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You are going to take a stand.

Decide who you agree with - Joseph, Paul or Antony.

In each case, either explain based on concepts or examples to convience others.

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connect your post to me if you agree with me! 

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connect your post to me if you agree with me! 

# data[].attributes.attachment_caption
connect your post to me if you agree with me! 

# data[].attributes.custom_properties.wC9Zh8UE
If n = 2
Paul 2+2=4
Antony 2*2=4
Joseph 2*2*2=8
If n = 3
Paul 2+3=5
Antony 2*3=6
Joseph 2*3*3=18
If n = 4
Paul 2+4=6
Antony 2*4=8
Joseph 2*4*4=32

So as you can see, the expression Joseph picked gave the largest answer in all the examples.

# data[].attributes.custom_properties.wC9Zh8UE
2+n= 2+n
2n=2xn
2n squared= 2xnx2xn
Thus 2n squared is the greatest

# data[].attributes.custom_properties.wC9Zh8UE
Joseph would be 
Let n be 2 
2n = 2x2 = 4
2+n = 2+2 = 4
2n^2 = 2x2^2 = 8 
But if n is either 1 or 0, Paul would be correct; if not, Joseph will be correct.
Let n be 0 
2n = 2x0 = 0
2+n = 2+0 = 2
2n^2 = 2x0^2 = 0
Therefore, Paul could be correct

# data[].attributes.custom_properties.wC9Zh8UE
For any value of n that is not 0 or 1, Joseph will have the largest value. For example if n=3:

Paul -> 2+3 = 5
Antony -> 2*3 = 6
Joseph 2*3^2 = 18
Thus, Joseph can be seen to have the greatest value, in this example

However for any value of n which has an exponent of 2 equal to n, Joseph will not have the greatest value.

# data[].attributes.custom_properties.wC9Zh8UE
2xn=2n
2n square = 2n x 2n
                  

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Let n = 4
Joseph 2n^2 = 2x4^2 = 32
Paul 2+n = 2+4 = 6
Antony 2n = 2x4 = 8

Thus, Joseph is correct as 2n^2 is the largest.

# data[].attributes.custom_properties.wC9Zh8UE
2n squared is 2n x2n 
                     =(2xn)(2xn)

# data[].attributes.custom_properties.wC9Zh8UE
If n is equal to or greater than 2, the answer would be the greatest.  However, if n is 1, the end result would be the same as Anthony's and less than Paul's. However, if the n < 1, n Joseph's answer would be the least. soooooo it would depen but most likely n > 1.



I think.

# data[].attributes.custom_properties.wC9Zh8UE
If n is 100,
2+n=102
2n square=20 000
2n=200
From this example, 2n square is the largest, thus making me agree with Joseph that 2n square is the largest.

# data[].attributes.custom_properties.wC9Zh8UE
let n= any prime or composite number

Example: n=7

2x7^2=2x49
           =98

2+n=2+7
       =9

thus as long as n is equal to any prime or composite number, 2n^2 is the equation with the largest value

# data[].attributes.custom_properties.wC9Zh8UE
2n squared will get you 2n x 2n which is (2 x n)(2 x n)

# data[].attributes.custom_properties.wC9Zh8UE
Lets say n=1 
Joseph:
2n^2=4
Paul:
2+n=3
Anthony:
2n=2
Thus using this example, Joseph's 2n^2 is the largest.

# data[].attributes.custom_properties.wC9Zh8UE
2n squared=2 x n x2 x n
2n=2 x n
2+n=2+n
2n squared has the largest value.
if n is negative, then 2 x (negative) x 2 x(negative), which can be interpreted as 2 x 2 x (negative) x (negative) =4 x (positive) and 2 n squared will still be the largest value.

# data[].attributes.custom_properties.wC9Zh8UE
Let n be 50.
2 + n = 52
2n = 100
2n square = 10 000

thus 2n square is the largest

# data[].attributes.custom_properties.wC9Zh8UE
If n is greater than 1, Joseph will always be correct. E.g. if n = 2 
Joseph: 2x2^2=16
Paul: 2+2=4
Antony: 2x2=4

But if n is equal to 1, Paul would be correct.
Joseph: 2x1^1=2
Paul: 2+1 = 3
Antony: 2x1=1

# data[].attributes.custom_properties.wC9Zh8UE
if n is more than 0 or 1 joseph will have the most.
e.g 

paul: 3+2 +5
antony:2x3 =6
joseph: 2x3squared = 18

so, joseph can be seen as the one with the most value. but is the number is -1, 0, or 1 then he will not have the most

# data[].attributes.custom_properties.wC9Zh8UE
even if its a negative number that is not -1, 0, 1, e.t.c around there for decimal points, it multiplies (itself times itself) by (itself times itself). which is basically taking it and notating it to the power of 4

# data[].attributes.custom_properties.wC9Zh8UE
If n > 1, Joseph’s expression 2n^2 will always give the largest value.

Example when n=2:
Joseph: 2 x 2^2 = 8
Paul: 2+2=4
Antony: 2×2=4

But when n=1, Paul’s expression 2+n is the largest.
Joseph: 2×1^2 =2
Paul: 2+1=3
Antony: 2×1=2