find three consecutive numbers if their sum is 3n

Answers

Answer 1

Answer:

n-1,n,n+1

Step-by-step explanation:

let the numbers be x,x+1,x+2

x+x+1+x+2=3n

3x+3=3n

3x=3n-3

x=1/3 (3n-3)=n-1

so mumbers are n-1,n,n+1

Answer 2

The three consecutive numbers are n-1, n, and n+1.

We have to determine, the three consecutive numbers if their sum is 3n.

According to the question,

The consecutive number series can be even as well as odd depending on the first term of the series and the difference between the numbers.

Let, the three consecutive numbers be x, x+1, and x+2.

The sum of three consecutive numbers is equal to 33.

[tex]\rm x + x+1+x+2=3n\\\\3x+3 = 3n \\\\Divided \ by \ 3 \ on \ both \ the \ sides \\\\x +1 = n\\\\x = n-1[/tex]

Therefore,

The first consecutive number is,

[tex]\rm x = n-1[/tex]

The second consecutive number is,

[tex]\rm x+1 = n-1+1 = n\\[/tex]

The third consecutive number is,

[tex]\rm x+2= n-1+2 = n+1\\[/tex]

Hence, The three consecutive numbers are n-1, n, and n+1.

For more details refer to the link given below.

https://brainly.com/question/16048559


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Answers

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Step-by-step explanation:

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Have attached the picture for explanation...

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What is a fraction?

A fraction is a figure that is not a whole number. A fraction usually has a numerator and a denominator. The numerator is the number above. While the denominator is the number below.

What is the product of the fraction?

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Answers

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Answers

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Answers

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Answers

Answer:

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Step-by-step explanation:

Part A

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Part B

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[tex]\huge \bf༆ Answer ༄[/tex]

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Answers

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Answers

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Answers

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Answers

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Answers

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Answers

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Answers

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Answers

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Answers

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(1, 5), (2, 7), (3, 10), (4, 12) solve for domain

D={5, 7, 10, 12}


D={1, 2, 3, 4}


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Cannot be determined

Answers

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Answers

Answer: See below

Step-by-step explanation:

Isolate x for x-2y+2z=9

3x=9+2y

[tex]x=\frac{9+2y}{3}[/tex]....(1)

Substitute (1) into the 2nd equation

[tex]\begin{bmatrix}-\frac{9+2y}{3}+3y=-4\\ 2\cdot \frac{9+2y}{3}-5y+3z=16\end{bmatrix}[/tex]

[tex]\frac{3\left(-9+7y\right)}{3}=3\left(-4\right)[/tex]

7y=-3

y=-3/7

Substitute y=-3/7

[tex]\begin{bmatrix}3z+\frac{18-11\left(-\frac{3}{7}\right)}{3}=16\end{bmatrix}[/tex]

[tex]\begin{bmatrix}3z+\frac{53}{7}=16\end{bmatrix}[/tex]

Isolate z by substituting it

[tex]3z+\frac{53}{7}=16[/tex]

[tex]\frac{3z}{3}=\frac{\frac{59}{7}}{3}[/tex]

[tex]z=\frac{59}{21}[/tex]

For x =9+2y/3 substitute z=59/21 and y=-3/7

[tex]x=\frac{9+2\left(-\frac{3}{7}\right)}{3}[/tex]

[tex]x=\frac{19}{7}[/tex]

[tex]x=\frac{19}{7},\:z=\frac{59}{21},\:y=-\frac{3}{7}[/tex]

Answer:

(1, - 1, 3 )

Step-by-step explanation:

x - 2y + 2z = 9 → (1)

- x + 3y = - 4 → (2)

2x - 5y + 3z = 16 → (3)

Add (1) and (2) term by term to eliminate x

y + 2z = 5 → (4)

Multiply (2) by 2

- 2x + 6y = - 8 → (5)

Add (3) and (5) term by term to eliminate x

y + 3z = 8 → (6)

Subtract (6) from (4) term by term to eliminate y

- z = - 3 ( multiply both sides by - 1 )

z = 3

Substitute z = 3 into (4)

y + 2(3) = 5

y + 6 = 5 ( subtract 6 from both sides )

y = - 1

Substitute y = - 1, z = 3 into (1) and solve for x

x - 2(- 1) + 2(3) = 9

x + 2 + 6 = 9

x + 8 = 9 ( subtract 8 from both sides )

x = 1

solution is (1, - 1, 3 )

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Answers

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