Memory Aid | Mathematics — Secondary 1 & 2

|\displaystyle​4\frac{1}{3} = \frac{4 \times 3+ 1}{3} = \frac{13}{3}|

|\color{blue}{\displaystyle\frac{8}{3}}|

|\color{red}{\displaystyle-0.625 = \frac{-625}{1000} = \frac{-5}{8}}|

|\color{green}{\displaystyle-80\%=\frac{-80}{100} = \frac{-4}{5}}|

|\color{fuchsia}{\displaystyle\left( \frac{-1}{2} \right)^2 = \frac{-1}{2} \times \frac{-1}{2} = \frac{1}{4}}|

|\displaystyle\color{orange}{\sqrt9 = 3}|

Use simplified fractional notation to represent each of the quantities.

|\displaystyle\frac{13}{3} = \frac{1040}{240}|

|\color{blue}{\displaystyle\frac{8}{3}=\frac{640}{240}}|

|\color{red}{\displaystyle\frac{-5}{8}=\frac{-150}{240}}|

|\color{green}{\displaystyle\frac{-4}{5}=\frac{-192}{240}}|

|\color{fuchsia}{\displaystyle\frac{1}{4}=\frac{60}{240}}|

|\color{orange}{\displaystyle3=\frac{3}{1} = \frac{720}{240}}|

Write all the numbers using a common denominator.

Since the denominators are all equal, the numbers can be ordered according to their numerator, as follows:

|\displaystyle \color{white}{=}\color{green}{\frac{-192}{240}} < \color{red}{\frac{-150}{240}} < \color{fuchsia}{\frac{60}{240}} < \color{blue}{\frac{640}{240}} < \color{orange}{\frac{720}{240}} < \frac{1040}{240}|
|\displaystyle \Rightarrow\ \color{green}{-80\%} < \color{red}{-0{.}625} < \color{fuchsia}{\left( \frac{-1}{2} \right)^2} < \color{blue}{\frac{8}{3}} < \color{orange}{\sqrt9} < 4\frac{1}{3}|

Total |=| Monday + Tuesday + Wednesday + Thursday

Write the chain of operations that will be used to solve the problem

|\color{white}{\text{Total}​}=\color{blue}{\displaystyle \left( \frac{50}{2}\right)} + \color{red}{50} + \color{orange}{\left(\left(50 + \displaystyle \frac{50}{2}\right) - 20\right)} + \color{fuchsia}{3 \times \displaystyle \left( \frac{50}{2}\right)}|​

Replace each day with its respective value.

|\color{white}{\text{Total}}=\color{blue}{25} + \color{red}{50} + \color{orange}{\left(\left(50 + \displaystyle \frac{50}{2}\right) - 20\right)} + \color{fuchsia}{3 \times \displaystyle \left(\frac{50}{2}\right)}|
|\color{white}{\text{Total}}=\color{blue}{25} + \color{red}{50} + \color{orange}{\big((50 + 25) - 20\big)} + \color{fuchsia}{3 \times \displaystyle \left( \frac{50}{2}\right)}|​​
|\color{white}{\text{Total}}=\color{blue}{25} + \color{red}{50} + \color{orange}{(75 - 20)} + \color{fuchsia}{3 \times \displaystyle \left(\frac{50}{2}\right)}|​​
|\color{white}{\text{Total}}=\color{blue}{25} + \color{red}{50} + \color{orange}{55} + \color{fuchsia}{3 \times \displaystyle \left(\frac{50}{2}\right)}|​​
|\color{white}{\text{Total}}=\color{blue}{25} + \color{red}{50} + \color{orange}{55} + \color{fuchsia}{3 \times 25}|​​
|\color{white}{\text{Total}}=\color{blue}{25} + \color{red}{50} + \color{orange}{55} + \color{fuchsia}{75}|​​
|\color{white}{\text{Total}} = 205 \ \text{min}|

Solve the problem by following the order of operations.

​​Simon spent |205| minutes preparing for his evaluation.

|264 \Rightarrow 2 + 6 + 4 = 12|
However, |12 \div 9 \neq | a whole number.

Determine if |264| is divisible by |9.|

|264 \Rightarrow 64 \div 4 = 16|
|264 \Rightarrow 4 \div 2 = 2|
Thus, |264| is divisible by |8.|

Find a divisor of |264| that is close to |9.|

Judith should let Justin invite |8| of his friends instead.

|\begin{align} \color{red}{\text{Quantity given}} &= \color{red}{\text{Its percentage}} \\ \color{red}{$14{.}85 \ } &= \color{red}{100\ \% - 45\ \%} \\ \color{red}{$14{.}85\ } &= \color{red}{55\ \%} \end{align}|

Identify the quantity given and associate a percentage with it.

|\begin{align} \color{blue}{\text{Quantity you are looking for}} &= \color{blue}{\text{Its percentage}}\\ \color{blue}{\text{Price before discount}} &= \color{blue}{100\ \%} \end{align}|

Identify the quantity you are looking for and its percentage.

|\Rightarrow \displaystyle \frac{\color{red}{$14{.}85 \ }}{\color{blue}{\text{Price before discount}}} = \frac{\color{red}{55\ \%}}{\color{blue}{100\ \%}}|

Build the proportion.

|\begin{align} \color{blue}{\text{Price before discount}} &= \displaystyle \frac{\color{blue}{100\ \%}\times \color{red}{$14{.}85\ \ }}{\color{red}{55\ \%}}​ \\ \color{blue}{\text{Price before discount}} &= \color{blue}{$27\ } \end{align}|

Solve the proportion.

Without the discount, she would have paid |$27| for her purchase.

|\color{red}{\text{Winner’s bursary}} : \color{blue}{\text{Finalist’s bursary}}|

Establish the order for the ratio.

|\begin{align} \color{red}{3 \ 500} &: \color{blue}{5 \ 000 - 3 \ 500} \\ \Rightarrow\ \color{red}{3 \ 500} &: \color{blue}{1 \ 500} \end{align}|

Determine the amount of each part.

|\begin{align} \color{red}{35} &: \color{blue}{15} &(\div 100) \\ \color{red}{7} &: \color{blue}{3} &(\div 5)​ \end{align}|

Simplify the ratio.

The bursary is shared using a ratio of |\color{red}{7}:\color{blue}{3}.|

|\color{red}{\text{Monday}: 476\ \text{km in} \ 6{.}5 \ \text{hrs} = 476 \ \text{km} / 6{.}5 \ \text{hrs}}|
|\color{blue}{\text{Tuesday}: 576\ \text{km in} \ 7{.}25 \ \text{hrs} = 576 \ \text{km} / 7{.}25 \ \text{hrs}}|
|\color{green}{\text{Wednesday}: 525\ \text{km in} \ 6{.}75 \ \text{hrs} = 525 \ \text{km} / 6{.}75 \ \text{hrs}}|
|\color{fuchsia}{\text{Thursday}: 712\ \text{km in} \ 9 \ \text{hrs} = 712 \ \text{km} / 9 \ \text{hrs}}|
|\color{orange}{\text{Friday}: 632\ \text{km in} \ 7{.}75 \ \text{hrs} = 632 \ \text{km} / 7{.}75 \ \text{hrs}}|

Convert each piece of information into a rate.

|\color{red}{\text{Monday}:476 \ \text{km} / 6{.}5 \ \text{hrs} \approx 73{.}23 \ \text{km/h}} \ (\div 6{.}5)|
|\color{blue}{\text{Tuesday}: 576 \ \text{km} / 7{.}25 \ \text{hrs} \approx 79{.}45 \ \text{km/h}}\ (\div 7{.}25)|
|\color{green}{\text{Wednesday}: 525 \ \text{km} / 6{.}75 \ \text{hrs} \approx 77{.}78 \ \text{km/h}}\ (\div 6{.}75)|
|\color{fuchsia}{\text{Thursday}: 712 \ \text{km} / 9 \ \text{hrs} \approx 79{.}11 \ \text{km/h}}\ (\div 9)|
|\color{orange}{\text{Friday}: 632 \ \text{km} / 7{.}75 \ \text{hrs} \approx 81{.}55 \ \text{km/h}}\ (\div 7{.}75)|

Determine the unit rates.

​His highest average speed was on |\color{orange}{\text{Friday}}| .

• The graphical representation of the situation passes through |(0,0).|

• The graphical representation forms a straight line.

|\begin{align} \text{Hours}\ \quad &\text{Cost} \\ \displaystyle \frac{\color{green}{1{.}4}}{\color{red}{125}} =\ &\frac{\color{green}{100{.}8}}{\color{red}{?}}​ \end{align}|

|\begin{align} \color{red}{?} &= \displaystyle \frac{\color{green}{100{.}8} \times \color{red}{125}}{\color{green}{1{.}4}} \\ &= \color{red}{9 \ 000\ $} \end{align}|

​The company's estimate should amount to |$9 \ 000.|

• The graph of the situation is a decreasing curved line.

• The graph does not touch the two axes.

|\begin{align} k &= \color{blue}{x y} \\ &= \color{blue}{2\times 3} \\ &= 6 \end{align}|

|\begin{align} k &= x y \\ 6 &= \color{green}{5} y \\ 1{.}2 &= y \end{align}|

​ In this situation, each winner would get the sum of |1{.}2| million dollars.

||\begin{align} &\underline{2(4x^2-6)} - \frac{1}{2}x^2 + (12x - 1) \div 4 \\\\ =\ &8x^2-12-\frac{1}{2}x^2 +\underline{(12x-1)\div 4} \\\\ =\ &\color{blue}{8x^2}\color{red}{-12}\color{blue}{-\frac{1}{2}x^2} + 3x \color{red}{- \frac{1}{4}} \\\\ =\ &\color{blue}{\frac{15}{2}x^2}+3x \color{red}{-\frac{49}{4}} \end{align}||

Simplify the algebraic expression.

||\begin{align} \text{Profit} &=\color{blue}{\frac{15}{2} (325)^2}+3 (325) \color{red}{-\frac{49}{4}} \\\\ &= \color{blue}{792 \ 187{.}5} + 975 \color{red}{-12{.}25} \\\\ &= $793 \ 150{.}25\ \end{align}||

Substitute |x| with |325| and calculate the final total.

​ The company can expect revenues of |$793 \ 150{.}25.|

Identify the unknowns and variables.

|36 =\color{orange}{ x} +\color{fuchsia}{(3x+12)} + \color{blue}{\displaystyle \frac{(3x+12)+x}{2}}|

Create the equation.

Simplify the equation.

Solve the equation by isolating the variable.

Validate by substituting the value of |x| in the initial equation.

Thus, the cost for the plates |= \color{orange}{x = $3},|

the cost for food |=\color{fuchsia}{3x + 12 = 3 \times3 + 12 = $21}|

and the cost for the gift |=\color{blue}{\displaystyle \frac{(3x+12)+x}{2} = \frac{(3 \times 3 + 12)+3}{2} = $12}.|

||\begin{align} \color{blue}{820 \ \text{cm}} &= (820 \div 100) \ \text{m} \\ &= \color{blue}{8{.}2 \ \text{m}} \\\\ \color{red}{1{.}2 \ \text{dam}} &= (1{.}2 \times 10) \ \text{m} \\ &= \color{red}{12 \ \text{m}} \end{align}||

||\begin{align} A_\text{triangle} &= \displaystyle \frac{\color{blue}{b} \times \color{red}{h}}{2} \\ &= \displaystyle \frac{\color{blue}{8{.}2} \times \color{red}{12}}{2} \\ &= 49{.}2 \ \text{m}^2 \end{align}||

The area for the triangle is |49{.}2 \ \text{m}^2.|

|\begin{align} P_\text{octagon} &= \color{red}{s} \times n \\ &= \color{red}{8} \times 8 \\ &= 64 \ \text{cm} \end{align}|

Calculate the perimeter of the regular octagon.

|\begin{align} P_\text{decagon} &= \color{blue}{s} \times n \\ &= \color{blue}{7} \times 10 \\ &= 70 \ \text{cm} \end{align}|

Calculate the perimeter of the regular decagon.

The statement is false because |70 > 64\ \Rightarrow\ P_\text{decagon} > P_\text{octagon}|

||\begin{align} \dfrac{\color{fuchsia}{\text{m arc of the circle}}}{\text{m central angle}} &= \dfrac{\text{Circumference}}{360^\circ} \\ \Rightarrow\ \dfrac{\color{fuchsia}{\text{m arc of the circle}}}{\color{red}{325^\circ}} &= \displaystyle \frac{2 \pi (\color{blue}{40}\div 2)}{360^\circ} \\\\ \Rightarrow\ \ \color{fuchsia}{\text{m arc of the circle}} &= \displaystyle \frac{\color{red}{325^\circ} \times 2 \pi (20)}{360^\circ} \\ &\approx \color{fuchsia}{113{.}45 \ \text{cm}} \end{align}||

Determine the size of the arc of the circle.

||\begin{align} \color{green}{\text{Initial length}} &= 8 \times \text{Circumference} \\ &= 8 \times 2 \pi (\color{blue}{40} \div 2) \\ &\approx \color{green}{1 \ 005{.}31 \ \text{cm}} \end{align}||

Determine the initial length of the lasso.

||\begin{align} \text{Final length} &= \color{green}{\text{Initial length}} - \color{fuchsia}{\text{Loop}} \\ &=\color{green}{1 \ 005{.}31} - \color{fuchsia}{113{.}45} \\ &= 891{.}86 \ \text{cm} \end{align}||

Determine the final length of the lasso.

By subtracting the size of the loop, the lasso’s length is |891{.}86 \ \text{cm}.|

Separate the figure into several shapes with known area formulas.

|\begin{align} A_\text{trapezoid} &= \dfrac{ (\color{fuchsia}{B}+\color{blue}{b}) \times h}{2} \\ &= \dfrac{ (\color{fuchsia}{12} + \color{blue}{8}) \times (\color{red}{15}-\color{green}{13})}{2} \\ &= \displaystyle \frac{ 20 \times 2}{2} \\ &= 20 \ \text{m}^2 \end{align}|

|\begin{align} A_\text{rectangle} &= \color{fuchsia}{b} \times \color{green}{h} \\ &= \color{fuchsia}{12} \times \color{green}{13} \\ &= 156 \ \text{m}^2 \end{align}|

|\begin{align} A_\text{half-circle} &= \dfrac{\pi r^2}{2}​ \\ &= \dfrac{\pi (\color{green}{13} \div 2)^2}{2}​ \\ &\approx 66{.}37 \ \text{m}^2 \end{align}|

​Calculate the area of ​​each of the sections.

Calculate the final bid.

The bid amount should be |$848{.}16.|

|\begin{align} \color{red}{A_\text{rectangles}} &= 4 \times (4,1 \times 2{.}4) \\ &= \color{red}{39{.}36 \ \text{dm}^2} \end{align}|

|\begin{align} A_\text{underneath} &= 4{.}1^2 \\ &= 16{.}81 \ \text{dm}^2 \end{align}|

|\begin{align} \color{orange}{A_\text{top}} &= A_\text{square} - A_\text{circle} \\ &= 4{.}1^2 - \pi (4{.}1 \div 2)^2 \\ &\approx \color{orange}{3{.}61 \ \text{dm}^2} \end{align}|

|\begin{align} \color{green}{A_\text{lateral of the cylinder}} &= 2 \pi (4{.}1\div 2) \times (3{.}5-2{.}4) \\ &\approx \color{green}{14{.}16 \ \text{dm}^2} \end{align}| |\begin{align} \color{blue}{A_\text{circle}} &= \pi (4{.}1\div 2)^2 \\ &\approx \color{blue}{13{.}20 \ \text{dm}^2} \end{align}|

Calculate the area of ​​each of the surfaces.

|\begin{align} A_\text{total} &= \color{red}{39{.}36} + 16{.}81 + \color{orange}{3{.}61} + \color{green}{14{.}16} + \color{blue}{13{.}20} \\ &= 87{.}14 \ \text{dm}^2 \end{align}|

​Add the areas of ​​each section.

You must use at least |87{.}14 \ \text{dm}^2| of duct tape.

Determine the formula to use.

Replace the known variables.

Isolate the desired variable |(\color{blue}{h}).|

The total height of the chest | \approx \color{blue}{h} + \color{fuchsia}{0{.}45} \approx \color{blue}{0{.}75}+\color{fuchsia}{0{.}45} \approx 1{.}2 \ \text{m}.|

|m\angle GJC = m\angle QJP=43^\circ|

​​

They are vertically opposite angles.

|m\angle BHG = m\angle MHG=90^\circ|

These are adjacent supplementary angles:

|\begin{align} m\angle BHG &= 180^\circ - m\angle GHM \\ &= 180^\circ - 90^\circ \\ &= 90^\circ \end{align}|

In any triangle, the sum of the interior angles is |180^\circ.| Thus, in the triangle |JCG,| we have:

|\begin{align} 180^\circ &= m\angle JGC + m\angle GJC + m\angle JCG \\ 180^\circ &= 90^\circ + 43^\circ + m\angle JCG \\47^\circ &= m\angle JCG \end{align}|

It is supplementary to |\angle JCG:|

|\begin{align} 180^\circ &= m\angle JCG + m\angle GCB \\ 180^\circ &= 47^\circ + m\angle GCB \\133^\circ &= m\angle GCB \end{align}|

Two parallel lines |(IK // ML)| intersected by a transversal |(QB)| form alternate-interior congruent angles.

Draw lines parallel to the translation arrow for each of the vertices.

​​​Measure the length of the translation arrow.

Use this same measurement to determine the location of each of the vertices in the image.

Connect the vertices in the image (same letters but accompanied by the symbol ') in the same order as the vertices of the initial image.

From the centre of rotation, draw a circle for each of the vertices using a compass.

​For each of the circles, measure the opening corresponding to the circular arc of each of the vertices.

​​Transpose this measurement from the vertex.

​​Repeat the previous two steps for each of the vertices.

Connect the vertices of the image in the same order as in the initial figure.

Draw lines perpendicular to the line of reflection from each of the vertices.

​Measure the distance between the line of reflection and one vertex.

Transfer the measurement to the same perpendicular line, but on the other side of the line of reflection.

​​Repeat the previous two steps for each of the vertices.

Connect vertices’ images in the same order as in the initial figure.

​​​Draw a straight line connecting the centre of dilation |O| to each of the vertices.

Measure the distance between the centre of dilation |O| and each of the vertices.

Apply the scale factor to each of the measured distances.

Identify the image’s vertex using the distance calculated from the centre of dilation |O| .

Connect each of the image vertices in the same order as in the initial figure.

Approach B) is the better strategy since the probability of winning is higher |\big(P(A) = 4\ \% > P(B) \approx 2{.}2\ \%\big).|

||\begin{align} \text{​​Mean} &= \dfrac{\color{green}{120}\color{red}{+124+\dots+148}\color{fuchsia}{+149}}{\color{blue}{20}} \\ &= 134{.}85 \ \text{cm} \end{align}||

Apply the formula.

||\begin{align} \text{Range} &= \color{fuchsia}{\text{max}} - \color{green}{\text{min}} \\ &= \color{fuchsia}{149}-\color{green}{120}​ \\&= 29\ \text{cm} \end{align}||

​Identify the maximum |(149)| and minimum |(120)| values to apply to the formula.

The average size is |134{.}85 \ \text{cm},| while the value of the range is |29\ \text{cm}.|

1) Build a distribution table.

2) Identify the title and axes of the graph.

3) ​Graduate the two axes appropriately.

4) Draw the horizontal or vertical bars (your choice).

1) ​​​Build the distribution table. ||\begin{align} \% = \displaystyle \frac{\text{Amount} \times 100}{3\ 500\ 000} \\ \text{Angle} = \displaystyle \frac{\text{Amount}\times 360}{3\ 500\ 000} \end{align}||