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3 years ago

.What is the nth term rule of the linear sequence below? 22 , 18 , 14 , 10 , 6 , . . .

Mathematics

2 answers:

Brrunno [24]3 years ago

8 0

Answer:

$a_n=26-4n$

Step-by-step explanation:

The nth term of an AP is

$a_n=a+(n-1)d$

where, a is the first term of the sequence and d is common difference.

The given linear sequence is

22 , 18 , 14 , 10 , 6 , . . .

Here, first term is 22 and common difference is

$d=a_2-a_1=18-22=-4$

The common difference of the sequence is -4. So, the nth term of given sequence is

$a_n=22+(n-1)(-4)$

$a_n=22-4n+4$

$a_n=26-4n$

Therefore, the nth term of the sequence is $a_n=26-4n$ .

kvv77 [185]3 years ago

7 0

N+4
how do I get it ?
put the 6 into n
(6)+4=10
so answer is n+4

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2 years ago

The graph of an equation is shown below:

ziro4ka [17]

Answer:

The point $(1,3)$ is a solution of the linear equation

Step-by-step explanation:

we have

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Find the equation of the line AB

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substitute the values

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$m=\frac{5}{4}$

$(x1,y1)=B(1,3)$

substitute

$y-3=\frac{5}{4}(x-1)$

we know that

If a point is a solution to the equation of the line

then

the point must be satisfy the equation and based on the graph the point must be on the line

Let's check every point.

case A) point $(-2,-3)$

Substitute the value of x and y in the equation of the line

$-3-3=\frac{5}{4}(-2-1)$

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The point $(-2,-3)$ is not a solution of the linear equation

See the attached figure-------> the point is not on the line

case B) point $(3,1)$

Substitute the value of x and y in the equation of the line

$1-3=\frac{5}{4}(3-1)$

$-2=\frac{10}{4}$ -------> is not true

The point $(3,1)$ is not a solution of the linear equation

See the attached figure-------> the point is not on the line

case C) point $(1,3)$

Substitute the value of x and y in the equation of the line

$3-3=\frac{5}{4}(1-1)$

$0=0$ -------> is true

The point $(1,3)$ is a solution of the linear equation

See the attached figure-------> the point is on the line

case D) point $(3,2)$

Substitute the value of x and y in the equation of the line

$2-3=\frac{5}{4}(3-1)$

$-1=\frac{10}{4}$ -------> is not true

The point $(3,2)$ is not a solution of the linear equation

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