We can use the fact that, for
,

Notice that
![\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1{1-x}\right]=\dfrac1{(1-x)^2}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Cdfrac1%7B1-x%7D%5Cright%5D%3D%5Cdfrac1%7B%281-x%29%5E2%7D)
so that
![f(x)=\displaystyle\frac5{(1-x)^2}=5\frac{\mathrm d}{\mathrm dx}\left[\sum_{n=0}^\infty x^n\right]](https://tex.z-dn.net/?f=f%28x%29%3D%5Cdisplaystyle%5Cfrac5%7B%281-x%29%5E2%7D%3D5%5Cfrac%7B%5Cmathrm%20d%7D%7B%5Cmathrm%20dx%7D%5Cleft%5B%5Csum_%7Bn%3D0%7D%5E%5Cinfty%20x%5En%5Cright%5D)



By the ratio test, this series converges if

so the series has radius of convergence
.
Answer:
1/sqrt10
Step-by-step explanation:
1) Find out cosA using formula (cosA)^2+(sinA)^2=1
The module of cosA= sqrt (1- (-3/5)^2)= sqrt 16/25=4/5
So cosA=-4/5 or cosA=4/5.
Due to the condition 270degrees< A<360 degrees, 0<cosA<1 that's why cosA=4/5.
2) Find sinA/2 using a formula cosA= 1-2sinA/2*sinA/2 where cosA=4/5.
(sinA/2)^2= 0.1
sinA= sqrt 0.1= 1/ sqrt10 or sinA= - sqrt 0.1= -1/sqrt10
But 270°< A< 360°, then 270/2°<A/2<360/2°
135°<A/2<180°, so sinA/2 must be positive and the only correct answer is
sin A/2= 1/sqrt10
Answer:
y = -x - 8
Step-by-step explanation:
Slope intercept form is y = mx + b, so subtract x from both sides to convert the equation to slope-intercept form.