updating precalculus problems for Tradler/Carley textbook

Contrib/CUNY/CityTech/CollegeAlgebra_Trig/ComplexNumbers/difference-complex.pg

@@ -122,11 +122,11 @@ END_PGML_HINT
 
 BEGIN_PGML_SOLUTION
 
->> [` \begin{align*}([$c1]) - ([$c2]) &= [$c1] - [$bR] - [$bI] i \\
+>> [` \begin{aligned}([$c1]) - ([$c2]) &= [$c1] - [$bR] - [$bI] i \\
 &= ([$aR] - [$bR]) + [$aI] i - [$bI] i \\
   &= [$ansR] + ([$aI] - [$bI]) i \\
  &= [$ansR] + [$ansI] i 
-\end{align*}
+\end{aligned}
 `] <<
 
 

Contrib/CUNY/CityTech/CollegeAlgebra_Trig/ComplexNumbers/divide-complex-numbers.pg

@@ -21,6 +21,7 @@ loadMacros(
    "MathObjects.pl",
    "contextLimitedComplex.pl",
    "contextFraction.pl",
+   "niceTables.pl",
    "PGML.pl"
 );
 
@@ -113,6 +114,19 @@ are conjugate pairs. This is significant because whenever you multiply conjugate
 
 We can take advantage of this when attempting to divide Complex numbers, because it is preferable to have a denominator that is a Real number instead of Complex.
 
+##### Example #####
+
+[@ LayoutTable(
+  [
+    [['\( \dfrac{5+6i}{3+4i}
+=\dfrac{(5+6i)\cdot (3-4i)}{(3+4i)\cdot (3-4i)}
+=\dfrac{15-20i+18i-24i^2}{9-12i+12i-16i^2} \)'] ],
+ ['\( \hspace{13mm}=\dfrac{15-2i-24\cdot (-1)}{9-16\cdot (-1)}=\dfrac{15-2i+24}{9+16}\)'],
+ ['\( \hspace{13mm}=\dfrac{39-2i}{25}=\dfrac{39}{25}-\dfrac{2}{25}i  \)'],
+  ],
+  align=>'l',
+  allcellcss=>'padding: 1pt;'
+) @]***
 
 >> ### Practice ### <<
 
@@ -185,10 +199,10 @@ _Last_: [` [$aI] i \cdot [$bIc] i = [$aIbIc] i^2 \quad\longrightarrow\quad [$aIb
 
 (b) Combine like terms: 
 
->> [` \begin{eqnarray*}
-\color{red}{[$aRbR] + [$aRbIc] i + [$aIbR] i - [$aIbIc]} &=& [$ansRp] + ([$aRbIc] + [$aIbR]) i \\
-&=&  \color{red}{[$ansRp] + [$ansIp] i} 
-\end{eqnarray*}
+>> [` \begin{aligned}
+\color{red}{[$aRbR] + [$aRbIc] i + [$aIbR] i - [$aIbIc]} &= [$ansRp] + ([$aRbIc] + [$aIbR]) i \\
+&=  \color{red}{[$ansRp] + [$ansIp] i} 
+\end{aligned}
 `]<<
 
 So our numerator reduces to [` \color{red}{[$ansRp] + [$ansIp] i}. `]
@@ -208,9 +222,9 @@ _Last_: [`[$bI] i \cdot [$bIc] i = -[$bI2] i^2 \quad\longrightarrow\quad
 
 (b)Combine like terms: 
 
->>[` \begin{eqnarray*}
-\color{blue}{[$bR2] + [$bRbIc] i + [$bRbI] i + [$bI2]} &=& [$denom] + ([$bRbIc] + [$bRbI]) i \\
-& =& [$denom] + 0\cdot i = \color{blue}{[$denom]}\end{eqnarray*} `]<<
+>>[` \begin{aligned}
+\color{blue}{[$bR2] + [$bRbIc] i + [$bRbI] i + [$bI2]} &= [$denom] + ([$bRbIc] + [$bRbI]) i \\
+&= [$denom] + 0\cdot i = \color{blue}{[$denom]}\end{aligned} `]<<
 
 And our denominator reduces to [`\color{blue}{[$denom]}`],
 
@@ -219,10 +233,10 @@ Recombine our fraction with the reduced numerator and denominator:
 Divide each term by the denominator and reduce: 
 
 >>[`
-\begin{align*}
+\begin{aligned}
 \displaystyle \frac{\color{red}{[$ansRp] + [$ansIp] i}}{\color{blue}{[$denom]}} &=& \frac{\color{red}{[$ansRp]}}{\color{blue}{[$denom]}} + \frac{\color{red}{[$ansIp]}}{\color{blue}{[$denom]}} \color{red}{i} \\
 &=& \frac{[$ansRr]}{[$ansRD]} + \frac{[$ansIr]}{[$ansID]} i 
-\end{align*}
+\end{aligned}
 `]<<
 
 

Contrib/CUNY/CityTech/CollegeAlgebra_Trig/ComplexNumbers/multiply-complex.pg

@@ -87,7 +87,7 @@ BEGIN_TEXT
 $PAR
 Simplify the expression using complex numbers:
 $PAR
-\[  ($c1) ($c2) \] 
+\[  ($c1) \cdot ($c2) \] 
 $PAR
 \{ans_rule(15)\}
 
@@ -153,10 +153,10 @@ _Last_:
 
 * Combine like terms: 
 
->> [`\begin{align*} 
+>> [`\begin{aligned} 
 \color{blue}{[$aRbR] + [$aRbI] i + [$aIbR] i - [$aIbI]} & = [$ansR] + ([$aRbI] + [$aIbR]) i \\
 & = [$ansR] + [$ansI] i \\
-\end{align*}
+\end{aligned}
 `]<<
 
 

Contrib/CUNY/CityTech/CollegeAlgebra_Trig/ComplexNumbers/sum-complex.pg

@@ -125,12 +125,12 @@ END_PGML_HINT
 
 BEGIN_PGML_SOLUTION
 
- [`\begin{align*}
+ [`\begin{aligned}
  ([$c1]) + ([$c2]) &= [$c1] + [$c2] \\
   &= [$aR] + [$bR] + [$aI] i + [$bI] i \\
  &= [$ansR] + ([$aI] + [$bI]) i \\
  &= [$ansR] + [$ansI] i \\
- \end{align*}`] 
+ \end{aligned}`] 
 
 END_PGML_SOLUTION
 Context()->normalStrings;

Contrib/CUNY/CityTech/CollegeAlgebra_Trig/ExponentialEquations-Calc/bacteria-word.pg

@@ -166,15 +166,15 @@ b. We want to know the population size [`A(t)`] when [`t= [$mins]`]. So we need
 c. The initial population is [`[$init]`].  The double of that is [`[@2*$init@]`]. We want to find [`t`] such that [`A(t)= [@2*$init@]`]:
 
 >>
-[`\begin{align*}
+[`\begin{aligned}
 [$At] & = [@2*$init@] & \qquad \text{divide both sides by } [$init] \\
 e^{[$rate] t}  & = 2  & \qquad \text{apply } \ln \text{to both sides}\\
 \ln\left(e^{[$rate] t}\right)  & = \ln 2  & \qquad \text{use the } \log \text{property for powers} \\
 [$rate] t \ln e & = \ln 2  & \qquad  \text{use that } \ln e = 1 \\
  [$rate] t  & = \ln 2  & \qquad  \text{divide both sides by } [$rate] \\
  t & = \dfrac{\ln 2}{[$rate]} &\\
  t & \approx [$time] &
-\end{align*}
+\end{aligned}
 `]<<
 
 The population will double in about [`[$time]`] minutes.

Contrib/CUNY/CityTech/CollegeAlgebra_Trig/ExponentialEquations-Calc/investment-word.pg

@@ -117,15 +117,15 @@ Hence [` A(t) = [$init] e^{[$rate]t} `]. We want to find [`t`] such that
 [`A(t) = [$double]`].
 
 >>
-[`\begin{align*}
+[`\begin{aligned}
 [$init] e^{[$rate]t} & = [@2*$init@] & \qquad \text{divide both sides by } [$init] \\
 e^{[$rate] t}  & = 2  & \qquad \text{apply } \ln \text{to both sides}\\
 \ln\left(e^{[$rate] t}\right)  & = \ln 2  & \qquad \text{use the } \log \text{property for powers} \\
 [$rate] t \ln e & = \ln 2  & \qquad  \text{use that } \ln e = 1 \\
  [$rate] t  & = \ln 2  & \qquad  \text{divide both sides by } [$rate] \\
  t & = \dfrac{\ln 2}{[$rate]} &\\
  t & \approx [$time] &
-\end{align*}
+\end{aligned}
 `]<<
 
 It will take about [`[$time]`] years for the investment to reach $[$double]. 

Contrib/CUNY/CityTech/CollegeAlgebra_Trig/ExponentialEquations-Calc/population-decline.pg

@@ -0,0 +1,65 @@
+##DESCRIPTION
+##  Algebra problem: exponential model of population growth
+##ENDDESCRIPTION
+
+
+########################################################################
+
+DOCUMENT();      
+
+loadMacros(
+   "PGstandard.pl",     # Standard macros for PG language
+   "MathObjects.pl",
+   "PGML.pl",
+   "contextFraction.pl"
+);
+
+# Print problem number and point value (weight) for the problem
+TEXT(beginproblem());
+
+# Show which answers are correct and which ones are incorrect
+$showPartialCorrectAnswers = 1;
+
+##############################################################
+#
+#  Setup
+#
+#
+Context("Numeric");
+
+$year = random(2013,2023,1);
+$initial_population = random(201000,699000,1000);
+$rate = random(1.6,6.3,0.1);
+$wait_time = random(17,65,1);
+$laterpopulation=$initial_population*exp(-$rate*$wait_time/100);
+$halftime=floor($year+100*ln(2)/$rate);
+
+##############################################################
+#
+#  Text
+#
+#
+
+BEGIN_PGML
+
+
+In [`[$year]`], the population of a city is [`[$initial_population]`], and is decreasing exponentially at [`[$rate]%`] per year.
+
+a. What is the population size after [`[$wait_time]`] years? [_________________________]{$laterpopulation}
+
+b. In what year will half of the population be left? [_________________________]{$halftime}
+
+
+
+END_PGML
+
+##############################################################
+#
+#  Answers
+#
+#
+
+
+COMMENT("Funded by US DoE Title V: Opening Gateways grant.");
+
+ENDDOCUMENT();        

Contrib/CUNY/CityTech/CollegeAlgebra_Trig/ExponentialEquations-Calc/population-growth.pg

@@ -0,0 +1,71 @@
+##DESCRIPTION
+##  Algebra problem: exponential model of population growth
+##ENDDESCRIPTION
+
+
+########################################################################
+
+DOCUMENT();      
+
+loadMacros(
+   "PGstandard.pl",     # Standard macros for PG language
+   "MathObjects.pl",
+   "PGML.pl",
+   "contextFraction.pl"
+);
+
+# Print problem number and point value (weight) for the problem
+TEXT(beginproblem());
+
+# Show which answers are correct and which ones are incorrect
+$showPartialCorrectAnswers = 1;
+
+##############################################################
+#
+#  Setup
+#
+#
+Context("Numeric");
+
+$year = random(2013,2023,1);
+$initial_population = random(501000,999000,1000);
+$rate = random(1.1,3.9,0.1);
+$wait_time = random(31,76,1);
+$wait_time_year = $year+$wait_time;
+$laterpopulation=$initial_population*exp($rate*$wait_time/100);
+$doubletime=floor($year+100*ln(2)/$rate);
+$tripletime=floor($year+100*ln(3)/$rate);
+
+##############################################################
+#
+#  Text
+#
+#
+
+BEGIN_PGML
+
+
+In [`[$year]`], the population of a country is [`[$initial_population]`], and is growing at [`[$rate]%`] per year.
+
+a. What will the population be in [`[$wait_time_year]`]? [_________________________]{$laterpopulation}
+
+b. In what year will the population be double? [_________________________]{$doubletime}
+
+c. In what year will the population be triple? [_________________________]{$tripletime}
+
+
+
+
+
+END_PGML
+
+##############################################################
+#
+#  Answers
+#
+#
+
+
+COMMENT("Funded by US DoE Title V: Opening Gateways grant.");
+
+ENDDOCUMENT();        

Contrib/CUNY/CityTech/CollegeAlgebra_Trig/LogarithmicProperties/condense-properties-powers.pg

@@ -125,7 +125,7 @@ END_PGML
 
 BEGIN_PGML_HINT
 
-* Do you have an expression like `\color{blue}{a} \log(\color{red}{b})`?  That is `\log(\color{red}{b}^\color{blue}{a})`.
+* Do you have an expression like `\color{blue}{a} \log(\color{red}{b})`?  That is `\log(\color{red}{b}^{\color{blue}{a}})`.
 
 * Do you have a sum of logarithms? Use the fact that `\log(\color{red}{a}) + \log(\color{blue}{b})  = \log(\color{red}{a}\color{blue}{b})`.
 
@@ -140,10 +140,10 @@ BEGIN_PGML_SOLUTION
 
 a.
 >>
-[`\begin{align*}
+[`\begin{aligned}
 &\color{magenta}{[$m] \log(x)}- \color{magenta}{[$p] \log(y)}+\color{magenta}{ [$n] \log(z)} & \text{three terms of the form } \color{blue}{a} \log(\color{red}{b})\\ \\
 \longrightarrow\quad&\color{blue}{[$m]} \log(\color{red}{x})- \color{blue}{[$p]} \log(\color{red}{y})+ \color{blue}{[$n]} \log(\color{red}{z}) & \text{identify the} \;\color{red}{a}\text{'s and the }\;\color{blue}{b}\text{'s} \\
-&&\text{ use that } \color{blue}{a}\log(\color{red}{b})= \log(\color{red}{b}^\color{blue}{a})\\ \\
+&&\text{ use that } \color{blue}{a}\log(\color{red}{b})= \log(\color{red}{b}^{\color{blue}{a}})\\ \\
 \longrightarrow\quad & \log(\color{red}{x}^{\color{blue}{[$m]}} )-  \log(\color{red}{y}^{\color{blue}{[$p]}})+  \log(\color{red}{z}^{\color{blue}{[$n]}}) & \text{the first two terms are a } \textbf{difference} \text{ of logs}\\ \\
 \longrightarrow\quad & \log(\color{red}{x^{[$m]}} )-  \log(\color{blue}{y^{[$p]}})+  \log(z^{[$n]})& \text{ use that }  \log (\color{red}{a}) - \log(\color{blue}{b}) = \log\left(\dfrac{\color{red}{a}}{\color{blue}{b}}\right) \\
 && \text{identify the } \color{red}{a} \text{ and the } \color{blue}{b} \\ \\
@@ -152,13 +152,13 @@ a.
 &&  \text{identify the } \color{red}{a} \text{ and the } \color{blue}{b} \\ \\
 \longrightarrow\quad & \log\left(\color{red}{\dfrac{x^{[$m]}}{y^{[$p]}}} \cdot\color{blue}{z^[$n]}\right)  &\\ \\
 \longrightarrow\quad &  \log\left(\dfrac{x^[$m]z^[$n]}{y^[$p]}\right) &\\
-\end{align*}`]<<
+\end{aligned}`]<<
 
 b.
-    >>[`\begin{align*} 
+    >>[`\begin{aligned} 
 &-\color{magenta}{[$m] \log(z)}+ \color{magenta}{[$p] \log(x)}-\color{magenta}{ [$n] \log(y)} & \text{three terms of the form } \color{blue}{a} \log(\color{red}{b})\\ \\
 \longrightarrow\quad&-\color{blue}{[$m]} \log(\color{red}{z})+ \color{blue}{[$p]} \log(\color{red}{x})- \color{blue}{[$n]} \log(\color{red}{y}) & \text{identify the} \;\color{red}{a}\text{'s and the }\;\color{blue}{b}\text{'s} \\
-&&\text{ use that } \color{blue}{a}\log(\color{red}{b})= \log(\color{red}{b}^\color{blue}{a})\\ \\
+&&\text{ use that } \color{blue}{a}\log(\color{red}{b})= \log(\color{red}{b}^{\color{blue}{a}})\\ \\
 \longrightarrow\quad & -\log(\color{red}{z}^{\color{blue}{[$m]}} )+ \log(\color{red}{x}^{\color{blue}{[$p]}})-  \log(\color{red}{y}^{\color{blue}{[$n]}}) & \text{the last two terms are a } \textbf{difference} \text{ of logs}\\ \\
 \longrightarrow\quad & -\log(z^{[$m]} )+  \log(\color{red}{x^{[$p]}})-  \log(\color{blue}{y^{[$n]}})& \text{ use that }  \log (\color{red}{a}) - \log(\color{blue}{b}) = \log\left(\dfrac{\color{red}{a}}{\color{blue}{b}}\right) \\
 && \text{identify the } \color{red}{a} \text{ and the } \color{blue}{b} \\ \\
@@ -170,7 +170,7 @@ b.
 \longrightarrow\quad & \log\left(\dfrac{\color{red}{\dfrac{x^{[$p]}}{y^{[$n]}}}}{\color{blue}{\dfrac{z^[$m]}{1}}}\right)  & \text{division of fractions}\\ \\
 \longrightarrow\quad & \log\left(\dfrac{x^{[$p]}}{y^{[$n]}}\cdot\dfrac{1}{z^{[$m]}}\right)  &\\ \\
 \longrightarrow\quad &  \log\left(\dfrac{x^[$p]}{y^[$n]z^[$m]} \right)  &\\
-\end{align*}`]<<
+\end{aligned}`]<<