Operations on real numbers and the laws of exponents

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).

When powers with the same base are multiplied, their exponents are added: \(2^3\cdot2^{-5}\cdot2^4=2^{3+(-5)+4}=2^2\). Therefore, option A is correct. Option B results from mishandling the negative exponent, while option D incorrectly multiplies the exponents. In an exam, retain the sign of a negative exponent while adding exponents.

For powers with the same base, add exponents during multiplication and subtract the denominator's exponent during division: \(2+(-4)-(-3)=1\). Thus, \(3^1=3\), so option A is correct. Option C, \(\frac{1}{3}\), results from incorrectly treating the final exponent as \(-1\). Exam tip: apply \(a^m\div a^n=a^{m-n}\) carefully, especially when the denominator has a negative exponent.

Using the power-of-a-power rule, \((x^3)^2=x^{3\times2}=x^6\). For the remaining product and division of powers with the same base, subtract the denominator exponent: \(x^6\cdot x^{-4}\div x=x^{6-4-1}=x\). Therefore, option A is correct. Exam tip: write the denominator as \(x^1\) before combining exponents.

For any non-zero real number \(a\), \(\frac{a^m}{a^m}=a^{m-m}=a^0\). Since the value on the left is \(1\), we get \(a^0=1\), where \(a\ne0\). Option D is a close distractor, but the zero-exponent law cannot be applied when \(a=0\); in school algebra, \(0^0\) is generally treated as undefined. Exam tip: Always remember the condition \(a\ne0\) with the zero-exponent rule.

Since \(16=2^4\), we get \(16^2=(2^4)^2=2^8\). Therefore, \(\frac{16^2}{2^5}=\frac{2^8}{2^5}=2^{8-5}=2^3=8\). Exam tip: when dividing powers with the same base, subtract their exponents.

\(0.2=\frac{1}{5}\), so \((0.2)^{-2}=\left(\frac{1}{5}\right)^{-2}=25\). Converting decimals to fractions helps.

\(\left(\frac{3}{2}\right)^{-2}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\). Hence \(\frac{4}{9}\cdot\frac{9}{4}=1\).