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3 years ago

Me ayudan a resolverlo por favor (3x+8y+8)+(4x+6y-6)

Mathematics

1 answer:

Anna007 [38]3 years ago

5 0

Answer:

7x + 14y + 2

Step-by-step explanation:

(3x+8y+8)+(4x+6y-6)

3x + 8y + 8 + 4x + 6y - 6

3x + 4x + 8y + 6y + 8 - 6

7x + 14y + 2

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Evgen [1.6K]

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Step-by-step explanation:

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3 years ago

Given the recursive formula for an arithmetic sequence find the common difference and the 52nd term

Mice21 [21]

Answer:

Common difference(d) $a_{52}$

(21) -10 -548

(22) -7 -323

(23) 10 547

(24) -100 -5118

Step-by-step explanation:

Let the common difference be denoted by 'd'.

Also the nth difference of an arithmetic sequence is given by: $a_{n}=a_{1}+(n-1)\times d$

(21)

We are given a recursive formula as:

$a_{n}=a_{n-1}-10$

The first term is given by:

$a_{1}=-38$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-10$

The nth term of an arithmetic sequence is given by:

$a_{n}=a_{1}+(n-1)\times d$

Here $a_{1}=-38$ and $d=-10$ .

Hence, $a_{52}=-38+(52-1)\times (-10)$

Hence, $a_{52}=-548$

(22)

$a_{n}=a_{n-1}-7$

$a_{1}=34$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-7$

Here $a_{1}=34$ and $d=-7$ .

Hence, $a_{52}=34+(52-1)\times (-7)$

Hence, $a_{52}=-323$

(23)

$a_{n}=a_{n-1}+10$

$a_{1}=37$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=10$

Here $a_{1}=37$ and $d=10$ .

Hence, $a_{52}=37+(52-1)\times (10)$

Hence, $a_{52}=547$

(24)

$a_{n}=a_{n-1}-100$

$a_{1}=-18$

The common difference for an arithmetic sequence is given by:

$a_{n}-a_{n-1}$

Hence, here we have the common difference as:

$d=-100$

Here $a_{1}=-18$ and $d=-100$ .

Hence, $a_{52}=-18+(52-1)\times (-100)$

Hence, $a_{52}=-5118$

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3 years ago

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