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Rasek [7]
3 years ago
11

U need question problem 2 to solve problem 3

Mathematics
2 answers:
garik1379 [7]3 years ago
7 0

The fraction is grather than 1 if the numerator is greater than denominator.

The fraction is less than 1 if the numerator is less than denominator.

The fraction is equal 1 if the numerator is equal the denominator.

----------------------------------------------------------------------------------

a.\ \text{greater than 1:}\ \dfrac{4}{3},\ \dfrac{7}{5};\ \dfrac{16}{10}\\\\b.\ \text{less than 1:}\ \dfrac{1}{3};\ \dfrac{12}{15};\ \dfrac{3}{10};\ \dfrac{3}{5}\\\\c.\ \text{equal to 1:}\ \dfrac{3}{3}


\dfrac{16}{10}=\dfrac{10+6}{10}=\dfrac{10}{10}+\dfrac{6}{10}=1+\dfrac{6}{10}=1\dfrac{6}{10}=1\dfrac{6:2}{10:2}=1\dfrac{3}{5}\\\\3\text{ is greater than half of 5.}\ \text{Therefore}\ \dfrac{3}{5}>\dfrac{1}{2}.\\\\\text{Therefore}\ \dfrac{16}{10}>1\dfrac{1}{2}.\\\\OTHER:\\\\\dfrac{16}{10}=1\dfrac{6}{10}\\\\1\dfrac{1}{2}=1\dfrac{1\cdot5}{2\cdot5}=1\dfrac{5}{10}1.5\Rightarrow\dfrac{16}{10}>1\dfrac{1}{2}

Sergio [31]3 years ago
3 0

Answer:

You are correct

Step-by-step explanation:

Start with 1 1/2. This can be made into an improper fraction which is 3/2

Now multiply both top and bottom of 3/2 by 5

(3*5)/(2 * 5) = 15 / 10

16/10 is just slightly bigger than 15/10

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5 0
2 years ago
Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

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

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

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

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

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.

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

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}

6 0
2 years ago
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