The answer is 7. hope this helps
Given:
The equation of the curve is:

To find:
The gradient of the given curve at the point where
.
Solution:
We have,

Differentiate with respect to x.
![[\because \dfrac{d}{dx}x^n=nx^{n-1}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%5Cdfrac%7Bd%7D%7Bdx%7Dx%5En%3Dnx%5E%7Bn-1%7D%5D)

Substituting
, we get


Using properties of exponents, we get
![[\because (a^m)^n=a^{mn}]](https://tex.z-dn.net/?f=%5B%5Cbecause%20%28a%5Em%29%5En%3Da%5E%7Bmn%7D%5D)



Therefore, the gradient of the given curve at the point where
is 20.
9514 1404 393
Answer:
14
Step-by-step explanation:
The first term of the sequence is 182, and the common difference is 176-182=-6. So, the general term is ...
an = a1 +d(n -1)
an = 182 -6(n -1)
The 29th term will be ...
a29 = 182 -6(29 -1)
a29 = 14
Sorry knoww the answer not