solve each of the following problems.

9. if variables $x$ and $y$ are inversely proportional, with $x_1 = 5$ and $y_1 = 4$, what is the constant of variation?

10. variables $x$ and $y$ are inversely proportional with a constant of variation of 2.5. if $x_1 = 6.25$, what is $y_1$?

11. if the fulcrum is at the midpoint of a lever and the weights $w_1$ and $w_2$ are inversely proportional to distances $s_1$ and $s_2$ from the fulcrum, respectively, then the lever is balanced. if $w_1$ equals 120 grams, $s_1$ is 9 centimeters, $w_2$ equals 135 grams, and the lever is balanced, what is the distance $s_2$?

12. in a transformer, the ratio of the currents is inversely proportional to the ratio of turns. use $n_1$ for the number of turns in the primary winding, $n_2$ for the number of turns in the secondary winding, $i_1$ for the primary current, and $i_2$ for the secondary current.
(a) write the proportion for this problem.
(b) suppose a transformer has 100 turns in its primary winding and 800 turns in its secondary winding. if the primary current is 5 a, determine the secondary current.

13. in a parallel circuit, the currents are inversely proportional to the resistances. use $i_1$ and $i_2$ for the currents and $r_1$ and $r_2$ for the resistances.
(a) write the proportion for this problem.
(b) if $i_1 = 0.5\text{ a}$, $i_2 = 0.35\text{ a}$, and $r_2 = 175\ \omega$, determine $r_1$.

14. whats the error? in a transformer, the currents in the primary and secondary windings are inversely proportional to the voltages in those windings. given that the primary voltage $e_p$ is 120 volts, the primary current $i_p$ is 5 amps, and the secondary voltage $e_s$ is 480 volts, a student solved a proportion for $i_s$ as shown below. describe the error and give the solution.
\\frac{e_p}{i_p} = \frac{e_s}{i_s}, \frac{120}{5} = \frac{480}{i_s}, 120 i_s = 5 \cdot 480, i_s = 20\text{ amps}\

Question

solve each of the following problems.

9. if variables $x$ and $y$ are inversely proportional, with $x_1 = 5$ and $y_1 = 4$, what is the constant of variation?

10. variables $x$ and $y$ are inversely proportional with a constant of variation of 2.5. if $x_1 = 6.25$, what is $y_1$?

11. if the fulcrum is at the midpoint of a lever and the weights $w_1$ and $w_2$ are inversely proportional to distances $s_1$ and $s_2$ from the fulcrum, respectively, then the lever is balanced. if $w_1$ equals 120 grams, $s_1$ is 9 centimeters, $w_2$ equals 135 grams, and the lever is balanced, what is the distance $s_2$?

12. in a transformer, the ratio of the currents is inversely proportional to the ratio of turns. use $n_1$ for the number of turns in the primary winding, $n_2$ for the number of turns in the secondary winding, $i_1$ for the primary current, and $i_2$ for the secondary current.
(a) write the proportion for this problem.
(b) suppose a transformer has 100 turns in its primary winding and 800 turns in its secondary winding. if the primary current is 5 a, determine the secondary current.

13. in a parallel circuit, the currents are inversely proportional to the resistances. use $i_1$ and $i_2$ for the currents and $r_1$ and $r_2$ for the resistances.
(a) write the proportion for this problem.
(b) if $i_1 = 0.5\text{ a}$, $i_2 = 0.35\text{ a}$, and $r_2 = 175\ \omega$, determine $r_1$.

14. whats the error? in a transformer, the currents in the primary and secondary windings are inversely proportional to the voltages in those windings. given that the primary voltage $e_p$ is 120 volts, the primary current $i_p$ is 5 amps, and the secondary voltage $e_s$ is 480 volts, a student solved a proportion for $i_s$ as shown below. describe the error and give the solution.
\\frac{e_p}{i_p} = \frac{e_s}{i_s}, \frac{120}{5} = \frac{480}{i_s}, 120 i_s = 5 \cdot 480, i_s = 20\text{ amps}\

Accepted Answer

{ "quality": "clear", "question_count": 6, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Inverse Proportion Equations", "Transformer Voltage Ratio" ], "new_concepts": [], "current_concepts": [ "Inverse Proportion Equations", "Transformer Voltage Ratio" ] } ## Solve inverse proportion problems 9, 10, and 11 $$ \begin{aligned} &\text{9. } y = \frac{k}{x} \implies k = x_1 y_1 = 5 \cdot 4 = 20\ &\text{10. } y = \frac{k}{x} \implies y = \frac{25}{6.25} = 4\ &\text{11. } w_1 s_1 = w_2 s_2 \implies 120 \cdot 9 = 135 \cdot s_2 \implies s_2 = \frac{1080}{135} = 8\text{ cm} \end{aligned} $$ ## Solve transformer and circuit proportion problems 12 and 13 $$ \begin{aligned} &\text{12. (a) } \frac{I_1}{I_2} = \frac{N_2}{N_1}\ &\text{12. (b) } \frac{5}{I_2} = \frac{800}{100} \implies 8 I_2 = 5 \implies I_2 = 0.625\text{ A}\ &\text{13. (a) } \frac{I_1}{I_2} = \frac{R_2}{R_1}\ &\text{13. (b) } \frac{0.5}{0.35} = \frac{175}{R_1} \implies R_1 = \frac{175 \cdot 0.35}{0.5} = 122.5\ \Omega \end{aligned} $$ ## Identify and correct the error in problem 14 $$ \begin{aligned} &\text{Error: The student set up a direct proportion } \frac{E_p}{I_s} = \frac{E_s}{I_p} \text{ instead of an inverse proportion } \frac{I_p}{I_s} = \frac{E_s}{E_p}.\ &\text{Correct setup: } \frac{5}{I_s} = \frac{480}{120} \implies 4 I_s = 5 \implies I_s = 1.25\text{ amps} \end{aligned} $$ ## Question 9 $20$ ## Question 10 $4$ ## Question 11 $8\text{ cm}$ ## Question 12 (a) $\frac{I_1}{I_2} = \frac{N_2}{N_1}$ (b) $0.625\text{ A}$ ## Question 13 (a) $\frac{I_1}{I_2} = \frac{R_2}{R_1}$ (b) $122.5\ \Omega$ ## Question 14 Error: The student set up a direct proportion $\frac{E_p}{I_s} = \frac{E_s}{I_p}$ instead of an inverse proportion $\frac{I_p}{I_s} = \frac{E_s}{E_p}$. Correct solution: $1.25\text{ amps}$ { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Algebra", "Inverse Proportion Equations" ] }

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QUESTION IMAGE

Question

Explanation:

Solve inverse proportion problems 9, 10, and 11

Solve transformer and circuit proportion problems 12 and 13

Identify and correct the error in problem 14

Answer:

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14