Operations on real numbers and the laws of exponents

The rule for a negative exponent is \(a^{-n}=\frac{1}{a^n}\), where \(a\ne0\). Hence, \(y^{-5}=\frac{1}{y^5}\), so option B is correct. Option A represents the positive exponent \(y^5\), while option D incorrectly treats 5 as a coefficient instead of an exponent. Exam tip: For a negative exponent, take the reciprocal of the base and make the exponent positive.

The binomial identity is \((a-b)^2=a^2-2ab+b^2\), so option B is correct. Option C is the expansion of \((a+b)^2\), while option A equals \((a-b)(a+b)\). In exams, remember that the middle term of the square of a difference is \(-2ab\).

When powers with the same base are multiplied, their exponents are added: \(x^m \times x^n=x^{m+n}\). Hence, \(x^4 \times x^3=x^{4+3}=x^7\), so option A is correct. Option B incorrectly multiplies the exponents. Exam tip: For multiplication of like bases, keep the base unchanged and add the exponents.

\(12a\) and \(9a\) are like terms, so their coefficients are added: \((12+9)a=21a\). The exponent of \(a\) does not change, so \(21a^2\) is incorrect. Exam tip: For like terms, add or subtract only the coefficients while keeping the variable part unchanged.

The governing concept is combining like terms. Both 13x and 5x contain the same variable x with the same power, so only their coefficients are subtracted: 13 - 5 = 8. Therefore, 13x - 5x = 8x, making option A correct. Option B adds the coefficients, option C changes the power of x incorrectly, and option D multiplies the coefficients instead of subtracting them.

Multiplying the coefficients gives \(6\times5=30\). For powers with the same base, \(x^m\cdot x^n=x^{m+n}\), so \(x\cdot x^2=x^{1+2}=x^3\). Hence, the product is \(30x^3\). Remember that exponents are added, not multiplied, when multiplying like bases.

When powers with the same nonzero base are divided, their exponents are subtracted. The numerical part and the variable part can therefore be simplified separately. In this expression, the coefficient becomes 5, while the power of x becomes \\(x^{6-3}=x^3\\). The complete simplified form is \\(5x^3\\), so the supplied answer A is correct.

Divide the coefficients first: \\(35/7=5\\). Then apply the quotient law: \\(x^6/x^3=x^{6-3}=x^3\\). Combining the two results gives \\(5x^3\\). The condition \\(x\\ne0\\) makes division by \\(x^3\\) valid. Adding the exponents would be incorrect because addition is used when multiplying like bases, not when dividing them.

Using the distributive law, \(6(2x+5)=6\times2x+6\times5=12x+30\). Therefore, option B is correct. Remember to multiply the factor outside the bracket by every term inside it; multiplying only \(2x\) is a common mistake.

Evaluate both powers and then add them: \(4^2+4^3=16+64=80\). Therefore, the correct answer is 80. The value 64 represents only \(4^3\), so it omits the first term. Exam tip: exponents are not added in addition; adding exponents applies to multiplication of powers with the same base, such as \(a^m\times a^n=a^{m+n}\).

In the left-hand side, \(p\) is multiplied separately by both terms inside the parentheses, \(q\) and \(r\): \(p(q+r)=pq+pr\). Therefore, it represents the distributive law. The associative law changes grouping, whereas this expression distributes multiplication over addition. Exam tip: whenever you see \(a(b+c)=ab+ac\) or \(a(b-c)=ab-ac\), identify the distributive law.

Changing the order of the addends does not change their sum; that is, m + n = n + m. This is the commutative law of addition. The distributive law involves multiplying across a sum or difference, such as a(b + c) = ab + ac. Exam tip: the commutative law changes the order, whereas the associative law changes the grouping.

In this equation, the order of the addends remains the same, but their grouping changes: \((m+n)\u0008 is grouped on the left and \((n+p)\u0008 on the right. Since regrouping does not change the sum, this is the associative law of addition. The distributive law involves multiplication over addition, while the commutative law changes the order of terms. Exam tip: To identify the associative law, look for a change in grouping, not a change in the order of terms.

Align the decimal points and add: 0.65 + 0.25 = 0.90. In terms of hundredths, 65 hundredths plus 25 hundredths equals 90 hundredths. Therefore, 0.90 is correct. The distractor 0.80 results from an incorrect addition. Exam tip: Always write decimal points in the same vertical column before adding decimals.

Using common denominator (12), (\frac{5}{6}=\frac{10}{12}). Hence (\frac{10}{12}+\frac{1}{12}=\frac{11}{12}).

Multiply and simplify: (\frac{5}{8}\cdot\frac{16}{15}=\frac{80}{120}=\frac{2}{3}). Cancelling first makes calculation easier.

According to the order of operations, multiplication is performed before subtraction. Thus, \(5\cdot3=15\), and then \(20-15=5\). Option B is only the intermediate product, not the final value. Exam tip: follow brackets, exponents, multiplication/division, and addition/subtraction in that order.

According to the order of operations, first calculate \(3^2=9\), then \(9\cdot5=45\), and finally \(7+45=52\). Therefore, the correct answer is 52. The value 45 is only an intermediate result after multiplication, not the final answer. Exam tip: evaluate exponents first, followed by multiplication or division, and then addition or subtraction.

\(6x^2\) and \(8x^2\) are like terms because they have the same variable with the same exponent. Add their coefficients: \(6+8=14\), while \(x^2\) remains unchanged. Therefore, the answer is \(14x^2\). Remember that exponents are added in multiplication, not when like terms are added.

The governing concept is subtraction of like terms. Since both terms have the identical variable part x⁵, their coefficients are subtracted while the variable part and its exponent remain unchanged: 14 - 9 = 5. Thus, 14x⁵ - 9x⁵ = 5x⁵, so option A is correct. The other choices add coefficients or incorrectly alter the exponent.

The expression 6x + 7y + 4 has three terms separated by plus signs: 6x, 7y, and 4. Therefore, the correct answer is 3. Choosing 2 is incorrect because the constant 4 is also a separate term. Exam tip: count the parts separated by plus or minus signs, including constant terms.

The relevant algebraic concept is the definition of like terms. Two terms are like terms when their variable parts, including every variable and its exponent, are exactly the same. The term z⁴ has variable part z⁴. In 9z⁴, the variable part is also z⁴; only the numerical coefficient changes from the implied 1 to 9, and that change is allowed. Therefore option A is correct. In 9z³, the variable is z but the exponent is 3 rather than 4, so it is unlike. In z⁵, the exponent is 5, so it is also unlike. In 9y⁴, the exponent matches but the variable is y instead of z. Thus the coefficient need not match, but the complete variable-and-exponent part must match exactly.

Multiplying the coefficients gives \(5 \times 2=10\). For the like base \(x\), the law of exponents gives \(x^4 \cdot x^3=x^{4+3}=x^7\). Therefore, the product is \(10x^7\). Option C is incorrect because the exponents are added, not multiplied, when powers with the same base are multiplied. Exam tip: multiply the coefficients and add the exponents of like bases.

Divide the numerical coefficients and then use the quotient law for powers: \(42\div6=7\) and \(a^7\div a^4=a^{7-4}=a^3\). Hence, the simplified form is \(7a^3\). Option B incorrectly adds the exponents. In exams, remember that division of powers with the same non-zero base gives \(a^m\div a^n=a^{m-n}\).

When powers with the same base are multiplied, their exponents are added: \(7^2\cdot7^0\cdot7^4=7^{2+0+4}=7^6\). Since \(7^0=1\), its exponent contributes 0 to the sum. Option B results from adding the exponents incorrectly. Exam tip: for multiplication of powers with the same base, keep the base unchanged and add the exponents.

First write (4^3=(2^2)^3=2^6) and (8^2=(2^3)^2=2^6). Thus (\frac{2^3\cdot2^6}{2^6}=2^{3+6-6}=2^3), so the correct option is (2^3).