chapter 10 review exercises

1. a resistance, $r$, decreased by $25\ \omega$ is $37\ \omega$.
(a) write an equation to describe this statement.
(b) determine the resistance.

2. two times the difference of a certain voltage and $12\text{ v}$ is $216\text{ v}$.
(a) write an equation to describe this statement.
(b) determine the voltage.

3. a power of $40\text{ w}$ is the square of the current in amps times a resistance of $250\ \omega$.
(a) write an equation to describe this statement.
(b) determine the voltage.

4. the diameter, $d$, in mils, of $240\text{ feet}$ of 10-gauge copper wire is given by $0.25 = \frac{(10.8)(240)}{d^2}$. solve this for $d$ to the nearest tenth.

5. solve the equation $i = \frac{e}{r+r}$ for $r$.

6. solve the equation $i = \frac{e_g - e_t}{r}$ for $e_g$.

7. if $r = \frac{kl}{d^2}$, find $l$ to the nearest tenth foot if $r = 2.5\ \omega$, $k = 10.8$, and $d = 85\text{ mil}$. (note, the $k$ is a constant and does not have any units.)

8. suppose $r = r_1 + \frac{r_2 r_3}{r_2 + r_3}$.
(a) solve this equation for $r_1$.
(b) determine $r_1$ if $r = 70\ \omega$, $r_2 = 20\ \omega$, and $r_3 = 30\ \omega$.

9. suppose $r = r_1 + \frac{r_2 r_3}{r_2 + r_3}$. solve this equation for $r_2$ if $r = 70\ \omega$, $r_1 = 58\ \omega$, and $r_3 = 20\ \omega$.

10. pavel is paid time-and-a-half for each hour he works over $40\text{ hours}$ a week. one week he worked $12\text{ hours}$ of overtime and made $\$962.80$.
(a) write an equation for this statement using $w$ as his hourly wage.
(b) what is his hourly wage?

11. write an inequality that describes the fact that it will take no more than $1,750\text{ feet}$ of cable to wire a certain house.

12. julio needed a transducer to replenish the supplies in his electrical shop. the cost of the supplies is $\$8,500$. he paid $\$1,245$ down and will pay $\$420.50$ a month until it is fully paid for.
(a) write an inequality to describe how many months it will take before he has fully paid for the transducer.
(b) how many months will it take before he has fully paid for the transducer?

Question

chapter 10 review exercises

1. a resistance, $r$, decreased by $25\ \omega$ is $37\ \omega$.
(a) write an equation to describe this statement.
(b) determine the resistance.

2. two times the difference of a certain voltage and $12\text{ v}$ is $216\text{ v}$.
(a) write an equation to describe this statement.
(b) determine the voltage.

3. a power of $40\text{ w}$ is the square of the current in amps times a resistance of $250\ \omega$.
(a) write an equation to describe this statement.
(b) determine the voltage.

4. the diameter, $d$, in mils, of $240\text{ feet}$ of 10-gauge copper wire is given by $0.25 = \frac{(10.8)(240)}{d^2}$. solve this for $d$ to the nearest tenth.

5. solve the equation $i = \frac{e}{r+r}$ for $r$.

6. solve the equation $i = \frac{e_g - e_t}{r}$ for $e_g$.

7. if $r = \frac{kl}{d^2}$, find $l$ to the nearest tenth foot if $r = 2.5\ \omega$, $k = 10.8$, and $d = 85\text{ mil}$. (note, the $k$ is a constant and does not have any units.)

8. suppose $r = r_1 + \frac{r_2 r_3}{r_2 + r_3}$.
(a) solve this equation for $r_1$.
(b) determine $r_1$ if $r = 70\ \omega$, $r_2 = 20\ \omega$, and $r_3 = 30\ \omega$.

9. suppose $r = r_1 + \frac{r_2 r_3}{r_2 + r_3}$. solve this equation for $r_2$ if $r = 70\ \omega$, $r_1 = 58\ \omega$, and $r_3 = 20\ \omega$.

10. pavel is paid time-and-a-half for each hour he works over $40\text{ hours}$ a week. one week he worked $12\text{ hours}$ of overtime and made $\$962.80$.
(a) write an equation for this statement using $w$ as his hourly wage.
(b) what is his hourly wage?

11. write an inequality that describes the fact that it will take no more than $1,750\text{ feet}$ of cable to wire a certain house.

12. julio needed a transducer to replenish the supplies in his electrical shop. the cost of the supplies is $\$8,500$. he paid $\$1,245$ down and will pay $\$420.50$ a month until it is fully paid for.
(a) write an inequality to describe how many months it will take before he has fully paid for the transducer.
(b) how many months will it take before he has fully paid for the transducer?

Accepted Answer

{ "quality": "clear", "question_count": 12, "skills_matched": [ "step_cot", "table_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Basic Arithmetic Operations", "Literal Equations", "Formula Rearrangement", "Linear Equations Solving", "Translating Word Problems", "Algebraic Modeling", "Overtime Pay Calculations" ], "new_concepts": [ "Electrical Formula Modeling", "Inequality Modeling" ], "current_concepts": [ "Basic Arithmetic Operations", "Literal Equations", "Formula Rearrangement", "Linear Equations Solving", "Translating Word Problems", "Algebraic Modeling", "Overtime Pay Calculations", "Electrical Formula Modeling", "Inequality Modeling" ] } ## Translate and solve Question 1 Using the *Translating Word Problems* and *Linear Equations Solving* knowledge points - (a) Let $R$ be the resistance. "A resistance, $R$, decreased by $25\ \Omega$ is $37\ \Omega$" translates to: $$R - 25 = 37$$ - (b) Solve for $R$: $$R = 37 + 25 = 62\ \Omega$$ ## Translate and solve Question 2 Using the *Translating Word Problems* and *Linear Equations Solving* knowledge points - (a) Let $V$ be the voltage. "Two times the difference of a certain voltage and $12\text{ V}$ is $216\text{ V}$" translates to: $$2(V - 12) = 216$$ - (b) Solve for $V$: $$V - 12 = 108 \implies V = 120\text{ V}$$ ## Translate and solve Question 3 Using the *Translating Word Problems* and *Formula Rearrangement* knowledge points - (a) Let $I$ be the current. "A power of $40\text{ W}$ is the square of the current in amps times a resistance of $250\ \Omega$" translates to: $$40 = I^2 \cdot 250$$ - (b) The question asks to "Determine the voltage" (which is likely a typo in the textbook for current $I$, or requires finding $V = I \cdot R$). Let's solve for both to be thorough. First, solve for current $I$: $$I^2 = \frac{40}{250} = 0.16 \implies I = 0.4\text{ A}$$ Using Ohm's law $V = I \cdot R$: $$V = 0.4\text{ A} \times 250\ \Omega = 100\text{ V}$$ ## Solve literal equations (Questions 4 to 9) Using the *Literal Equations* and *Formula Rearrangement* knowledge points - **Question 4**: Solve $0.25 = \frac{10.8 \times 240}{d^2}$ for $d$: $$d^2 = \frac{2592}{0.25} = 10368 \implies d = \sqrt{10368} \approx 101.8\text{ mils}$$ - **Question 5**: Solve $I = \frac{E}{R + r}$ for $r$: $$I(R + r) = E \implies R + r = \frac{E}{I} \implies r = \frac{E}{I} - R$$ - **Question 6**: Solve $I = \frac{E_s - E_g}{R}$ for $E_s$: $$I \cdot R = E_s - E_g \implies E_s = I \cdot R + E_g$$ - **Question 7**: Solve $R = \frac{KL}{d^2}$ for $L$: $$L = \frac{R \cdot d^2}{K}$$ Substitute $R = 2.5$, $K = 10.8$, $d = 85$: $$L = \frac{2.5 \times 85^2}{10.8} = \frac{2.5 \times 7225}{10.8} \approx 1672.5\text{ feet}$$ - **Question 8**: - (a) Solve $R = R_1 + \frac{R_2 R_3}{R_2 + R_3}$ for $R_1$: $$R_1 = R - \frac{R_2 R_3}{R_2 + R_3}$$ - (b) Substitute $R = 70$, $R_2 = 20$, $R_3 = 30$: $$R_1 = 70 - \frac{20 \times 30}{20 + 30} = 70 - \frac{600}{50} = 70 - 12 = 58\ \Omega$$ - **Question 9**: Solve $R = R_1 + \frac{R_2 R_3}{R_2 + R_3}$ for $R_2$ given $R = 70$, $R_1 = 58$, $R_3 = 20$: $$70 = 58 + \frac{20 R_2}{R_2 + 20} \implies 12 = \frac{20 R_2}{R_2 + 20}$$ $$12(R_2 + 20) = 20 R_2 \implies 12 R_2 + 240 = 20 R_2 \implies 8 R_2 = 240 \implies R_2 = 30\ \Omega$$ ## Translate and solve real-world scenarios (Questions 10 to 12) Using the *Overtime Pay Calculations* and *Algebraic Modeling* knowledge points - **Question 10**: Pavel works 40 hours regular time plus 12 hours overtime. - (a) Regular pay is $40w$. Overtime pay rate is $1.5w$. Overtime pay is $12 \times 1.5w = 18w$. $$40w + 1.5w(12) = 962.80 \implies 40w + 18w = 962.80 \implies 58w = 962.80$$ - (b) Solve for $w$: $$w = \frac{962.80}{58} = \$16.60\text{ per hour}$$ - **Question 11**: Let $c$ be the length of cable in feet. "No more than 1,750 feet" means: $$c \le 1750$$ - **Question 12**: Let $m$ be the number of months. - (a) Total paid after $m$ months is $1245 + 420.50m$. To fully pay for the transducer, this total must be at least the cost of \$8,500: $$1245 + 420.50m \ge 8500$$ - (b) Solve for $m$: $$420.50m \ge 7255 \implies m \ge \frac{7255}{420.50} \approx 17.25$$ Since payments are made monthly, it will take $18$ months to fully pay it off. ## Question 1 - (a) $R - 25 = 37$ - (b) $R = 62\ \Omega$ ## Question 2 - (a) $2(V - 12) = 216$ - (b) $V = 120\text{ V}$ ## Question 3 - (a) $40 = 250 I^2$ - (b) $I = 0.4\text{ A}$ (or $V = 100\text{ V}$) ## Question 4 $d \approx 101.8\text{ mils}$ ## Question 5 $r = \frac{E}{I} - R$ ## Question 6 $E_s = IR + E_g$ ## Question 7 $L \approx 1672.5\text{ feet}$ ## Question 8 - (a) $R_1 = R - \frac{R_2 R_3}{R_2 + R_3}$ - (b) $R_1 = 58\ \Omega$ ## Question 9 $R_2 = 30\ \Omega$ ## Question 10 - (a) $40w + 18w = 962.80$ - (b) $w = \$16.60\text{/hr}$ ## Question 11 $c \le 1750\text{ feet}$ ## Question 12 - (a) $1245 + 420.50m \ge 8500$ - (b) $18\text{ months}$ { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Algebra", "Formula Rearrangement" ] }

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