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ipn [44]
3 years ago
15

How can you tell that a fraction corresponds to a repating decimal and not a terminating decimal

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
8 0
If the primes divisors of denominator are only 2 and 5, then a fraction corresponds terminating decimal. If not, then a fraction corresponds repeating decimal.

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Find the sum or difference. a. -121 2 + 41 2 b. -0.35 - (-0.25)
s344n2d4d5 [400]

Answer:

2

Step-by-step explanation:

The reason an infinite sum like 1 + 1/2 + 1/4 + · · · can have a definite value is that one is really looking at the sequence of numbers

1

1 + 1/2 = 3/2

1 + 1/2 + 1/4 = 7/4

1 + 1/2 + 1/4 + 1/8 = 15/8

etc.,

and this sequence of numbers (1, 3/2, 7/4, 15/8, . . . ) is converging to a limit. It is this limit which we call the "value" of the infinite sum.

How do we find this value?

If we assume it exists and just want to find what it is, let's call it S. Now

S = 1 + 1/2 + 1/4 + 1/8 + · · ·

so, if we multiply it by 1/2, we get

(1/2) S = 1/2 + 1/4 + 1/8 + 1/16 + · · ·

Now, if we subtract the second equation from the first, the 1/2, 1/4, 1/8, etc. all cancel, and we get S - (1/2)S = 1 which means S/2 = 1 and so S = 2.

This same technique can be used to find the sum of any "geometric series", that it, a series where each term is some number r times the previous term. If the first term is a, then the series is

S = a + a r + a r^2 + a r^3 + · · ·

so, multiplying both sides by r,

r S = a r + a r^2 + a r^3 + a r^4 + · · ·

and, subtracting the second equation from the first, you get S - r S = a which you can solve to get S = a/(1-r). Your example was the case a = 1, r = 1/2.

In using this technique, we have assumed that the infinite sum exists, then found the value. But we can also use it to tell whether the sum exists or not: if you look at the finite sum

S = a + a r + a r^2 + a r^3 + · · · + a r^n

then multiply by r to get

rS = a r + a r^2 + a r^3 + a r^4 + · · · + a r^(n+1)

and subtract the second from the first, the terms a r, a r^2, . . . , a r^n all cancel and you are left with S - r S = a - a r^(n+1), so

(IMAGE)

As long as |r| < 1, the term r^(n+1) will go to zero as n goes to infinity, so the finite sum S will approach a / (1-r) as n goes to infinity. Thus the value of the infinite sum is a / (1-r), and this also proves that the infinite sum exists, as long as |r| < 1.

In your example, the finite sums were

1 = 2 - 1/1

3/2 = 2 - 1/2

7/4 = 2 - 1/4

15/8 = 2 - 1/8

and so on; the nth finite sum is 2 - 1/2^n. This converges to 2 as n goes to infinity, so 2 is the value of the infinite sum.

8 0
3 years ago
Please help me please​
baherus [9]

Answer:

The answer is C

4 0
3 years ago
Solve the equation 0.3 = -15z.
melamori03 [73]

Answer:

z = - 0.02

Step-by-step explanation:

0.3 = -15z

0.3 / -15 = z

-0.02 = z

6 0
3 years ago
Read 2 more answers
8a + 12b = 92 6a – 4b = 4 -- a =? b = ?
Yanka [14]

Answer:

a = 4 b = 5

Step-by-step explanation:

8a + 12b = 92......eqn 1

6a - 4b = 4......eqn 2

multiply eqn 2 by 3

18a - 12b = 12.....eqn 2

add eqn 1 and 2 to eliminate b

26a = 104

a = 4

put a = 4 into eqn 2

18(4) - 12b = 12

72 - 12b = 12

72 - 12 = 12b

60 = 12b

5 = b

6 0
2 years ago
Help me on question 24 please and explain it thanks
Papessa [141]
You know that the sum of the interior angles in a triangle is 180
we will add all the angles,
(x+8)+(2x-8)+(3x-12)=180
x+8+2x-8+3x-12=180    8 and -8 cancel out 
6x-12=180
6x=180+12
6x=192
x=192/6=32
x=32

8 0
3 years ago
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