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Kryger [21]
3 years ago
5

Which table best classifies the following numbers as rational and irrational?

Mathematics
1 answer:
g100num [7]3 years ago
4 0
The rational negative 6 over 5.,3.5,0 point 5 bar,. square root of 4.
Hope this helps :)
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a computer and printer a totoal cost $1132 the cost of the computer is three times the cost of the ptinter
zavuch27 [327]
Use a system of equations

C+P=1132
3P=C

Substitute C in first equation as
3P+P=1132
Simplify
4P=1132
Solve
P=1132/4
P=283

NOW SOLVE FOR C SUBSTITUTING P VALUE IN FIRST EQUATION

C+283=1132
C=1132-283
C=849


Printer = 283$
Computer = 849$
3 0
2 years ago
Tori sells snow cones. His profit in dollars and high temperature in degrees Fahrenheit for two weeks are shown in the table bel
masya89 [10]
Sorry i cant answer if there is no table <3
6 0
3 years ago
Can sum1 answer asap and pls give explanation :)
Tju [1.3M]

Answer:

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3 years ago
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Do I divide or multiply not sure
Makovka662 [10]

Answer:

x=40

Step-by-step explanation:

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5 0
3 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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