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MakcuM [25]
3 years ago
6

Fill in the blanks to rewrite the following statement with variables: Is there an integer with a remainder of 1 when it is divid

ed by 4 and a remainder of 3 when it is divided by 7? (a) Is there an integer n such that n has (b) Does there exist such that if n is divided by 4 the remainder is 1 and if
Mathematics
1 answer:
bearhunter [10]3 years ago
8 0

Answer:

See explanation below

Step-by-step explanation:

a) Is there an integer n such that n has...

Is there an integer n such that n has a remainder of 1 when n is divided by 4 and a remainder of 3 when n is divided by 7?

b) Does there exist such that if n is divided by 4 the remainder is 1 and if...

Does there exist an integer n such that if n is divided by 4 the remainder is 1 and if n is divided by 7 the remainder is 3?

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12-A
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28-18=10
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fourth of 100% is 25%
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(3x - 1)(2 {x}^{2}  + 4x - 3)

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3 years ago
HELP!Please solve! this!!
Angelina_Jolie [31]

Answer:

Recursive formula for geometric sequence

a_{n}=a_{n-1}\times r

is a_{n}=a_{n-1}\times 6

and explicit formula for geometric sequence a_{n}=a_{1}^{r-1} is

a_{n}=(\frac{1}{2})^{6-1}

Step-by-step explanation:

Given sequence is \frac{1}{2},3,18,108,648,...

To find the recursive and explicit formula for this sequence:

Let a_{1}=\frac{1}{2},a_{2}=3,a_{3}=18,a_{4}=108,a_{5}=648,...

To find the common ratio r:

r=\frac{a_{2}}{a_{1}}

=\frac{3}{\frac{1}{2}}

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Therefore r=6

r=\frac{a_{3}}{a_{2}}

=\frac{18}{3}

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Therefore r=6

Therefore the common ration r=6

Therefore the given sequence is geometric sequence

Recursive formula for geometric sequence is a_{n}=a_{n-1}\times r

a_{n}=a_{n-1}\times 6

and explicit formula is a_{n}=a_{1}^{r-1}

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=(\frac{1}{2})^{5}

=\frac{1}{32}

Therefore a_{n}=\frac{1}{32}

8 0
3 years ago
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Answer:

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Step-by-step explanation:

Step 1: Plug In--> <u>8-(-13)</u>

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Step 2: Simplify: <u>21</u>

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13m-22=9m-6

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3 years ago
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