Question

1
2
1
13. Simiplify the following by rationalising the denominators To+V5 15+17
√7 +16​

Answer 1

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Answer 2

Answer:

(i)

13

5

Since, 5=20×51

⇒ The denominator is in the form of 2m×5n, where m and n are non-negative integers.

So, the decimal form of

13

5

will be terminating type.

(ii)

2

11

Since, 11=20×50×111

⇒ The denominator is not in the form of 2m×5n, where m and n are non-negative integers.

So, the decimal form of

2

11

will be non-terminating recurring type.

(iii)

29

16

Since, 16=24×50

⇒ The denominator is in the form of 2m×5n, where m and n are non-negative integers.

So, the decimal form of

29

16

will be terminating type.

(iv)

17

125

Since, 125=20×53

⇒ The denominator is in the form of 2m×5n, where m and n are non-negative integers.

So, the decimal form of

17

125

will be terminating type.

(v)

11

6

Since, 6=21×50×31

⇒ The denominator is not in the form of 2m×5n, where m and n are non-negative integers.

So, the decimal form of

11

6

will be non-terminating recurring type.

Step-by-step explanation:

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