Question

The first sequence rule is multiply by 3 starting from 4. The second sequence rule is add 8 starting from 20. What is the first number that appears in both sequences?

12
20
28
36

Answer 1

help?

Answer 2

ill help

Answer 3

oop

Answer 4

well i tried

Answer 5

it's ok. i will give you fan and medal.

Answer 6

ty

Answer 7

Rule 1: Multiply by 2, then add one third starting from 1. Rule 2: Add one half, then multiply by 4 starting from 0. What is the fourth ordered pair using the two sequences?

(5, 10)
(four and two thirds, 42)
(21, 170)
(two and one third, 2)

Answer 8

I'M NOW RED!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 9

bruh

Answer 10

it could be B or D

Answer 11

Which one

Answer 12

question

Answer 13

the second one

Answer 14

This one?

Answer 15

@candyplays321 dm'd the answer.

Answer 16

for first
\[y=4(3)^{n-1}\]
for second
\[y=20+8(n-1)\]
for n=1
\[4(3)^{1-1}=20+8(1-1)\]
4=20
for n=2
\[4(3)^{2-1}=20+8(2-1)\]
\[4(3)=20+8\]
12=28
not true
for n=3
\[4(3)^{3-1}=20+8(3-1)\]
\[4(3)^2=20+8(2)\]
36=36
so third term =36
and first term equal=36