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# Principles of Math MPM1D

This course enables students to develop an understanding of mathematical concepts related to algebra, analytic geometry and measurement and geometry through investigation, the effective use of technology, and abstract reasoning. Students will investigate relationships, which they will then generalize as equations of lines, and will determine the connections between different representations of a linear relation. They will also explore relationships that emerge from the measurement of thee dimensional figures and two dimensional shapes. Students will reason mathematically and communicate their thinking as they solve multi step problems.

xa × xb = xa + b

xa ÷ xb = xa - b

(xa)b = xa × b

# Linear Relations

## Given the equations...

\begin{align} Slope & = m \\ & = \frac{\Delta y}{\Delta x} \\ & = \frac{y_2\ -\ y_1}{x_2\ -\ x_1} \\ \end{align}

## A cyclist records their distance travelled as they ride their bicycle.

 Time, x (hours) Distance, y (km) 1 10 2 20 3 30 4 40

y = mx + b

## Amanda is working hard to save up to buy herself a nice iPhone from Apple. She knows it will be expensive and that money doesn't grow on trees so she wants to compare the different plan options A - D.

 Plan Phone Cost of Phone Cost of Plan (per month) A Latest iPhone $230$50 B Latest iPhone $700$35 C Previous Generation iPhone $280$50 D Previous Generation iPhone $450$35

# Linear Analytic Geometry

## Answer the questions for the equation:

$3x + 6y + 2 = 0$

# Measurement and Geometry

## This large Roman window arch is made of a semicircle (half a circle) and a rectangle. The base (b) of the window is 2.0 m and the total height, h is 4.0 m

Acircle = $\pi$r2

## A square-based prism has a surface area equal to 96 cm2.

### Determine the maximum volume of this square-based prism, and the dimensions of the base and height. Solution Hint Clear Info Incorrect Attempts: CHECK cm3 Hint Unavailable \begin{align} height & = \dfrac{48 - b^2}{2b} \\ \\ & = \dfrac{48 - 1^2}{2(1)} = 23.5\ cm \\ \\ & = \dfrac{48 - 2^2}{2(2)} = 11\ cm \\ \\ & = \dfrac{48 - 3^2}{2(3)} = 6.5\ cm \\ \\ & = \dfrac{48 - 4^2}{2(4)} = 4\ cm \\ \\ & = \dfrac{48 - 5^2}{2(5)} = 2.3\ cm \\ \\ & = \dfrac{48 - 6^2}{2(6)} = 1\ cm \\ \\ \end{align} Base (cm)Height (cm)Volume (cm3) = b2 × h 123.5= 12 × (23.5) =    23.5 211= 22 × (11) =    44 36.5= 32 × (6.5) =    58.5 44= 42 × (4) =    64 52.3= 52 × (2.3) =    57.5 61= 62 × (1) =    36 Therefore the maximum volume (64 cm3) occurs when the base equals 4cm, and the height equals 4 cm.

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