# data[].attributes.body
Joseph is correct as 2n square is probably 4n, which is bigger that 2+n or 2n.

# data[].attributes.body
joseph is correct as long as n is not below 1

# data[].attributes.body
I only will agree if n is less than 1, because any number 1 or higher would result in another expression larger

# data[].attributes.body
I only agree with Paul if n=0.

# data[].attributes.body
I agree with Joseph. However, it depends on the value of n. If n=0 or n=1, the other numbers will be larger as 2n2 would be 0 or 2

# data[].attributes.body
I only agree if n is less than 1 because if n is less than 1 then it will be 2 while 2n2 will be 0

# data[].attributes.body
I agree with Joseph but only if the number is more than 1 as if n equals to 0 then it will be larger as it will be 2 while 2n2 is 0

# data[].attributes.body
If it is 0 or 1, it would be smaller than 2+n.

# data[].attributes.body
This also depends. If n is one or zero, this would be the largest, but, however, if other values are used, Joseph and Antony would be correct.

# data[].attributes.body
It kind of depends. If n=0, the answer would be 0, or if n=1, the others would be bigger.

# data[].attributes.body
[DONE] 1 | ALEXIS LIM EN YU
[DONE] 2 | LIN LEI LEI PHYO
3 | NGO SEAN GINN
[DONE] 4 | VICTORIA TU JIA WEI
[DONE] 5 | VIJAYAGANAPATHY PAVITHRAA
[DONE] 6 | YOSHIOKA LILI
[DONE] 7 | ANG SEN LUEN MATTHIAS
8 | ARJUN NAVNEET
9 | BALASARAVANAN DHANWIN BASIL
10 | CHEW SING HUN
[DONE] 11 | DAYAN SOH
12 | DEAN CHAN
[DONE] 13 | DEIAN WEE SHUO XIAN
14 | ISAAC LIM ZU-LE
[DONE] 15 | JOAB LEE JIE EN
16 | JOVAN KEE BING LE
17 | KELVIN SIAH MENG HAN
18 | KESHAV MICHAEL VIKNESH
[DONE] 19 | LAU KAI RUI GAVIN
[DONE] 20 | PRAJAN MANIVANNAN
21 | PRAKASH DHANVIN
22 | TAN JIT FONG, ANSEL
[DONE] 23 | TEH TAO DER
[DONE] 24 | TEO ZHI EN IAN
25 | YIO TI CHI

# data[].attributes.body
It depends on the value of n. If n is lesser than 1, this option will be the biggest. This is because if n is 0, the answer will be 2 which will make it smaller than Joseph's answer. But if the n is bigger than one, Paul is wrong. This is because his answer will equate to 2, which will be smaller than Joseph's answer.

# data[].attributes.body
It depends on the value of n. If n is lesser than 1, this option will not be the biggest. This is because if n is 0, the answer will be 0 which will make it smaller than Paul's answer. But if the n is bigger than one, Joseph is correct. This is because (2x1)square, the answer will be 4, making it the biggest.

# data[].attributes.body
if n = 3,
2n^2 = 36
2n = 6
2+n = 5

so 2n^2 is the largest unless n = 0 (2+n would then be the largest)

# data[].attributes.body
If n=0 I agree

# data[].attributes.body
I only agree if n≠0

# data[].attributes.body
If n = 0, then yes; Paul is correct

# data[].attributes.body
I agree with Joseph. If n is 1, 2n square will be 4. 2+n will be 3 and 2n will only be 2

# data[].attributes.body
Unless n=0, 2+n would not be the largest

# data[].attributes.body
I agree with Joseph. If value of n is 5, then 2n square would be 100 and 2+n would be 7. 100>7. However if the value of n is 0, then 2n square would be 0 and 2+n would be 2. 0<2

# data[].attributes.body
I agree with Joseph because I like his name and also because N squared will be bigger than 2+n  and 2xn  as long as n is not 0 or 1 because if n = 1 then 1 squared x 2 = 2 while 2 + 1 (Paul)
will get 3

# data[].attributes.body
I agree with Joseph. Assume That n=2,
2n square = 16
2+n = 4
2n = 4

2n square is the largest.
However, if n=0, then 2+n would be the largest.

# data[].attributes.body
I agree with Joseph. However, if the value is 0 or 1, then 2n square will be lesser than 2+n. (e.g. If n = 1, 2n Square = 2. While 2+n = 3.) If the value of n is greater than 1, 2n square will be greater than 2+n and 2n.
(e.g. if n = 6, 2n square = 72. While 2+n = 8 and 2n = 12. )

# data[].attributes.body
I only agree with Paul if the value of n is 0.

# data[].attributes.body
I agree with Joseph. If the value of n is 5, then 2n square would be 100 which is greater than 2+n (which would give you 7) and 2n (which would give you 10). However, if the value of n is 0, then 2n square would give you 0 which is lesser than 2+n (which would give you 2).

# data[].attributes.body
josep is correct as n is the same value in all scenarios and n squared is larger than 2n ans 2+n in the case that n is not equal to 2, even so 2n squared is essentially (n+n) squared, which would provide an answer larger than n squared assuming that n is not 0.

# data[].attributes.body
2n can only bigger than 2nsquared if n is >0 and <1. But this will cause it to be smaller than 2+n.

# data[].attributes.headline
I agree with Joseph only if the value of n is bigger than 1. This is because 2*1squared is 2, which is smaller than 2+1. But let's say the value of n is 2, 2*2squared is 8, which is bigger than 2+2.

# data[].attributes.subject
I agree with Joseph only if the value of n is bigger than 1. This is because 2*1squared is 2, which is smaller than 2+1. But let's say the value of n is 2, 2*2squared is 8, which is bigger than 2+2.

# data[].attributes.body
I only agree if n is less than 1. My explanation is below Joseph.

# data[].attributes.body
refer to Joseph (2+0=2)

# data[].attributes.body
it depends, as if n=0, 2n=0, while 2+n would be 2, however, if n=13, 2*169=388, which is more than 15 and 26

# data[].attributes.body
Submit your response as a new post

# data[].attributes.body
Submit your response as a new post

# data[].attributes.body
Submit your response as a new post