Test your knowledge with AI-generated multiple choice questions
Divide coefficients: $6.0\div1.5=4.0$. Subtract exponents: $10^{8}/10^{3}=10^{5}$. So the result is $4.0\times10^{5}$.
Square numerator: $25x^4y^2$. Square denominator: $100z^2$. Reduce $25/100$ to $1/4$ to get $x^4y^2/(4z^2)$.
Negative exponents create reciprocals: $2^{-4}=\frac{1}{2^{4}}$. Since $2^{4}=16$, the value is $\frac{1}{16}$.
Apply the power to each factor: $3^2=9$, $(x^2)^2=x^4$, and $y^2$. Multiply them to get $9x^4y^2$.
Move the decimal 4 places right to get $5.4$, so the exponent is $-4$. Proper scientific notation needs a number between 1 and 10 times a power of 10. Thus $5.4\times10^{-4}$ is correct.
Multiply coefficients: $2.5\times4=10$. Add exponents: $3+(-2)=1$. So $10\times10^{1}=10^{2}$, which is $1.0\times10^{2}$ in standard form. B is not standard; C and D give the wrong value.
Multiply the coefficients and add exponents: $3.2\times 4=12.8$ and $10^{5}\cdot10^{3}=10^{8}$. Rewrite $12.8\times10^{8}$ as $1.28\times10^{9}$ to keep the coefficient between 1 and 10.
Move the decimal 4 places to the right to get 8.76, so the power is $10^{-4}$. The coefficient is between 1 and 10, so $8.76\times10^{-4}$ is the correct scientific notation.
Use $(ab)^{2}=a^{2}b^{2}$. So $4^{2}=16$, $(k^{3})^{2}=k^{6}$, and $(m^{2})^{2}=m^{4}$. Combine to get $16k^{6}m^{4}$.
Subtract exponents when dividing: $a^{-4}/a^{-1}=a^{-4-(-1)}=a^{-3}$. Rewrite $a^{-3}$ as $\dfrac{1}{a^3}$ to use positive exponents.
A negative exponent inverts the base: $(1/2)^{-3}=(2)^{3}=8$. Options A and D give the reciprocal, and C has an incorrect negative sign.
Distribute the exponent: $3^{2}=9$, $(x^{2})^{2}=x^{4}$, and $y^{2}=y^{2}$. Multiply to get $9x^{4}y^{2}$. The other options misuse exponent distribution or coefficients.
Apply the power to numerator and denominator: $(2a)^{3}=8a^{3}$ and $(5b)^{3}=125b^{3}$. Option B treats the power as multiplication by 3. C and D only cube part of the expression.
Use the quotient rule: subtract exponents. $5-(-2)=7$, so $y^{5}/y^{-2}=y^{7}$. The other options subtract in the wrong order or keep the negative sign incorrectly.
A negative exponent means reciprocal: $a^{-n}=\frac{1}{a^{n}}$. So $5^{-2}=\frac{1}{25}$ and $a^{-1}=\frac{1}{a}$, giving $\frac{1}{25a}$.
Move the decimal 4 places to the right to get 4.56, so use $10^{-4}$ to keep the value. Standard form requires $1\le a<10$. Options C and D are not in standard form, and B has the wrong sign on the exponent.
For the same base in a quotient, subtract exponents: $10-4=6$. So $\frac{m^{10}}{m^{4}}=m^{6}$. The other options misapply the rules.
First apply power of a power: $(p^4)^2=p^{8}$. Then use the quotient rule: $p^{8}/p^{3}=p^{8-3}=p^{5}$.
Distribute the exponent: $3^{2}=9$, $(a^{2})^{2}=a^{4}$, and $b^{2}$. Thus $(3a^{2}b)^{2}=9a^{4}b^{2}$.
$39{,}000{,}000=3.9\times10^{7}$. The coefficient must be between 1 and 10 in standard form. Option C has the correct value but not standard form; B and D use incorrect exponents.
For the same base, add exponents: $x^3\cdot x^5=x^{3+5}=x^8$. This is the product rule.
Move factors with negative exponents to the denominator: $a^{-2}=1/a^{2}$ and $b^{-1}=1/b$, giving $1/(a^{2}b)$. The other options misplace exponents or change signs without reason.
A negative exponent means a reciprocal: $5^{-3}=1/5^{3}=1/125$. The result is positive because the base is positive. The other options have incorrect sign or magnitude.
Use $a^{-n}=\dfrac{1}{a^n}$. Since $5^3=125$, $5^{-3}=\dfrac{1}{125}$.
Subtract exponents on the same base: $x^{-2}/x^{-5}=x^{-2-(-5)}=x^{3}$. The $y^{3}$ is unaffected. Final answer: $x^{3}y^{3}$.
Convert km to m: $742\text{ km}=742{,}000\text{ m}$. Writing $742{,}000$ with coefficient between 1 and 10 gives $7.42\times10^{5}\text{ m}$.
Power of a power means multiply exponents: $(p^{4})^{3}=p^{4\cdot3}=p^{12}$. The other options add or otherwise mis-handle exponents.
For the same base, add exponents: $3+5=8$. So $x^{3}\cdot x^{5}=x^{8}$. The other options use incorrect exponent operations.
A negative exponent inverts the fraction: $(\frac{3}{x})^{-2}=(\frac{x}{3})^{2}$. Then square to get $x^{2}/9$ with positive exponents.
For division with the same base, subtract exponents: $a^7/a^3=a^{7-3}=a^4$. This keeps the base and reduces the power.
Raise numerator and denominator: $(2a)^3=8a^3$ and $(5b)^3=125b^3$. Power of a quotient distributes to both parts.
Distribute the exponent: $(3)^{2}=9$, $(x^{2})^{2}=x^{4}$, and $(y)^{2}=y^{2}$ in the denominator. The other options misapply the exponent to coefficients or variables.
Apply $(\tfrac{a}{b})^{n}=\tfrac{a^{n}}{b^{n}}$ and $(ab)^{n}=a^{n}b^{n}$. Then $2^{3}=8$, $5^{3}=125$, giving $\frac{8x^{3}}{125y^{3}}$.
Use the product rule: add exponents with the same base. $7+(-3)=4$, so $x^{7}x^{-3}=x^{4}$. The other options come from incorrect addition or sign errors.
Write $4.8\times10^{4}$ as $0.48\times10^{5}$. Then add mantissas: $3.2+0.48=3.68$ and keep $\times10^{5}$. Answer: $3.68\times10^{5}$.
First apply power of a power: $(p^{4})^{3}=p^{12}$. Then use the quotient rule: $p^{12}/p^{5}=p^{12-5}=p^{7}$. The other options result from incorrect exponent arithmetic.
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