Respuesta :
Answer:
1) Neither
2) Exponential
3) Exponential
4) Linear
Step-by-step explanation:
1) x-values [tex]{}[/tex] y-values
0 [tex]{}[/tex] 0
1 [tex]{}[/tex] 1
2 [tex]{}[/tex] 4
3 [tex]{}[/tex] 9
4 [tex]{}[/tex] 16
Therefore, we have, the relationship of the function given as_f(x) = x² Which is neither a linear function nor an exponential function
2) x-values [tex]{}[/tex] y-values
0 [tex]{}[/tex] 1/3
1 [tex]{}[/tex] 1
2 [tex]{}[/tex] 3
3 [tex]{}[/tex] 9
4 [tex]{}[/tex] 27
The relationship between 'x', and 'y' is given as follows;
f(x) = 3⁽ˣ⁻¹⁾
Therefore, the relationship between 'x', and 'y', is an exponential relationship, and the function is an exponential function
3) x-values [tex]{}[/tex] y-values
0 [tex]{}[/tex] 5
1 [tex]{}[/tex] 5/2
2 [tex]{}[/tex] 5/4
3 [tex]{}[/tex] 5/8
4 [tex]{}[/tex] 5/16
The relationship between 'x', and 'y' is given as follows;
f(x) = 5×(1/2)ˣ
Therefore, the relationship between 'x', and 'y', is an exponential relationship, and the function is an exponential function
4) x-values [tex]{}[/tex] y-values
0 [tex]{}[/tex] 4.5
1 [tex]{}[/tex] 4
2 [tex]{}[/tex] 3.5
3 [tex]{}[/tex] 3
4 [tex]{}[/tex] 2.5
The given data for the y-values has a constant first common difference of 4 - 4.5 = -0.5, therefore, the relationship between the 'x', and 'y' values is a linear relationship.