I’m pretty sure it’s B because you can put in any number and it could still have a difference of 1
Method 1:
The recursive formula states that every next term is three times the previous one, plus two.
So, we have:
Method 2:
You can write the expression for
, and use the recursive formula until you get to the starting point:
![a_5 = 3a_4+2 = 3(3a_3+2)+2 = 9a_3+8 = 9(3a_2+2)+8 = 27a_2+26 = 27(3a_1+2)+26 = 81a_1+80](https://tex.z-dn.net/?f=a_5%20%3D%203a_4%2B2%20%3D%203%283a_3%2B2%29%2B2%20%3D%209a_3%2B8%20%3D%209%283a_2%2B2%29%2B8%20%3D%2027a_2%2B26%20%3D%2027%283a_1%2B2%29%2B26%20%3D%2081a_1%2B80)
Now use the fact that
to get the conclusion
![a_5 = 81+80 = 161](https://tex.z-dn.net/?f=a_5%20%3D%2081%2B80%20%3D%20161)
Answer:
41.47%
Step-by-step explanation:
618÷(745+618+127)×100
1) 8p+3f=$10.39
2) 5p+4f=$10.51
where p represents pencil & f represents folder.
multiply the first equation by 5 and the second equation by 8.
40p+15f=$51.95
40p+32f=$84.08
Subtract
17f=$32.13
f=$32.13/17 =$1.89
Statement A is needed to complete this syllogism. its diagonals are perpendicular. Therefore, if a <u>quadrilateral has 4 congruent sides</u> then its diagonals are perpendicular.
<h3>What is a syllogism?</h3>
A logical structure for a formal argument that includes a major, minor, and conclusion.
This syllogism cannot be concluded without Statement A. The diagonals of it are parallel. As a result, a quadrilateral's diagonals are perpendicular if it has four congruent sides.
If a quadrilateral has 4 congruent sides, then it is a rhombus. If it is a rhombus, then its diagonals are perpendicular.
Therefore, if a <u>quadrilateral has 4 congruent sides</u> then its diagonals are perpendicular.
Hence option A is correct.
To learn more about the syllogism refer;
brainly.com/question/21635821
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