QUESTION IMAGE
Question
algebraic
for the following exercises, simplify the given expression. write answers with positive exponents
- $\frac{a^{3}a^{2}}{a}$
- $\frac{mn^{2}}{m^{-2}}$
- $(b^{3}c^{4})^{2}$
- $(\frac{x^{-3}}{y^{2}})^{-5}$
- $ab^{2}div d^{-3}$
- $(w^{0}x^{5})^{-1}$
- $\frac{m^{4}}{n^{0}}$
- $y^{-4}(y^{2})^{2}$
- $\frac{p^{-4}q^{2}}{p^{2}q^{-3}}$
- $(l\times w)^{2}$
- $(y^{7})^{3}div x^{14}$
- $(\frac{a}{2^{3}})^{2}$
- $(25m)div(\frac{5}{6}m)$
- $\frac{(16sqrt{x})^{2}}{y^{-1}}$
- $\frac{z^{3}}{(3u)^{-2}}$
- $(ma^{6})^{2}\frac{1}{m^{3}a^{2}}$
- $(b^{-3}c)^{3}$
- $(x^{2}y^{13}div y^{0})^{2}$
Response
- Exercise 25:
- Explanation:
- Step 1: Use the rule \(a^m\times a^n=a^{m + n}\) in the numerator
- \(a^3a^2=a^{3 + 2}=a^5\). So the expression becomes \(\frac{a^5}{a}\).
- Step 2: Use the rule \(\frac{a^m}{a^n}=a^{m - n}\)
- \(\frac{a^5}{a}=a^{5-1}=a^4\).
- Exercise 26:
- Explanation:
- Step 1: Use the rule \(\frac{a^m}{a^n}=a^{m - n}\) for the \(m\) - terms
- \(\frac{mn^2}{m^{-2}}=m^{1-(-2)}n^2\).
- Step 2: Simplify the exponent of \(m\)
- \(m^{1 + 2}n^2=m^3n^2\).
- Exercise 27:
- Explanation:
- Step 1: Use the power - of - a - product rule \((ab)^n=a^nb^n\)
- \((b^3c^4)^2=(b^3)^2(c^4)^2\).
- Step 2: Use the power - of - a - power rule \((a^m)^n=a^{mn}\)
- \((b^3)^2=b^{3\times2}=b^6\) and \((c^4)^2=c^{4\times2}=c^8\). So the result is \(b^6c^8\).
- Exercise 28:
- Explanation:
- Step 1: Use the power - of - a - quotient rule \((\frac{a}{b})^n=\frac{a^n}{b^n}\)
- \((\frac{x^{-3}}{y^2})^{-5}=\frac{(x^{-3})^{-5}}{(y^2)^{-5}}\).
- Step 2: Use the power - of - a - power rule \((a^m)^n=a^{mn}\)
- \((x^{-3})^{-5}=x^{(-3)\times(-5)} = x^{15}\) and \((y^2)^{-5}=y^{2\times(-5)}=y^{-10}\).
- Step 3: Rewrite with positive exponents
- \(\frac{x^{15}}{y^{-10}}=x^{15}y^{10}\).
- Exercise 29:
- Explanation:
- Step 1: Rewrite the division as multiplication by the reciprocal
- \(ab^2\div d^{-3}=ab^2\times d^3=ab^2d^3\).
- Exercise 30:
- Explanation:
- **Step 1: Use the power - of - a - product rule \((ab)^n=a^nb^n\) and \(a^0 = 1\) (\(a
eq0\))**
- \((w^0x^5)^{-1}=(1\times x^5)^{-1}=x^{-5}\).
- Step 2: Rewrite with positive exponents
- \(\frac{1}{x^5}\).
- Exercise 31:
- Explanation:
- **Step 1: Use the rule \(a^0 = 1\) (\(a
eq0\))**
- \(\frac{m^4}{n^0}=\frac{m^4}{1}=m^4\).
- Exercise 32:
- Explanation:
- Step 1: Use the power - of - a - power rule \((a^m)^n=a^{mn}\) for \((y^2)^2\)
- \((y^2)^2=y^{2\times2}=y^4\). So the expression is \(y^{-4}\times y^4\).
- Step 2: Use the rule \(a^m\times a^n=a^{m + n}\)
- \(y^{-4 + 4}=y^0 = 1\).
- Exercise 33:
- Explanation:
- Step 1: Use the rule \(\frac{a^m}{a^n}=a^{m - n}\) for \(p\) and \(q\) terms separately
- \(\frac{p^{-4}q^2}{p^2q^{-3}}=p^{-4-2}q^{2-(-3)}\).
- Step 2: Simplify the exponents
- \(p^{-6}q^5\).
- Step 3: Rewrite with positive exponents
- \(\frac{q^5}{p^6}\).
- Exercise 34:
- Explanation:
- Step 1: Use the power - of - a - product rule \((ab)^n=a^nb^n\) and \(1^2 = 1\)
- \((l\times w)^2=l^2w^2\).
- Exercise 35:
- Explanation:
- Step 1: Use the power - of - a - power rule \((a^m)^n=a^{mn}\) for \((y^7)^3\)
- \((y^7)^3=y^{7\times3}=y^{21}\). So the expression is \(y^{21}\div x^{14}=\frac{y^{21}}{x^{14}}\).
- Exercise 36:
- Explanation:
- Step 1: Use the power - of - a - quotient rule \((\frac{a}{b})^n=\frac{a^n}{b^n}\)
- \((\frac{a}{z^3})^2=\frac{a^2}{z^{3\times2}}=\frac{a^2}{z^6}\).
- Exercise 37:
- Explanation:
- Step 1: Recall that \(_n^nC = 1\) and rewrite the division
- \((25m)\div(_{5}^5m)=\frac{25m}{1\times m}=25\).
- Exercise 38:
- Explanation:
- Step 1: First, square the numerator \((16\sqrt{x})^2=(16)^2\times(\sqrt{x})^2\)
- \((16)^2 = 256\) and \((\sqrt{x})^2=x\…
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