Answer:
Power series representation of given f(x) is ![\sum^{\infty}_{n=0}(-1)^{n}(\frac{x^{6n+3}}{a^{6n+6}})](https://tex.z-dn.net/?f=%5Csum%5E%7B%5Cinfty%7D_%7Bn%3D0%7D%28-1%29%5E%7Bn%7D%28%5Cfrac%7Bx%5E%7B6n%2B3%7D%7D%7Ba%5E%7B6n%2B6%7D%7D%29)
Step-by-step explanation:
Given that:
--- (1)
writing it in form
for which power series is:
---(2)
--- (3)
consider:
![u=\frac{x^{6}}{a^{6}}](https://tex.z-dn.net/?f=u%3D%5Cfrac%7Bx%5E%7B6%7D%7D%7Ba%5E%7B6%7D%7D)
substituting above eq. in (3)
![f(x)=\frac{x^{3}}{a^{6}}\frac{1}{1-u}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7Bx%5E%7B3%7D%7D%7Ba%5E%7B6%7D%7D%5Cfrac%7B1%7D%7B1-u%7D)
Expanding
using (2)
![=\frac{x^{3}}{a^{6}}\sum^{\infty}_{n=0}(-1)^{n}u^{n}](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bx%5E%7B3%7D%7D%7Ba%5E%7B6%7D%7D%5Csum%5E%7B%5Cinfty%7D_%7Bn%3D0%7D%28-1%29%5E%7Bn%7Du%5E%7Bn%7D)
Back-substituting value of u in above
![=\frac{x^{3}}{a^{6}}\sum^{\infty}_{n=0}(-1)^{n}(\frac{x^{6}}{a^{6}})^{n}\\=\sum^{\infty}_{n=0}(-1)^{n}(\frac{x^{6n+3}}{a^{6n+6}})](https://tex.z-dn.net/?f=%3D%5Cfrac%7Bx%5E%7B3%7D%7D%7Ba%5E%7B6%7D%7D%5Csum%5E%7B%5Cinfty%7D_%7Bn%3D0%7D%28-1%29%5E%7Bn%7D%28%5Cfrac%7Bx%5E%7B6%7D%7D%7Ba%5E%7B6%7D%7D%29%5E%7Bn%7D%5C%5C%3D%5Csum%5E%7B%5Cinfty%7D_%7Bn%3D0%7D%28-1%29%5E%7Bn%7D%28%5Cfrac%7Bx%5E%7B6n%2B3%7D%7D%7Ba%5E%7B6n%2B6%7D%7D%29)
Answer:
8x - 14 = 5x + 34 (alternate exterior angles)
8x - 5x = 34 + 14
3x = 48
x = 48/3
x = 16
5y + 16 + 8x - 14 = 180 (linear pair)
5y + 16 + 8(16) - 14 = 180
5y + 16 + 128 - 14 = 180
5y + 130 = 180
5y = 180 - 130
5y = 50
y = 50/5
y = 10
∠ (8x - 14) = 114 [(8 × 16 -14)=114]
∠ (5x + 34) = 114 [(5 × 16 +34)=114]
∠ (5y + 16) = 66 [(5 × 10 +16)=66]
hope this helps you!
The question is wrong can you explain.
Answer:
(-1, 2) will be point A'
Step-by-step explanation:
Answer:
5
Step-by-step explanation:
2x + 4 = 14 ( - 4 )
2x + 10 ( /2)
x= 5