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disa [49]
3 years ago
6

Consider the expression 35.86 divided by 2.2 .

Mathematics
1 answer:
Anastasy [175]3 years ago
3 0

Answer:

  • a = 100, b = 100
  • 16.3

Step-by-step explanation:

<h3>Part A</h3>
  • a = 100 will convert the number on the numerator to whole number
  • b = 10 will convert the number on the denominator to whole number

But we need a = b in order not to change the value of the fraction, therefore we choose

  • a = b = 100
<h3>Part B</h3>

Multiplying both numerator and denominator by 100 will make the fraction:

  • 35.86*100/2.2*100 =
  • 3586/220 =
  • 16 66/220 =
  • 16 3/10
  • 16.3
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Determine formula of the nth term 2, 6, 12 20 30,42​
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Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

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Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

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and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

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Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

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cluponka [151]

Answer:

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