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igomit [66]
3 years ago
5

Find the 75th term of the arithmetic sequence 16, 15, 14,

Mathematics
1 answer:
slega [8]3 years ago
3 0

Answer:

a_{75}=-58

Step-by-step explanation:

This is an arithmetic sequence:

a_n=a_1+(n-1)d

where d is the common difference and n is the index of any given term.

For the given sequence, the common difference is -1:

15-16=-1\\14-15=-1

Knowing both the common difference and the first term, you can write the equation for this sequence:

a_n=16+(n-1)(-1)

Then you can use that equation to find the 75th term:

a_{75}=16+(75-1)(-1)\\a_{75}=16+(74)(-1)\\a_{75}=16-74\\a_{75}=-58

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Write as a product of 2 binomials and a monomial (factor out as much as possible from each binomial). (14x+21y)(6ab–3a)
mihalych1998 [28]

Answer: The answer is 21a(2x+3y)(2b-1).


Step-by-step explanation: Given expression is as follows

E=(14x+21y)(6ab-3a).

We are to write the above expression "E" as a product of two binomials and a monomial.

A monomial is an expression with a single term and a binomial is an expression with two terms.

So, let us start as follows -

E\\\\=(14x+21y)(6ab-3a)\\\\=7(2x+3y)3a(2b-1)\\\\=21a(2x+3y)(2b-1).

Here,

21a~\textup{is a monomial and}~(2x+3y),~(2b-1)~\textup{are binomials}.

Thus, the required factorisation is

E=21a(2x+3y)(2b-1).


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thank you for your time today

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