From what I am looking at, it seems as if it is. :)
Given:
The sequence is 25, 20, 15, 10, 5.
To find:
The ninth term of the given sequence.
Solution:
We have,
25, 20, 15, 10, 5
It is an AP because the difference between two consecutive terms are same.
Here,
First term (a) = 25
Common difference (d) = 20-25
= -5
The nth terms of an AP is

Where, a is the first term and d is the common difference.
Putting a=25, n=9 and d=-5 to get the 9th term.




Therefore, the ninth term of the given sequence is -15.
Answer:
well 126/9=14 but is the other stuff factor families or something? like whats the question asking
Step-by-step explanation:
Answer:
0.0005 I believe
Step-by-step explanation:
Answer:

Step-by-step explanation:
![f(x)=4\sqrt{2x^3-1}=4\left(2x^3-1\right)^\frac{1}{2}\\\\f'(x)=4\cdot\dfrac{1}{2}(2x^3-1)^{-\frac{1}{2}}\cdot3\cdot2x^2=\dfrac{12x^2}{(2x^3-1)^\frac{1}{2}}=\dfrac{12x^2}{\sqrt{2x^3-1}}\\\\\text{used}\\\\\sqrt{a}=a^\frac{1}{2}\\\\\bigg[f\left(g(x)\right)\bigg]'=f'(g(x))\cdot g'(x)\\\\\bigg[nf(x)\bigg]'=nf'(x)\\\\(x^n)'=nx^{n-1}](https://tex.z-dn.net/?f=f%28x%29%3D4%5Csqrt%7B2x%5E3-1%7D%3D4%5Cleft%282x%5E3-1%5Cright%29%5E%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5Cf%27%28x%29%3D4%5Ccdot%5Cdfrac%7B1%7D%7B2%7D%282x%5E3-1%29%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%5Ccdot3%5Ccdot2x%5E2%3D%5Cdfrac%7B12x%5E2%7D%7B%282x%5E3-1%29%5E%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cdfrac%7B12x%5E2%7D%7B%5Csqrt%7B2x%5E3-1%7D%7D%5C%5C%5C%5C%5Ctext%7Bused%7D%5C%5C%5C%5C%5Csqrt%7Ba%7D%3Da%5E%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%5Cbigg%5Bf%5Cleft%28g%28x%29%5Cright%29%5Cbigg%5D%27%3Df%27%28g%28x%29%29%5Ccdot%20g%27%28x%29%5C%5C%5C%5C%5Cbigg%5Bnf%28x%29%5Cbigg%5D%27%3Dnf%27%28x%29%5C%5C%5C%5C%28x%5En%29%27%3Dnx%5E%7Bn-1%7D)