Rational numbers

The governing idea is interpreting an open interval. The multiples of 7 from this range are 14,21,28,35. Because “between 14 and 35” excludes both endpoints, only 21 and 28 remain. Therefore option B is correct. Option A includes both endpoints, C includes 7 and 14, and D includes the excluded endpoint 35. The wording would need “inclusive” or “from 14 to 35” to include the endpoints.

Solving x²=9 over the integers requires considering both square roots of 9. Since 3²=9 and (-3)²=9, the integer solutions are x=3 and x=-3. A set lists both distinct solutions, so S={-3,3}. Option A and option B each omit one valid solution, while option D contains values that do not all satisfy the equation. Therefore C is correct.

Direct answer: Option B is correct. To find the value of an algebraic expression, substitute the given value of the variable first and then follow the operations. Here, the variable is x and its value is 3. Replace x by 3: 2x+5 becomes 2(3)+5. Multiply before adding: 2(3)=6. Then add 5: 6+5=11. Thus the answer is 11, which is option B. Option A, 8, would result from an incorrect operation such as adding 3 and 5 while forgetting the coefficient 2. Option B is correct because it follows direct substitution and multiplication. Option C, 13, may come from adding incorrectly or using an unsupported operation; it is not the value of the expression. Option D, 15, is also obtained by an incorrect calculation and does not follow the stated expression. Exam cue: replace the letter, use brackets, multiply first, and add last.

A rational number can be written as ( \frac{p}{q} ) where the denominator is not zero. Always check the denominator in exams.

The governing concept is the definition of a rational number. A number is rational if it can be written in the form p/q, where p and q are integers and q is not zero. The fraction 3/5 already has this form: the numerator 3 is an integer, the denominator 5 is an integer, and 5 is not zero. Therefore, option B is correct. It is not irrational because irrational numbers cannot be expressed as a ratio of two integers. Although 3 is a natural and whole number, the fraction 3/5 is not itself a whole number or a natural number. Its decimal form, 0.6, also terminates, which provides another familiar indication that it is rational.

( \frac{-7}{4} ) is a ratio of two integers with non-zero denominator. The given square-root numbers are not in simple rational form.

A number is rational if it can be written as \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q\neq 0\). Thus, \(\frac{0}{5}=0\) is a valid representation. Options B and D are undefined because their denominators are zero, and option C, \(\frac{0}{0}\), is also undefined. Exam tip: the denominator of a rational-number fraction can never be zero.

Dividing both (6) and (8) by (2) gives ( \frac{3}{4} ). Remove common factors for simplest form.

The governing concept is the standard form of a rational number. In standard form, the denominator must be positive and the numerator and denominator must have no common factor other than 1. Starting with -9/12, the greatest common divisor of 9 and 12 is 3. Dividing both numerator and denominator by 3 gives -9/12 = -3/4. The denominator 4 is positive, and 3 and 4 are coprime, so option A is correct. Option B has the same numerical value but leaves the denominator negative, so it is not the standard form. Option C is not obtained by reducing the fraction, and option D changes the sign and is not equivalent to the original negative fraction.

Multiplying ( \frac{2}{3} ) by (4) gives ( \frac{8}{12} ). For equivalent fractions multiply numerator and denominator by the same number.

The additive inverse of a number is the number that gives zero when added to the original number. Therefore, the additive inverse of \(\frac{5}{7}\) is \(-\frac{5}{7}\), because \(\frac{5}{7}+\left(-\frac{5}{7}\right)=0\). Note that \(\frac{7}{5}\) is its reciprocal, not its additive inverse. Exam tip: to find the additive inverse, change only the sign.

The governing concept is the multiplicative inverse, or reciprocal, of a nonzero rational number. The multiplicative inverse of a/b is b/a, provided a is not zero, because their product must equal 1. For -4/9, interchange the numerator and denominator while retaining the negative sign: the inverse is -9/4. Verification gives (-4/9) × (-9/4) = 36/36 = 1, so option B is correct. Option A loses the negative sign and its product with -4/9 is -1. Option C is the original number, not its inverse. Option D is equivalent to 4/9 because both signs are negative, so it also loses the required negative sign. The sign and reciprocal operation must both be handled correctly.

Taking common denominator (6), ( \frac{3}{6}+\frac{2}{6}=\frac{5}{6} ). Use LCM while adding fractions.

Both fractions have the same denominator, 10. When fractions with an equal denominator are subtracted, the denominator remains unchanged and only the numerators are subtracted. Thus, \\(\frac{7}{10}-\frac{2}{10}=\frac{7-2}{10}=\frac{5}{10}\\). This fraction can be reduced by dividing its numerator and denominator by 5, giving \\(\frac{1}{2}\\), but the unsimplified result \\(\frac{5}{10}\\) is exactly the form listed in option A.

Option A is therefore correct. The value \\(\frac{9}{10}\\) would result from adding the numerators, not subtracting them. The denominator 20 is unnecessary because the original denominators are already equal, and \\(\frac{1}{10}\\) would represent a difference of one tenth rather than five tenths. Hence the subtraction gives five tenths, or one half.

Direct answer: Option B, \(\frac{3}{10}\). To multiply fractions, multiply the numerators together and multiply the denominators together. The numerators are 3 and 2, so the numerator becomes \(3\times2=6\). The denominators are 4 and 5, so the denominator becomes \(4\times5=20\). Thus the product is \(\frac{6}{20}\). Divide numerator and denominator by their common factor 2: \(\frac{6\div2}{20\div2}=\frac{3}{10}\). Option A, \(\frac{5}{9}\), comes from adding or combining numbers incorrectly and is not the product. Option B is correct because it is the simplified product. Option C, \(\frac{6}{9}\), uses neither the correct denominator nor correct simplification. Option D, \(\frac{1}{20}\), loses the factor 3 and is therefore too small. A safe check is that \(\frac34\) is less than 1 and \(\frac25\) is less than 1, so the answer should be positive and less than either factor; \(\frac3{10}\) fits. Remember: fraction multiplication is numerator times numerator and denominator times denominator.

Change division into multiplication by the reciprocal of the second fraction. ( \frac{5}{6}\times\frac{3}{10}=\frac{1}{4} ).

Both fractions have the same denominator, 9. When denominators are equal, compare the numerators; the fraction with the greater numerator is larger. Since 5 > 4, \(\frac{5}{9} > \frac{4}{9}\), so option C is correct. Exam tip: For fractions with equal denominators, compare only their numerators.

( \frac{1}{2}=\frac{4}{8} ), so ( \frac{3}{8} ) is smaller. Making common denominators helps in comparison.

( \frac{-1}{5} ) lies between ( \frac{-2}{5} ) and ( \frac{1}{5} ) on the number line. Be careful with the direction of negative numbers.

( \frac{1}{2}=\frac{2}{4} ), which lies between ( \frac{1}{4} ) and ( \frac{3}{4} ). Use common denominators to find a middle number.

The direct answer is D, \\(\frac{4}{0}\\). A fraction is defined only when its denominator is not zero, because division asks how many times the denominator fits into the numerator, and division by zero has no defined value. Check the options carefully. In A, \\(\frac{8}{2}\\) has denominator 2, so it is defined and equals 4. In B, \\(\frac{0}{6}\\) has denominator 6, so it is defined and equals 0. In C, \\(\frac{-3}{7}\\) has denominator 7, so it is defined, even though its value is negative. In D, the denominator is 0, so \\(\frac{4}{0}\\) is undefined; it cannot be treated as an ordinary rational number. Thus D is correct. Remember: a rational number has the form \\(\frac{p}{q}\\) with integers p and q and the condition \\(q\ne0\\).

The governing concept is the definition of a rational number: it is any number that can be written as p/q, where p and q are integers and q is not zero. The integer -6 can therefore be represented by placing it over 1: -6/1 = -6. Option A is correct because its denominator is non-zero and its value is exactly -6. Option B equals -1/6, not -6. Option C is invalid because division by zero is undefined. Option D equals 0, so it also does not represent -6. Thus every integer, including a negative integer, is rational because it can be divided by 1.

Any integer can be written in the form \(\frac{p}{1}\), where \(p\) is an integer and the denominator is non-zero. Therefore, every integer is a rational number. Option B is false because rational numbers such as \(\frac{1}{2}\) and \(-3\) are not natural numbers. A fraction with a zero denominator is undefined, not rational. Exam tip: the denominator of a rational number must always be non-zero.

The direct answer is B, \\(\frac{1}{3}\\). The bar over 3 means that 3 repeats forever: \\(0.\overline{3}=0.3333\ldots\\). Let \\(x=0.\overline{3}\\). Multiplying by 10 gives \\(10x=3.\overline{3}\\). Subtract the first equation: \\(10x-x=3.\overline{3}-0.\overline{3}\\), so \\(9x=3\\). Dividing by 9 gives \\(x=\frac{3}{9}=\frac{1}{3}\\). Option A, \\(\frac{3}{10}\\), represents 0.3 with only one 3, not an endlessly repeating 3. Option B is correct because its decimal expansion is 0.3333... . Option C, \\(\frac{3}{100}\\), represents 0.03, not 0.333... . Option D, \\(\frac{10}{3}\\), is greater than 3 and has decimal value 3.333..., so it is not the given number. Memory cue: one repeating digit can be converted by placing that digit over 9, so repeating 3 gives \\(\frac{3}{9}=\frac{1}{3}\\).

( \frac{-3}{2}=-1.5 ), which lies between (-2) and (-1). Place negative decimals carefully on the number line.