a specific frequency (f) passes through it. </u>This is measured in ohms (Ω) and its formula is:


X_{C}=\frac{1}{2\pi fC}   (1)



Where:


X_{C} is the Reactance in ohms (Ω)

f is the frequency of the alternating current in Hertz (Hz)

C is the Capacitance of the Capacitor in Farad (F)



In this case we have a 25 mF capacitor with an applied frequency of 400 Hz.


Note 1mF=1*10^{-3}F=0.001F



Well, with the given data we have to solve (1):


X_{C}=\frac{1}{2\pi(400Hz)(25*10^{-3}F)}    

Then:


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3 years ago
Compute the sum:
Nady [450]
You could use perturbation method to calculate this sum. Let's start from:

S_n=\sum\limits_{k=0}^nk!\\\\\\\(1)\qquad\boxed{S_{n+1}=S_n+(n+1)!}

On the other hand, we have:

S_{n+1}=\sum\limits_{k=0}^{n+1}k!=0!+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=1}^{n+1}k!=1+\sum\limits_{k=0}^{n}(k+1)!=\\\\\\=1+\sum\limits_{k=0}^{n}k!(k+1)=1+\sum\limits_{k=0}^{n}(k\cdot k!+k!)=1+\sum\limits_{k=0}^{n}k\cdot k!+\sum\limits_{k=0}^{n}k!\\\\\\(2)\qquad \boxed{S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n}

So from (1) and (2) we have:

\begin{cases}S_{n+1}=S_n+(n+1)!\\\\S_{n+1}=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\end{cases}\\\\\\&#10;S_n+(n+1)!=1+\sum\limits_{k=0}^{n}k\cdot k!+S_n\\\\\\&#10;(\star)\qquad\boxed{\sum\limits_{k=0}^{n}k\cdot k!=(n+1)!-1}

Now, let's try to calculate sum \sum\limits_{k=0}^{n}k\cdot k!, but this time we use perturbation method.

S_n=\sum\limits_{k=0}^nk\cdot k!\\\\\\&#10;\boxed{S_{n+1}=S_n+(n+1)(n+1)!}\\\\\\&#10;

but:

S_{n+1}=\sum\limits_{k=0}^{n+1}k\cdot k!=0\cdot0!+\sum\limits_{k=1}^{n+1}k\cdot k!=0+\sum\limits_{k=0}^{n}(k+1)(k+1)!=\\\\\\=&#10;\sum\limits_{k=0}^{n}(k+1)(k+1)k!=\sum\limits_{k=0}^{n}(k^2+2k+1)k!=\\\\\\=&#10;\sum\limits_{k=0}^{n}\left[(k^2+1)k!+2k\cdot k!\right]=\sum\limits_{k=0}^{n}(k^2+1)k!+\sum\limits_{k=0}^n2k\cdot k!=\\\\\\=\sum\limits_{k=0}^{n}(k^2+1)k!+2\sum\limits_{k=0}^nk\cdot k!=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n\\\\\\&#10;\boxed{S_{n+1}=\sum\limits_{k=0}^{n}(k^2+1)k!+2S_n}

When we join both equation there will be:

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\boxed{\sum\limits_{k=0}^{n}(1+k^2)k!=n(n+1)!+1}

Sorry for my bad english, but i hope it won't be a big problem :)
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With the fewest terms possible is

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✉️ If any further questions, inbox me! ✉️

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