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forsale [732]
2 years ago
9

A student borrows a total of $5400 in student loan from two lenders. Bank One charges 4.3% simple

Mathematics
1 answer:
g100num [7]2 years ago
8 0

9514 1404 393

Answer:

  $1400

Step-by-step explanation:

Let x represent the amount borrowed from Bank 1. For each loan, the interest is given by I=Prt, where P is the principal borrowed at rate r for t years.

Then the total interest over 3 years is ...

  3(4.3%·x +5.9%(5400 -x)) = 888.60

  -1.6%·x +318.60 = 296.20 . . . . . . . . . . . divide by 3, simplify

  -1.6%·x = -22.40 . . . . . . . . . . . subtract 318.60

  x = 22.40/0.016 = 1400

The student borrowed $1400 from Bank 1.

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Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

\large\underline{\sf{Solution-}}

Given expression is

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

On substituting directly x = 1, we get,

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Consider again,

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]

can be rewritten as

\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]

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\rule{190pt}{2pt}

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