No circle can never be the perfect squares
To find f(3), plug x = 3 in f(x),
∴ f(3) = 3(3)+1 = 9+1 = 10
Therefore, the ordered pair is (3,10)
For second one,
plug x = 10 in f⁻¹(x),
f⁻¹(10) = 10-1/3 = 9/3 = 3
Therefore, the ordered pair is (10,3)
Step-by-step explanation:
y = x + 2
X. <u>Y</u>
3. 5
4. 6
5. 7
6. 8
Based on this table, the graph that best represents the table is the one on the right.
I believe it would be the first choice
Hope that helps!
A I am pretty sure. I apologize if I am incorrect.
