Memory Aid | Mathematics — Secondary 4 (CST)

|\begin{align} \color{blue}{f(x)} &= a \color{green}{x}^2 \\ \color{blue}{-6{.}75} &= a \color{green}{(-1{.}5)}^2 \\ -3 &= a \end{align}|

​Substitute the values ​​of |x| and |y| with a point of the curve to find the value of |a|.

The equation of the parabola is |f(x) = -3 x^2|.

|x:| Number of years elapsed since 2005

|f(x):| Number of toads

Define the variables.

|​\begin{align} f(x) &= a (c)^{x} \\ f(x) &= 500 (c)^{x} \end{align}|

​Replace |a| by the initial value, i.e., the number of toads in 2005.

|​\begin{align} f(x) &= 500 (c)^{x} \\ f(x) &= 500 (0{.}95)^{x} \end{align}|

Replace |c| with |0{.}95,| since |c = 1 - 5\ \text{​%} = 1 - 0{.}05 = 0{.}95.|

|​\begin{align} f(18) &= 500 (0{.}95)^{6} \\ ​f(18) &\approx 368 \end{align}|

​By replacing |x| with different values and using trial-and-error, we deduce that |x=6.|

If |x=6,| then in |6| years there will be |368| toads. Since the study took place in 2005, we can deduce that in 2011 (2005 + 6) there will be approximately |368| toads.

Identify 6 p.m. on the |x|-axis.

Select the solid point which is included in the function at that moment and mark its height on the |y|-axis.

​​According to the graph, there were 7 000 people visiting the Videotron Centre at 6 p.m..

Identify the places where the speed is 16 km/h on the first cycle.

Identify the points on each of the lines where the desired minimum speed is located.

|a = \dfrac{\Delta y}{\Delta x} = \dfrac{24-12}{4-3} = 12|

So, |y = 12x + b|

By substituting in x and y,
|\begin{align} 24 &= 12 (4) + b \\ -24 &= b \end{align}|

Finally, |\color{green}{y=12x - 24}|

​Find the equation of |\color{green}{\text{the green line}}.| To do so, use the points |(3,12)| and |(4,24)|, since they are on |\color{green}{\text{the green line}}.|

​​|\begin{align} 16 &= 12x - 24 \\ 40 &= 12x \\ \color{green}{3{.}33} &\approx x \end{align}|

​Replace |y|with |16| to find the time, in minutes, of the |\color{green}{\text{green point}}.|

|a = \dfrac{\Delta y}{\Delta x} = \dfrac{24-0}{4-6} = -12|

So, |y = -12x + b|

From substituting x and y,
|\begin{align} 24 &= -12 (4) + b \\ 72 &= b \end{align}|

Finally, |\color{blue}{y=-12x + 72}|

​Find the equation of |\color{blue}{\text{the blue line}}.| To do so, use the points |(4,24)| and |(6,0)|, since they are on |\color{blue}{\text{the blue line}}.|

|\begin{align} 16 &= -12x + 72 \\ -56 &= -12x \\ \color{blue}{4{.}67} &\approx x\end{align}|

Replace |y| with |16| to find the time, in minutes, of the |\color{blue}{\text{blue point}}.|

|\color{blue}{4{.}67} - \color{green}{3{.}33} = \color{red}{1{.}34 \ \text{min}}|

​Determine the duration between |\color{green}{\text{the green point}}| and |\color{blue}{\text{the blue point}}.|

​|6 - 0 = 6 \ \text{min}|

​Determine the period (length) of the repeated cycle.

​|\begin{align} \text{No. of cycles} &= \dfrac{\text{Total trip duration}} {\text{Duration of One Cycle}} \\ \text{No. of cycles} &= \dfrac{45}{6} \\ \text{No. of cycles} &=7{.}5 \end{align}|

​Determine the number of complete cycles in the route.

​|7{.}5| equals |7| complete cycles |(7 \times 6 = 42\ \text{min})| and another half a cycle.

So, the |1{.}34| minute interval repeats exactly seven times.

​Analyze the incomplete cycle to determine if the portion meets the requirements of the initial question (minimum speed of 16 km / h).

|\begin{align} \text{Total duration} &= \color{red}{\text{Duration of One Cycle}} \times \text{No. of periods} \\ \text{Total duration} &= \color{red}{1{.}34} \times 7 \\ \text{Total duration} &= 9{.}38 \end{align}|

​Determine the total duration of the analyzed interval.

​By the end of her training, Marie-Claude will have spent a total of |9{.}38| minutes pedaling at a speed of at least 16 km / h.​

​Domain: |[0, 12]|​

The smallest value on the |x|-axis is |0| and the largest is |12.|

​Range: |[-5, 20]|

​The smallest value on the |y|-axis is |-5| and the largest is |20.|

​Increase: |[0, 4] \cup [9, 12]|

When analyzing the |x| values, these two portions of the graph go up or remain constant.

​Decrease: |[4, 10]|

​When analyzing the |x| values, this is the only portion of the graph that goes down or remains constant.

​Maximum: |\left\{20 \right\}|

​Among the |y| values, this is the largest value reached by the graph.

​Minimum: |\left\{-5\right\}|

Among the |y| values, this is the smallest value reached by the graph.

​Zeros of the Function: |(0,0), (8,0), (11,0)|

​These are the coordinates of the points where the graph intercepts the |x|-axis.

​Y-intercept: |(0,0)|

​It is the coordinate of the point where the graph intercepts the |y|-axis.

​Positive: | [0,8] \cup [11,12]|

Among the |x| values, these portions of the graph are above or equal to the |x|-axis.

|\begin{align} &a\; = 6{.}5 \\ &m = 4{.}1 \\ &c \ (m\overline {AB}) =\ ? \\ &b \ (m\overline {BC}) =\ ? \end{align}|

Associate all known and desired measurements with one of the measurements in the reference drawing.

|\begin{align} a^2 &= m c \\ 6{.}5^2 &= 4{.}1 c \end{align}|

Choose the theorem where there will be only one unknown:

In a right angle triangle, the measure of each side of the right angle is the geometric mean between its projection onto the hypotenuse and the hypotenuse itself.

|\begin{align} 6{.}5^2 &= 4{.}1 c \\ 42{.}25 &= 4{.}1 c \\ \dfrac{42{.}25}{\color{red}{4{.}1}} &= \dfrac{4{.}1c}{\color{red}{4{.}1}} \\ 10{.}3 &\approx c \end{align}|

​Solve the equation.

|\begin{align} a^2 + b^2 &= c^2 \\ 6{.}5^2 + m\overline {BC}^2 &= 10{.}3^2 \\ 42{.}25 + m\overline {BC}^2 &\approx 106{.}09 \\ m\overline {BC}^2 &\approx 63{.}84 \\ m\overline {BC} &\approx 8 \end{align}|

​Apply the Pythagorean theorem on the large green right triangle to find the measure of the missing leg.

​ So, |m \overline {AB} \approx 10{.}3 \ \text{m}| and |m \overline {BC} \approx 8 \ \text{m}.|

​|\sin \color{red}{35^\circ} = \dfrac{\color{red}{c}}{\color{green}{13}}|

​Identify the appropriate trigonometric ratio: |\sin \theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}|

|\begin{align} \sin \color{red}{35^\circ} &= \dfrac{\color{red}{c}}{\color{green}{13}} \\ \color{green}{13}\sin \color{red}{35^\circ} &= \color{red}{c} \\ 7{.}46 &\approx \color{red}{c} \end{align}|

​Solve the equation.

|\cos \color{red}{35^\circ} = \dfrac{\color{blue}{b}}{\color{green}{13}}|

Identify the correct trigonometric ratio: |\cos \theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}|

​|\begin{align} \cos \color{red}{35^\circ} &= \dfrac{\color{blue}{b}}{\color{green}{13}} \\ \color{green}{13} \cos \color{red}{35^\circ} &= \color{blue}{b} \\ 10{.}65 &\approx \color{blue}{b} \end{align}|

​Solve the equation.

​ So, |\color{blue}{m \overline {AC}} \approx 10{.}65 \ \text{m}| and |\color{red}{m \overline {AB}} \approx 7{.}46 \ \text{m}.|

​|\tan ? = \dfrac{\color{red}{18{.}5}}{\color{blue}{12}}|

Identify the correct trigonometric ratio: |\tan \theta = \dfrac{\text{Opposite}}{\text{Adjacent}}|

|\begin{align} \tan ? &= \dfrac{\color{red}{18{.}5}}{\color{blue}{12}} \\ \tan ? &\approx 1{.}54 \\ ? &= \tan^{-1}(1{.}54) \\ ? &\approx 57^\circ \end{align}|

​Solve the equation.

​ ​The helicopter's orientation angle should be |57^\circ.|

Identify the vertices and edges of the triangle.

​If possible, deduce the other measurements of the triangle (using the sum of interior angles of a triangle and the properties of the isosceles triangle).

|\begin{align} \dfrac{\color{green}{a}}{\sin 40^\circ} &= \displaystyle \dfrac{\color{blue}{20}}{\sin \color{blue}{70^\circ}} \\\\ \Rightarrow\ \color{green}{a} &= \dfrac {\color{blue}{20} \sin 40^\circ}{\sin \color{blue}{70^\circ}} \\ \color{green}{a} &\approx 13{.}68 \ \text{m} \end{align}|

Apply the law of sines and isolate the variable.

​So, |m \overline{AB} = m \overline {AC} = 20 \ \text{m}| and |m \overline {BC} \approx 13{.}68 \ \text{m}|

​Identify the vertices and edges of the triangle.

|\begin{align} \dfrac{\color{blue}{1{.}18}}{\sin \color{blue}{15}} &= \dfrac {\color{red}{3{.}39}}{\sin \color{red}{B}} \\\\ \Rightarrow\ \sin \color{red}{B} &= \dfrac{\color{red}{3{.}39} \sin \color{blue}{15}}{\color{blue}{1{.}18}} \\ \sin \color{red}{B} &\approx 0{.}744 \end{align}|

​Use the law of sines and isolate the sine of the desired angle.

|\begin{align} \sin \color{red}{B} &\approx 0{.}744 \\ \color{red}{ B} &\approx \sin^{-1} (0{.}744) \\ \color{red}{B} &\approx 48{.}1^\circ \end{align}|

​Calculate the value of the variable by using |\sin^{-1}.|

|\begin{align} \color{red}{m\angle B} &\approx 180^\circ - 48{.}1^\circ \\ \color{red}{m\angle B} &\approx 131{.}9^\circ \end{align}|

Find the value of the obtuse angle.

​ ​The measure of the angle is |131{.}9^\circ.|

Identify the vertices and edges of the triangle.

|\begin{align} p &= \dfrac{\color{blue}{a}+\color{red}{b}+\color{green}{c}}{2} \\ &= \dfrac{\color{blue}{22}+\color{red}{24}+\color{green}{21}}{2} \\ &= 33{.}5\ \text{m} \end{align}|

​Calculate the value of the half-perimeter |(p)| of the figure.

|\begin{align} A &=\sqrt{p(p-\color{blue}{a})(p-\color{red}{b})(p-\color{green}{c})}\\ &=\sqrt{33{.}5(33{.}5-\color{blue}{22})(33{.}5-\color{red}{24})(33{.}5-\color{green}{21})} \end{align}|

​Substitute the values ​​into the formula to calculate the area of ​​the triangle.

|\begin{align} A &=\sqrt{33{.}5(33{.}5-\color{blue}{22})(33{.}5-\color{red}{24})(33{,}5-\color{green}{21})} \\ &= \sqrt{33{.}5(11{.}5)(9{.}5)(12{.}5)} \\ &\approx 213{.}89\ \text{m}^2 \end{align}|

Solve the equation while respecting the order of operations.

​​The area of the floor that is covered by surveillance camera is |213{.}89\ \text{m}^2.|

|\angle \color{red}{BAC} \cong \angle \color{green}{EFG}|

​|m \angle \color{green}{EFG} = 180^\circ - 130^\circ - 17^\circ = 33^\circ = m \angle \color{red}{BAC}|

|\color{red}{\overline{AB}} \cong \color{green}{\overline{EF}}|

By hypothesis

​|\angle \color{red}{ABC} \cong \angle \color{green}{DEF}|

​By hypothesis

​​ ​|\Delta \color{red}{ABC} \cong \Delta \color{green}{DEF}| by A-S-A

|\dfrac{m\color{green}{\overline{AC}}}{m \color{blue}{\overline {DF}}} = \dfrac{m\color{green}{\overline{AB}}}{m \color{blue}{\overline {DE}}}= \dfrac{m\color{green}{\overline{BC}}}{m \color{blue}{\overline {EF}}}|

​|\begin{align} \dfrac{\color{green}{1{.}4}}{\color{blue}{3{.}5}} &= \dfrac{\color{green}{1{.}1}}{\color{blue}{2{.}75}}= \dfrac{\color{green}{0{.}8}}{\color{blue}{2}} \\ \dfrac{2}{5}\ \ &=\ \ \dfrac{2}{5}\ \ \ =\ \ \dfrac{2}{5}\end{align}|

​​​​Therefore, |\Delta \color{green}{ABC} \sim \Delta \color{blue}{DEF}| by S-S-S.

​Montreal |= (\color{blue}{x_1}, \color{red}{y_1}) = (\color{blue}{512}, \color{red}{647})|
Paris | = (\color{green}{x_2}, y_2) = ( \color{green}{5936}, 1603)|

​Identify the points |(x_1, y_1)| and |(x_2, y_2).|

|\begin{align} \text{Distance} &= \sqrt{(y_2 - \color{red}{y_1})^2 + (\color{green}{x_2} - \color{blue}{x_1})^2} \\ \text{Distance} &= \sqrt{(1\ 603 - \color{red}{647})^2 + (\color{green}{5\ 936} - \color{blue}{512})^2} \end{align}|

​Substitute the values ​​into the formula.

|​\begin{align} ​\text{Distance} &= \sqrt{(1\ 603 - \color{red}{647})^2 + (\color{green}{5\ 936} - \color{blue}{512})^2} \\ \text{Distance} &= \sqrt{ 956^2 + 5\ 424^2} \\ \text{Distance} &\approx 5\ 507{.}6 \ \text{km} \end{align}|

​Solve the equation.

​ ​The distance between Montreal and Paris is approximately |5\ 507{.}6 \ \text{km}.|

​|a = \dfrac{\Delta y}{\Delta x} = \dfrac{0{.}7 - 2{.}5}{1{.}5 - 0} = -1{.}2|

​Find the slope of |\overline{AB}.|

​|\begin{align} y &= -1{.}2x + b \\ 2{.}25 &= -1{.}2 (1{.}55) + b \\ 2{.}25 &= -1{.}86 + b \\ 4{.}11 &= b \end{align}|

​Find the equation of the line passing through |C (1{.}55, 2{.}25)| in function form |y = ax + b.|

In this case, the value of |a| of the desired equation is the same as that of |\overline{AB}|, since the lines are parallel.

​​The equation of the line that is parallel to |\overline{AB}| and passes through point | C (1{.}55, 2{.}25)| is |y = -1{.}2x + 4{.}11.|

|a_1 = \dfrac{\Delta y}{\Delta x} = \dfrac{0{.}7 - 2{.}5}{1{.}5 - 0} = -1{.}2|

Find the slope of |\overline{AB}.|

|\begin{align} a_1 \times a_2 &= -1 \\-1{.}2 \times a_2 &= -1 \\ a_2 &= \dfrac{-1}{-1{.}2} \\ a_2 &= 0{.}8\overline{3} \end{align}|

​Keeping in mind that the product of the slopes of two perpendicular lines is equal to |-1| , find the value of |a_2| of the line passing through C.

​|\begin{align} y &= a_2 x + b \\ y &= 0{.}8\overline{3}x + b \\ 2{.}25 &= 0{.}8\overline{3} (1{.}55) + b \\ 2{.}25 &\approx 1{.}29 + b \\ 0{.}96 &\approx b \end{align}|

​Find the rule of the line passing through |C (1{.}55, 2{.}25)|
in function form |y = a_2x + b.|

​​​Finally, the equation of the line which is perpendicular to |\overline{AB}| and which passes through the point |C(1{.}55, 2{.}25)| is |y=0{.}8\overline{3}x + 0{.}96.|

|\dfrac{156\ 700 + 158\ 900 + \dots + 179\ 000 + 183\ 000}{11} = \color{green}{\$168\ 700}|

​Calculate the mean given by |\frac {\sum x_i}{n}.|

|\begin{align} \mid \color{blue}{156\ 700} - \color{green}{168\ 700}{\mid} &= 12\ 000 \\ \mid \color{red}{158\ 900} - \color{green}{168\ 700}{\mid} &= 9\ 800 \\ \mid 159\ 000 - \color{green}{168\ 700}{\mid} &= 9\ 700 \\ \mid 162\ 500 - \color{green}{168\ 700}{\mid} &= 6\ 200 \\ \mid 164\ 100 - \color{green}{168\ 700}{\mid} &= 4\ 600 \\ \mid 167\ 400 - \color{green}{168\ 700}{\mid} &= 1\ 300 \\ \mid 172\ 000 - \color{green}{168\ 700}{\mid} &= 3\ 300 \\ \mid 175\ 000 - \color{green}{168\ 700}{\mid} &= 6\ 300 \\ \mid 178\ 100 - \color{green}{168\ 700}{\mid} &= 9\ 400 \\ \mid 179\ 000 - \color{green}{168\ 700}{\mid} &= 10\ 300 \\ \mid 183\ 000 - \color{green}{168\ 700}{\mid} &= 14\ 300 \end{align}|

​Calculate the deviations from the mean of each of the data.

|MD = \dfrac{12\ 000+9\ 800+ \dots + 10\ 300 + 14\ 300}{11} \approx \$7\ 927{.}27|

Calculate the average of the deviations from the mean (mean deviation).

​|R_{100}84 = \displaystyle \frac{\color{blue}{21} + \frac{\color{green}{3}}{2}}{\color{red}{33}} \times 100|

Looking at the table, we find: ||​\begin{align} \color{blue}{21} &\color{blue}{= \text{number of data less than the value in question}} \\ \color{green}{3} &\color{green}{= \text{number of data equal}} \\ \color{red}{33} &\color{red}{= \text{total number of date}} \end{align}||

​|R_{100}84 \approx 68{.}18|

This result was found using the order of operations.

​|R_{100}84 = 69|

If the answer is not a whole number, the percentile rank always corresponds to the next whole number (round up).

​Despite your result of |84\ \%,| your application will not be kept for the rest of the process , because you have a percentile rank of |69.|

|\begin{align} \text{Percentile Rank of the data} &= \frac{\color{blue}{82}}{100} \times \color{red}{33} \\ &\approx 27{.}06\end{align}|

According to the data from the question, we get:

|​\begin{align} \color{blue}{\text{percentile rank} = 82} \\ \color{red}{\text{total number of data} = 33} \end{align}|

|\text{Position of the data} = 27|

The position of the data always corresponds to the lower integer if the answer is not a whole number.

Since the data located at the |27^{th}| position is |92|, everyone who received a result greater than |92\ \%| will be retained.​

Draw a rectangle, as small as possible, to frame the scatter plot.

Using a ruler, measure the length |(L)| and width |(l)| of the rectangle. In this case,

|\color{blue}{L = 13{.}5 \ \text{cm}}|

|\color{red}{l = ​2{.}8 \ \text{cm}}|

|\begin{align} r &\approx \pm \left(1 - \dfrac{\color{red}{2{.}8}}{\color{blue}{13{.}5}}\right) \\ r &\approx \pm 0{.}79 \end{align}|

Replace |L| and |l| in the formula |r \approx \pm \left(1 - \frac{\color{red}{l}}{\color{blue}{L}}\right).|

​|r \approx - 0{.}79|

​Since the rectangle is oriented downwards (decreasing), the correlation coefficient is negative.

The correlation coefficient between the number of hours absent and the final result in percentage is approximately |-0{.}79,| which means it is a moderate negative correlation.

Place the points in ascending order according to the value of the independent variable |(x)|, taking care not to “undo” the initial ordered pairs.

​Separate the points into three equal groups. If this is not possible, make sure that the first and the last group have the same amount of data.

|M_1 = (1, 29)|

|M_2 = \left(\dfrac{2+3}{2}, \dfrac{54+58}{2}\right) = (2{.}5 , 56)|

|M_3 = (4, 90)|

​Calculate the median coordinate of each group.

|\begin{align} \color{green}{P} &= \left(\frac{1 + 2{.}5 + 4}{3} , \frac{29 + 56 + 90}{3}\right) \\ &= \color{green}{(2{.}5 , 58{.}33)} \end{align}|

Calculate the mean point by taking the average of the |x|’s and |y|’s of the three midpoints.

​|\color{blue}{a} = \dfrac{\Delta y}{\Delta x} = \dfrac{90 - 29}{4 - 1} \approx 20{.}33|

​According to the equation of the regression line |y = \color{blue}{a}x + \color{red}{b}|, determine the value of |\color{blue}{a}| from the points |M_1| and |M_3.|

​

|\begin{align} y &= \color{blue}{20{.}33} x + \color{red}{b}\\ \color{green}{58{.}33} &= \color{blue}{20{.}33}(\color{green}{2{.}5}) + \color{red}{b} \\ \color{red}{7{.}503} &= \color{red}{b} \end{align}|

So, |y = \color{blue}{20{.}33} x + \color{red}{7{.}503}|

​Find the value of parameter |\color{red}{b}| by substituting |x| and |y| with the coordinates of the point |\color{green}{P}.|

|\begin{align} y &= \color{blue}{20{.}33} x + \color{red}{7{.}503} \\ y &= \color{blue}{20{.}33} (20) + \color{red}{7{.}503} \\ y &= 414{.}103 \end{align}|

​Since we want to know the height of the trees after |20| years, substitute |x| with |20.|

After |20| years, the trees will be about |414{.}103\ \text{cm}.| Thus, the first balconies must have a minimum height of |414{.}103\ \text{cm}.|

First, place the points in ascending order according to the value of the independent variable, without undoing the initial ordered pairs.

If possible, separate the distribution into two equal groups.

|\begin{align} P_1 &= \left(\dfrac{\color{red}{1+1+... +2+2}}{7}, \dfrac{\color{blue}{21+23+...+42+54}}{7}\right) \\ &\approx (\color{red}{1{.}43}, \color{blue}{34{.}29}) \\\\ P_2 &= \left(\dfrac{\color{green}{​​3+3...+4+4}}{7}, \dfrac{58+59+...+90+97}{7}\right) \\ &\approx (\color{green}{3{.}43},; 76) \end{align}|

​Calculate the mean points of each of the groups.

|a = \dfrac{\Delta y}{\Delta x} = \dfrac{76 - \color{blue}{34{.}29}}{\color{green}{3{.}43} - \color{red}{1{.}43}} \approx 20{.}86|

Thereby, |y = 20{.}86x + b.|

By substituting the coordinates of |P_1:| ||\begin{align} y &= 20{.}86x + b \\ 34{.}29 &= 20{.}86 (1{.}43) + b \\ 34{.}29 &= 29{.}83 + b \\ 4{.}46 &= b \end{align}||

Thereby, |y = 20{.}86x + 4{.}46.|

Find the equation of the regression line in the form |y = ax + b| according to the points |P_1| and |P_2.|

|\begin{align} y &= 20{.}86x + 4{.}46 \\ y &= 20{.}86 (20) +4{.}46 \\ y &= 417{.}2 + 4{.}46 \\ y &= 421{.}66 \end{align}|

​Since we want to know the height of the trees after |20| years, substitute |x| with |20.|

​After |20| years, the trees will be about |421{.}66\ \text{cm}.| Thus, the first balconies must be of a minimum height of |421{.}66\ \text{cm}.|