Decimal expansion of rational numbers

(0.254545\ldots=\frac{252}{990}=\frac{14}{55}), so the denominator is (55). In such questions, reduce the final fraction fully.

The non-repeating part (2) and repeating part (54) give (\frac{252}{990}), which reduces to (\frac{14}{55}). In exams, identify repeating and non-repeating digits separately.

Since (147=3\cdot 7^2), the reduced denominator is (2\cdot 5^4). The larger exponent is (4), so the decimal terminates after (4) places.

For exactly (7) places, the larger exponent of (2) and (5) must be (7). Only (2^7\cdot 5^3) satisfies this.

A rational number has a terminating decimal exactly when its denominator (in lowest terms) is of the form \(2^m5^n\). The number of decimal places needed equals max(m,n). Here 29 is coprime to 2 and 5, so the denominator remains \(2^a5^b\) after simplification. Given \(a>b\), the maximum exponent is \(a\); for exactly 8 decimal places we must have \(a=8\). Option 7 is wrong because it would give only 7 decimal digits; \(a+b\) is irrelevant since the decimal length depends on the maximum exponent, not the sum. Exam tip: always reduce the fraction first and then take the larger of the exponents of 2 and 5 to get the number of decimal places.

(0.00096=\frac{96}{100000}), and reducing by (32) gives (\frac{3}{3125}). Even for small decimals, check the greatest common factor carefully.

(0.00096=\frac{96}{100000}=\frac{3}{3125}), so the denominator is (3125). Convert the decimal to a fraction and reduce fully.

The factor (13) makes the decimal recurring, and the larger exponent among (2) and (5) is (5), giving the initial non-repeating part. Understand recurrence and delay separately.

Since (121=11^2), the reduced denominator is (2^3\cdot 5^2). Therefore the reason correctly explains the terminating decimal rule.

(0.04\overline{6}) has a fixed repeating digit, so it is rational but not terminating. A decimal is terminating only when zeros continue after some point.

The direct answer is C, 6 decimal places. First cancel common factors: \\(5^4\\) cancels from numerator and denominator, and 7 also cancels. The fraction becomes \\(1/(2^6\\cdot5^3)\\). A rational number has a terminating decimal when its reduced denominator contains only factors 2 and 5. To convert the denominator into a power of 10, match the smaller exponent 3 by multiplying conceptually by \\(2^3\\); then the denominator has \\(2^6\\), giving \\(10^3\\cdot2^3=2^6\\cdot5^3\\). Therefore six places are needed. Option A, 3, is only the exponent of 5 and ignores the larger exponent of 2. Option B, 5, is not the larger required exponent. Option C, 6, is correct. Option D, 7, has no basis after cancellation. Always reduce first, then use the larger exponent of 2 and 5.

Since (q) has only (2) and (5), the decimal terminates. Not dividing (10^7) means the larger exponent is (8) or (9).

(0.00\overline{63}=\frac{63}{9900}=\frac{7}{1100}), so the denominator is (1100). In recurring decimals, the first denominator formed may not be final.

Two non-repeating zeros and two repeating digits give (\frac{63}{9900}). Reducing it gives (\frac{7}{1100}).

The direct answer is B, 5 decimal places. Simplify the powers separately: \\(2^4/2^9=1/2^5\\) and \\(5^3/5^5=1/5^2\\). Thus the fraction becomes \\(1/(2^5\\cdot5^2)\\). The reduced denominator has only 2 and 5, so the decimal terminates. The number of places is the larger exponent, 5, because \\(2^5\\cdot5^2\\) can be completed to \\(10^5\\) by supplying three more factors of 5. Option A, 2, uses only the exponent of 5 and ignores the exponent of 2. Option B, 5, is correct. Option C, 7, incorrectly adds the exponents instead of taking the larger one. Option D, 9, uses the original denominator exponents without cancellation. Reduce powers before deciding. Memory cue: for a reduced denominator made only of 2s and 5s, the larger exponent gives the terminating length.

(10^k) has only prime factors (2) and (5), so any divisor gives a terminating decimal. The other statements are not always true because extra factors may occur.

The answer is B, 0.38. A decimal ending in infinitely many 9s has an equivalent terminating form because 0.00999… = 0.01000…, just as 0.37999… = 0.38000… = 0.38. Step by step, the 9s fill the hundredths place up to the next hundredth: 0.37 + 0.00999… = 0.37 + 0.01 = 0.38. A, 0.37, is too small because the repeating 9s add more than zero. B is correct because it is the equal terminating decimal. C, 379/999, is not the value; that fraction represents 0.379379… after division. D, 3799/10000, equals 0.3799, which stops and is also smaller. Memory cue: an ending block of 9s carries one unit to the preceding digit.

The factors (3^3) and (23) must be removed from the reduced denominator, so the minimum factor is (27\cdot 23=621). Factors (2) and (5) may remain.

The direct answer is C, non-terminating recurring. Factor the numerator: \\(175=5^2\\cdot7\\). The denominator is \\(2^2\\cdot5^3\\cdot7^2\\). Cancel \\(5^2\\) and one factor of 7. The reduced denominator is \\(2^2\\cdot5\\cdot7\\). A rational decimal terminates only when the reduced denominator has no prime factors other than 2 and 5. The factor 7 remains, so the decimal cannot terminate; because the number is rational, its non-terminating decimal is recurring. Option A, terminating after 3 places, ignores the remaining factor 7. Option B, terminating after 2 places, is also impossible because 7 remains. Option C is correct. Option D, non-terminating non-recurring, describes irrational decimals, not a rational fraction. Always reduce before applying the rule.

This decimal does not end, and the number of zeros keeps changing. Since there is no fixed repeating block, it is non-terminating non-recurring.

Since (10^7=2^7\cdot 5^7), the denominator lacks (2^4). Thus (N=13\cdot 16=208), so the correct value is not listed.

Key idea: \(10^7=2^7\cdot 5^7\). The given denominator has only \(2^3\), so multiply numerator and denominator by \(2^4\) to raise the power of 2 to 7. Thus \(N=13\cdot 2^4=13\cdot16=208\). Option C (416) would be \(13\cdot2^5\) — one extra factor of 2 — and A (104) corresponds to \(13\cdot2^3\), so both are incorrect. Exam tip: match prime powers of 2 and 5 to form \(10^n\) quickly by comparing exponents.

The reduced denominator is (10^6), so the decimal terminates exactly after (6) places. If the denominator is reduced, do not assume further cancellation.

Since (81=3^4), the reduced denominator is (2^4\cdot 5^6). The larger exponent is (6), so the decimal terminates after (6) places.

At most (7) decimal places means the fraction can be written with denominator (10^7). Therefore the reduced denominator must divide (10^7).