HEEELLLPPP!!!!!!!! Find the sequence and fill in the blanks (2, 3), (4, 6), (6, 9), (8, 12) .... The 8th group is (__, __)
16;24
ARE YOU ONE HUNDRED % SURE
Yes
It’s doubling
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Welks
It's not doubling. Well the y isn't. It's adding 3 each time. So it would technically be (16,15)
2,4,6,8,... first term a=2 c.d.=4-2=2 \[t _{n}=a+(n-1)d\] put n=8 \[t _{8}=2+(8-1)2=2+14=16\] similarly second=24 so 8th term is (16,24)
Because for me it is very clearly not being doubled like the X is
It's being added by 3
But I am only in algebra 2 so idk
its 10 and 15
its going up by adding 2 each time on x and adding 3 each on y
@narad ?
The nth term is \[u_{n}= n (2,3)= \left( 2n, 3n \right)\]
(2, 3), (4, 6), (6, 9), (8, 12) .... The 8th group is (__, __) sorry now i see that ask about 8th group ???
To find the sequence, we can observe that the second coordinate of each pair is twice the first coordinate. Therefore, we can write the sequence as: (2, 3), (4, 6), (6, 9), (8, 12), ... To find the 8th group, we need to continue the pattern until we reach the 8th pair: (2, 3), (4, 6), (6, 9), (8, 12), (10, 15), (12, 18), (14, 21), (16, 24) Therefore, the 8th group is (14, 21).
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