1. 9 Academic
Review

2. Number Sense Order of Operations Unit Price Proportion Fractions
Percent
Ratios

3. B E D M A S Order of Operation: RACKETS XPONENTS IVISION LEFT TO RIGHT
ULTIPLICATION A
DDITION S
LEFT TO RIGHT
UBTRACTION

4. Example: Solve
6×3÷2−(4−9)
=6×3÷2−(−5)
=6×3÷2+5
=18÷2+5
=9+5
=14

5. Unit Price MONEY ÷ ITEM Jennifer Ahmad $0.08/g + $0.02/g = $0.10/g
Ahmad bought nuts from a local market and paid $40 for 500 g. Jennifer went to the grocery store to purchase her apples and paid $0.02/g more. If Jennifer bought 350 g of apples, how much did she pay?
Jennifer
Ahmad
$0.08/g + $0.02/g = $0.10/g
$40 500g =$0.08/g
$0.10/g ×350g = $350
∴ Jennifer paid $350 for her apples.

6. Fractions Change all mixed fractions to improper
Multiplication/Division → NO COMMON DENOMINATOR
To multiply invert (flip) the fraction after the division sign and multiply
i.e. ÷ 𝟐 𝟑 is the same as× 𝟑 𝟐
Follow BEDMAS
To convert to a decimal: NUMERATOR ÷ DENOMINATOR
If you’re stuck use the fraction button on your calculator.

7. Percent Calculating Percentage 21 25 ×100 =0.84×100 =84%
Robbie got a run 21 out of his last 25 times at bats, calculate his batting average.
21 25 ×100
=0.84×100
=84%
∴ Robbie’s batting average is 84%.

8. Percent Percent OF 24÷100=0.24 0.24×250=60
24% of Grade 9 students have decided to take drama next year. If there are 250 grade 9 students, how many have signed up for drama?
24÷100=0.24
0.24×250=60
∴ 60 grade 9 students signed up to take drama next year.

9. Percent Percent IS 30% off →she is paying 70%
Becky paid $60 before tax for a Champion Sweater. If the sweater was on sale for 30% off what was the original price?
30% off →she is paying 70%
$60 𝐢𝐬 70% of the original price
70÷100=0.70
$60÷0.70=$85.71
∴ the original price of the sweater was $85.71.

10. Ratios Jordan Kim Mike TOTAL 3 5 4 12 𝑥 𝑦 𝑧 $245,862
Jordan, Kim and Mike share a business and split their profits in the ratio of 3:5:4 respectively. If they earned $245,862 last year, how much did each person make?
Jordan
Kim
Mike
TOTAL 3 5 4 12 𝑥 𝑦 𝑧
$245,862
3 𝑥 = ,862
5 𝑦 = ,862
4 𝑧 = ,862
𝑥=$61,465.50
𝑦=$102,442.50
𝑧=$81,954.00

11. Algebra Simplify Expressions Follow BEDMAS
Expand using the distributive property
Collect Like terms
Adding/Subtracting – Do NOT change exponents
Multiplying – ADD exponents
Dividing – SUBTRACT exponents
Solve equations
Isolate the variable by using inverse operations.
Move “letters” to one side, “numbers” to the other and divide.

12. Example: Simplify (4𝑥)(3 𝑥 2 )−2𝑥(5 𝑥 2 −9𝑥)
=12 𝑥 3 −10 𝑥 𝑥 2
=2 𝑥 𝑥 2

13. Example: Solve 15 𝑥−3 =3𝑥−9 15 𝑥−3 =3𝑥−9 15𝑥−45=3𝑥−9 15𝑥−3𝑥=−9+45
12𝑥=36 12 12
𝑥=3

14. Example: Application
Determine a simplified expression for the perimeter of the rectangle shown below.
𝑥+1
3𝑥−5
3𝑥−5
𝑥+1 𝑃= 8𝑥 −8
8𝑥−8=112
8𝑥=112+8
Solve for 𝑥 when the perimeter is 112 cm.
8𝑥=120 8 8
𝑥=15 cm

15. Linear Relations SLOPE = RATE (OF CHANGE) 𝑦−intercept = INITIAL VALUE
Table of Values
Graph
Equation
Description

16. Linear Relations – Table of Values𝒙 𝒚 16 97 18
106 22
124 30
160 32
169
Rate:
∆𝑦 ∆𝑥
Initial Value:
4.5×16=72
= 106−97 18−16
97−72=25
= 9 2
Partial Variation
=4.5
Equation:
𝑦= 9 2 𝑥+25
Note:
This is linear as there is a constant rate of change i.e. ( )/(30-22)=4.5.

17. Linear Relations – Graph
Rate:
∆𝑦 ∆𝑥
Initial Value:
= 75−0 3−0
Direct Variation
= 75 3
=25 mi/h
Equation:
𝑑=25𝑡
Note:
This is linear as there is a constant rate of change i.e. it’s a straight line.

18. Linear Relations – Equation
5𝑥−3𝑦+45=0
Rate:
5 3
−3𝑦=−5𝑥−45 −3 −3 −3
Initial Value: 15
𝑦= 5 3 𝑥+15
Partial Variation

19. Linear Relations – Equation from 2 points
Determine the equation of the line that passes through the points 𝐴 2,−5 and 𝐵 −4,7 .
(𝑥,𝑦)
𝑚= 7−(−5) −4−2
𝑦=−2𝑥+𝑏
−5=−2(2)+𝑏
= 12 −6
−5=−4+𝑏
−5+4=𝑏
∴𝑦=−2𝑥−1
=−2
𝑏=−1

20. Linear Relations – Equation Example 2
Determine the equation of the line that is perpendicular to the line 5𝑥−2𝑦+10=0 and has the same 𝑥−intercept as the line 3𝑥+5𝑦−30=0.
𝑦=− 2 5 𝑥+𝑏
slope
𝑥−intercept
5𝑥−2𝑦+10=0
3𝑥+5 0 −30=0
0=− 2 5 (10)+𝑏
3𝑥−30=0
−2𝑦=−5𝑥−10
0=−4+𝑏
3𝑥=30
𝑦= 5 2 𝑥+5
𝑥=10
𝑏=4
∴𝑦=− 2 5 𝑥+4
𝑚=− 2 5
point:(10,0)

21. Scatter Plots & Motion Scatter Plot
Know how to read and interpret points.
Draw a line of best fit.
Create the equation of the line of best fit.
Distance/Speed/Time Graphs
Remember to state 4 details for every section:
Direction
Change in distance
Change in time
Speed (Distance ÷ Time)

22. Example: Scatter Plot
𝑚= 1300−100 55−0 𝑚=
𝑚=24 𝑏
∴𝑙=24ℎ+100

23. Example: Distance/Speed/Time
Distance Away From Home
Away from home
150 m
100 mins.
Distance Away From Home (m)
1.5 m/min.
Stopped
0 m
100 mins.
0 m/min.
Towards home
150 m
50 mins.
3 m/min.

24. Geometry 2D Shapes: Perimeter and Area
Don’t use the formula page for the perimeter just add the outside (Recall: 𝐶=𝜋𝑑).
For the area of a composite shape, use the formula page to find the area of each shape and then add and/or subtract.
3D Shapes: Surface Area and Volume
Remember for Surface Area just find the Area of each side and add them together (lateral surface – curved surface).
For the volume of a composite shape, use the formula page to find the volume of each shape then add and/or subtract.

25. Angles Straight Angles add to 1800. Right Angles add to 900 .
Opposite Angles are equal (X-Pattern)
Parallel lines (Z-Pattern, F-Pattern, C-Pattern)
The sum of the angles in a triangle equals 1800.
The sum of the angles in a quadrilateral equals 3600.
The sum of the interior angles in a polygon is 𝑛−2 × 180 𝑜 .
The sum of the exterior angles in a polygon is 360 𝑜 .

26. Example: Interior Angles in a Polygon
The sum of the interior angles of a polygon is How many sides does the polygon have?
𝑛−2 × =
𝑛−2 = ÷ 180 0
𝑛−2=26
𝑛=26+2
𝑛=28
∴ the polygon has 28 sides.

27. Example: Exterior Angles of a Polygon
Determine the value of on this regular nonagon.
𝑥=360 0 ÷9 𝑥
𝑥= 40 0