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https://imgur.com/a/aK0UKl0

Posted a picture of the problem on here. Can't seem to figure it out thanks in advance.

 Dec 13, 2019
 #1
avatar+106539 
0

When t  = 0     we have this equation

 

n  +  a  =  2       (1)

 

When t  = 1   we have this equation 

 

n + m   =   3       (2)

 

When  t  = 2   we have this equation

 

n + 2m  - ab^2   = 20      (3)

 

When  t = 3  we have this equation

 

n + 3m  =  1     (4)

 

And  when  t  =4  we have this

 

n + 4m +  ab^t  =   -162     (5)

 

Using (2)  and (4) we have that

 

n + m  = 3

n  + 3m  =  1        subtract the first equation from the second and we have that

 

2m  =  -2

m  = -1

So....  n =  4

 

And using (1)

 

4  +  a  = 2

a  = -2

 

And using (3)  we have that

 

n+ 2m  -ab^2  = 20

 

Subbing we have that

 

4  + 2(-1)  - (-2)b^2  = 20

2  + 2b^2  = 20

1  + b^2  =  10

b^2  = 9

b =   3  or  b  = -3

 

Assuming that  b  is positive

 

 

So  using 5...if   n = 4, m= -1, a =-2  and  b  = 3    and  t  = 4  we have that

 

4 - 4  + (-2)(3)^4   =

 

-2 * 81  = 162

 

So  

 

n = 4

m = -1

a = -2

b = 3

 

 

Here's a graph using  x  instead of  t  : https://www.desmos.com/calculator/b677bwjlwf

 

 

cool cool cool

 Dec 14, 2019

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