Given:
The equation of the curve is:

To find:
The gradient (slope) of the given curve at point (2,7).
Solution:
We have,

Differentiate the given equation with respect to x.


Now we need to find the value of this derivative at (2,7).




Therefore, the gradient (slope) of the given curve at point (2,7) is 19.
1.556302501 or 1.556
hope this helps
Answer:
first y then x because y represents 22 and x represents 4 a week.
Answer:
2.50*10 350+d
Step-by-step explanation:
- thats the answer up below