Memory Aid | Mathematics — Secondary 5 (TS)

|\color{white}{=}\dfrac{(27 a^3 b)^{\frac{1}{2}}}{27^{\frac{1}{3}}a^3}|

|= \dfrac{(3^3 a^3 b)^{\frac{1}{2}}}{(3^{3})^{\frac{1}{3}}a^3}|​​

Put the coefficients on the same base, if possible.

|\color{white}{=} \dfrac{\sqrt{3^3 a^3 b}}{3^1 a^3}|​

|= \dfrac{\sqrt{\color{blue}{3^2} \times 3^1 \times\color{red}{a^2}\times a^1 \times b}}{3^1 a^3}|

|= \dfrac{\color{blue}{3} \times \color{red}{a} \sqrt{3ab}}{3 a^3}|

|= ​\dfrac{\sqrt{3ab}}{a^2}|

​Use the laws and properties of exponents to simplify as much as possible.

​ The simplified expression is |\dfrac{\sqrt{3ab}}{a^2}.|

​|\sqrt{45} = \sqrt{\color{blue}{9} \times 5}|

Factor the radicand with a square number.

​|\sqrt {\color{blue}{9} \times 5}|
|= \sqrt{\color{blue}{9}} \times \sqrt {5}|

Use the law of the product of radicals |\rightarrow \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}.|

​|\sqrt {\color{blue}{9}} \times \sqrt {5}|
| = \color{blue}{3} \times \sqrt {5}|

Calculate the square roots.

​So, |\sqrt{45} = 3 \sqrt{5}.|

||\begin{align}=\,&(\color{#3b87cd}{\log_4 3x^2 + \log_4 4y} - \log_4 6x)^4\\=\,&(\color{#3b87cd}{\log_4 (3x^2 \times 4y)} - \log_4 6x)^4\\ =\,&(\color{#ec0000}{\log_4 12x^2y - \log_4 6x})^4\\ =\,&\left(\color{#ec0000}{\log_4 \left(\dfrac{12x^2y}{6x}\right)}\right)^4\\ =\,&4 \log_4(2xy)\end{align}||

Use the laws of logarithms with those who have the same base.

​ So, the simplified expression is |4 \log_4 (2xy).|

||​\begin{align}f(x) &= a (x-h)^2+k\\f(x)&= a (x-6)^2 + 1.5\\0.69 &= a (3-6)^2 + 1.5\\ 0.69 &= a \times 9 + 1.5\\ -0.09 &= a\\[4pt] \Rightarrow f(x) &= -0.09 (x-6)^2 + 1.5\end{align}||

Find the equation for the parabola in the appropriate form: ||f(x) = a(x-h)^2+k||

||​\begin{align}f(x) &= -0.09 (x^2 - 12x + 36) + 1.5\\f(x) &= -0.09x^2 + 1.08x - 1.74\end{align}||

Transform the rule into its general form: ||f(x) = Ax^2 + Bx + C||

||0 = -0.09x^2 + 1.08x - 1.74||
||​\begin{align}\{x_1, x_2\} &= \dfrac{-(1.08) \pm \sqrt{(1.08)^2 - 4(-0.09)(-1.74)}}{2(-0.09)}\\&= \{1.92, 10.08\}\end{align}||

Find the zeros of the function by replacing |f(x) = 0| and using the quadratic formula.

So, the length of the jump is |10.08 - 1.92 = 8.16| meters.

|h=3|

Find the value of |h| along the vertical asymptote.

||​\begin{align}\color{blue}{\text{function's zero}} &= \dfrac{1}{b} + h\\[3pt] \color{blue}{\dfrac{13}{4}}&= \dfrac{1}{b} + 3\\[3pt] \dfrac{1}{4} &= ​\dfrac{1}{b}\\[3pt] b &= 4\end{align}||

Use the zero of a function to find the value of the parameter |b|.
Zero of a function | =\dfrac{1}{b} + h.|

||​\begin{align}\color{red}{1.79} &= \log_c\big(4(\color{red}{6}-3)\big)\\ \color{red}{1.79} &= \log_c(12)\\ c^{1.79} &= 12\\
c&\approx 4\end{align}||Therefore, |f(x) = \log_4\big(4(x-3)\big).|

Find the value of the parameter |c| using the coordinates of another point: |\color{red}{(6, 1.79)}.|

​||\begin{align}3& = \log_4\big(4(x-3)\big)\\4^3 &= 4(x-3)\\19 &= x\end{align}||

Replace |f(x)| by |3.|

When |y= 3, x = 19.|

Minimum investment | = \$3\ 000|

|\Rightarrow k=3\ 000|

Find the value of parameter |k.|

​|\begin{align}c &= 1 \pm 5 \%\\
c& = 1 + 0.05\\
c& = 1.05\end{align}|

|\Rightarrow f(x) = a(1.05)^{bx}+ 3\ 000|

Find the value of parameter |c.|

capitalized every 2 years |\Rightarrow b = \dfrac{1}{2}|

Therefore, |f(x) = a(1.05)^{\frac{1}{2}x}+ 3\ 000.|

Find the value of parameter |b| according to the context.

​|\begin{align}5\ 000 &= a(1.05)^{\frac{1}{2}(0)}+3\ 000\\
5\ 000 &= a(1) + 3\ 000\\
2\ 000 &= a\end{align}|

​Replace |(x,y)| by the given initial value |(0, 5\ 000).|

|\begin{align}8\ 000 &= 2000(1.05)^{\frac{1}{2}x}+3\ 000\\[3pt]
2.5 &= (1.05)^{\frac{1}{2}x}\\[3pt]
\log_{1.05}(2.5)& = \dfrac{1}{2}x\\[3pt]
18.78 &\approx \dfrac{1}{2}x\\[3pt]
37.56 &\approx x\end{align}|

Replace |f(x)| by |\$8\ 000.|

The capital invested will be worth at least |\$8\ 000| after |37.56| years.

​|f(x) = \pm a \sqrt{\pm 1(x-h)} + k|

Determine the model to be used.

|f(x) = a \sqrt{1(x-h)} + k|
|f(x) = a \sqrt{x-h} + k|

Determine the sign of |a| and |b| depending on the orientation of the graph (both are positive).

​|\begin{align} \color{green}{8} &= a \sqrt{\color{green}{12} - h} + 3\\
25& = a^2 (\color{green}{12}- h)\\
\dfrac{25}{\color{green}{12}-h}&=a^2\\\\
\color{red}{10{.}5}& = a \sqrt{\color{red}{17} - h} + 3\\
56{.}25 &= a^2 (\color{red}{17}- h)\\
\dfrac{56{.}25}{\color{red}{17}-h}& = a^2\end{align}​|

Create two equations using the points provided.

​|\begin{align}\dfrac{56{.}25}{17 - h}& = \dfrac{25}{12-h}\\
56{.}25(12-h)& = 25(17-h)\\ 675 - 56{.}25h& = 425 - 25h\\
250 &= 31{.}25h\\
8 &= h\end{align}|​

Compare the two values for |a^2.|

​|\begin{align}f(x) &= a \sqrt{x-8} + 3\\
\color{green}{8}&= a \sqrt{\color{green}{12} - 8} + 3\\
\color{green}{8}& = a(2) + 3\\
2{.}5 &= a\end{align}|

Use one of the points to find the value of |a.|

​​|\begin{align} f(x)& = 2{.}5 \sqrt{x-8} + 3\\
50& = 2{.}5 \sqrt{x-8}+3\\
18{,}8 &= \sqrt{x-8}\\
353{.}44& = x-8\\
361{.}44& = x\end{align}|

The bird will be found at a horizontal distance of |361{.}44| metres.

​|x =| number of jackets
|y =| number of shirts

Identify the variables. The combination of |x| and |y| is generally done randomly.

​|Z = 32x + 17y|

Find the function to be optimised.

​|\color{blue}{y \le 21 - 4x}|
|\color{green}{x \ge 8 - 3y}|
|\color{red}{3x - 2x \ge 2}|
|x \ge 0| et |y \ge 0|

Identify the inequalities without forgetting the non-negativity constraints.

​|\color{blue}{y \le 21 - 4x}|
|\color{green}{y \ge -\dfrac{1}{3}x + \dfrac{8}{3}}|
|\color{red}{y \le \dfrac{3}{2}x - 1}|

Isolate |y| in each of the inequalities in order to write them in functional form.

Trace the boundary lines of each of the inequalities in a Cartesian plane.

Find the constraint polygon that satisfies all the inequalities.

Find the coordinates of each vertex using the Comparison Method, Substitution Method or Simplification Method.

According to  |A (4,5),| 
|\Rightarrow Z = 32 \times 4 + 17 \times 5 = 213.|
According to |B (5,1),|
|\Rightarrow Z = 32 \times 5 + 17 \times 1 = 177.|
According to |C (2,2),|
|\Rightarrow Z = 32 \times 2 + 17 \times 2 = 98.|

Calculate the profit for each of the points using the function to be optimised.

To maximize profits, the manager would have to sell |4| jackets and |5| shirts for a maximum profit of |$213\.|

​|\color{red}{A_\text{Ancien}} = \color{blue}{A_\text{Nouveau}}|

The two figures are equivalent.

​|\begin{align} \color{red}{A_\text{Ancien}} &= \color{blue}{A_\text{Nouveau}} \\ \color{red}{b\times h} &= \color{blue}{b\times h} \\ \color{red}{8 \times 12} &= \color{blue}{b \times 10} \\ \color{red}{96} &= \color{blue}{10b} \\ 9{.}6\ \text{m} &= \color{blue}{b} \end{align}|

Create an equation using the area formulae and solve it.

The width of the new parking area must be |9{.}6\ \text{m}.|

​|\color{blue}{V_\text{Prisme}} = \color{red}{V_\text{Demi-boule}}|

The two solids are equivalent.

​|\begin{align} \color{blue}{A_b \times h} &= \color{red}{\dfrac{4 \pi r^3}{3} \div 2} \\ \color{blue}{\dfrac{1{.}8 \times 1{.}7}{2} \times 2{.}1} &= \color{red}{\dfrac{4 \pi r^3}{6}} \\ \color{blue}{3{.}21} &\approx \color{red}{\frac{4 \pi r^3}{6}} \\ 1{.}53 &\approx \color{red}{r^3} \\ 1{.}15\ \text{m} &\approx \color{red}{r} \end{align}|

Create an equation with the respective volume formulae and solve.

The radius of the half-ball tent should be approximately |1{.}15\ \text{m}.|

​|\begin{align}\dfrac{\text{Angle measure in degrees}}{180^\circ} &= \frac{\text{Angle measure in radians}}{\pi\ \text{rad}}\\ \dfrac{\color{red}{227^\circ}}{180^\circ} &= \dfrac{\text{Angle measure in radians}}{\pi \ \text{rad}}\end{align}|​​

Use the proportion identified higher.

​|\begin{align}\color{red}{227^\circ} \times \pi \div 180^\circ &= \text{Angle measure in radians}\\
\color{white}{\Rightarrow 227^\circ \cdot \pi}3{.}96 \ \text{rad} &= \text{Angle measure in radians}\end{align}|

Solve using the cross product.

The angle at the centre measures |3{.}96 \ \text{rad}.|

Identify the vertices and edges of the triangle.

If possible, deduce other measures of the triangle (sum of the interior angles of a triangle and properties of an isosceles triangle).

|\begin{align} \dfrac{\color{green}{a}}{\sin 40^\circ} &= \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}| et |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 to isolate the sine of the angle you are looking for.

|\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 doing |\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 for the obtuse angle.

In this situation, the angle measure is |131{.}9^\circ.|

Identify the vertices and edges of the triangle.

​|\begin{align} \color{red}{a}^2 &= \color{blue}{b}^2 + \color{green}{c}^2 - 2\color{blue}{b} \color{green}{c} \cos \color{red}{A} \\ \color{red}{a}^2 &= \color{blue}{92}^2 + \color{green}{125}^2 - 2 \color{blue}{(92)} \color{green}{(125)} \cos \color{red}{81^\circ} \end{align}|

Apply the appropriate formula to ensure that there is only one unknown measurement.

|\begin{align} \color{red}{a}^2 &= \color{blue}{92}^2 + \color{green}{125}^2 - 2 \color{blue}{(92)} \color{green}{(125)} \cos \color{red}{81^\circ}​ \\ \color{red}{a}^2 &\approx 8\ 464 +  15\ 625 - 3\ 598 \\ \color{red}{a}^2 &\approx 20\ 491 \\ \color{red}{a}\ &\approx 143{,}15 \end{align}|

Solve the equation by isolating the variable.

Moose can walk over a |\color{red}{\text{distance}}| of approximately |143{,}15\ \text{m}.|

Identify the vertices and edges of the triangle.

​|\begin{align} \color{blue}{a^2} &= \color{red}{b^2} + \color{green}{c^2} - 2 \color{red}{b} \color{green}{c} \cos \color{blue}{A} \\ \color{blue}{22^2} &= \color{red}{24^2} + \color{green}{21^2} - 2 \color{red}{(24)} \color{green}{(21)} \cos \color{blue}{A} \end{align}|

Substitute the values in the formula. Here, we use |a^2 = b^2 + c^2 - 2bc \cos A| since it's the measure of the angle |A| that we're looking for.

|\begin{align} \color{blue}{22^2} &= \color{red}{24^2} + \color{green}{21^2} - 2 \color{red}{(24)} \color{green}{(21)} \cos \color{blue}{A} \\ \color{blue}{484} &=576+441 - 1 \ 008 \cos \color{blue}{A}\\\\ \Rightarrow\ \dfrac{484 - 576-441}{- 1 \ 008} &= \cos \color{blue}{A} \\ 0{,}529 &\approx \cos \color{blue}{A} \\ 58^\circ &\approx \color{blue}{m\angle A} \end{align}|

Isolate the variable.

To ensure that there are no blind spots, the camera should describe rotations at an angle of approximately |58^\circ.|

​|\text{m} \ \overline{HC} = 2{,}5 = \text{m} \ \overline{CG}|

Since |\overline{BF}| is a diameter, it cuts |\overline{GH}| into two equal parts.

||\begin{align}\text{m} \ \overline{DH} \times \text{m} \ \overline{DG} &=\text{m}\ \overline{DI} \times \text{m} \ \overline{DF}\\
1{.}62 \times \big(2{.}5+(2{.}5-1{.}62)\big)& = 1{.}76 \times x\\
5{.}48& \approx 1{.}76x\\
3{.}11 &\approx x\end{align}||

||\color{fuchsia}{\overline{FI}}=1{.}76+3{.}11=4{.}87||

Use the 3^rd formula above.

The measure of |\color{fuchsia}{\overline{FI}}| is |4{.}87.|

|\begin{align}​\color{green}{\text{m} \ \angle BFD} &= \dfrac{\color{fuchsia}{\text{m} \ \angle BED} - \color{red}{\text{m} \ \angle AEC}}{2}\\
\color{green}{42^\circ} &=\dfrac{\color{fuchsia}{m \ \angle BED} - \color{red}{58^\circ}}{2}\\
\color{fuchsia}{142^\circ} &= \color{fuchsia}{m \ \angle BED}\end{align}|

Find |\color{fuchsia}{\text{m} \ \angle BED}| according to the 3^rd formula.

||\begin{align}​\color{blue}{\text{m} \ \angle BGD} &=\dfrac{\color{fuchsia}{\text{m} \ \angle BED}}{2}\\
\color{blue}{\text{m} \ \angle BGD} &=  \dfrac{\color{fuchsia}{142^\circ}}{2}\\
\color{blue}{\text{m} \ \angle BGD} &=\color{blue}{ 71^\circ}\end{align}||

Find |\color{blue}{\text{m} \ \angle BGD}| according to the 1^st formula.

So,  ​|\color{blue}{\text{m} \ \angle BGD = 71^\circ}.|

​|\sec \theta - \cos \theta =​ \dfrac{1}{\cos \theta} - \cos \theta|

Transform terms into |\sec \theta| and |\cos \theta.|

|\begin{align} \phantom{\sec \theta - \cos \theta​}​ &= \dfrac{1}{\cos \theta} - \frac{\cos ^2 \theta}{\cos \theta}\\ &= \dfrac{1 - \cos ^2 \theta}{\cos \theta} \end{align}|

Find a common denominator for the subtraction.

|\phantom{\sec \theta - \cos \theta​}​ =​ \dfrac{\sin ^2 \theta}{\cos \theta}|

Use |\sin ^2 \theta + \cos ^2 \theta = 1 \Rightarrow \sin ^2 \theta = 1 - \cos ^2 \theta.|

|\phantom{\sec \theta - \cos \theta​}​ = ​\dfrac{\color{blue}{\sin \theta} \ \sin \theta}{\color{blue}{\cos \theta}}​|

Decompose |\sin ^2 \theta = \sin \theta \ \sin \theta.|

|\phantom{\sec \theta - \cos \theta​}​ = \color{blue}{\tan \theta} \ \sin \theta|

Use |\dfrac{\sin \theta}{\cos \theta} = \tan \theta.|

The identity is verified, as the result is what was initially indicated.

Use the components of the vector, starting from the origin in the Cartesian plane.

Trace a right triangle to deduce its norm. ||\begin{align} \mid\mid \color{red}{\overrightarrow u} \mid \mid &= \sqrt{\color{blue}{3}^2 + \color{green}{8}^2} \\ &\approx 8{.}54 \end{align}||

Use trigonometric ratios to find the angle measure associated with the orientation of |\color{red}{\overrightarrow u}.|

||\begin{align} \text{Orientation of}\ \color{red}{\overrightarrow u} &= \color{orange}{m\angle ABC} \\ &= 180^\circ - \color{fuchsia}{m\angle DBC} \\ &= 180^\circ - \tan^{-1} \left(\frac{\color{green}{8}}{\color{blue}{3}}\right) \\ &\approx 180^\circ - 69 ^\circ \\ &\approx 111^\circ \end{align}||

So, |\mid \mid  \color{red}{\overrightarrow u} \mid \mid\ \approx 8{.}54| and its orientation is approximately |111^\circ.|

​|\begin{align} \color{blue}{\overrightarrow w} &= k_1 \color{red}{\overrightarrow u} + k_2 \color{green}{\overrightarrow v} \\ \color{blue}{(4,-12)} &= k_1 \color{red}{(-1,4)} + k_2 \color{green}{(2,5)}\end{align}|

|\Rightarrow\begin{cases}\color{blue}{\ \ \ \ \ 4} = k_1 \color{red}{(-1)} + k_2 (\color{green}{2})\\\color{blue}{-12} = k_1 (\color{red}{4}) + k_2 (\color{green}{5})\end{cases}|

Create two equations using the definition of linear combination, one for the |x| component and one for the |y.| component.

​|\begin{align} \color{blue}{-16} &= \color{red}{4}k_1\color{green}{-8}k_2 \\
\color{blue}{-12} &=\color{red}{4}k_1 +\color{green}{5}k_2 \\\hline -4 &=\ \ \ \ -13k_2 \\
\color{fuchsia}{\dfrac{4}{13}} &= \color{fuchsia}{k_2} \end{align}|

|\begin{align} \color{blue}{-12} &=\color{red}{4}k_1 +\color{green}{5}\left(\color{fuchsia}{\dfrac{4}{13}}\right) \\ \color{orange}{-\dfrac{44}{13}} &= \color{orange}{k_1} \end{align}|

Solve the system of equations using the simplification method by multiplying the first equation by |-4.|.

​So, |\color{blue}{\overrightarrow w} = \color{orange}{-\dfrac{44}{13}}\color{red}{\overrightarrow u} + \color{fuchsia}{\dfrac{4}{13}}\color{green}{\overrightarrow v}.|

​|\begin{align}W &= \color{red}{150}\cos \color{blue}{21^\circ} \times \color{green}{12}\\&\approx 1\ 680 \ J\end{align}|

Use the formula to calculate the work. ||W = \color{red}{F}\cos \color{blue}{\theta} \times \color{green}{\Delta x}||

Julien will have to do a work of |1\ 680 \ J.|

​​|A(3,2)|
|B(-1,5)|
|C(4,-1)|

|r_{(0,270^\circ)} : (\color{blue}{x},\color{red}{y}) \mapsto (\color{red}{y},\color{blue}{-x})|

​|A(\color{blue}{3},\color{red}{2}) \mapsto A' (\color{red}{2}, \color{blue}{-3})|

​|B(\color{blue}{-1},\color{red}{5}) \mapsto B' (\color{red}{5}, \color{blue}{1})|

​|C(\color{blue}{4},\color{red}{-1}) \mapsto C' (\color{red}{-1}, \color{blue}{-4})|

The coordinates of the figure which is the image of this rotation are|A'(2,-3),| |B'(5,1)| and |C'(-1,-4).|

​​|A(-3,-5)|
|B(-2,1)|
|C(4,3)|
|D(2,-2)|

Determine the coordinates of each vertex.

|s_y : (\color{blue}{x}, \color{red}{y}) \mapsto (\color{blue}{-x}, \color{red}{y})|

Identify the relation to be used.

|A(\color{blue}{-3},\color{red}{-5}) \mapsto A' (\color{blue}{3}, \color{red}{-5})|
​|B(\color{blue}{-2},\color{red}{1}) \mapsto B' (\color{blue}{2}, \color{red}{-1})|
​|C(\color{blue}{4},\color{red}{3}) \mapsto C' (\color{blue}{-4}, \color{red}{3})|
|D(\color{blue}{2}, \color{red}{-2}) \mapsto D'(\color{blue}{-2}, \color{red}{-2})|

Apply the identified relation to each of the vertex coordinates.

​​|A(2,3)|
|B(4,7)|
|C(8,-2)|
|D(-3,12)|

Determine the coordinates of each of the initial vertices.

|t_{(5,-4)} : (x,y) \mapsto (x + 5, y - 4)|

Identify the relation to be used.

​|A(\color{blue}{2},\color{red}{3}) \mapsto A' (\color{blue}{7}, \color{red}{-1})|
​|B(\color{blue}{4},\color{red}{7}) \mapsto B' (\color{blue}{9}, \color{red}{3})|
​|C(\color{blue}{8},\color{red}{-2}) \mapsto C' (\color{blue}{13}, \color{red}{-6})|
|D(\color{blue}{-3},\color{red}{12}) \mapsto D'(\color{blue}{2}, \color{red}{8})|

Apply the identified relation to each of the vertex coordinates.

In the statement, |C‘=(12,-6)| and |D’=(2,-8)| whereas the calculations show that |C‘=(\color{blue}{13}, \color{red}{-6})| and |D’=(\color{blue}{2}, \color{red}{8}).| Since there is an error in the coordinates given in the statement, the image and initial figures will not be isometric.

|A(1,2)|
|B(2,3)|
|C(4,0)|
|D(3,-2)|
|E(-1,-2)|

Determine the coordinates of each vertex.

​|h_{(0,12)} : (\color{blue}{x},\color{red}{y}) \mapsto (\color{blue}{12x},\color{red}{12y})|

Identify the relation to be used.

​|A(\color{blue}{1},\color{red}{2}) \mapsto A' (\color{blue}{12}, \color{red}{24})|
​|B(\color{blue}{2},\color{red}{3}) \mapsto B' (\color{blue}{24}, \color{red}{36})|
​|C(\color{blue}{4},\color{red}{0}) \mapsto C' (\color{blue}{48}, \color{red}{0})|
|D(\color{blue}{3}, \color{red}{-2}) \mapsto D'(\color{blue}{36}, \color{red}{-24})|
|E(\color{blue}{-1}, \color{red}{-2}) \mapsto E'(\color{blue}{-12}, \color{red}{-24})|

Apply the identified relation to each of the vertex coordinates.

The new coordinates would be |A'(12,24),| |B'(24,36),| |C'(48,0),| |D'(36,-24)| and |E'(-12,-24).|

|\begin{align}(\color{green}{6{.}35} - h)^2 + (\color{green}{10{.}92} -\color{blue}{4})^2 &= r^2\\\\
(\color{red}{8{.}35} -h)^2 + (\color{red}{-1{.}98} - \color{blue}{4})^2 &= r^2\end{align}|

Create two equations using the circle equation.

|\begin{align}(\color{green}{6{.}35} - h)^2 + (\color{green}{10{.}92} -\color{blue}{4})^2& = (\color{red}{8{.}35} - h)^2 + (\color{red}{-1{.}98} - \color{blue}{4})^2\\
h^2 - 12{.}7h + 88{.}21&=h^2 -16{.}7h +105{.}46\\
17{.}27& = 4h\\
4{.}32&=h\end{align}|

Compare the equations and solve the one obtained.

|\begin{align}(\color{green}{x} -h)^2 + (\color{green}{y} -\color{blue}{k})^2 &= r^2\\
(\color{green}{6{.}35} -4{.}32)^2 + (\color{green}{10{.}92} -\color{blue}{4})^2 &= r^2\\
52{.}01& \approx r^2\\
7{.}21& \approx r
\end{align}|

Use a point to find the radius value.

Since the radius measures |7{.}21\ \text{km},| then the coverage of Gitane's sonar is |pi \times 7{.}21^2 \approx 163{.}31 \text{km}^2.|

​|\begin{align}\text{m}\overline{\color{fuchsia}{F_1} \color{blue}{P}} &= \sqrt{(\color{fuchsia}{1{.}5} - \color{blue}{2{,}01})^2 + (\color{fuchsia}{1{.}2} - \color{blue}{2})^2}\\ &= \sqrt {0{.}9}\\&\approx 0{.}95\\​
\text{m}\overline{\color{red}{F_2} \color{blue}{P}} &= \sqrt{(\color{red}{1{.}5} - \color{blue}{2{.}01})^2 + (\color{red}{4{.}8} - \color{blue}{2})^2}\\
&= \sqrt {8{.}1}\\& \approx 2{.}85\end{align}|

Find |\text{m}\overline{\color{fuchsia}{F_1} \color{blue}{P}}| and |\text{m}\overline{\color{red}{F_2} \color{blue}{P}}.|

|2a:| sum of the distance between the foci and one point.
​|\begin{align}2a& = 0.95 + 2.85\\2a&= 3.8\\a &= 1.9\\\end{align}|

Use the definition of the ellipse to find the measure of the longest axis.

​|\begin{align}1 &=\dfrac{(\color{blue}{x}-\color{green}{h})^2}{a^2} + \frac{(\color{blue}{y}-\color{green}{k})^2}{b^2}\\
1& =\dfrac{(\color{blue}{2}-\color{green}{3})^2}{1{.}9^2} + \dfrac{(\color{blue}{2{.}01}-\color{green}{1{.}5})^2}{b^2}\\
1& =\dfrac{1}{3{.}61} + \dfrac{0{.}26}{b^2}\\
0{.}723&=\dfrac{0{.}26}{b^2}\\
b^2 &= \dfrac{0{.}26}{0{.}723}\\
b &= 0{.}6\end{align}|

Replace |\color{blue}{(x,y)}| with a point on the ellipse.

The canoe has the following dimensions.
Maximum length: |2a = 2 \times 1{.}9 = 3{.}8\ \text{m}|
Maximum width: |2b = 2 \times 0{.}6 = 1{.}2\ \text{m}|

​|\begin{align} d(\color{red}{F}, \color{blue}{S}) &= \color{blue}{y_2} - \color{red}{y_1} \\ &= \color{blue}{2{.}8} - \color{red}{1{.}3} \\ &= \color{green}{1{.}5}\end{align}|

​Calculate the distance between |\color{red}{F}| and |\color{blue}{S}.|

​|(x-\color{blue}{h})^2 = -4 \color{green}{c} (y-\color{blue}{k})|

Determine the appropriate model of the equation for the parabola.

|\begin{align} \color{green}{c} &= \color{blue}{2{.}8} - \color{red}{1{,}3} \\ &= \color{green}{1{.}5} \end{align}|

Calculate the parameter value |\color{green}{c}.|

​|(x  \color{blue}{+ 3})^2 = -4 \color{green}{(1{.}5)}(y- \color{blue}{2{.}8})|

Replace the parameters with their respective values.

Finally, Gitane can model this situation by the equation ​|(x  \color{blue}{+ 3})^2 = -6(y- \color{blue}{2{.}8}).|

|\begin{align}\text{m}\overline{\color{red}{F_1} P} &= \sqrt{(\color{red}{-8{.}21} - 10{.}63)^2 + (\color{red}{5} - 0)^2}\\​
& = 19{.}49\\​
​​\text{m}\overline{\color{fuchsia}{F_2} P}& = \sqrt{(\color{fuchsia}{16{.}21} - 10{.}63)^2 + (\color{fuchsia}{5} - 0)^2}\\
&= 7{.}49\end{align}|​​

​Calculate |d(\color{red}{F_1},P)| and |d(\color{fuchsia}{F_2}, P).|

​|\begin{align}​2a &=\left \vert \text{m}\overline{\color{red}{F_1}P} - \text{m} \overline{\color{fuchsia}{F_2} P}\right \vert\\
&=\left \vert 19{.}49 - 7{.}49\right \vert\\
&= 12\\
a& = 6\\\\
\Rightarrow a^2 &= 36\end{align}|

Use the definition to find the measure of |a.|

​|\begin{align}​\color{blue}{k} &= 5\\\\
\color{blue}{h}& = \dfrac{\color{red}{-8{.}21} +\color{fuchsia}{16{.}21}}{2}\\
& = 4\end{align}|

Determine the vertex coordinates |\color{blue}{(h,k)}| according to the properties of the hyperbola.

​|\begin{align}​c &= \color{blue}{4} - \color{red}{(-8{.}21)}\\
c&= 12{.}21
\end{align}|

​Deduct the value of |c.|

|​\begin{align}​c^2 &= a^2 + b^2\\
12{.}21^2 &= 6^2 + b^2\\
113 &\approx b^2
\end{align}|

Find the value for |b^2| using the relation |c^2 = a^2 + b^2.|

The equation that defines the hyperbola of the swells that will meet is |\dfrac{(x-\color{blue}{4})^2}{36} - \dfrac{(y-\color{blue}{5})^2}{113} = 1.|

​|\begin{align} \color{red}{1} &= \color{red}{\dfrac{x^2}{196} + \dfrac{y^2}{25} } \\ \color{blue}{y} &= \color{blue}{\dfrac{2}{5}x - 1} \end{align}​|

​​|\color{red}{1 =\dfrac{x^2}{196}} + \dfrac{\left(\color{blue}{\frac{2}{5}x - 1}\right)^2}{\color{red}{25}}|

Substitute the |\color{red}{y}| in the ellipse equation with the |\color{blue}{y}| in the linear function.

​|\begin{align} 1 &= \dfrac{x^2}{196}+\dfrac{0{.}16x^2-0{.}8x+1}{25} \\
4\ 900 &= 25x^2 +31{.}36x^2 - 156{.}8x + 196 \\
0 &= 56{.}36x^2 - 156{.}8x - 4\ 704 \\\\
\Rightarrow \{\color{fuchsia}{x_1}, \color{green}{x_2}\} &= \dfrac{-(-156{.}8)\pm \sqrt{(-156{.}8)^2 - 4 (56{.}36)(-4\ 704)}}{2(56{.}36)} \\
\color{fuchsia}{x_1}  &\color{fuchsia}{\approx}\color{fuchsia}{-7{.}85}\quad \text{et}\quad \color{green}{x_2 \approx 10{.}63} \end{align}|

Solve the equation to find the values for |\color{fuchsia}{x_1}| and |\color{green}{x_2}.|

​|\begin{align} \color{fuchsia}{y_1} &= \color{blue}{\dfrac{2}{5}} \color{fuchsia}{(-7{.}85)} \color{blue}{-1} \\&\approx \color{fuchsia}{-4{.}14} \\\\ \color{green}{y_2} &= \color{blue}{ \dfrac{2}{5}} \color{green}{(10{.}63)} \color{blue}{-1} \\ &\approx \color{green}{3{.}25}\end{align}|​

Calculate the values for |\color{fuchsia}{y_1}| and |\color{green}{y_2}.|

​|\begin{align} d(\color{fuchsia}{A}, \color{green}{B}) &= \sqrt{\big(\color{green}{3{.}25} - \color{fuchsia}{(-4{.}14)}\big)^2 + \big(\color{green}{10{.}63} - \color{fuchsia}{(-7{.}85)}\big)^2} \\ &\approx 19{.}90 \ \text{m} \end{align}|

Calculate the distance between |\color{fuchsia}{A (-7{.}85, -4{.}14)}| and |\color{green}{B(10{.}63, 3{.}25)}.|

|\begin{align} \dfrac{5\ \text{m}}{19{.}90\ \text{m}} &= \dfrac{1\ \text{sec}}{?\ \text{sec}} \\\\ \Rightarrow\ ? &= 1 \times 19{.}90 \div 5 \\ &\approx 3{.}98\ \text{sec} \end{align}|

Determine how long the sharks spend under the boat.

​​ ​​The sharks remained under the boat for around |3{.}98| seconds.

|\begin{align}\dfrac{-17\pi}{4} &+ 2\pi \\= \dfrac{-17\pi}{4} &+ \dfrac{8\pi}{4} = \dfrac{-9\pi}{4}\\\\
\dfrac{-9\pi}{4} &+ 2\pi \\= \dfrac{-9\pi}{4} &+ \dfrac{8\pi}{4} = \dfrac{-\pi}{4}\\\\\dfrac{-\pi}{4} &+ 2\pi \\= \dfrac{-\pi}{4} &+ \dfrac{8\pi}{4} = \dfrac{7\pi}{4}\end{align}|​​

Add or subtract one turn |(2\pi)| until you reach the interval |[0, 2\pi].|

| \begin{align}P\left(\dfrac{-17\pi}{4}\right) &= P\left(\dfrac{7\pi}{4}\right) \\&= \left(\dfrac{\sqrt{2}}{2}, \dfrac{-\sqrt{2}}{2}\right)\end{align}|

Match the correct coordinates to the angle found.

The coordinates of the point are |\left(\dfrac{\sqrt{2}}{2}, \dfrac{-\sqrt{2}}{2}\right).|