Download Math Book - Riverside Secondary School

Chapter 1: Rational Numbers
In the rational numbers chapter, we learnt how to compare and order rational
numbers, to solve problems involving rational numbers, to determine the square
root of a perfect square rational number, and to determine the approximate
square root of a non-perfect 8
Terminology I learnt…
Rational number: A
number that can be
expressed as the quotient
of two integers, where the
divisor is not zero. Example;
0.75, ¾, and -2 are all
rational numbers.
Key Words
Non-perfect square: A rational
number that cannot be expressed as
the product of two equal rational
factors. For example, you cannot
multiply any rational number by itself
and get an answer of 3, 5, 1.5, or 7/8.
The square root of a non-perfect square
is a non-repeating, non-terminating
decimal.
Perfect Square: A
number that is the product
of two identical factors.
2 X 2 = 4, so 4 is a perfect
square
6 X 6 = 36, so 36 is a
perfect square.
Compare Rational Numbers
Which fraction is bigger 3/4 or 2/3?
1. Find a common denominator: 3/4 and 2/3 share 12 as a common denominator
2. Whatever you do to the top, you do to the bottom: We multiplied 4 by 3 so now we
have 12 and we multiply 3 by 3 to get nine. New fraction equals 9/12. We multiplied 3
by 4 to get 12 so we multiply 2 by 4 to get 8. New fractions equals 8/12.
9/12 is larger than 8/12.
Option 2; use decimals
1. Convert fractions into decimals and compare.
3/4= 0.75
2/3= 0.6’
0.75 > 0.6’
¾ is the greater fraction.
*Now that you can compare fractions in different ways you can order them from least to greatest.*
Problem Solving with Rational Numbers, as decimals and fractions.
a) 2.93+ (-3.87)= ?
b) 6.42 - 7.83 =?
C) 5.32 X 2.14=?
Estimate and Calculate
a)3+(-4)= (-1) estimation
2.93+(-3.87)= (-0.94) calculation
b) 6.42 – 7.83 *add the opposite* 6.42 + (-7.83) = (-1.41)
c) 5 X 2 = 10
5.32 X 2.14= 11.38
Adding and Subtracting Fractions
1. Find a common denominator
2. Keep the denominator the same
3. Add or subtract the numerators
2/3 + 5/6
4/6 + 5/6= 9/6 reduced to 3/2 or 1 ½
Multiplying and Dividing Fractions
Multiplying
1. Multiply the numerators and multiply the denominators
¾ x ¼ = 3/12 = ¼
Dividing
1. Multiply by the reciprocal
*Change only second fraction to its reciprocal*
2/3
3/5 =
2/3 X 5/3 = 10/9 = 1 1/9
Determining Square Roots of Rational Numbers:
Determine whether each of the following numbers is a perfect square;
A) 25/49
B) 0.4
A) Both the denominator and numerator are a perfect square because it is expressed
as the product of two equal rational factors. 25= 5X5 , 49= 7X7
B) 0.4 can be expressed as 4/10 , the numerator is a perfect square (2X2) but the
numerator is not… so 0.4 is not a perfect square
Area= 0.64 units squared, is it perfect square?
0.64= 16/25
The square root of 16 is 4 and the square root of 25 is 5, therefore a
*use square roots to determine perfect squares and non-perfect squares.
Chapter 2: Exponents
In the exponent unit we learnt how to represent repeated multiplication exponents, how to
describe how powers represent repeated multiplication, demonstrate the difference
between the exponent and the base by building models of a given power. We
demonstrated the difference between two given powers in which the exponent and the
base are interchanged by using repeated multiplication. We also evaluated powers with
integral bases and whole exponents. We learnt that if the base is negative, use brackets (-3)7
means -3 multiplied 7 times. But -37 means -1 time 37. We also learnt how to model addition
and subtraction of polynomial expressions concretely, pictorially, and record the process
symbolically. Also, how to solve problems using addition and subtraction of polynomials and
how to simplify the expressions. Terminology I learnt…
Key words
Power; An expression
Base; the number you
Exponent: the number
Exponential form; a
made up of a base
and exponent
multiply by
of times you multiply
the base in a power
shorter way to write
repeated
multiplication
Examples of learning:
How do you write 4x4x4x4x4x4x4x4x4x4x4 in exponential form? =
411
What is the difference between 43 and 34? The difference is, for 43 you would multiply 4 three
times; 4x4x4 but for 34 you would multiply 3 four times; 3x3x3x3.
Use a model to represent 52,
25
5
5
The Exponent Laws
A) When you multiply the same bases, add exponents; 74 x 73 = 77
B) When you divide the same bases subtract exponents; 69 x 64 = 65
C) When you raise a power to a power, multiply exponents; (45)3= 415
D) Anything to the power of 0 is 1; 80= 1
*Last tip*- When using BEDMAS, Exponents are done second*
Chapter 3: Polynomials
In this chapter I learnt how to write expressions represented by algebra tiles,
how to write expressions for a given situation and solve them. How to find the
perimeter of shapes containing polynomials, how to add and subtract
polynomials and simplify the expressions.
Terminology I learnt…
Degree of a
term: The sum
of exponents
in the
variable in a
single term
(monomial)
Degree of a
polynomial:
The degree
of the
highestdegree in a
polynomial.
Number of
variables
Coefficients:
The number
with a
variable
attached to
the end.
Eg; 7x and 5y
are
coefficients
but 3-x is not
a coefficient.
Constant:
The number
with no
variable
attached to
the end.
Like terms:
Terms that
differ only by
their
numerical
coefficients.
Eg; 3x2+7x2
Zero Pairs:
Algebra tiles
that have a
combined
value of
zero.
x2
-x2 =0
What polynomial expression is represented by the set of algebra tiles
= -X2+2x+1
Write an expression for this statement
Twice the sum of 5 times a number and 8= 2(5x+8)
Seven and a number are added together= 7+x
The product of a number and 8 is decreased by 3= 8x-3
A triangles base and height are equal. Find an expression for the area. A= X2
,
2
Tickets for the basketball game are $7 for adults and $4 for students. Write an expression that
shows the total income. I=7a+4s
Write an expression represented by the algebra tiles after you’ve gotten rid of
zero pairs.
= X2 –X
Adding and subtracting polynomials.
Adding; *To add, just find zero pairs and simplify
(2x2 – 2x +1) + (-x2+ 2x+2) = (Find zero pairs) X2-x+3
Subtraction; *To subtract a polynomial, you add the opposite terms and then simplify
Chapter 4:Multiplying and Dividing Polynomials.
In this short chapter, I learnt how to model multiplication of a monomial by
monomial and a polynomial by a monomial and record the process
algebraically. How to model division of a monomial by a monomial and a
polynomial by a monomial and record the process and then to simplify
polynomial expressions.
Legend:
= -x2
=x2
=
x
=-x
1
 1
=y
= -y
=xy2
=xy2
Examples of learning:
= 2x * -2= -4x
=2y * (-2x+2) = -4xy+4y
Or, -4x
or, -4xy+4y
2x= -2
2y= -2x+2
You can also do these questions without the algebra tiles, this is a more efficient and faster
way to do it.
a) 2xy(3xy-4)= 6xy2 – 8xy
a) 3x2 – 6x
, 3x = x-2
b) 3x(-x+5y-3) = -3x2 + 15 xy- 9x
b) 2x2+3x
-x
= -2x - 3
Using this process in certain situations.
4y
3x
A=24x2
4x
3x * 2 *4y = volume
2
24x2
4x= 6x
y-2
y-2 – y-2
=y2 +4
5y * 5y=25y2-y2+4
5y
25y2+ y2 -4
27y2 - 4
Chapter 5: Surface Area
In this chapter, we learnt how to the find surface area of 1 dimensional to 3
dimensional shapes. We learnt how to find the area of a circle, trapezoid,
triangle, rectangle, parallelogram, and a square. We learnt that symmetry can
help us find surface area of a 3 dimensional shape because there are 2 of
every side so we only need to find half of the dimensions and then multiply by 2
Vocab and property’s I learnt throughout this chapter
 Area=the number of square units contained in a two-dimensional region.
 Circumference=The boundary of or distance around a circle. This is a
linear measurement.
 Diameter=the distance across a circle through its centre.
 Perimeter=the distance around the outside of a two dimensional shape
 Polygon = a two dimensional closed figure made of three or more line
segments, triangles, squares, octagon, etc.
 Radius= a line segment joining the centre of a circle to the outside edge.
Half the circumference.
 Line symmetry=A line running through the centre of an object or design
such that the halves on each side are mirror images.
 Surface Area=The number of square units needed to cover a 3-D object.
The sum of all the faces of an object.
Area
Area of a circle = pi * radius2
Area of a rectangle = base * height
Area of a parallelogram= base * height
Area of a trapezoid= ½ (b1 +b2)
Area of a triangle= ½ base * height
Area of a square= s2
How to apply this to finding the surface area of shapes
Top= 15X5=75cm2 (so bottom equals 75cm2)=150cm2
10cm Front= 15X10=150cm2(so back equals 150cm2)=300cm2
Sides= 5X10= 50cm2 (other side equals 50cm2)=100cm2
15cm
5cm
Add all sides together, 150+300+100=550cm2
Find the surface area of the cylinder (Height=18cm, radius=11cm)
18x22=396cm2
18cm
C=2pi r, C=22cm
Area of a circle= pi x r2
3.141592653589793… X 112= 69.1cm2
Both circles equal 69.1, so add them together (138.2) Add this number and the rectangle
side together, 132.2 + 396 = 534.2cm2
Chapter 6: Linear Relations
In this chapter I learnt how to interpret graphs and apply it to real life
circumstances, how to find patterns in figures and/or graphs and write an
equation for it, read and make tables relating to linear relations and make an
equation about it, and how to graph the linear relations.
Some terminology I learnt is…
Key Words
Linear
patterns: a
sequence
of numbers
in which
the
pattern inly
involves
addition or
subtraction
Common
difference:
the
difference
between
any two
consecutive
numbers in
a linear
pattern
Dependent
variable:
graphed on
the vertical
axis
e.g. cost
Independent
variable
Graphed on
the
horizontal
axis
e.g. time
Interpolate:
Estimate a
value
between
two given
values
Extrapolate:
estimate a
value
beyond a
given set of
values
Slope:
describes
the
steepness
of a line
y-intercept:
Where the
graph
crosses the
y-axis. The x
coordinate
is zero.
Finding relations in patterns
Fig.1=3 fig.2=5 fig.3=7
fig.4=9
The relation is, you add 2 squares each time,
The equation is, the difference multiplied by the fig. number and then add or
subtract to get the right number of squares needed. 2xf+1
How many squares might the next three figures contain?
Figure 5, 2x5+1= 11
Figure 6, 2x6+1= 13
Figure 7, 2x7+1= 15
Putting the numbers into a graph…
Figure #
#of squares
1
3
2
5
3
7
4
9
5
11
6
13
And now you can look at the table and figure out an equation that applies to
this series of numbers. (2xf+1=) (difference x figure # plus or minus a number.)
Interpreting graphs and relating them to real life.
40
30
20
10
0
2
4
6
8
This graph represents how much money Briana gets in her bank each day. She gets exactly
$10 every day. put the information in a table.
Days
Money
1
$5
2
$15
3
$25
4
$35
5
$45
6
$55
7
$65
8
$75
We are able to extrapolate this graph because she makes exactly $10 every day so we
know what she will make in 17 days.
An equation we can use is, 10d-5=p (d representing the day and p representing profit.
We can use this equation to find out how much money she will make in 15 days,
10x15-5= $145
*If you can extrapolate on the graph, you put an arrow through the dots. Because it will
always go up or down the same amount.*
Graphing linear relations.
I learnt, if you are given an equation you can graph it.
Y=-3x-4
X
Y
1
-7
2
-10
3
-13
4
-16
And now you can graph it,
0
-1
-3
-5
-7
-9
-11
-13
-15
-17
1
2
3
4
Chapter 7: Solving Linear Equations
In this chapter I learnt how solve equations algebraically using inverse/opposite
operations. We reviewed that addition and subtraction and opposite and
multiplying and dividing are opposite operations. I learnt how to check my
answer by putting the answer I got into the equation. I found this chapter super
useful and know I will continue to use it throughout my high-school career.
To solve linear equations we use inverse operations, for example…
3x+4=-5
3x=-9 (we subtract 4 from each side.)
X=-3 (we divide 3 by each side because 3 is being multiplied by x)
*Another property we learnt was distributive property*
2(-2+x)=18
-4+2x=18 (distributive property, multiply -2 and x by 2)
2x=22 (add 4 to each side to isolate the variable.)
X=11
*GOAL! TO ISOLATE THE VARIABLE!!*
“Balance the scale”
To check you answer, just put the answer you got into the equation.
2(-2+11)=18
2x9=18
18=18
*Yup it works!*
Chapter 8: Linear Inequalities
I learnt how to solve a linear inequality algebraically and explain how to do it.
How to graph the answer of linear inequality on a number line. I learnt how to
translate an equation into a single variable linear inequality using the symbols <,
>, <, >, or =. I learnt the difference between solving a linear equation and linear
inequality, which is if you multiply or divide by a negative while solving a linear
inequality you must switch the direction of the sign to make it true. And lastly
how to check if your answer is correct.
Inequality: a
mathematical
statement comparing
expressions that may
not be equal
Key Words
Boundary Point:
Open circle boundary
separates the values
point: the boundary
less than and great
point is not included in
than a specified value. the solution.
Might be the solution
<,>
Closed circle
boundary point:
the boundary
point is included
in the solution.
<,>
3<5, 6>4, 8>4, 19<22, -10<-3, 12<13, 5>1, -5>-17, 15>-18, 17<18, 3>1, 1<6, 9=9, 1<2
Translating verbal statements into inequalities on a number line
X is larger than or equal to 2, x>2
-2 -1 0 1 2
Up above is the three ways you can express inequalities. The first way is verbally,
describing the inequality in words. The second way is algebraically, using <,>,
<,>. The last way is graphically, using visuals and in this case a number line.
Solving Inequalities.
Solving inequalities is the same as solving equations in the previous chapters.
The differences though, are if you multiply or divide by a negative when
working with inequalities you must switch the direction of the sign to be true.
The other difference is equations we use = sign and inequalities use <,>, <,>.
Example
-3x -10 < 5x+38
-3x<5x +48 (added 10 to each side.)
-8x< 48 (subtracted 5 to each side.)
X>-6 (divided each side by -8 and then switched the sign.)
To check if the answer is correct put any number equal or greater than -6 into
the equation.
-3*-4 < 5*-4 +38
12< -20+38
12< 18
It is correct. 12 is smaller than 18.
Chapter 9: Scale Factor and Similarity
In this chapter, I learnt that scale factor is when you take an image and divide
by the actual object length or the scale factor and that will equal the actual
diagram measurements. I learnt how to represent an enlargement and
reduction by drawing them. I also learnt how to determine if a diagram is
proportional or not.
When a polygon is enlarged or reduced its corresponding angle measurements
will stay the same but the corresponding side lengths are enlarged or reduced.
When a triangle is enlarged or reduced its corresponding angle measurements
will remain the same. And then the corresponding side lengths are proportional.
11
8
4
5.5
4.5
The triangles are similar because the side lengths are proportional.
That means the angles are similar as well because it is a triangle.
9
The black triangle is the original one and the blue one is the image. The scale
factor is… 4/8= 0.5. The scale factor is 0.5. Since it’s less than 1 it is a reduction.
26cm
16
17
9
Although the angles are the same, the side lengths are not proportional
therefore this polygon is not similar.
Key Words
Enlargement: An
increase in the
dimensions of an
object by a scale
factor
Reduction: A
decrease in the
dimensions of an
object by a scale
factor
Scale factor: The
constant amount by
which all dimensions
of an object are
multiplying to draw
an enlargement or
reduction
Corresponding
angles/sides: Have
the same relative
position in a
geometric figure
Similar figures: Have
the same shape but
different size
Scale: A comparison
between the actual
size of an object and
the size of its
diagram. Expressed
as a ratio, fraction,
percent, words, or a
diagram
Scale Diagram: A
drawing similar to the
actual figure or
object. Might be
smaller or larger but
is the same
proportion.
Proportion: Is a
relationship that
shows two ratios are
equal. Written as a
fraction or ratio
Scale Factor=Image length/object length
Scale= Diagram measurement/actual measurement.
Fraction
Ratios
Percent
1 /4
1:4
25%
2/5
2:5
20%
6/7
6:7
85%
3 /4
3:4
75%
Chapter 10: Circle Geometry
This chapter was completely new to me and I found it really interesting. I learnt
how to find angles within circles, the different properties to use. I learnt how to
solve problems involving using these circle properties.
Key words and properties I learnt.
Key Words
Inscribed Angle:
An angle formed
by two chords
that share a
common
endpoint.
Central Angle:
An angle
formed by two
radii of a circle
Chord: A line
between two
points on a
circle.
Arc: A section of
the
circumference of
the circle.
Tangent: A line
outside the
circle touching
the circle at
one point
Important Properties, to find angles within circles.
 An inscribed angle subtended to the same arc of a central angle is half
the angle of the central angle. (Central angle is double the inscribed.)
 When two or more inscribed angles are connected to the same arc they
feature the same angles.
 Complementary <’s add to 90o
 Supplementary <’s add to 180o
 Angles on a line add to 1800
 Right angles are 90o
 Straight angles are 180o
 Angles at a point add to 360o
 Vertically opposite angles are equal
I learnt how to use Pythagorean theorem to find the side lengths of
RIGHT TRIANGLES in a circle.
The formula is, a2+b2+=c2
Example,
52+62=c2
25+36=c2
61=c2
7.8cm
X=7.8cm
Perpendicular Bisector of a Chord
This was an important property I learnt. You use this when you want to find the
center of a circle. It is a line that passes through the centre of a circle and is
perpendicular to a chord bisects the chord.
And lastly, we learnt about tangents. At the top of this chapter it explains that I
tangent is a line outside of a circle that touches the circle and precisely one
point. We can use these tangents to find side lengths, angles, and bisectors just
like how we chords inside the circle. If a tangent is perpendicular to the point of
tangency, that angle made is 90o.