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ioda
3 years ago
11

Write the first five terms of the arithmetic sequence. a1 = 9; d = -2

Mathematics
1 answer:
asambeis [7]3 years ago
4 0

Answer:

nth term = 1 1/2n -1

Step-by-step explanation:

The arithmetic sequence formula is:

a n = a 1 + ( n – 1 ) d

Where: a n

is the nth term in the sequence

a 1

is the first term in the sequence

n

is the term you are solving for

d

is the common difference for any pair of consecutive numbers in the sequence.

First Term or  

n = 1 :

This is given in the problem.

a 1  =  9

Second Term or  

n = 2 :

Substitute  

2  for  n

in the formula and substitute the values from the problem giving:

a 2 = 9 + ( ( 2 – 1 ) × -2 )

a 2 = 9 + ( 1 × -2 )

a 2 = 9 +-2

a 2 = 7

Fifth Term or  n = 5 :

Substitute      in the formula and substitute the values from the problem giving:

a 5 = 9+ ( ( 5– 1 ) × -2 )

a 5 = 9 + ( 4 × -2 )

a 2 = 9 + -8

a 2 = 1  

Using this same process you should be able to determiner the

Third Term or  n = 3

: and Fourth Term or  n = 4 :

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Find the length of the third side. If necessary, write in simplest radical form.
Arlecino [84]
Answer:

9

Explanation:

9 root 2 = root 81 x root 2 = root 162
Root 162 = 12.7279220614
9^2 + b^2 = 12.7279220614^2
81 + b^2 = 162
b^2 = 81
Root both
B = 9

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2 years ago
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V125BC [204]

Answer:

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Step-by-step explanation:

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3 years ago
Read 2 more answers
1. Let a; b; c; d; n belong to Z with n > 0. Suppose a congruent b (mod n) and c congruent d (mod n). Use the definition
lukranit [14]

Answer:

Proofs are in the explantion.

Step-by-step explanation:

We are given the following:

1) a \equi b (mod n) \rightarrow a-b=kn for integer k.

1) c \equi  d (mod n) \rightarrow c-d=mn for integer m.

a)

Proof:

We want to show a+c \equiv b+d (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(a+c)-(b+d)=rn. If we do that we would have shown that a+c \equiv b+d (mod n).

kn+mn   =  (a-b)+(c-d)

(k+m)n   =   a-b+ c-d

(k+m)n   =   (a+c)+(-b-d)

(k+m)n  =    (a+c)-(b+d)

k+m is is just an integer

So we found integer r such that (a+c)-(b+d)=rn.

Therefore, a+c \equiv b+d (mod n).

//

b) Proof:

We want to show ac \equiv bd (mod n).

So we have the two equations:

a-b=kn and c-d=mn and we want to show for some integer r that we have

(ac)-(bd)=tn. If we do that we would have shown that ac \equiv bd (mod n).

If a-b=kn, then a=b+kn.

If c-d=mn, then c=d+mn.

ac-bd  =  (b+kn)(d+mn)-bd

          =    bd+bmn+dkn+kmn^2-bd

          =           bmn+dkn+kmn^2

          =            n(bm+dk+kmn)

So the integer t such that (ac)-(bd)=tn is bm+dk+kmn.  

Therefore, ac \equiv bd (mod n).

//

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3 years ago
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kati45 [8]

Answer:

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Novay_Z [31]

Answer:

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