11 college math

11 College Math

Final Exam Review

Topics

• Part 1: Trigonometry• Part 2: Quadratic Relations• Part 3: Probability • Part 4: Statistics• Part 5: Exponents• Part 6: Personal Finance• Part 7: Measurement

AgendaThis week …

Monday: Right-angle Trigonometry (SohCahToa)Tuesday: Sine Law & Cosine LawWednesday: Quadratics (Algebra & The Parabola)Thursday: Quadratics (3 Forms & Applications)Friday: Probability & Statistics

Next week …

Monday: ExponentsTuesday: Personal FinanceWednesday: Geometry

Part 1: Trigonometry

Monday:

Tuesday:

• Labelling: Opposite, Adjacent, Hypotenuse• SOHCAHTOA: Opposite, Adjacent, Hypotenuse• Right-angle Applications

• The Sine Law• The Cosine Law• Applications

Right-angle Trigonometry

Non-Right Angle Trigonometry

Right-Angle TrigonometryPrimary Trigonometric Ratios: Sine, Cosine & Tangent

Recall: When solving for an angle we must use the inverse functions!

Example 1: Evaluate to 3 decimal places.

sin 650 = cos 1240 = tan 3410 =

Example 2: Solve for the angle to the nearest degree.sin R = 0.25 cos B = 0.92 tan Q = 1.54

-0.5590.906 -0.344

ANGLES

R = sin-10.25 = 14o

B = cos-10.92 = 23o

Q = tan-11.54 = 57o

When do we use these?When we want to solve for an angle or a side in a Right-angle Triangle.

What is the first thing that we must do when solving a right angle triangle?

LABLE THE SIDES: OPPOSITE, ADJACENT, HYPOTENUSE

OPPOSITE

ADJACENT

HYPOTENUSE

Recall: Labelling depends upon the reference angle!

HYPOTENUSE

OPPOSITE

ADJACENT

Recall: SOHCAHTOA

Hence, the primary trig ratios for angle A below are:

A

BC

𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒h𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒sinA =

cosA =𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡h𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

tanA = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡OPPOSITE

ADJACENTHYPOTENUSE

Example 1: Determine the length of side b to one decimal place.

42o

11.5 m

b

O

H

A

SOHCAHTOACOSINE

cos (𝑎𝑛𝑔𝑙𝑒)=𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡h𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

cos 420= 𝑏11.5

𝑏=11.5 cos420

Variable on top→ MULTIPLY

𝑏=8.5𝑚

Example 2: Determine the length of side b to one decimal place.

37o

7 cm

b

O

HA

SOHCAHTOATANGENT

t an(𝑎𝑛𝑔𝑙𝑒)=𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

tan 370=7𝑏

𝑏=7÷ tan 370

Variable on bottom→ DIVIDE

𝑏=9.3𝑐𝑚

Example 3: Determine the length of angle A to the nearest degree.

1.2 in

2.5 in

O

HA

SOHCAHTOASINE

sin (𝑎𝑛𝑔𝑙𝑒)=𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒h𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒

s inA=1.22.5

A=sin−1( 1.22.5 )Solving for an angle→ INVERSE

A=29°

A

BC

2nd sin (1.22.5)

Example 4:

OH

A

SOHCAHTOA

t an(𝑎𝑛𝑔𝑙𝑒)=𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡

t an29 °= 𝑥150

𝑥=150 tan 29°

Variable on top→ MULTIPLY

𝑥=83 𝑓𝑡

Joey is standing 150 ft from the base of a building. The angle of elevation to the top of the building is 29o. How tall is the building.

Diagram:

BUIL

DIN

G

JOEY150 ft

29o

x

Right Angle Triangle

the building is 83 feet tall.

By the end of today …I can label the sides a right angle triangle OPPOSITE, ADJACET AND HYPOTENUSE in reference to a certain angle.

I can use SOHCAHTOA to select the appropriate Primary Trig Ratio based on the given information.

I can solve for the unknown side or angle in a right-angle triangle using SOHCAHTOA.

I can create a diagram from the words of a typical right-angle trig application question and apply SOHCAHTOA to solve the problem.

Non Right-Angle Trigonometry

Solving triangle that are NOT right angles

Can we use the Pythagorean Theorem?

Can we use SOHCAHTOA? NO

NO

Then ... what can we use?

SINE LAW COSINE LAW

Before we begin let's recall a few things about triangles:

#1 Sum of the angle in a triangle:

#2 Labelling Conventions:

· Angles are labelled with UPPER CASE letters.

· Sides are labelled with LOWER CASE letters.

· OPPOSITE sides and angles correspond.

R

Q

P

B

A C

r

q

p

A + B + C = 180o

The Sine Law

Recall: Look for the OPPOSITE SIDE-ANGLE PAIR!i.e. Are you given A & a, B & b or C & c?

If you are given two angles, always start by finding the third angle (i.e. 1800 minus the other two)

B

A C

a

b

c

Solving for a side:

Solving for an angle:

asinA=

bsinB=

csinC

sinAa =

sinBb =

sinCc

Example 1: Determine the length of side b to one decimal place.

B

A C

12 cm

b

c

32o65o

asinA=

bsinB=

csinC83o

180 ° −32 °−65 °

bsin 83 °=

12sin 65 °

Only use 2 fractions …

b=12sin 83 °sin 65 °

12sin83o = sin65o

b=13.1cm

Given: A & a → THE SINE LAW

The size of opposite side-anglepairs should correspond (i.e. thebiggest side should be across from the biggest angle).

REFLECT

Example 2: Determine the measure of angle A to the nearest degree.

B

A C

3.1 in

b

2.3 in

47o sinA3.1 =

sin 47 °2.3

Note: angles are on top!

s inA=3.1 sin 47 °2.3

3.1sin47o = 2.32nd sin ANS

A=80 °

Given: C & c → THE SINE LAW

sinAa =

sinBb =

sinCc

A=sin−1( 3.1 sin 47 °2.3 )

The Cosine Law

Recall:

Use cosine law if you can’t use sine law (i.e. no Side-Angle pair)

B

A C

a

b

c

Solving for a side:

Solving for a angle:

a2=b2+𝑐2−2bc ∙ cosA

cosA=b2+𝑐2−a22bc

Solving for a side → look for a side-angle-side “sandwich”Solving for an angle → look for all 3 sides

Example 3: Determine the length of side b to one decimal place.

B

A C

12 m

b

11 m

75o Change letters to solve for b…

= ANS

b=14.0m

Side-angle-side sandwich

→ THE COSINE LAW

a2=b2+𝑐2−2bc ∙ cosA

b2=a2+𝑐2−2ac ∙ cosBb2=112+122−2(11)(12)cos75 °

b=√112+122−2(11)(12)cos75 °

Example 4: Determine the measure of angle A to the nearest degree.

B

A C

26 m

15 m

19 m

2nd cos (-90 570)

A=99 °

Given: All 3 sides

→ THE COSINE LAW

cosA=b2+𝑐2−a22bc

Always subtract the side across from the angle you are solving for.

cosA=192+152−262

2(19)(15)

cosA=−90570

A=cos−1(−90570 )

Example 5:

Cosine Law

𝑥=574.9Km

Two planes took off from Pearson International Airport at the same time. The first plane is travelling due west at a speed of 168 Km/hr. The second plane is travelling due east at a speed of 156 Km/hr. How far apart are the planes after 2 hours? Assume the angle between them is 125o.

Diagram: Non Right-Angle Triangle

the planes are 574.9 Km apart after 2 hours.

Side-Angle-Side Sandwich

AIRPORT

Plane 1

Plane 2

336 Km(168 x 2)

312 Km(156 x 2)125 °

𝑥 𝑥2=3362+3122−2(336)(312)cos 125 °

𝑥=√3362+3122−2(336)(312)cos 125 °  

By the end of today …I can identify when to use the Sine Law and apply it to solve for the unknown side or angle in a triangle.

I can identify when to use the Cosine Law and apply it to solve for the unknown side or angle in a triangle.

I can create a diagram from the words of a typical non right-angle trig application question and apply the Sine Law or the Cosine Law to solve the problem.

Part 2: Quadratics

Wednesday:

Thursday:

• Expanding from Factored Form to Standard Form• Expanding from Vertex Form to Standard Form• Factoring: Common Factoring and Simple Trinomials

• Identifying the key features of a parabola from the three forms.

• Applications (projectile & revenue)

Algebra & The Parabola

Graphing, Three Forms & Applications

• Key features of the Parabola• Vertex form and transformations from

• Graphing from Vertex Form

Algebra - ExpandingExample 1: Convert the following equations form Factored Form to Standard Form.

Example 2: Convert the following equations form Vertex Form to Standard Form.

a) b)𝑦=(𝑥−2)(𝑥+7) 𝑦=4 (𝑥−3)(𝑥−5)𝑦=𝑥2+7 𝑥−2 𝑥−14𝑦=𝑥2+5𝑥−14

𝑦=4 (𝑥2−5𝑥−3 𝑥+15 )𝑦=4 (𝑥2−8 𝑥+15 )𝑦=4 𝑥2−32 𝑥+60

a) b)𝑦=(𝑥+2 )2−7 𝑦=2 (𝑥−3 )2+5𝑦=(𝑥+2 ) (𝑥+2 )−7𝑦=𝑥2+2𝑥+2 𝑥+4−7

+5

𝑦=2 (𝑥2−3𝑥−3 𝑥+9 )+5𝑦=2 𝑥2−6 𝑥−6 𝑥+18+5𝑦=𝑥2+4 𝑥−3𝑦=2 𝑥2−12 𝑥+23

Algebra - FactoringExample: Convert the following equations form Standard Form to Factored Form.

a)

b)

𝑦=𝑥2+10𝑥+21

𝑦=3 𝑥2−21𝑥−54

𝑦=(𝑥+3 ) (𝑥+7 )

𝑦=3 (𝑥2−7 𝑥−18 )

×21 +10121 22 x37 10

3 33×−18+−7

+1−18−17 x+2−9 −7

𝑦=3 (𝑥+2 ) (𝑥−9 )

The Parabola

Axis of Symmetry

-intercept -intercept

-intercept

Vetex (Max/Min)

Transformations from

Vertex Form:

𝑦=𝑎 (𝑥−h )2+𝑘

• vertical stretch/compression by a factor of

• horizontal translation• vertical translation

• reflection in the -axis.Step Pattern: 1, ,

v

Transformations from

Example 2:

Example 1: State the parabola represented by has been transformed from the graph of .

Determine the equation of the parabola that has been transformed from the graph of as follows:

• Translated 3 units right.• Translated 7 units up.• Vertical stretch by a factor of 2. • Reflected in the x-axis.

• Translated 3 units left.• Translated 7 units down.• Vertical stretch by a factor of 5.

𝑦=5 (𝑥+3 )2−7

By the end of today …I can convert a Quadratic Relation from Factored Form to Standard Form by Expanding and collecting like terms.I can convert a Quadratic Relation from Vertex Form to Standard Form by Expanding and collecting like terms.

I can convert a Quadratic Relation from Standard Form to Factored Form by common factoring and/or simple trinomial factoring.

I know and can identify the key features of a parabola.

I can identify the transformations of a parabola form the graph of when given the equation in vertex form.

I can write the equation of a Quadratic Relation in Vertex Form when given the transformations from the graph of .

Graphing from Vertex Form

Recall from yesterday …

𝑦=𝑎 (𝑥−h )2+𝑘

• vertical stretch/compression by a factor of

• horizontal translation• vertical translation

• reflection in the -axis.Step Pattern: 1, ,

v

Graphing from Vertex FormExample: Graph the equation, on the graph below.

Vertex: (−5 ,7 )

Step Pattern:

−2 ,−6 ,−10

Recall From yesterday ….

Axis of Symmetry

-intercept -intercept

-intercept

Vetex (Max/Min)

Three Forms

Vertex Form: 𝑦=𝑎 (𝑥−h )2+𝑘

𝑎 d

step pattern: 1, ,

All forms:

+ + - -

vertex: (h ,𝑘 )

Standard Form: 𝑦=𝑎𝑥2+𝑏𝑥+𝑐-intercept: 𝑐

Factored Form: 𝑦=𝑎 (𝑥−𝑠 ) (𝑥−𝑡 )-intercept(s): 𝑠 , 𝑡

Three FormsExample: Complete the chart of the key features of the following

quadratic relation.

Standard Form

Factored Form

Vertex Form

Direction of Opening

Step Pattern

y-intercept

x-intercepts

Vertex

DOWN

-3, -9, -15

-36

-3, -7

(-5, 12)

Application - ProjectileA football is kicked upwards. Its height is described by the equation where is the height measured in metres and is the times measured in seconds.

a) What is the height of the ball after 3 seconds?

b) When does the ball hit the ground?

𝑡=3

h=20𝑚 the ball is 20 metres high after 3 seconds.

-intercept(s): 2 ,8

the ball hits the ground after 8 seconds.

Application - ProjectileA football is kicked upwards. Its height is described by the equation where is the height measured in metres and is the times measured in seconds.

c) When does the ball reach its maximum height?

d) What is the balls maximum height?

-intercept(s): 2 ,8

the ball reaches its maximum height at 5 seconds.

Middle: 2+8  2 ¿

10  2 ¿5

h=36𝑚 the ball reaches a maximum height of 36 metres.

𝑡=5

Application - ProjectileA football is kicked upwards. Its height is described by the equation where is the height measured in metres and is the times measured in seconds.

e) Sketch a graph of the projectile of the football.

Application - RevenueThe owner of a waste management company is given a graph that shows the relation between profit and mass of garbage his company recycles. The graph is shown below:

Profit Versus Mass of Garbage Recycled

-400,000

-300,000

-200,000

-100,000

0

100,000

200,000

300,000

0 200 400 600 800 1000

Mass (tonne)

Prof

it ($

)

Application - RevenueThe owner of a waste management company is given a graph that shows the relation between profit and mass of garbage his company recycles. The graph is shown below:

How many tonnes of garbage must be recycled to produce the maximum profit possible?

a) What is the maximum profit possible?

b)

A company is said to “break even” when revenue equals expenses, or when profit equals zero. How many tonnes of garbage must be recycled to break even?

c)

$180,000

500 tonnes

200 tonnes OR 800 tonnes

By the end of today …I can graph a Quadratic Relation from Vertex Form.

I can identify the y-intercept and step-pattern of a parabola given the equation in Standard Form.

I can identify the x-intercepts and step-pattern of a parabola given the equation in Factored Form.

I can identify the vertex and step-pattern of a parabola given the equation in Vertex Form.

I can create a diagram from the words of a typical application question (i.e. projectile, revenue) and select the appropriate information form the equation to solve the problem.

Probability

Experimental ProbabilityTheoretical Probability VS• The likelihood (odds) of

an event occuring.• The number of times an event

actually happens out of a certain number of trials.

• The Theoretical Probability of rolling a 5 on a dice is ALWAYS …

• The Experimental Probability of rolling a 5 on a dice is …

For example …. Whereas….

16

The¿ of× youroll a5 ¿The¿ of× youroll the dice ¿

A jar contains 5 red balls, 8 green balls and 2 black balls. If you select on ball from the jar, what is the probability of NOT getting a green ball? Leave your answer as a fraction in lowest terms.

Theoretical ProbabilityExample 1:

Example 2:

A card is drawn from a standard deck of cards. What is the probability that it is a 9 of any suit or a black face card? Leave your answer as a fraction in lowest terms.

# of 9’s: 4# of black face cards: 6Total # of cards:

52

Probability: 𝟏𝟎𝟓𝟐

𝟓𝟐𝟔¿

Lowest Terms←10 a

the probability of drawing a 9 or a black face card is .

# of green balls: 8Total # of balls: 15

Probability:

𝟏𝟓−𝟖𝟏𝟓

𝟕𝟏𝟓¿

Already in Lowest Terms←

the probability of NOT getting a 9 green ball is .

Statistics

Mean:

Measures of Central Tendency:

Measures of Spread:

Median:

Mode:

Standard Deviation:

Interquartile Range:

Range:

Quartiles:

The “average”.• Add them all up, press equal

and then divide by how many there are.

The middle.• Line up the data and find the

middle # or the average of the 2 middle #’s.

The most frequent.• Note: there can be more than

one mode.

Biggest - Smallest

• The typical distance a particular value is from the mean.

• 3 #’s that divide the data into 4 equal parts.

• The 2nd quartile is the median.

• 3rd Quartile – 1st Quartile

Example: Mackenzie recorded the following daily temperatures for 8 days in June:

23 25 17 16 27 25 32 12

a) Calculate the Mean, Median, Mode and Range:

Mean: 23+25+17+16+27+25+32+12=¿154154÷8=¿19.25

Median: (23+25)÷ 2=¿24

Mode: 25

Range: 32−12¿20

12

16

17

23

25

25

27

32

Middle

Example: Mackenzie recorded the following daily temperatures for 8 days in June:

23 25 17 16 27 25 32 12

b) Find the three quartiles and state what they are:

12

16

17

23

25

25

27

32

1st Quartile:

3rd Quartile:

2nd Quartile:

c) Determine the Interquartile Range:

Middle(23+25)÷ 2=¿24

→ Note: this is the same as the Median.

Middle of 1st half(16+17)÷ 2=¿16.5

Middle of 2nd half(25+27)÷2=¿26

26 9.5−16.5¿

Standard DeviationAn engine part is manufactured with a mean weight of 900g and a standard deviation of 9g. Parts are rejected if they are not within one standard deviation of the mean. How many of the parts with the following weights would be rejected?

890g 899g 907g 905g 911g 895g 905g

900g

+9g-9g

891g 909g

x x

3 of the parts would be rejected: 890g, 911g & 895g.

By the end of today …I understand and can recognize the difference between experimental and theoretical probability.

I can use the respective formula to calculate the experimental or theoretical probability of various situations (i.e. rolling a die, drawing cards from a deck, etc.)

I can calculate the mean, median, mode and range of a given set of data.

I can identify the THREE quartiles in a given set of data.

ExponentsExponent Laws:

1. Multiplication 𝑏𝑚 ∙𝑏𝑛=¿

2. Division 𝑏𝑚

𝑏𝑛 =¿

3. Power of a power (𝑏𝑚 )𝑛=¿

𝑏𝑚+𝑛

𝑏𝑚−𝑛

𝑏𝑚𝑛

(𝑏𝑐 )𝑛=¿𝑏𝑛𝑐𝑛

(𝑏𝑐 )𝑛=¿ 𝑏

𝑛

𝑐𝑛

4. Negative exponents 𝑏−𝑛=¿1𝑏𝑛

(𝑏𝑐 )−𝑛

=¿𝑐𝑛

𝑏𝑛(𝑐𝑏 )𝑛=¿

𝑏0=¿1

Don’t forget …

𝑏1=¿𝑏

ExponentsExamples:

a) b) c)

d) e)

¿ 𝑥10 ¿32𝑥101

¿9 𝑥10¿3 𝑥7

Simplify by expressing as a single power. Write all answers with positive exponents where required.

¿1

Exponential Curve

𝑦=𝑎𝑥

0<𝑎<1Decreasing for …

𝑎>1Increasing for …

Always crosses the -axis at 0.

→The smaller the value of a the steeper the slope.

→The larger the value of a the steeper the slope.

Exponential Curve

Example:Given the equations and the graphs, match each equation with its graph. Write the matching number under each equation.

x-4 -2 2 4

y

5

10

15

20

25

1  

2  

3  

4  

a)

b)

c)

d)

¿0.29𝑥

¿3.5𝑥

Largest increasing →

Smallest increasing →

Largest decreasing →

Smallest decreasing →

3

2

4

1

Exponential Growth & Decay

𝑦=𝑎𝑏𝑥

• , starting value• , growth/decay factor• , # of growth/decay periods

Example 1:

Example 2:

The value of a rare coin increases by 16% per year. If the coin is worth $52 now, how much will it be worth in 5 years?

The half-life of Radium-52 is 20 min. How much of a 6000 mg sample will be left in 4 hours?

𝑦=(52)(1.16)5𝑏=¿1.16

¿ $109.22the coin will be worth $109.22 in 5 years.

∴

𝑦=(6000)(0.5)12

𝑏=¿0.5

mg

1.46 mg of the sample will be left in 4 hours.

∴x=¿3×4=12

Comparing Linear, Quadratic & Exponential Models

Linear Quadratic Exponential

Equation

Graph

Table of Values

𝑦=𝑎𝑥 2+𝑏𝑥+𝑐𝑦=𝑚𝑥+𝑏 𝑦=𝑎𝑏𝑥

Straight line

1st differences are the same.

2nd differences are the same.

Ratios are the same.

Parabola Exponential Curve

Comparing Linear, Quadratic & Exponential Models

Example: Classify each of the following relations as linear, quadratic or exponential.

c ) 𝑦=3 𝑥2−2 𝑥a ) b )

__________________ ____________________ c )¿LinearExponential Quadratic

4

4

4

4

By the end of today …ALGEBRA: I can use the Exponent Rules to simplify expressions leaving only positive exponents..

GRAPHING: I can identify the graph of an exponential relation and match the equations of increasing and decreasing functions from the equations.

APPLICATIONS: I can determine the growth/decay factor from a word problem and apply this in the formula to find a solution..

I can identify and differentiate between linear, quadratic and exponential relations from their Table of Values, Graphs and Equations.

Personal Finance

Compound InterestSimple Interest VS

Compare two $1000 loans both collecting 5% annual interest over a 4 year period. The first one is collecting simple interest and the other is being compounded annually.

Year Interest Amount

0 - $1000

1

2

3

4

Year Interest Amount

0 - $1000

1

2

3

4

$50

$50

$50

$50

$50.00

$52.50

$55.13

$57.88

$1050$1100$1150

$1200

$1102.50

$1050

$1157.63

$1215.51

Simple Interest

𝐼=𝑃𝑟𝑡

𝐴=𝑃+ 𝐼

• the total interest earned over years.

• the amount of the investment after years.

NOTE: With the SIMPLE INTEREST FORMULA you are calculating the INTEREST on the investment/loan NOT the AMOUNT.

To determine the AMOUNT of the investment/loan after years you must ADD the INTEREST to the PRINCIPAL amount invested.

• the principal amount invested.• the annual interest rate as a decimal.• the time in years.

Simple InterestExample 1: Determine the amount of a $5,000 investment making 3.2% simple

interest annually after 4 months.

Example 2: A $2000 investment made $400 in simple interest in 10 years, what was the interest rate?

𝐼=𝑃𝑟𝑡.032)

𝐼=$53.33

𝐴=𝑃+ 𝐼𝐴=$5000+$ 53.33𝐴=$5053.33

the amount of the investment after 4 months is .

𝑃𝐼𝑟𝑡

𝑟=𝐼𝑃𝑡

𝑟=400

(2000)(10)

𝟒𝟎𝟎÷𝟐𝟎𝟎𝟎 ÷𝟏𝟎𝑟=0.02𝑟=2%

the interest rate was 2%.

Compound Interest𝐴=𝑃 (1+𝑖 )𝑛

𝐼=𝐴−𝑃 • the total interest earned over years.

• the amount of the investment after years.

NOTE: With the COMPOUND INTEREST FORMULA you are calculating the TOTAL AMOUNT on the investment/loan NOT the INTEREST.

To determine the INTEREST earned you must SUBTRACT the PRINCIPAL from the AMOUNT.

• the principal amount invested.• the interest rate per compounding period.• the total number of compounding periods.

→ .→ .

is the # of compounding periods per year

Compound Interest

Example: You invest $8,000 at 3.6%, compounded quarterly for 5 years.

(a) What is your investment worth in five years?

(b) How much interest did you make?

𝑁=¿𝑃=¿𝑖=¿𝑛=¿

480000.036÷ 45

¿0.009×4 ¿20

𝐴=𝑃 (1+𝑖 )𝑛

𝐴=8000 (1+0.009 )20

𝐴=9570.03

your investment is worth in 5 years.

𝐼=𝐴−𝑃𝐼=9570.03−8000

𝐼=1570.03 you made in interest.

Present Value

𝑃=𝐴 (1+𝑖 )−𝑛

• the amount of the investment after years.• the principal amount invested.• the interest rate per compounding period.• the total number of compounding periods.

NOTE: This is still compound interest → everything works the same!

The only difference is that we know the future AMOUNT of the loan/investment and we want to know the PRINCIPAL (Present Value).

HINT: Notice the , use this as a reminder to use this formula when you are working backwards in time.

Present ValueExample: You would like to have $4,000 saved to travel in 2 years.

How much must you invest today at 5.4%, compounded semi-annually, in order to save the required amount?

𝑁=¿𝐴=¿𝑖=¿𝑛=¿

240000.054÷22

¿0.027×2 ¿ 4

𝑃=𝐴 (1+𝑖 )−𝑛

𝑃=4000 (1+0.027 )− 4

𝑃=3595.66

you must invest today to have saved in 2 years.

By the end of today …

I can use the Simple Interest formula to determine the Interest, Amount, Principle, time or interest rate of an investment or loan from a word problem

I can use the Compound Interest formula to solve for the Amount and/or interest of an investment or loan from a word problem.

I can use the Present Value formula to solve for the Principal of an investment or loan from a word problem.

Unit ConversionsExamples: Convert each measurement. Show your work.

a) 2.7 L = _________ mL

d) 2.7 L = _________ pints

b) 12.4 cm = _________ inches

c) 5.4 yards = _________ ft

Formula Page

100 mL = 1 L

L → mL we expect our # to get bigger.

10002700

2.54 cm = 1 inch

cm → in we expect our # to get smaller.

2.544.88

3 feet = 1 yard

yrds → ft we expect our # to get bigger.

3

0.473 L = 1 pint

L → pints we expect our # to get bigger.

0.47316.2 5.7

Geometry – Perimeter & Area

15 cm12 cm

9 cm

𝑃=12𝜋2 +15+9𝐶=𝜋 𝑑÷2

𝑃=42.8𝑐𝑚

𝐴=𝜋 (6 )2

2+

(9 ) (12 )2

𝐴=110.5 𝑐𝑚2

Example: Calculate the perimeter & area of the following composite shape.

Geometry - Volume

1

2

3

V12 cm

V1

V2

V2

V3

V3

V

= 9 cm15 cm – 6 cm

9 cm – 4 cm= 5 cm

12cm - 3cm – 5cm= 4cm

Example: Calculate the volume of the following composite figure.

By the end of today …I can convert from Metric to Metric measurements using a conversion chart.

I can convert from Imperial to Imperial measurements using a conversion chart.

I can calculate the Perimeter of a composite shape.

I can convert from Imperial to Metric measurements using a conversion chart.

I can calculate the Area of a composite shape.

I can calculate the Volume of a composite figure.

Formula Page

Formula Page

BACK

Formula Page

Supplies

• Pencils – at least two.

• Eraser

• Ruler

• Scientific Calculator

→ Make sure that you get this early and have practiced using it before the EXAM.

• Water bottle

→ preferably one that works in the correct order (i.e. sin 30 NOT 30 sin)

→ CELL PHONES may NOT be used as calculators for the exam.

Study Tips• Use the Learning Goals checklist to write out a study sheet for the exam.

Be sure to include diagrams and examples.• Attempt all review booklet questions and ask for extra-help immediately

on questions you do not understand.• When doing the above questions, be sure to use the calculator you will be

using for the exam.• Review old tests.• Use the Learning Goals checklist to reflect and keep track of skills you have

attained and those that still need practice.• GET A GOOD NIGHTS SLEEP.• EAT A GOOD BREAKFAST.• Double check that you have packed all necessary supplies.• Arrive to school at least 15 minutes early.• Relax! You are going to do great!

See you next …

Tuesday, June 24th at 8:15 am

GOOD LUCK!