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elena55 [62]
3 years ago
15

Your credit card has a balance of $3300 and an annual interest rate of 14%. You decided to pay off the balance over two years. I

f there are no further purchases charged to the card, you must pay $158.40 each month, and you will pay a total interest of $501.60. Assume you decided to pay off the balance over one year rather than two. How much more must you pay each month and how much less will you pay in total interest?
Mathematics
1 answer:
NARA [144]3 years ago
3 0

9514 1404 393

Answer:

  • $137.90 more each month
  • $246.00 less total interest

Step-by-step explanation:

The amortization formula is ...

  A = P(r/12)/(1 -(1 +r/12)^(-12t))

for the monthly payment on principal P at annual rate r for t years. Here, we have P=3300, r = 0.14, and t=1, so the monthly payment is ...

  A = $3300(0.14/12)/(1 -(1 +0.14/12)^-12) ≈ $296.30

The payment of $296.30 is ...

  $295.30 -158.40 = $137.90 . . . more each month

The total amount paid is 12×$296.30 = $3555.60, so 255.60 in interest. This amount is ...

  $501.60 -255.60 = $246.00 . . . less total interest

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The length of a rectangle is 1 ft less than twice the width, the area of the rectangle is 21 ft squared. What’s the length
guapka [62]

Answer:

The length of the rectangle is 6 feet

Step-by-step explanation:

Given as :

The length of rectangle is 1 feet less than twice the width

So, let The width of rectangle = w feet

The length of rectangle = ( 2 w - 1 )  feet

The area of the rectangle = 21 feet²

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∵ The area of the rectangle = length × width

Or, 21 feet² =  ( 2 w - 1 ) feet × w feet

or, 21 = 2 w² - w

or, 2 w² - w - 21 = 0

or, 2 w² + 6 w - 7 w - 21 = 0

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or, ( w + 3 ) ( 2 w - 7 ) = 0

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here we consider only positive value of w i.e  \frac{7}{2} feet

∴ The length of rectangle = ( 2 w - 1 )  feet

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3 0
3 years ago
Y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′′(0) = 0
Snowcat [4.5K]

Answer:

y(t) =  3u₂(t) [ e^{-2t+4}  - e^{-5t + 10)} ] - 4u₅(t) [ e^{-2t+10)}  - e^{-5t + 25)} ]

Step-by-step explanation:

To find - y″+ 5y′ + 6y = 3δ(t − 2) − 4δ(t −5); y(0) = y′(0) = 0

Formula used -

L{δ(t − c)} = e^{-cs}

L{f''(t) = s²F(s) - sf(0) - f'(0)

L{f'(t) = sF(s) - f(0)

Solution -

By Applying Laplace transform, we get

L{y″+ 5y′ + 6y} = L{3δ(t − 2) − 4δ(t −5)}

⇒L{y''} + 5L{y'} + 6L{y} = 3L{δ(t − 2)}  − 4L{δ(t −5)}

⇒s²Y(s) - sy(0) - y'(0) + 5[sY(s) - y(0)] + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒s²Y(s) - 0 - 0 + 5[sY(s) - 0] + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒s²Y(s) + 5sY(s) + 6Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s² + 5s + 6] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s² + 3s + 2s + 6] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[s(s + 3) + 2(s + 3)] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒[(s + 2)(s + 3)] Y(s) = 3e^{-2s} - 4e^{-5s}

⇒Y(s) = \frac{3e^{-2s} }{(s + 2)(s + 3)} -  \frac{4e^{-5s} }{(s + 2)(s + 3)}

Now,

Let

\frac{1}{(s+2)(s+3)} = \frac{A}{s+2}  + \frac{B}{s+3} \\\frac{1}{(s+2)(s+3)} = \frac{A(s + 3) + B(s+2)}{(s+2)(s+3)}\\1 = As + 3A + Bs + 2B\\1 = (A+B)s + (3A + 2B)

By Comparing, we get

A + B = 0 and 3A + 2B = 1

⇒A = -B

and

3(-B) + 2B = 1

⇒-B = 1

⇒B = -1

So,

A = 1

∴ we get

\frac{1}{(s+2)(s+3)} = \frac{1}{s+2}  + \frac{-1}{s+3}

So,

Y(s) = 3e^{-2s}[ \frac{1}{(s + 2)} -    \frac{1}{(s + 3)}] - 4e^{-5s}[ \frac{1}{(s + 2)} -    \frac{1}{(s + 3)}]

⇒Y(s) = 3e^{-2s} \frac{1}{(s + 2)} -    3e^{-2s} \frac{1}{(s + 3)} - 4e^{-5s}\frac{1}{(s + 2)} + 4e^{-5s}\frac{1}{(s + 3)}

By applying inverse Laplace , we get

y(t) = 3u₂(t) [ e^{-2(t-2)}  - e^{-5(t - 2)} ] - 4u₅(t) [ e^{-2(t-5)}  - e^{-5(t - 5)} ]

⇒y(t) =  3u₂(t) [ e^{-2t+4}  - e^{-5t + 10)} ] - 4u₅(t) [ e^{-2t+10)}  - e^{-5t + 25)} ]

It is the required solution.

3 0
3 years ago
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