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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}

$2416.6

step-by-step explanation:

per hour = $ 25

total work time = 35

commission = 15%

total she has to make = $ 4,500

let she sold x dollar products

so her commission will be 15 % of x

now

total = weekly basic pay + commission

4500= (25*35) + 15 % of x

4500=875 + * x

4500-875 =   * x

3625= * x

multiplying wwith 100 on both sides gives

36250 = 15 x

now

x =  

x= $2416.6

0.038

step-by-step explanation:

the answer for this question is a= 13,622.84

step-by-step explanation: