Operations on real numbers and the laws of exponents

The whole number ((-6)) is squared, so the answer is (36). Parentheses include the negative sign in the power.

The governing concept is the order of operations, especially the precedence of exponents over a unary minus. In the expression -7^2, there are no parentheses around -7, so the exponent applies to 7 first: 7^2 = 49. The negative sign then remains in front, giving -(49) = -49. Therefore, option B is correct. Option A, 49, would be obtained from (-7)^2, where the negative number is explicitly included in parentheses. Options C and D incorrectly treat the exponent as if it meant multiplication by 2 rather than squaring. Thus, careful attention to parentheses and exponent precedence resolves the question.

For the product of two powers with the same exponent, multiply the bases and retain the common exponent: \(a^n\cdot b^n=(ab)^n\). Thus, \(6^2\cdot2^2=(6\cdot2)^2=12^2\), so option B is correct. In option C, the exponents have been added incorrectly; exponents are added when the bases are the same. Exam tip: with equal exponents, multiply the bases and keep the exponent unchanged.

With equal exponents, \(\frac{18^2}{6^2}=\left(\frac{18}{6}\right)^2=3^2\). Simplify equal powers together.

When powers with the same base are multiplied, their exponents are added: \(v^m\cdot v^n=v^{m+n}\). Hence, \(v^4\cdot v^5=v^{4+5}=v^9\). Option B incorrectly multiplies the exponents instead of adding them. Exam tip: for multiplication of like bases, add the exponents.

When powers with the same non-zero base are divided, their exponents are subtracted: \(\frac{h^{12}}{h^7}=h^{12-7}=h^5\). Therefore, option A is correct. In option B, the exponents have been added, which is appropriate for multiplication, not division. Exam tip: remember \(a^m\div a^n=a^{m-n}\) for division of powers with the same base.

When powers with the same non-zero base are divided, their exponents are subtracted, so \(a^m/a^n=a^{m-n}\). Option B is the product rule, while D incorrectly divides exponents. Exam tip: for division, subtract exponents.

When powers with the same base are multiplied, their exponents are added; hence \(a^m\times a^n=a^{m+n}\). The form \(a^{mn}\) applies to a power raised to another power. In exams, first check whether the bases are identical.

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.