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Anestetic [448]
3 years ago
10

How do i find the nth sequence in this sequen 7cm 10cm 13cm 16cm

Mathematics
2 answers:
Butoxors [25]3 years ago
4 0

Answer:

a_{n} = 3n + 4

Step-by-step explanation:

There is a common difference between consecutive terms in the sequence

10 - 7 = 13 - 10 = 16 - 13 = 3

This indicates the sequence is arithmetic with nth term

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Here a₁ = 7 and d = 3 , then

a_{n} = 7 + 3(n - 1) = 7 + 3n - 3 = 3n + 4

german3 years ago
3 0

Answer:

This is a sequence of 3 added to digits which start from 7 so 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, etc.

If u wanna find the 9th sequence the answer is 31

Step-by-step explanation:

Hope it helps :D

Brainliest?

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2X+2Y=48<br> 3X+Y=40 <br><br> What is X and Y
marshall27 [118]

Answer:

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Step-by-step explanation:

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Ther is many ways to solve this, one of them is clearance:

WE CLEAR AN UNLOCKED (X) AND REPLACE IT IN THE OTHER

x= (40-y)/3

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4 0
3 years ago
Find x(3x+72) triangle
Alinara [238K]

Answer: x = -4 ; Angle = 60°

Concept:

The given figure is a triangle with 3 arc signs on each angle. This <u>arc sign </u>stands for the corresponding angles are congruent, which in this question, it shows that all three angles are congruent. Sometimes, if the figure has multiple angles and there are different groupings of congruent angles, then we use different numbers of arcs or symbols.

Solve:

<u>Given information</u>

An angle = 3x + 72

Total measure of angles = 180 (triangle angle sum theorem)

Total number of congruent angles = 3

<u>Given expression</u>

Total measure = Total number of congruent angles × An angle measure

<u>Substitute values into the expression</u>

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<u>Divide 3 on both sides</u>

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<u>Subtract 72 on both sides</u>

60 - 72 = 3x + 72 - 72

-12 = 3x

<u>Divide 3 on both sides</u>

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\boxed{x=-4}

<u />

<u>Find the angle measure</u>

3x + 72 = 3 (-4) + 72 = \boxed{60}

Hope this helps!! :)

Please let me know if you have any questions

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