Answer:
Radius of the convergence is R = 
and,
The Interval of convergence is 
Step-by-step explanation:
Given function : Σ(21x)^k
Now,
Using the ratio test, we have
R = 
or
R = 
or
R = 
now,
for convergence R|x| < 1
Therefore,
< 1
or
and,
Radius of the convergence is R =
and,
The Interval of convergence is
Answer:
x = 348
Step-by-step explanation:
The measure of angle ABD is (0.15x+65) and the measure of angle CBD is(0.1x+28) . Find the value of x.
0.15x + 65 + 0.1x + 28 = 180
0.25x + 93 = 180
0.25x = 180 - 93
0.25x = 87
x = 87/0.25
x = 348
Answer:
(B) Write Negative 3 and one-third as Negative StartFraction 10 over 3 EndFraction, and find the reciprocal of StartFraction 4 over 9 EndFraction as StartFraction 9 over 4 EndFraction. Then, rewrite Negative 3 and one-third divided by StartFraction 4 over 9 EndFraction as Negative StartFraction 10 over 3 EndFraction times StartFraction 9 over 4 EndFraction. The quotient is Negative 7 and StartFraction 6 over 12 EndFraction = Negative 7 and one-half.
Step-by-step explanation:
To find the quotient of the division:
Step 1: 
Step 2: Find the reciprocal of 
see the attached figure with the letters
1) find m(x) in the interval A,BA (0,100) B(50,40) -------------- > p=(y2-y1(/(x2-x1)=(40-100)/(50-0)=-6/5
m=px+b---------- > 100=(-6/5)*0 +b------------- > b=100
mAB=(-6/5)x+100
2) find m(x) in the interval B,CB(50,40) C(100,100) -------------- > p=(y2-y1(/(x2-x1)=(100-40)/(100-50)=6/5
m=px+b---------- > 40=(6/5)*50 +b------------- > b=-20
mBC=(6/5)x-20
3)
find n(x) in the interval A,BA (0,0) B(50,60) -------------- > p=(y2-y1(/(x2-x1)=(60)/(50)=6/5
n=px+b---------- > 0=(6/5)*0 +b------------- > b=0
nAB=(6/5)x
4) find n(x) in the interval B,CB(50,60) C(100,90) -------------- > p=(y2-y1(/(x2-x1)=(90-60)/(100-50)=3/5
n=px+b---------- > 60=(3/5)*50 +b------------- > b=30
nBC=(3/5)x+30
5) find h(x) = n(m(x)) in the interval A,B
mAB=(-6/5)x+100
nAB=(6/5)x
then
n(m(x))=(6/5)*[(-6/5)x+100]=(-36/25)x+120
h(x)=(-36/25)x+120
find <span>h'(x)
</span>h'(x)=-36/25=-1.44
6) find h(x) = n(m(x)) in the interval B,C
mBC=(6/5)x-20
nBC=(3/5)x+30
then
n(m(x))=(3/5)*[(6/5)x-20]+30 =(18/25)x-12+30=(18/25)x+18
h(x)=(18/25)x+18
find h'(x)
h'(x)=18/25=0.72
for the interval (A,B) h'(x)=-1.44
for the interval (B,C) h'(x)= 0.72
<span> h'(x) = 1.44 ------------ > not exist</span>
Answer:
x=-17.......................
Step-by-step explanation:
x+17=0
