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prohojiy [21]
2 years ago
11

Identify the expicit function for the sequence in the table.

Mathematics
1 answer:
marysya [2.9K]2 years ago
4 0
It is B. Because the formula for sequences in in the form of a+(n-1)d where a is the first number in the sequence, n is the nth term(position) and d is the difference between terms. So if a=9 and d=5 the answer is 9 + (n-1)*5
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Wich equation shows y + 1/5 = 3x in standard form
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15x - 5y = 1 it's B in standard form



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Giveaway time!!! Get your free 50 points
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The sum of 2 positive integers is 3. the sum of their squares is 5. find the 2 numbers
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Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.
Lelechka [254]

Answer:

1) \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}} = 32

Step-by-step explanation:

1) \frac{(-2)^{-5}}{(-2)^{-10}}

Solving using exponent rule: a^{-m}=\frac{1}{a^m}

\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32

So, \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4}

Using the exponent rule: a^m.a^n=a^{m+n}

We have:

2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}

We also know that: a^{-m}=\frac{1}{a^m}

Using this rule:

2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}

So, 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2

Solving:

(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}

So, (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}}

We know that: a^{-m}=\frac{1}{a^m}

\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32

So, \frac{2}{2^{-4}} = 32

3 0
3 years ago
In the partially shown sequence: …, 17711, A, B, 75025, …, each new term is the sum of the two previous terms. Find the whole nu
yanalaym [24]

Answer:

24513

Step-by-step explanation:

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2 years ago
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