Question

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: (i) 13/3125 (ii) 17/8 (iii) 64/455 (iv) 15/1600 (v) 29/343 (vi) 23/(2352) (vii) 129/(225775) (viii) 6/15 (ix) 35/50 (x) 77/210​

Answer 1

Step-by-step explanation:

If the denominator has only factors of 2 and 5 or in the form of 2m ×5n then it has terminating decimal expansion.

If the denominator has factors other than 2 and 5 then it has a non-terminating decimal expansion.

(i) 13/3125

Factorizing the denominator, we get,

3125 = 5 × 5 × 5 = 55

Since, the denominator has only 5 as its factor, 13/3125 has a terminating decimal expansion.

(ii) 17/8

Factorizing the denominator, we get,

8 = 2×2×2 = 23

Since, the denominator has only 2 as its factor, 17/8 has a terminating decimal expansion.

(iii) 64/455

Factorizing the denominator, we get,

455 = 5×7×13

Since, the denominator is not in the form of 2m × 5n, thus 64/455 has a non-terminating decimal expansion.

(iv) 15/ 1600

Factorizing the denominator, we get,

1600 = 2652

Since, the denominator is in the form of 2m × 5n, thus 15/1600 has a terminating decimal expansion.

(v) 29/343

Factorizing the denominator, we get,

343 = 7×7×7 = 73 Since, the denominator is not in the form of 2m × 5n thus 29/343 has a non-terminating decimal expansion.

(vi)23/(2352)

Clearly, the denominator is in the form of 2m × 5n.

Hence, 23/ (2352) has a terminating decimal expansion.

(vii) 129/(225775)

As you can see, the denominator is not in the form of 2m × 5n.

Hence, 129/ (225775) has a non-terminating decimal expansion.

(viii) 6/15

6/15 = 2/5

Since, the denominator has only 5 as its factor, thus, 6/15 has a terminating decimal expansion.

(ix) 35/50

35/50 = 7/10

Factorising the denominator, we get,

10 = 2 5

Since, the denominator is in the form of 2m × 5n thus, 35/50 has a terminating decimal expansion.

(x) 77/210

77/210 = (7× 11)/ (30 × 7) = 11/30

Factorising the denominator, we get,

30 = 2 × 3 × 5

As you can see, the denominator is not in the form of 2m × 5n .Hence, 77/210 has a non-terminating decimal expansion.

hope it helps you please make me as brainist

Answer 2

Answer:

Theorem: Let x=

qp

be a rational number, such that the prime factorisation of q is of the form 2

n

5

m

, where n, m are non-negative integers. Then, x has a decimal expansion which terminates.

(i) 312/513

Factorise the denominator, we get

3125=5×5×5×5×5=5

5

So, denominator is in form of 5

m

so, 312513

is terminating.

(ii) 8/17

Factorise the denominator, we get

8=2×2×2=2

3

So, denominator is in form of 2

n

so,

8/17 is terminating.

(iii) 455/64

Factorise the denominator, we get

455=5×7×13

So, denominator is not in form of 2

n 5 m so, 455/64

is not terminating.

(iv)

1600

15

Factorise the denominator, we get

1600=2×2×2×2×2×2×5×5=2 6 5 2

So, denominator is in form of 2

n 5m

so, 160015

is terminating.

(v) 343/29

Factorise the denominator, we get

343=7×7×7=7

3

So, denominator is not in form of 2

n 5 m

so, 34329

is not terminating.

(vi) 2 3 5 2

2३

Here, the denominator is in form of 2

n 5 m so, 2 3 5 2

23

is terminating.

(vii)

22 5 7 7 5 129

Here, the denominator is not in form of 2

n 5 m

so, 2 2 5 7 75

129

is not terminating.

(viii) 15/6

Divide nominator and denominator both by 3 we get 15/3

So, denominator is in form of 5

m so, 15/6

is terminating.

(ix) 50/35

Divide nominator and denominator both by 5 we get 10/7

Factorise the denominator, we get

10=2×5

So, denominator is in form of 2

n 5 m

so, 50/35

is terminating.

(x)

210/77

Divide nominator and denominator both by 7 we get

30/11

Factorise the denominator, we get

30=2×3×5

So, denominator is not in the form of 2 n

5 m so 15/6

is not terminating.

♡──━━━━━━⊱✿⊰━━━━━━──♡

$\huge \pink{✿} \red {C} \green {u} \blue {t} \orange {e} \pink {/} \red {Q} \blue {u} \pink {e} \red {e} \green {n} \pink {♡}$