Show that 43\2^4×5^3 will terminate after how many places of the decimal​

4 places after the decimal.

the result is 0.0215

Step-by-step explanation:

I assume the expression is really

43 / (2⁴ × 5³)

this is the same as

43 / 2) / 2) / 2) / 2) / 5) / 5) / 5)

since the starting value is an odd number, the first division by 2 creates a first position after the decimal point, and it must be a 5, as the result is xx.5

the second division by 2 splits again the uneven end .5 in half, creating a second position after the decimal point again ending in 5, as the result is now xx.x5

the third division by 2 does the same thing with that last 5 and creates a third position after the decimal point ending again in 5, as the result is now xx.xx5

the fourth division by 2 does again the same thing, a fourth position after the decimal point is created ending in 5. now xx.xxx5

in essence, every division of the 0.5 part by 2 is the same as a multiplication by 0.5, which squares 0.5 leading to 0.5². the next division did the same thing leading to 0.5³.

and finally the fourth division to 0.5⁴.

0.5⁴ = (5/10)⁴ = 5⁴/10⁴

so, now we start to divide this result by 5. since the positions after the decimal point are divisible by 5 without remainder, as we have 5⁴ to work with.

every divisible by 5 takes one of these powers away.

so, we go from 5⁴/10⁴ to 5³/10⁴ to 5²/10⁴ to 5/10⁴.

all the time we maintain the 10⁴ in the denominator of the fraction. and that determines the positions after the decimal point.

so, after all the individual divisions we come to and end and are still limited to the 4 positions after the decimal point.

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}