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fenix001 [56]
1 year ago
8

2. Each year a certain amount of money is deposited in an account which pays an

Mathematics
1 answer:
sergiy2304 [10]1 year ago
8 0

Answer:

  2a.  ((500x +200)x +600)x

  2b.  $1350.68

  3.  -1

Step-by-step explanation:

2. You want to know the balance in an account at the end of 3 years with deposits at the beginning of successive years being $500, $200, and $600, and with deposits having an annual growth factor of x. You want (a) an expression for the balance in terms of x, and (b) the balance when x=1+2%.

3. You want the value of p(-2) for p(x) = 5x³ +8x² -3x + 1.

<h2>2. Account Balance</h2><h3>a. Expression</h3>

$500 is deposited at the beginning of the first year. The problem statement tells us that this has been multiplied by growth factor x by the end of the year, so the balance at that point is 500x.

At the beginning of the second year, $200 is added to the account, and the entire amount is multiplied by the growth factor for the year. That makes the balance at the end of the second year be ...

  (500x +200)x

Then $600 is added, and again the account grows by a factor of x. At the end of the third year, the balance is ...

  balance = ((500x +200)x +600)x

<h3>b. Balance for r=2%</h3>

The growth factor is given as x = 1 +r. When r = 2%, this becomes x = 1.02. Then the balance at the end of year 3 is ...

  ((500·1.02 +200)·1.02 +600)·1.02 = 1350.68

The amount in the account at the end of three years is $1350.68.

__

<h2>3. Polynomial value</h2>

The value of p(x) = 5x³ +8x² -3x +1 for x=-2 is found by substituting -2 where x is found. Evaluation can be easier by rewriting the polynomial to Horner form:

  p(x) = ((5x +8)x -3)x +1

  p(-2) = ((5(-2) +8)(-2) -3)(-2) +1 = ((-2)(-2) -3)(-2) +1 = (1)(-2) +1 = -1

The value of the function for x=-2 is -1.

__

<em>Additional comment</em>

You will note that the expression in problem 2 is also written in Horner form. If it were expanded, it would be 500x³ +200x² +600x. Evaluation takes fewer steps when Horner form is used.

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