PATTERNING �AND ALGEBRA
TERM 1
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PATTERNING AND ALGEBRA
Represent, through investigation with concrete materials, the general term of a linear pattern, using one or more algebraic expressions.
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CURRICULUM
PATTERNING AND ALGEBRA
Determine a term, given its term number, in a linear pattern that is represented by a graph or an algebraic equation.
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CURRICULUM
DO IT NOW
# Linear Patterns – Algebraic Expressions PA
Linear Patterns – Algebraic Expressions
PA
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ALGEBRAIC EXPRESSIONS
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EXPRESSIONS
There are 2 types of expressions
5 + 84 - Numerical
3x + 10 - Algebraic
Notice there are no equal signs in these expressions so they are not equations!
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Contains only
EXAMPLE
5 x 3 + 8
NUMERICAL EXPRESSIONS
ALGEBRAIC EXPRESSIONS
Contains only
EXAMPLE
8y
j + 12
k - 6
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VARIABLE
A variable represents an unknown value.
4 + x
10 - ?
5y
20 ÷
A variable can be any letter of the alphabet since it represents an unknown
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ALGEBRAIC EXPRESSIONS
The product of 3 and x
The sum of m and 8
The difference of r and 2
3x
m + 8
r - 2
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Write an algebraic expression �for this statement:
Three more than four times a number
3
4
?
+
x
3 + 4 x n
3 + 4n
n
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Write an algebraic EQUATION�for this statement:
A number divided by four is 5
n
4
÷
=
n ÷ 4 = 5
5
n = 5
4
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WHERE DO I PUT THE VARIABLE?
The product of 5 and a number c
Seven times a number t
6 multiplied by a number d
5c
7t
6d
Always write the variable AFTER the number.
(COEFFICIENT)
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EVALUATE
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LET
THEN
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EVALUATE
Evaluate the expression if a = 3 and b = 4
1) 3a + b
2) 2b − a
LET
THEN
LET
a = 3 and b= 4
THEN
3a + b
3(3) + 4
9 + 4
13
LET
a = 3 and b= 4
THEN
2b - a
2(4) - 3
8 - 3
5
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THEN:
LINE IT UP (Line up your equal signs)
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TRY IT
Evaluate the expression if a = 3 and b = 4
3) 2ab
Evaluate the expression if x = 5 and y = 3
5) 3x
y
4) 5 + ab
6) x2 + y2
LET a = 3 and b= 4
THEN
2ab
2(3)(4)
6 x 4
24
LET a = 3 and b= 4
THEN
5 + ab
5 + (3)(4)
5 + 12
17
LET x = 5 and y = 3
THEN
3x
y
3(5)
3
15
3
5
LET x = 5 and y = 3
THEN
x2 + y2
52 + 32
(5x5) + (3*3)
25 + 9
34
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LINEAR PATTERNS
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A pattern is a group of numbers, shapes, or objects that follow a rule while repeating or changing.
To extend a pattern you can use a table or a pattern rule that relates the term number to the pattern rule.
LINEAR PATTERNS
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LINEAR PATTERNS
A relationship between 2 variables that form a straight line when graphed
Linear patterns can be written as algebraic expressions to learn their “PATTERN RULE”
LINEAR PATTERNS
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8, 12, 16, 20…
2 VERY VEEEERY IMPORANT TERMS:
LINEAR PATTERNS
1
2
3
4
8
12
16
20
Term Number (n)
Term Value (v)
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8, 12, 16, 20…
2 VERY VEEEERY IMPORANT TERMS:
LINEAR PATTERNS
1
2
3
4
8
12
16
20
Term Number (n)
Term Value (v)
n | v |
| |
| |
| |
| |
| |
1
8
2
12
3
16
4
20
5
?
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PATTERN RULE of a LINEAR PATTERN
v=2n+3
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Determine the first 4 terms of this linear pattern given this pattern rule: v= 2n+3
v=2n+3
__, __, __, __…
n | v |
| |
| |
| |
| |
1
?
2
?
3
?
4
?
1
2
3
4
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Determine the first 4 terms of this linear pattern given this pattern rule: v= 2n+3
__, __, __, __…
n | v |
| |
| |
| |
| |
1
?
2
?
3
?
4
?
1
2
3
4
LET n= 1
THEN
v=2n+3
v=2(1)+3
v=2+3
V=5
5
5
LET n= 2
THEN
v=2n+3
v=2(2)+3
v=4+3
V=7
7
7
LET n= 3
THEN
v=2n+3
v=2(3)+3
v=6+3
V=9
9
9
LET n= 4
THEN
v=2n+3
v=2(4)+3
v=8+3
V=11
11
11
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Determine the first 4 terms of this linear pattern given this pattern rule: v= 8n-6
__, __, __, __…
n | v |
| |
| |
| |
| |
1
?
2
?
3
?
4
?
1
2
3
4
LET n= 1
THEN
v=8n-6
v=8(1)-6
v=8-6
V=2
2
2
LET n= 2
THEN
v=8n-6
v=8(2)-6
v=16-6
V=10
10
10
LET n= 3
THEN
v=8n-6
v=8(3)-6
v=24-6
V=18
18
18
LET n= 4
THEN
v=8n-6
v=8(4)-6
v=32-6
V=26
26
26
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Determine the first 4 terms of this linear pattern given this pattern rule: v= 8n-6
__, __, __, __…
n | v |
| |
| |
| |
| |
1
?
2
?
3
?
4
?
1
2
3
4
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LINEAR PATTERNS
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Area of a circle’s formula is π r to the power of 2. The pi symbol stands for 3.14159 which you have to multiply it with the radius which is the r. But because we are finding the area of the circle, it has to be the radius to the power of 2. The power of 2 means to multiply the number by itself. For example, if the radius of the circle you wanted to find the area of was 14, you would have it to the power of 2 which to find the answer of the power, you would do 14 x 14. For example, here is how you would find the area of a circle.
Circle’s radius is 8 cm.
a= πr²
a=(3.14159) 8²
a= (3.14159) 8 x 8
a= (3.14159) 64
a= 201.06 cm 2
You always have to keep the equal signs in a straight line down. On the first line, you right the formula, which is pi times radius, to the power of 2. Right below that you right out what the formula’s numbers are. Which in this case the radius is 8 so you would but 8 to the power of 2. Pi is always the same for every circle, so it’s 3.14159. The third line you put the pi in brackets, but because the radius has to be to the power of 2, you multiply the radius by itself so it would be 8 x 8. The fourth line you put the pi in brackets once again and you just multiply to two numbers together, which is radius. So 8 x 8 equals 64, so you would write 64. The last line you write down the answer of 3.14159 x 64, which will add up to 201.06. But on the last line, after you put the answer you write the units and squared. If the circle's information only gives you the diameter, you only can do it with the radius, so you would have to add divided by 2 at the end of the formula. We divided the diameter into 2 because the radius is half of the diameter.
For example:
The Circle's diameter is 20 m.
a= πr² ÷ 2
a= (3.14159) 20² ÷ 2
a= (3.14159) 20 x 20 ÷ 2
a= (3.14159) 400 ÷ 2
a= 1,256.63 m² ÷2
a= 628.31 m² �
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Area of a circle’s formula is π r to the power of 2. The pi symbol stands for 3.14159 which you have to multiply it with the radius which is the r. But because we are finding the area of the circle, it has to be the radius to the power of 2. The power of 2 means to multiply the number by itself. For example, if the radius of the circle you wanted to find the area of was 14, you would have it to the power of 2 which to find the answer of the power, you would do 14 x 14. For example, here is how you would find the area of a circle.
Circle’s radius is 8 cm.
a= πr²
a=(3.14159) 8²
a= (3.14159) 8 x 8
a= (3.14159) 64
a= 201.06 cm 2
MRSTAV
You always have to keep the equal signs in a straight line down. On the first line, you right the formula, which is pi times radius, to the power of 2. Right below that you right out what the formula’s numbers are. Which in this case the radius is 8 so you would but 8 to the power of 2. Pi is always the same for every circle, so it’s 3.14159. The third line you put the pi in brackets, but because the radius has to be to the power of 2, you multiply the radius by itself so it would be 8 x 8. The fourth line you put the pi in brackets once again and you just multiply to two numbers together, which is radius. So 8 x 8 equals 64, so you would write 64. The last line you write down the answer of 3.14159 x 64, which will add up to 201.06. But on the last line, after you put the answer you write the units and squared. If the circle's information only gives you the diameter, you only can do it with the radius, so you would have to add divided by 2 at the end of the formula. We divided the diameter into 2 because the radius is half of the diameter. For example:
The Circle's diameter is 20 m.
a= πr² ÷ 2
a= (3.14159) 20² ÷ 2
a= (3.14159) 20 x 20 ÷ 2
a= (3.14159) 400 ÷ 2
a= 1,256.63 m² ÷2
a= 628.31 m² �
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ALGEBRAIC �EXPRESSION
5, 7, 9, 11, ?...
| |
| |
| |
| |
| |
| |
1
2
3
5
7
9
n
v
4
11
100
?
1
2
3
4
100
2
2
2
FIRST
DIFFERENCE
V=
FIRST�DIF
n
+or- #
5 =
2
(1)
+or-?
5 =
2 +or- ?
7-5=
9-7=
11-9=
5 =
2 + 3
5 =
5
v = 2n + 3
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