Please help me!! Thank you

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

use formula for the circle a=3.14

step-by-step explanation:

none.

step-by-step explanation:

m

1

=

5

m1=5

b

=

−

1

b=-1

find the slope and y-intercept of the second equation.

tap for more

m

2

=

0

m2=0

b

=

−

6

b=-6

compare the slopes

m

m of the two equations.

m

1

=

5

,

m

2

=

0

m1=5,m2=0

compare the two slopes in the decimal form. if the slopes are equal, then the lines are parallel. if the slopes are not equal, then the lines are not parallel.

m

1

=

5

,

m

2

=

0

m1=5,m2=0