Question

hw04b fall25: problem 5
(1 point)
$(x^{61}cdot y^{126})cdot(x^{24}) = x^{p}cdot y^{q}$ when $p = \square$ and $q = \square$
$(e^{61x}cdot(\sin(11x))^{126})cdot(e^{24x}) = e^{px}cdot(\sin(11x))^{q}$ when $p = \square$ and $q = \square$
$\frac{x^{99}cdot y^{121}}{x^{24}} = x^{p}cdot y^{q}$ when $p = \square$ and $q = \square$
$\frac{x^{24}cdot y^{121}}{x^{99}} = x^{p}cdot y^{q}$ when $p = \square$ and $q = \square$
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Explanation:

Step1: Use exponent - multiplication rule

For \((x^{61}\cdot y^{126})\cdot(x^{24})\), when multiplying terms with the same base \(x\), we add the exponents. So \(x^{61}\cdot x^{24}=x^{61 + 24}=x^{85}\), and \(y\) remains \(y^{126}\). So \(P = 85\) and \(Q=126\).

Step2: Simplify exponential - function expression

For \((e^{61x}\cdot(\sin(11x))^{126})\cdot(e^{24x})\), when multiplying terms with the same base \(e\), we add the exponents of \(e\). So \(e^{61x}\cdot e^{24x}=e^{(61 + 24)x}=e^{85x}\), and \((\sin(11x))\) remains \((\sin(11x))^{126}\). So \(P = 85\) and \(Q = 126\).

Step3: Use exponent - division rule

For \(\frac{x^{99}\cdot y^{121}}{x^{24}}\), when dividing terms with the same base \(x\), we subtract the exponents. So \(\frac{x^{99}}{x^{24}}=x^{99-24}=x^{75}\), and \(y\) remains \(y^{121}\). So \(P = 75\) and \(Q = 121\).

Step4: Simplify another division expression

For \(\frac{x^{24}\cdot y^{121}}{x^{99}}\), when dividing terms with the same base \(x\), we subtract the exponents. So \(\frac{x^{24}}{x^{99}}=x^{24 - 99}=x^{-75}\), and \(y\) remains \(y^{121}\). So \(P=-75\) and \(Q = 121\).

Answer:

For \((x^{61}\cdot y^{126})\cdot(x^{24})=x^{P}\cdot y^{Q}\), \(P = 85\), \(Q = 126\)
For \((e^{61x}\cdot(\sin(11x))^{126})\cdot(e^{24x})=e^{P x}\cdot(\sin(11x))^{Q}\), \(P = 85\), \(Q = 126\)
For \(\frac{x^{99}\cdot y^{121}}{x^{24}}=x^{P}\cdot y^{Q}\), \(P = 75\), \(Q = 121\)
For \(\frac{x^{24}\cdot y^{121}}{x^{99}}=x^{P}\cdot y^{Q}\), \(P=-75\), \(Q = 121\)