SlideShare a Scribd company logo
Order of Operations
If we have two $5-bills and two $10-bills,
Order of Operations
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars.
Order of Operations
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
Order of Operations
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
Order of Operations
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
If I have three $10-bills and you have four $10-bills,
Order of Operations
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
If I have three $10-bills and you have four $10-bills, we have
3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $.
Order of Operations
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
If I have three $10-bills and you have four $10-bills, we have
3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $.
In this case, we group the 3 + 4 in the “( )” to indicate that we
are to add them first,
Order of Operations
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
If I have three $10-bills and you have four $10-bills, we have
3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $.
In this case, we group the 3 + 4 in the “( )” to indicate that we
are to add them first, then multiply the sum to 10.
Order of Operations
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
If I have three $10-bills and you have four $10-bills, we have
3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $.
In this case, we group the 3 + 4 in the “( )” to indicate that we
are to add them first, then multiply the sum to 10.
Order of Operations
This motivates us to set the rules for the order of operations.
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
If I have three $10-bills and you have four $10-bills, we have
3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $.
In this case, we group the 3 + 4 in the “( )” to indicate that we
are to add them first, then multiply the sum to 10.
Order of Operations
Order of Operations (excluding raising power)
This motivates us to set the rules for the order of operations.
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
If I have three $10-bills and you have four $10-bills, we have
3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $.
In this case, we group the 3 + 4 in the “( )” to indicate that we
are to add them first, then multiply the sum to 10.
Order of Operations
Order of Operations (excluding raising power)
Given an arithmetic expression, we perform the operations in
the following order .
This motivates us to set the rules for the order of operations.
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
If I have three $10-bills and you have four $10-bills, we have
3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $.
In this case, we group the 3 + 4 in the “( )” to indicate that we
are to add them first, then multiply the sum to 10.
Order of Operations
Order of Operations (excluding raising power)
Given an arithmetic expression, we perform the operations in
the following order .
1st . Do the operations within grouping symbols, starting with
the innermost grouping symbol.
This motivates us to set the rules for the order of operations.
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
If I have three $10-bills and you have four $10-bills, we have
3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $.
In this case, we group the 3 + 4 in the “( )” to indicate that we
are to add them first, then multiply the sum to 10.
Order of Operations
Order of Operations (excluding raising power)
Given an arithmetic expression, we perform the operations in
the following order .
1st . Do the operations within grouping symbols, starting with
the innermost grouping symbol.
2nd. Do multiplications and divisions (from left to right).
This motivates us to set the rules for the order of operations.
If we have two $5-bills and two $10-bills, we have the total of
2(5) + 2(10) = 30 dollars. To get the correct answer 30,
we multiply the 2 and the 5 and multiply the 2 and the10 first,
then we add the products 10 and 20.
If I have three $10-bills and you have four $10-bills, we have
3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $.
In this case, we group the 3 + 4 in the “( )” to indicate that we
are to add them first, then multiply the sum to 10.
Order of Operations
Order of Operations (excluding raising power)
Given an arithmetic expression, we perform the operations in
the following order .
1st . Do the operations within grouping symbols, starting with
the innermost grouping symbol.
2nd. Do multiplications and divisions (from left to right).
3rd. Do additions and subtractions (from left to right).
This motivates us to set the rules for the order of operations.
Example A.
a. 4(–8) + 3(5)
Order of Operations
Example A.
a. 4(–8) + 3(5)
Order of Operations
Example A.
a. 4(–8) + 3(5)
= –32 + 15
Order of Operations
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
Order of Operations
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
Order of Operations
b. 4 + 3(5 + 2)
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
Order of Operations
b. 4 + 3(5 + 2)
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
= 25
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
c. 9 – 2[7 – 3(6 + 1)]
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
= 25
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
c. 9 – 2[7 – 3(6 + 1)]
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
= 25
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
c. 9 – 2[7 – 3(6 + 1)]
= 9 – 2[7 – 3(7)]
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
= 25
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
c. 9 – 2[7 – 3(6 + 1)]
= 9 – 2[7 – 3(7)]
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
= 25
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
c. 9 – 2[7 – 3(6 + 1)]
= 9 – 2[7 – 3(7)]
= 9 – 2[7 – 21]
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
= 25
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
c. 9 – 2[7 – 3(6 + 1)]
= 9 – 2[7 – 3(7)]
= 9 – 2[7 – 21]
= 9 – 2[ –14 ]
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
= 25
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
c. 9 – 2[7 – 3(6 + 1)]
= 9 – 2[7 – 3(7)]
= 9 – 2[7 – 21]
= 9 – 2[ –14 ]
= 9 + 28
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
= 25
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
c. 9 – 2[7 – 3(6 + 1)]
= 9 – 2[7 – 3(7)]
= 9 – 2[7 – 21]
= 9 – 2[ –14 ]
= 9 + 28
= 37
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
= 25
Example A.
a. 4(–8) + 3(5)
= –32 + 15
= –17
c. 9 – 2[7 – 3(6 + 1)]
= 9 – 2[7 – 3(7)]
= 9 – 2[7 – 21]
= 9 – 2[ –14 ]
= 9 + 28
= 37
(Don’t perform “4 + 3” or “9 – 2” in the above problems!!)
Order of Operations
b. 4 + 3(5 + 2)
= 4 + 3(7)
= 4 + 21
= 25
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Coefficients
Starting with 0, adding N copies
of x’s to 0 is written as Nx:
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Coefficients
Starting with 0, adding N copies
of x’s to 0 is written as Nx:
0 + x + x + x .. + x = Nx
N copies added
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Coefficients
Starting with 0, adding N copies
of x’s to 0 is written as Nx:
So 0 = 0x
0 + x + x + x .. + x = Nx
N copies added
0 + x = 1x
0 + x + x = 2x
0 + x + x + x = 3x
0 + x + x + x + x = 4x
.
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Coefficients
Starting with 0, adding N copies
of x’s to 0 is written as Nx:
So 0 = 0x
0 + x + x + x .. + x = Nx
N copies added
0 + x = 1x
0 + x + x = 2x
0 + x + x + x = 3x
The number N of added copies is called the coefficient.
0 + x + x + x + x = 4x
.
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Coefficients
Starting with 0, adding N copies
of x’s to 0 is written as Nx:
So 0 = 0x
0 + x + x + x .. + x = Nx
N copies added
0 + x = 1x
0 + x + x = 2x
0 + x + x + x = 3x
The number N of added copies is called the coefficient.
So the coefficient of 3x is 3.
0 + x + x + x + x = 4x
.
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Coefficients
Starting with 0, adding N copies
of x’s to 0 is written as Nx:
So 0 = 0x
0 + x + x + x .. + x = Nx
N copies added
0 + x = 1x
0 + x + x = 2x
0 + x + x + x = 3x
The number N of added copies is called the coefficient.
So the coefficient of 3x is 3.
Similarly ab + ab + ab + ab is 4ab, with coefficient 4,
0 + x + x + x + x = 4x
.
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Coefficients
Starting with 0, adding N copies
of x’s to 0 is written as Nx:
So 0 = 0x
0 + x + x + x .. + x = Nx
N copies added
0 + x = 1x
0 + x + x = 2x
0 + x + x + x = 3x
The number N of added copies is called the coefficient.
So the coefficient of 3x is 3.
Similarly ab + ab + ab + ab is 4ab, with coefficient 4,
and that 3(x + y) is (x + y) + (x + y) + (x + y).
0 + x + x + x + x = 4x
.
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Coefficients
Starting with 0, adding N copies
of x’s to 0 is written as Nx:
So 0 = 0x
0 + x + x + x .. + x = Nx
N copies added
0 + x = 1x
0 + x + x = 2x
0 + x + x + x = 3x
The number N of added copies is called the coefficient.
So the coefficient of 3x is 3.
Similarly ab + ab + ab + ab is 4ab, with coefficient 4,
and that 3(x + y) is (x + y) + (x + y) + (x + y). Note that 0x = 0.
0 + x + x + x + x = 4x
.
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Exponents
Starting with 1, multiplying N copies
of x’s to 1 is written as xN.
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Exponents
Starting with 1, multiplying N copies
of x’s to 1 is written as xN.
1* x * x * x…* x as xN
N copies of x’s
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Exponents
Starting with 1, multiplying N copies
of x’s to 1 is written as xN.
1* x * x * x…* x as xN
N copies of x’s
1 * x = x1
1 * x * x = x2
1 * x * x * x = x3
1 * x * x * x * x = x4
.
So
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Exponents
Starting with 1, multiplying N copies
of x’s to 1 is written as xN.
1* x * x * x…* x as xN
N copies of x’s
1 * x = x1
1 * x * x = x2
1 * x * x * x = x3
1 * x * x * x * x = x4
.
So 1 = x0 (x ≠ 0)
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Exponents
Starting with 1, multiplying N copies
of x’s to 1 is written as xN.
1* x * x * x…* x as xN
N copies of x’s
1 * x = x1
1 * x * x = x2
1 * x * x * x = x3
The number of multiplied copies N of xN is called the exponent.
1 * x * x * x * x = x4
.
So 1 = x0 (x ≠ 0)
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Exponents
Starting with 1, multiplying N copies
of x’s to 1 is written as xN.
1* x * x * x…* x as xN
N copies of x’s
1 * x = x1
1 * x * x = x2
1 * x * x * x = x3
The number of multiplied copies N of xN is called the exponent.
So the exponent of x3 is 3.
1 * x * x * x * x = x4
.
So 1 = x0 (x ≠ 0)
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Exponents
Starting with 1, multiplying N copies
of x’s to 1 is written as xN.
1* x * x * x…* x as xN
N copies of x’s
1 * x = x1
1 * x * x = x2
1 * x * x * x = x3
The number of multiplied copies N of xN is called the exponent.
So the exponent of x3 is 3.
An exponent applies only to the quantity directly under it.
1 * x * x * x * x = x4
.
So 1 = x0 (x ≠ 0)
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Exponents
Starting with 1, multiplying N copies
of x’s to 1 is written as xN.
So 1 = x0 (x ≠ 0)
1* x * x * x…* x as xN
N copies of x’s
1 * x = x1
1 * x * x = x2
1 * x * x * x = x3
The number of multiplied copies N of xN is called the exponent.
So the exponent of x3 is 3.
An exponent applies only to the quantity directly under it.
So ab3 = a*b*b*b and that (ab)3 = ab*ab*ab.
1 * x * x * x * x = x4
.
Exercise: Don’t do the part that you shouldn’t do!
1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4)
3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)]
Order of Operations
Ans: a. 18 b. 18 c. 23 4. 15
Exponents
Starting with 1, multiplying N copies
of x’s to 1 is written as xN.
So 1 = x0 (x ≠ 0)
1* x * x * x…* x as xN
N copies of x’s
1 * x = x1
1 * x * x = x2
1 * x * x * x = x3
The number of multiplied copies N of xN is called the exponent.
So the exponent of x3 is 3.
An exponent applies only to the quantity directly under it.
So ab3 = a*b*b*b and that (ab)3 = ab*ab*ab. Note that x0 =1.
1 * x * x * x * x = x4
.
Order of Operations
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
Order of Operations
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Order of Operations
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3)
Order of Operations
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
Order of Operations
b. Expand – 32
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
Order of Operations
b. Expand – 32
The base of the 2nd power is 3.
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
Order of Operations
b. Expand – 32
The base of the 2nd power is 3.
Hence – 32 means – (3*3)
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
Order of Operations
b. Expand – 32
The base of the 2nd power is 3.
Hence – 32 means – (3*3) = – 9
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
c. Expand (3*2)2 and simplify the answer.
Order of Operations
b. Expand – 32
The base of the 2nd power is 3.
Hence – 32 means – (3*3) = – 9
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
c. Expand (3*2)2 and simplify the answer.
The base for the 2nd power is (3*2).
Order of Operations
b. Expand – 32
The base of the 2nd power is 3.
Hence – 32 means – (3*3) = – 9
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
c. Expand (3*2)2 and simplify the answer.
The base for the 2nd power is (3*2).
Hence(3*2)2 is (3*2)(3*2)
Order of Operations
b. Expand – 32
The base of the 2nd power is 3.
Hence – 32 means – (3*3) = – 9
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
c. Expand (3*2)2 and simplify the answer.
The base for the 2nd power is (3*2).
Hence(3*2)2 is (3*2)(3*2) = (6)(6) = 36
Order of Operations
b. Expand – 32
The base of the 2nd power is 3.
Hence – 32 means – (3*3) = – 9
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
c. Expand (3*2)2 and simplify the answer.
The base for the 2nd power is (3*2).
Hence(3*2)2 is (3*2)(3*2) = (6)(6) = 36
d. Expand 3*22 and simplify the answer.
Order of Operations
b. Expand – 32
The base of the 2nd power is 3.
Hence – 32 means – (3*3) = – 9
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
c. Expand (3*2)2 and simplify the answer.
The base for the 2nd power is (3*2).
Hence(3*2)2 is (3*2)(3*2) = (6)(6) = 36
d. Expand 3*22 and simplify the answer.
The base for the 2nd power is 2.
Order of Operations
b. Expand – 32
The base of the 2nd power is 3.
Hence – 32 means – (3*3) = – 9
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
c. Expand (3*2)2 and simplify the answer.
The base for the 2nd power is (3*2).
Hence(3*2)2 is (3*2)(3*2) = (6)(6) = 36
d. Expand 3*22 and simplify the answer.
The base for the 2nd power is 2.
Hence 3*22 means 3*2*2
Order of Operations
b. Expand – 32
The base of the 2nd power is 3.
Hence – 32 means – (3*3) = – 9
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
c. Expand (3*2)2 and simplify the answer.
The base for the 2nd power is (3*2).
Hence(3*2)2 is (3*2)(3*2) = (6)(6) = 36
d. Expand 3*22 and simplify the answer.
The base for the 2nd power is 2.
Hence 3*22 means 3*2*2 = 12
Order of Operations
b. Expand – 32
The base of the 2nd power is 3.
Hence – 32 means – (3*3) = – 9
Example B. (Exponential Notation)
a. Expand (–3)2 and simplify the answer.
The base is (–3).
Hence (–3)2 is (–3)(–3) = 9
Order of Operations
e. Expand (–3y)3 and simplify the answer.
Order of Operations
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y)
Order of Operations
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
Order of Operations
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
= –(3)(3)(3)(y)(y)(y)
Order of Operations
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
= –(3)(3)(3)(y)(y)(y)
= –27y3
Order of Operations
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
= –(3)(3)(3)(y)(y)(y)
= –27y3
From part b above, we see that the power is to be carried out
before multiplication.
Order of Operations
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
= –(3)(3)(3)(y)(y)(y)
= –27y3
From part b above, we see that the power is to be carried out
before multiplication. Below is the complete rules of order of
operations.
Order of Operations
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
= –(3)(3)(3)(y)(y)(y)
= –27y3
Order of Operations (PEMDAS)
From part b above, we see that the power is to be carried out
before multiplication. Below is the complete rules of order of
operations.
Order of Operations
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
= –(3)(3)(3)(y)(y)(y)
= –27y3
Order of Operations (PEMDAS)
1st. (Parenthesis) Do the operations within grouping symbols,
starting with the innermost one.
From part b above, we see that the power is to be carried out
before multiplication. Below is the complete rules of order of
operations.
Order of Operations
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
= –(3)(3)(3)(y)(y)(y)
= –27y3
Order of Operations (PEMDAS)
1st. (Parenthesis) Do the operations within grouping symbols,
starting with the innermost one.
2nd. (Exponents) Do the exponentiation
From part b above, we see that the power is to be carried out
before multiplication. Below is the complete rules of order of
operations.
Order of Operations
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
= –(3)(3)(3)(y)(y)(y)
= –27y3
Order of Operations (PEMDAS)
1st. (Parenthesis) Do the operations within grouping symbols,
starting with the innermost one.
2nd. (Exponents) Do the exponentiation
3rd. (Multiplication and Division) Do multiplications and
divisions in order from left to right.
From part b above, we see that the power is to be carried out
before multiplication. Below is the complete rules of order of
operations.
Order of Operations
e. Expand (–3y)3 and simplify the answer.
(–3y)3
= (–3y)(–3y)(–3y) (the product of three negatives number is negative)
= –(3)(3)(3)(y)(y)(y)
= –27y3
Order of Operations (PEMDAS)
1st. (Parenthesis) Do the operations within grouping symbols,
starting with the innermost one.
2nd. (Exponents) Do the exponentiation
3rd. (Multiplication and Division) Do multiplications and
divisions in order from left to right.
4th. (Addition and Subtraction) Do additions and
subtractions in order from left to right.
From part b above, we see that the power is to be carried out
before multiplication. Below is the complete rules of order of
operations.
Example C. Order of Operations
a. 52 – 32
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
= – 4
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
= – 4
c. –2*32 + (2*3)2
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
= – 4
c. –2*32 + (2*3)2
= –2*9 + (6)2
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
= – 4
c. –2*32 + (2*3)2
= –2*9 + (6)2
= –18 + 36
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
= – 4
c. –2*32 + (2*3)2
= –2*9 + (6)2
= –18 + 36
= 18
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
= – 4
c. –2*32 + (2*3)2
= –2*9 + (6)2
= –18 + 36
= 18
d. –32 – 5(3 – 6)2
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
= – 4
c. –2*32 + (2*3)2
= –2*9 + (6)2
= –18 + 36
= 18
d. –32 – 5(3 – 6)2
= –9 – 5(–3)2
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
= – 4
c. –2*32 + (2*3)2
= –2*9 + (6)2
= –18 + 36
= 18
d. –32 – 5(3 – 6)2
= –9 – 5(–3)2
= –9 – 5(9)
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
= – 4
c. –2*32 + (2*3)2
= –2*9 + (6)2
= –18 + 36
= 18
d. –32 – 5(3 – 6)2
= –9 – 5(–3)2
= –9 – 5(9)
= –9 – 45
Order of Operations
Example C. Order of Operations
a. 52 – 32
= 25 – 9
= 16
b. – (5 – 3)2
= – (2)2
= – 4
c. –2*32 + (2*3)2
= –2*9 + (6)2
= –18 + 36
= 18
d. –32 – 5(3 – 6)2
= –9 – 5(–3)2
= –9 – 5(9)
= –9 – 45 = –54
Order of Operations
Make sure that you interpret the operations correctly.
Exercise A. Calculate the following expressions.
Order of Operations
7. 1 + 2(3) 8. 4 – 5(6) 9. 7 – 8(–9)
1. 3(–3) 2. (3) – 3 3. 3 – 3(3) 4. 3(–3) + 3
5. +3(–3)(+3) 6. 3 + (–3)(+3)
B.Make sure that you don’t do the ± too early.
10. 1 + 2(3 – 4) 11. 5 – 6(7 – 8) 12. (4 – 3)2 + 1
13. [1 – 2(3 – 4)] – 2 14. 6 + [5 + 6(7 – 8)](+5)
15. 1 + 2[1 – 2(3 + 4)] 16. 5 – 6[5 – 6(7 – 8)]
17. 1 – 2[1 – 2(3 – 4)] 18. 5 + 6[5 + 6(7 – 8)]
19. (1 + 2)[1 – 2(3 + 4)] 20. (5 – 6)[5 – 6(7 – 8)]
C.Make sure that you apply the powers to the correct bases.
23. (–2)2 and –22 24 (–2)3 and –23 25. (–2)4 and –24
26. (–2)5 and –25 27. 2*32 28. (2*3)2
21. 1 – 2(–3)(–4) 22. (–5)(–6) – (–7)(–8)
Order of Operations
D.Make sure that you apply the powers to the correct bases.
29. (2)2 – 3(2) + 1 30. 3(–2)2 + 4(–2) – 1
31. –2(3)2 + 3(3) – 5 32. –3(–1)2 + 4(–1) – 4
33. 3(–2)3 – 4(–2)2 – 1 34. (2)3 – 3(2)2 + 4(2) – 1
35. 2(–1)3 – 3(–1)2 + 4(–1) – 1 36. –3(–2)3 – 4(–2)2 – 4(–2) – 3
37. (6 + 3)2 38. 62 + 32 39. (–4 + 2)3 40. (–4)3 + (2)3
E. Calculate.
41. 72 – 42 42. (7 + 4)(7 – 4 )
43. (– 5)2 – 32 44. (–5 + 3)(–5 – 3 )
45. 53 – 33 46. (5 – 3) (52 + 5*3 + 32)
47. 43 + 23 48. (4 + 2)(42 – 4*2 + 22)
7 – (–5)
5 – 353.
8 – 2
–6 – (–2)
54.
49. (3)2 – 4(2)(3) 50. (3)2 – 4(1)(– 4)
51. (–3)2 – 4(–2)(3) 52. (–2)2 – 4(–1)(– 4)
(–4) – (–8)
(–5) – 3
55.
(–7) – (–2)
(–3) – (–6)
56.

More Related Content

What's hot

Integers slidecast
Integers slidecastIntegers slidecast
Integers slidecast
millergjtime
 
1 s2 addition and subtraction of signed numbers
1 s2 addition and subtraction of signed numbers1 s2 addition and subtraction of signed numbers
1 s2 addition and subtraction of signed numbers
math123a
 
Ch 1 Final 10Math.pdf
Ch 1 Final 10Math.pdfCh 1 Final 10Math.pdf
Ch 1 Final 10Math.pdf
HabibDawar3
 
Tugas 5.3 kalkulus integral
Tugas 5.3 kalkulus integralTugas 5.3 kalkulus integral
Tugas 5.3 kalkulus integral
Nurkhalifah Anwar
 
Mathematical Operational Order
Mathematical Operational OrderMathematical Operational Order
Mathematical Operational Order
ficklenm
 
Prefix Sum Algorithm | Prefix Sum Array Implementation | EP2
Prefix Sum Algorithm | Prefix Sum Array Implementation | EP2Prefix Sum Algorithm | Prefix Sum Array Implementation | EP2
Prefix Sum Algorithm | Prefix Sum Array Implementation | EP2
Kanahaiya Gupta
 
PM5006 Week 6
PM5006 Week 6PM5006 Week 6
PM5006 Week 6
Carlos Vázquez
 
01. integral fungsi aljabar
01. integral fungsi aljabar01. integral fungsi aljabar
01. integral fungsi aljabar
Hirwanto Iwan
 
BSC_Computer Science_Discrete Mathematics_Unit-I
BSC_Computer Science_Discrete Mathematics_Unit-IBSC_Computer Science_Discrete Mathematics_Unit-I
BSC_Computer Science_Discrete Mathematics_Unit-I
Rai University
 

What's hot (9)

Integers slidecast
Integers slidecastIntegers slidecast
Integers slidecast
 
1 s2 addition and subtraction of signed numbers
1 s2 addition and subtraction of signed numbers1 s2 addition and subtraction of signed numbers
1 s2 addition and subtraction of signed numbers
 
Ch 1 Final 10Math.pdf
Ch 1 Final 10Math.pdfCh 1 Final 10Math.pdf
Ch 1 Final 10Math.pdf
 
Tugas 5.3 kalkulus integral
Tugas 5.3 kalkulus integralTugas 5.3 kalkulus integral
Tugas 5.3 kalkulus integral
 
Mathematical Operational Order
Mathematical Operational OrderMathematical Operational Order
Mathematical Operational Order
 
Prefix Sum Algorithm | Prefix Sum Array Implementation | EP2
Prefix Sum Algorithm | Prefix Sum Array Implementation | EP2Prefix Sum Algorithm | Prefix Sum Array Implementation | EP2
Prefix Sum Algorithm | Prefix Sum Array Implementation | EP2
 
PM5006 Week 6
PM5006 Week 6PM5006 Week 6
PM5006 Week 6
 
01. integral fungsi aljabar
01. integral fungsi aljabar01. integral fungsi aljabar
01. integral fungsi aljabar
 
BSC_Computer Science_Discrete Mathematics_Unit-I
BSC_Computer Science_Discrete Mathematics_Unit-IBSC_Computer Science_Discrete Mathematics_Unit-I
BSC_Computer Science_Discrete Mathematics_Unit-I
 

Similar to 4 order of operations 125s

9 order of operations
9 order of operations9 order of operations
9 order of operations
elem-alg-sample
 
2.1 order of operations w
2.1 order of operations w2.1 order of operations w
2.1 order of operations w
Tzenma
 
Ns 1.2 add-subtracting integers
Ns 1.2  add-subtracting integersNs 1.2  add-subtracting integers
Ns 1.2 add-subtracting integers
david19961
 
order of operations
order of operationsorder of operations
order of operations
Mary Ann Joy Cedillo
 
GCSE-CompletingTheSquare.pptx
GCSE-CompletingTheSquare.pptxGCSE-CompletingTheSquare.pptx
GCSE-CompletingTheSquare.pptx
MitaDurenSawit
 
2.2 add real numbers day 1-2
2.2 add real numbers   day 1-22.2 add real numbers   day 1-2
2.2 add real numbers day 1-2
bweldon
 
math m1
math m1math m1
math m1
ZoRo Lossless
 
Intro adding integres
Intro adding integresIntro adding integres
Intro adding integres
Julie Ann Pelayo
 
Integers And Order of Operations
Integers And Order of OperationsIntegers And Order of Operations
Integers And Order of Operations
nickromero76
 
Add and subtract pos and neg numbers 4 parts
Add and subtract pos and neg numbers 4 partsAdd and subtract pos and neg numbers 4 parts
Add and subtract pos and neg numbers 4 parts
Melanie_Anderson
 
BIDMAS-4-Demonstration.pptx
BIDMAS-4-Demonstration.pptxBIDMAS-4-Demonstration.pptx
BIDMAS-4-Demonstration.pptx
GovindSharma606333
 
BIDMAS-4-Demonstration.pptx
BIDMAS-4-Demonstration.pptxBIDMAS-4-Demonstration.pptx
BIDMAS-4-Demonstration.pptx
sabreenTaj10
 
Order of operations
Order of operationsOrder of operations
Order of operations
sonnynexus
 
Reshma digital textbook
Reshma digital textbookReshma digital textbook
Reshma digital textbook
reshmatettikuzhi
 
Chapter 2
Chapter 2Chapter 2
Chapter 2
jenpenbrad
 
Ejercicios john rangel
Ejercicios john rangelEjercicios john rangel
Ejercicios john rangel
johndaddy
 
ppt math6 qt3-wk3.pptx
ppt math6 qt3-wk3.pptxppt math6 qt3-wk3.pptx
ppt math6 qt3-wk3.pptx
casafelicemcat04
 
Basic math akash
Basic math akashBasic math akash
Basic math akash
Akash Varaiya
 
Vii ch 1 integers
Vii  ch 1 integersVii  ch 1 integers
Vii ch 1 integers
AmruthaKB2
 
Factoring common monomial
Factoring common monomialFactoring common monomial
Factoring common monomial
AjayQuines
 

Similar to 4 order of operations 125s (20)

9 order of operations
9 order of operations9 order of operations
9 order of operations
 
2.1 order of operations w
2.1 order of operations w2.1 order of operations w
2.1 order of operations w
 
Ns 1.2 add-subtracting integers
Ns 1.2  add-subtracting integersNs 1.2  add-subtracting integers
Ns 1.2 add-subtracting integers
 
order of operations
order of operationsorder of operations
order of operations
 
GCSE-CompletingTheSquare.pptx
GCSE-CompletingTheSquare.pptxGCSE-CompletingTheSquare.pptx
GCSE-CompletingTheSquare.pptx
 
2.2 add real numbers day 1-2
2.2 add real numbers   day 1-22.2 add real numbers   day 1-2
2.2 add real numbers day 1-2
 
math m1
math m1math m1
math m1
 
Intro adding integres
Intro adding integresIntro adding integres
Intro adding integres
 
Integers And Order of Operations
Integers And Order of OperationsIntegers And Order of Operations
Integers And Order of Operations
 
Add and subtract pos and neg numbers 4 parts
Add and subtract pos and neg numbers 4 partsAdd and subtract pos and neg numbers 4 parts
Add and subtract pos and neg numbers 4 parts
 
BIDMAS-4-Demonstration.pptx
BIDMAS-4-Demonstration.pptxBIDMAS-4-Demonstration.pptx
BIDMAS-4-Demonstration.pptx
 
BIDMAS-4-Demonstration.pptx
BIDMAS-4-Demonstration.pptxBIDMAS-4-Demonstration.pptx
BIDMAS-4-Demonstration.pptx
 
Order of operations
Order of operationsOrder of operations
Order of operations
 
Reshma digital textbook
Reshma digital textbookReshma digital textbook
Reshma digital textbook
 
Chapter 2
Chapter 2Chapter 2
Chapter 2
 
Ejercicios john rangel
Ejercicios john rangelEjercicios john rangel
Ejercicios john rangel
 
ppt math6 qt3-wk3.pptx
ppt math6 qt3-wk3.pptxppt math6 qt3-wk3.pptx
ppt math6 qt3-wk3.pptx
 
Basic math akash
Basic math akashBasic math akash
Basic math akash
 
Vii ch 1 integers
Vii  ch 1 integersVii  ch 1 integers
Vii ch 1 integers
 
Factoring common monomial
Factoring common monomialFactoring common monomial
Factoring common monomial
 

More from Tzenma

6 slopes and difference quotient x
6 slopes and difference quotient x6 slopes and difference quotient x
6 slopes and difference quotient x
Tzenma
 
5 algebra of functions
5 algebra of functions5 algebra of functions
5 algebra of functions
Tzenma
 
4 graphs of equations conic sections-circles
4 graphs of equations conic sections-circles4 graphs of equations conic sections-circles
4 graphs of equations conic sections-circles
Tzenma
 
3 graphs of second degree functions x
3 graphs of second degree functions x3 graphs of second degree functions x
3 graphs of second degree functions x
Tzenma
 
2 graphs of first degree functions x
2 graphs of first degree functions x2 graphs of first degree functions x
2 graphs of first degree functions x
Tzenma
 
1 functions
1 functions1 functions
1 functions
Tzenma
 
9 rational equations word problems-x
9 rational equations word problems-x9 rational equations word problems-x
9 rational equations word problems-x
Tzenma
 
9 rational equations word problems-x
9 rational equations word problems-x9 rational equations word problems-x
9 rational equations word problems-x
Tzenma
 
7 proportions x
7 proportions x7 proportions x
7 proportions x
Tzenma
 
10 complex fractions x
10 complex fractions x10 complex fractions x
10 complex fractions x
Tzenma
 
6 addition and subtraction ii x
6 addition and subtraction ii x6 addition and subtraction ii x
6 addition and subtraction ii x
Tzenma
 
5 addition and subtraction i x
5 addition and subtraction i x5 addition and subtraction i x
5 addition and subtraction i x
Tzenma
 
4 the lcm and clearing denominators x
4 the lcm and clearing denominators x4 the lcm and clearing denominators x
4 the lcm and clearing denominators x
Tzenma
 
3 multiplication and division of rational expressions x
3 multiplication and division of rational expressions x3 multiplication and division of rational expressions x
3 multiplication and division of rational expressions x
Tzenma
 
2 cancellation x
2 cancellation x2 cancellation x
2 cancellation x
Tzenma
 
1 rational expressions x
1 rational expressions x1 rational expressions x
1 rational expressions x
Tzenma
 
8 linear word problems in x&y x
8 linear word problems in x&y x8 linear word problems in x&y x
8 linear word problems in x&y x
Tzenma
 
7 system of linear equations ii x
7 system of linear equations ii x7 system of linear equations ii x
7 system of linear equations ii x
Tzenma
 
6 system of linear equations i x
6 system of linear equations i x6 system of linear equations i x
6 system of linear equations i x
Tzenma
 
5 equations of lines x
5 equations of lines x5 equations of lines x
5 equations of lines x
Tzenma
 

More from Tzenma (20)

6 slopes and difference quotient x
6 slopes and difference quotient x6 slopes and difference quotient x
6 slopes and difference quotient x
 
5 algebra of functions
5 algebra of functions5 algebra of functions
5 algebra of functions
 
4 graphs of equations conic sections-circles
4 graphs of equations conic sections-circles4 graphs of equations conic sections-circles
4 graphs of equations conic sections-circles
 
3 graphs of second degree functions x
3 graphs of second degree functions x3 graphs of second degree functions x
3 graphs of second degree functions x
 
2 graphs of first degree functions x
2 graphs of first degree functions x2 graphs of first degree functions x
2 graphs of first degree functions x
 
1 functions
1 functions1 functions
1 functions
 
9 rational equations word problems-x
9 rational equations word problems-x9 rational equations word problems-x
9 rational equations word problems-x
 
9 rational equations word problems-x
9 rational equations word problems-x9 rational equations word problems-x
9 rational equations word problems-x
 
7 proportions x
7 proportions x7 proportions x
7 proportions x
 
10 complex fractions x
10 complex fractions x10 complex fractions x
10 complex fractions x
 
6 addition and subtraction ii x
6 addition and subtraction ii x6 addition and subtraction ii x
6 addition and subtraction ii x
 
5 addition and subtraction i x
5 addition and subtraction i x5 addition and subtraction i x
5 addition and subtraction i x
 
4 the lcm and clearing denominators x
4 the lcm and clearing denominators x4 the lcm and clearing denominators x
4 the lcm and clearing denominators x
 
3 multiplication and division of rational expressions x
3 multiplication and division of rational expressions x3 multiplication and division of rational expressions x
3 multiplication and division of rational expressions x
 
2 cancellation x
2 cancellation x2 cancellation x
2 cancellation x
 
1 rational expressions x
1 rational expressions x1 rational expressions x
1 rational expressions x
 
8 linear word problems in x&y x
8 linear word problems in x&y x8 linear word problems in x&y x
8 linear word problems in x&y x
 
7 system of linear equations ii x
7 system of linear equations ii x7 system of linear equations ii x
7 system of linear equations ii x
 
6 system of linear equations i x
6 system of linear equations i x6 system of linear equations i x
6 system of linear equations i x
 
5 equations of lines x
5 equations of lines x5 equations of lines x
5 equations of lines x
 

Recently uploaded

A Strategic Approach: GenAI in Education
A Strategic Approach: GenAI in EducationA Strategic Approach: GenAI in Education
A Strategic Approach: GenAI in Education
Peter Windle
 
PIMS Job Advertisement 2024.pdf Islamabad
PIMS Job Advertisement 2024.pdf IslamabadPIMS Job Advertisement 2024.pdf Islamabad
PIMS Job Advertisement 2024.pdf Islamabad
AyyanKhan40
 
Digital Artifact 1 - 10VCD Environments Unit
Digital Artifact 1 - 10VCD Environments UnitDigital Artifact 1 - 10VCD Environments Unit
Digital Artifact 1 - 10VCD Environments Unit
chanes7
 
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
National Information Standards Organization (NISO)
 
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
Levi Shapiro
 
A Survey of Techniques for Maximizing LLM Performance.pptx
A Survey of Techniques for Maximizing LLM Performance.pptxA Survey of Techniques for Maximizing LLM Performance.pptx
A Survey of Techniques for Maximizing LLM Performance.pptx
thanhdowork
 
PCOS corelations and management through Ayurveda.
PCOS corelations and management through Ayurveda.PCOS corelations and management through Ayurveda.
PCOS corelations and management through Ayurveda.
Dr. Shivangi Singh Parihar
 
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Dr. Vinod Kumar Kanvaria
 
Natural birth techniques - Mrs.Akanksha Trivedi Rama University
Natural birth techniques - Mrs.Akanksha Trivedi Rama UniversityNatural birth techniques - Mrs.Akanksha Trivedi Rama University
Natural birth techniques - Mrs.Akanksha Trivedi Rama University
Akanksha trivedi rama nursing college kanpur.
 
S1-Introduction-Biopesticides in ICM.pptx
S1-Introduction-Biopesticides in ICM.pptxS1-Introduction-Biopesticides in ICM.pptx
S1-Introduction-Biopesticides in ICM.pptx
tarandeep35
 
MARY JANE WILSON, A “BOA MÃE” .
MARY JANE WILSON, A “BOA MÃE”           .MARY JANE WILSON, A “BOA MÃE”           .
MARY JANE WILSON, A “BOA MÃE” .
Colégio Santa Teresinha
 
Top five deadliest dog breeds in America
Top five deadliest dog breeds in AmericaTop five deadliest dog breeds in America
Top five deadliest dog breeds in America
Bisnar Chase Personal Injury Attorneys
 
What is the purpose of studying mathematics.pptx
What is the purpose of studying mathematics.pptxWhat is the purpose of studying mathematics.pptx
What is the purpose of studying mathematics.pptx
christianmathematics
 
Thesis Statement for students diagnonsed withADHD.ppt
Thesis Statement for students diagnonsed withADHD.pptThesis Statement for students diagnonsed withADHD.ppt
Thesis Statement for students diagnonsed withADHD.ppt
EverAndrsGuerraGuerr
 
Lapbook sobre os Regimes Totalitários.pdf
Lapbook sobre os Regimes Totalitários.pdfLapbook sobre os Regimes Totalitários.pdf
Lapbook sobre os Regimes Totalitários.pdf
Jean Carlos Nunes Paixão
 
CACJapan - GROUP Presentation 1- Wk 4.pdf
CACJapan - GROUP Presentation 1- Wk 4.pdfCACJapan - GROUP Presentation 1- Wk 4.pdf
CACJapan - GROUP Presentation 1- Wk 4.pdf
camakaiclarkmusic
 
Advanced Java[Extra Concepts, Not Difficult].docx
Advanced Java[Extra Concepts, Not Difficult].docxAdvanced Java[Extra Concepts, Not Difficult].docx
Advanced Java[Extra Concepts, Not Difficult].docx
adhitya5119
 
MATATAG CURRICULUM: ASSESSING THE READINESS OF ELEM. PUBLIC SCHOOL TEACHERS I...
MATATAG CURRICULUM: ASSESSING THE READINESS OF ELEM. PUBLIC SCHOOL TEACHERS I...MATATAG CURRICULUM: ASSESSING THE READINESS OF ELEM. PUBLIC SCHOOL TEACHERS I...
MATATAG CURRICULUM: ASSESSING THE READINESS OF ELEM. PUBLIC SCHOOL TEACHERS I...
NelTorrente
 
RPMS TEMPLATE FOR SCHOOL YEAR 2023-2024 FOR TEACHER 1 TO TEACHER 3
RPMS TEMPLATE FOR SCHOOL YEAR 2023-2024 FOR TEACHER 1 TO TEACHER 3RPMS TEMPLATE FOR SCHOOL YEAR 2023-2024 FOR TEACHER 1 TO TEACHER 3
RPMS TEMPLATE FOR SCHOOL YEAR 2023-2024 FOR TEACHER 1 TO TEACHER 3
IreneSebastianRueco1
 
DRUGS AND ITS classification slide share
DRUGS AND ITS classification slide shareDRUGS AND ITS classification slide share
DRUGS AND ITS classification slide share
taiba qazi
 

Recently uploaded (20)

A Strategic Approach: GenAI in Education
A Strategic Approach: GenAI in EducationA Strategic Approach: GenAI in Education
A Strategic Approach: GenAI in Education
 
PIMS Job Advertisement 2024.pdf Islamabad
PIMS Job Advertisement 2024.pdf IslamabadPIMS Job Advertisement 2024.pdf Islamabad
PIMS Job Advertisement 2024.pdf Islamabad
 
Digital Artifact 1 - 10VCD Environments Unit
Digital Artifact 1 - 10VCD Environments UnitDigital Artifact 1 - 10VCD Environments Unit
Digital Artifact 1 - 10VCD Environments Unit
 
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
Pollock and Snow "DEIA in the Scholarly Landscape, Session One: Setting Expec...
 
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
 
A Survey of Techniques for Maximizing LLM Performance.pptx
A Survey of Techniques for Maximizing LLM Performance.pptxA Survey of Techniques for Maximizing LLM Performance.pptx
A Survey of Techniques for Maximizing LLM Performance.pptx
 
PCOS corelations and management through Ayurveda.
PCOS corelations and management through Ayurveda.PCOS corelations and management through Ayurveda.
PCOS corelations and management through Ayurveda.
 
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
Exploiting Artificial Intelligence for Empowering Researchers and Faculty, In...
 
Natural birth techniques - Mrs.Akanksha Trivedi Rama University
Natural birth techniques - Mrs.Akanksha Trivedi Rama UniversityNatural birth techniques - Mrs.Akanksha Trivedi Rama University
Natural birth techniques - Mrs.Akanksha Trivedi Rama University
 
S1-Introduction-Biopesticides in ICM.pptx
S1-Introduction-Biopesticides in ICM.pptxS1-Introduction-Biopesticides in ICM.pptx
S1-Introduction-Biopesticides in ICM.pptx
 
MARY JANE WILSON, A “BOA MÃE” .
MARY JANE WILSON, A “BOA MÃE”           .MARY JANE WILSON, A “BOA MÃE”           .
MARY JANE WILSON, A “BOA MÃE” .
 
Top five deadliest dog breeds in America
Top five deadliest dog breeds in AmericaTop five deadliest dog breeds in America
Top five deadliest dog breeds in America
 
What is the purpose of studying mathematics.pptx
What is the purpose of studying mathematics.pptxWhat is the purpose of studying mathematics.pptx
What is the purpose of studying mathematics.pptx
 
Thesis Statement for students diagnonsed withADHD.ppt
Thesis Statement for students diagnonsed withADHD.pptThesis Statement for students diagnonsed withADHD.ppt
Thesis Statement for students diagnonsed withADHD.ppt
 
Lapbook sobre os Regimes Totalitários.pdf
Lapbook sobre os Regimes Totalitários.pdfLapbook sobre os Regimes Totalitários.pdf
Lapbook sobre os Regimes Totalitários.pdf
 
CACJapan - GROUP Presentation 1- Wk 4.pdf
CACJapan - GROUP Presentation 1- Wk 4.pdfCACJapan - GROUP Presentation 1- Wk 4.pdf
CACJapan - GROUP Presentation 1- Wk 4.pdf
 
Advanced Java[Extra Concepts, Not Difficult].docx
Advanced Java[Extra Concepts, Not Difficult].docxAdvanced Java[Extra Concepts, Not Difficult].docx
Advanced Java[Extra Concepts, Not Difficult].docx
 
MATATAG CURRICULUM: ASSESSING THE READINESS OF ELEM. PUBLIC SCHOOL TEACHERS I...
MATATAG CURRICULUM: ASSESSING THE READINESS OF ELEM. PUBLIC SCHOOL TEACHERS I...MATATAG CURRICULUM: ASSESSING THE READINESS OF ELEM. PUBLIC SCHOOL TEACHERS I...
MATATAG CURRICULUM: ASSESSING THE READINESS OF ELEM. PUBLIC SCHOOL TEACHERS I...
 
RPMS TEMPLATE FOR SCHOOL YEAR 2023-2024 FOR TEACHER 1 TO TEACHER 3
RPMS TEMPLATE FOR SCHOOL YEAR 2023-2024 FOR TEACHER 1 TO TEACHER 3RPMS TEMPLATE FOR SCHOOL YEAR 2023-2024 FOR TEACHER 1 TO TEACHER 3
RPMS TEMPLATE FOR SCHOOL YEAR 2023-2024 FOR TEACHER 1 TO TEACHER 3
 
DRUGS AND ITS classification slide share
DRUGS AND ITS classification slide shareDRUGS AND ITS classification slide share
DRUGS AND ITS classification slide share
 

4 order of operations 125s

  • 2. If we have two $5-bills and two $10-bills, Order of Operations
  • 3. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. Order of Operations
  • 4. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, Order of Operations
  • 5. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, then we add the products 10 and 20. Order of Operations
  • 6. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, then we add the products 10 and 20. If I have three $10-bills and you have four $10-bills, Order of Operations
  • 7. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, then we add the products 10 and 20. If I have three $10-bills and you have four $10-bills, we have 3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $. Order of Operations
  • 8. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, then we add the products 10 and 20. If I have three $10-bills and you have four $10-bills, we have 3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $. In this case, we group the 3 + 4 in the “( )” to indicate that we are to add them first, Order of Operations
  • 9. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, then we add the products 10 and 20. If I have three $10-bills and you have four $10-bills, we have 3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $. In this case, we group the 3 + 4 in the “( )” to indicate that we are to add them first, then multiply the sum to 10. Order of Operations
  • 10. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, then we add the products 10 and 20. If I have three $10-bills and you have four $10-bills, we have 3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $. In this case, we group the 3 + 4 in the “( )” to indicate that we are to add them first, then multiply the sum to 10. Order of Operations This motivates us to set the rules for the order of operations.
  • 11. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, then we add the products 10 and 20. If I have three $10-bills and you have four $10-bills, we have 3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $. In this case, we group the 3 + 4 in the “( )” to indicate that we are to add them first, then multiply the sum to 10. Order of Operations Order of Operations (excluding raising power) This motivates us to set the rules for the order of operations.
  • 12. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, then we add the products 10 and 20. If I have three $10-bills and you have four $10-bills, we have 3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $. In this case, we group the 3 + 4 in the “( )” to indicate that we are to add them first, then multiply the sum to 10. Order of Operations Order of Operations (excluding raising power) Given an arithmetic expression, we perform the operations in the following order . This motivates us to set the rules for the order of operations.
  • 13. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, then we add the products 10 and 20. If I have three $10-bills and you have four $10-bills, we have 3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $. In this case, we group the 3 + 4 in the “( )” to indicate that we are to add them first, then multiply the sum to 10. Order of Operations Order of Operations (excluding raising power) Given an arithmetic expression, we perform the operations in the following order . 1st . Do the operations within grouping symbols, starting with the innermost grouping symbol. This motivates us to set the rules for the order of operations.
  • 14. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, then we add the products 10 and 20. If I have three $10-bills and you have four $10-bills, we have 3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $. In this case, we group the 3 + 4 in the “( )” to indicate that we are to add them first, then multiply the sum to 10. Order of Operations Order of Operations (excluding raising power) Given an arithmetic expression, we perform the operations in the following order . 1st . Do the operations within grouping symbols, starting with the innermost grouping symbol. 2nd. Do multiplications and divisions (from left to right). This motivates us to set the rules for the order of operations.
  • 15. If we have two $5-bills and two $10-bills, we have the total of 2(5) + 2(10) = 30 dollars. To get the correct answer 30, we multiply the 2 and the 5 and multiply the 2 and the10 first, then we add the products 10 and 20. If I have three $10-bills and you have four $10-bills, we have 3 + 4 = 7 $10-bills, and we have a total of (3 + 4)10 = 70 $. In this case, we group the 3 + 4 in the “( )” to indicate that we are to add them first, then multiply the sum to 10. Order of Operations Order of Operations (excluding raising power) Given an arithmetic expression, we perform the operations in the following order . 1st . Do the operations within grouping symbols, starting with the innermost grouping symbol. 2nd. Do multiplications and divisions (from left to right). 3rd. Do additions and subtractions (from left to right). This motivates us to set the rules for the order of operations.
  • 16. Example A. a. 4(–8) + 3(5) Order of Operations
  • 17. Example A. a. 4(–8) + 3(5) Order of Operations
  • 18. Example A. a. 4(–8) + 3(5) = –32 + 15 Order of Operations
  • 19. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 Order of Operations
  • 20. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 Order of Operations b. 4 + 3(5 + 2)
  • 21. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 Order of Operations b. 4 + 3(5 + 2)
  • 22. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7)
  • 23. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7) = 4 + 21
  • 24. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7) = 4 + 21 = 25
  • 25. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 c. 9 – 2[7 – 3(6 + 1)] Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7) = 4 + 21 = 25
  • 26. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 c. 9 – 2[7 – 3(6 + 1)] Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7) = 4 + 21 = 25
  • 27. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 c. 9 – 2[7 – 3(6 + 1)] = 9 – 2[7 – 3(7)] Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7) = 4 + 21 = 25
  • 28. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 c. 9 – 2[7 – 3(6 + 1)] = 9 – 2[7 – 3(7)] Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7) = 4 + 21 = 25
  • 29. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 c. 9 – 2[7 – 3(6 + 1)] = 9 – 2[7 – 3(7)] = 9 – 2[7 – 21] Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7) = 4 + 21 = 25
  • 30. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 c. 9 – 2[7 – 3(6 + 1)] = 9 – 2[7 – 3(7)] = 9 – 2[7 – 21] = 9 – 2[ –14 ] Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7) = 4 + 21 = 25
  • 31. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 c. 9 – 2[7 – 3(6 + 1)] = 9 – 2[7 – 3(7)] = 9 – 2[7 – 21] = 9 – 2[ –14 ] = 9 + 28 Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7) = 4 + 21 = 25
  • 32. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 c. 9 – 2[7 – 3(6 + 1)] = 9 – 2[7 – 3(7)] = 9 – 2[7 – 21] = 9 – 2[ –14 ] = 9 + 28 = 37 Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7) = 4 + 21 = 25
  • 33. Example A. a. 4(–8) + 3(5) = –32 + 15 = –17 c. 9 – 2[7 – 3(6 + 1)] = 9 – 2[7 – 3(7)] = 9 – 2[7 – 21] = 9 – 2[ –14 ] = 9 + 28 = 37 (Don’t perform “4 + 3” or “9 – 2” in the above problems!!) Order of Operations b. 4 + 3(5 + 2) = 4 + 3(7) = 4 + 21 = 25
  • 34. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Coefficients Starting with 0, adding N copies of x’s to 0 is written as Nx:
  • 35. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Coefficients Starting with 0, adding N copies of x’s to 0 is written as Nx: 0 + x + x + x .. + x = Nx N copies added
  • 36. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Coefficients Starting with 0, adding N copies of x’s to 0 is written as Nx: So 0 = 0x 0 + x + x + x .. + x = Nx N copies added 0 + x = 1x 0 + x + x = 2x 0 + x + x + x = 3x 0 + x + x + x + x = 4x .
  • 37. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Coefficients Starting with 0, adding N copies of x’s to 0 is written as Nx: So 0 = 0x 0 + x + x + x .. + x = Nx N copies added 0 + x = 1x 0 + x + x = 2x 0 + x + x + x = 3x The number N of added copies is called the coefficient. 0 + x + x + x + x = 4x .
  • 38. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Coefficients Starting with 0, adding N copies of x’s to 0 is written as Nx: So 0 = 0x 0 + x + x + x .. + x = Nx N copies added 0 + x = 1x 0 + x + x = 2x 0 + x + x + x = 3x The number N of added copies is called the coefficient. So the coefficient of 3x is 3. 0 + x + x + x + x = 4x .
  • 39. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Coefficients Starting with 0, adding N copies of x’s to 0 is written as Nx: So 0 = 0x 0 + x + x + x .. + x = Nx N copies added 0 + x = 1x 0 + x + x = 2x 0 + x + x + x = 3x The number N of added copies is called the coefficient. So the coefficient of 3x is 3. Similarly ab + ab + ab + ab is 4ab, with coefficient 4, 0 + x + x + x + x = 4x .
  • 40. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Coefficients Starting with 0, adding N copies of x’s to 0 is written as Nx: So 0 = 0x 0 + x + x + x .. + x = Nx N copies added 0 + x = 1x 0 + x + x = 2x 0 + x + x + x = 3x The number N of added copies is called the coefficient. So the coefficient of 3x is 3. Similarly ab + ab + ab + ab is 4ab, with coefficient 4, and that 3(x + y) is (x + y) + (x + y) + (x + y). 0 + x + x + x + x = 4x .
  • 41. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Coefficients Starting with 0, adding N copies of x’s to 0 is written as Nx: So 0 = 0x 0 + x + x + x .. + x = Nx N copies added 0 + x = 1x 0 + x + x = 2x 0 + x + x + x = 3x The number N of added copies is called the coefficient. So the coefficient of 3x is 3. Similarly ab + ab + ab + ab is 4ab, with coefficient 4, and that 3(x + y) is (x + y) + (x + y) + (x + y). Note that 0x = 0. 0 + x + x + x + x = 4x .
  • 42. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Exponents Starting with 1, multiplying N copies of x’s to 1 is written as xN.
  • 43. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Exponents Starting with 1, multiplying N copies of x’s to 1 is written as xN. 1* x * x * x…* x as xN N copies of x’s
  • 44. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Exponents Starting with 1, multiplying N copies of x’s to 1 is written as xN. 1* x * x * x…* x as xN N copies of x’s 1 * x = x1 1 * x * x = x2 1 * x * x * x = x3 1 * x * x * x * x = x4 . So
  • 45. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Exponents Starting with 1, multiplying N copies of x’s to 1 is written as xN. 1* x * x * x…* x as xN N copies of x’s 1 * x = x1 1 * x * x = x2 1 * x * x * x = x3 1 * x * x * x * x = x4 . So 1 = x0 (x ≠ 0)
  • 46. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Exponents Starting with 1, multiplying N copies of x’s to 1 is written as xN. 1* x * x * x…* x as xN N copies of x’s 1 * x = x1 1 * x * x = x2 1 * x * x * x = x3 The number of multiplied copies N of xN is called the exponent. 1 * x * x * x * x = x4 . So 1 = x0 (x ≠ 0)
  • 47. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Exponents Starting with 1, multiplying N copies of x’s to 1 is written as xN. 1* x * x * x…* x as xN N copies of x’s 1 * x = x1 1 * x * x = x2 1 * x * x * x = x3 The number of multiplied copies N of xN is called the exponent. So the exponent of x3 is 3. 1 * x * x * x * x = x4 . So 1 = x0 (x ≠ 0)
  • 48. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Exponents Starting with 1, multiplying N copies of x’s to 1 is written as xN. 1* x * x * x…* x as xN N copies of x’s 1 * x = x1 1 * x * x = x2 1 * x * x * x = x3 The number of multiplied copies N of xN is called the exponent. So the exponent of x3 is 3. An exponent applies only to the quantity directly under it. 1 * x * x * x * x = x4 . So 1 = x0 (x ≠ 0)
  • 49. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Exponents Starting with 1, multiplying N copies of x’s to 1 is written as xN. So 1 = x0 (x ≠ 0) 1* x * x * x…* x as xN N copies of x’s 1 * x = x1 1 * x * x = x2 1 * x * x * x = x3 The number of multiplied copies N of xN is called the exponent. So the exponent of x3 is 3. An exponent applies only to the quantity directly under it. So ab3 = a*b*b*b and that (ab)3 = ab*ab*ab. 1 * x * x * x * x = x4 .
  • 50. Exercise: Don’t do the part that you shouldn’t do! 1. 6 + 3(3 + 1) 2. 10 – 4(2 – 4) 3. 5 + 2[3 + 2(1 + 2)] 4. 5 – 2[3 + 2(5 – 9)] Order of Operations Ans: a. 18 b. 18 c. 23 4. 15 Exponents Starting with 1, multiplying N copies of x’s to 1 is written as xN. So 1 = x0 (x ≠ 0) 1* x * x * x…* x as xN N copies of x’s 1 * x = x1 1 * x * x = x2 1 * x * x * x = x3 The number of multiplied copies N of xN is called the exponent. So the exponent of x3 is 3. An exponent applies only to the quantity directly under it. So ab3 = a*b*b*b and that (ab)3 = ab*ab*ab. Note that x0 =1. 1 * x * x * x * x = x4 .
  • 51. Order of Operations Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer.
  • 52. Order of Operations Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3).
  • 53. Order of Operations Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3)
  • 54. Order of Operations Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 55. Order of Operations b. Expand – 32 Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 56. Order of Operations b. Expand – 32 The base of the 2nd power is 3. Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 57. Order of Operations b. Expand – 32 The base of the 2nd power is 3. Hence – 32 means – (3*3) Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 58. Order of Operations b. Expand – 32 The base of the 2nd power is 3. Hence – 32 means – (3*3) = – 9 Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 59. c. Expand (3*2)2 and simplify the answer. Order of Operations b. Expand – 32 The base of the 2nd power is 3. Hence – 32 means – (3*3) = – 9 Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 60. c. Expand (3*2)2 and simplify the answer. The base for the 2nd power is (3*2). Order of Operations b. Expand – 32 The base of the 2nd power is 3. Hence – 32 means – (3*3) = – 9 Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 61. c. Expand (3*2)2 and simplify the answer. The base for the 2nd power is (3*2). Hence(3*2)2 is (3*2)(3*2) Order of Operations b. Expand – 32 The base of the 2nd power is 3. Hence – 32 means – (3*3) = – 9 Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 62. c. Expand (3*2)2 and simplify the answer. The base for the 2nd power is (3*2). Hence(3*2)2 is (3*2)(3*2) = (6)(6) = 36 Order of Operations b. Expand – 32 The base of the 2nd power is 3. Hence – 32 means – (3*3) = – 9 Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 63. c. Expand (3*2)2 and simplify the answer. The base for the 2nd power is (3*2). Hence(3*2)2 is (3*2)(3*2) = (6)(6) = 36 d. Expand 3*22 and simplify the answer. Order of Operations b. Expand – 32 The base of the 2nd power is 3. Hence – 32 means – (3*3) = – 9 Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 64. c. Expand (3*2)2 and simplify the answer. The base for the 2nd power is (3*2). Hence(3*2)2 is (3*2)(3*2) = (6)(6) = 36 d. Expand 3*22 and simplify the answer. The base for the 2nd power is 2. Order of Operations b. Expand – 32 The base of the 2nd power is 3. Hence – 32 means – (3*3) = – 9 Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 65. c. Expand (3*2)2 and simplify the answer. The base for the 2nd power is (3*2). Hence(3*2)2 is (3*2)(3*2) = (6)(6) = 36 d. Expand 3*22 and simplify the answer. The base for the 2nd power is 2. Hence 3*22 means 3*2*2 Order of Operations b. Expand – 32 The base of the 2nd power is 3. Hence – 32 means – (3*3) = – 9 Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 66. c. Expand (3*2)2 and simplify the answer. The base for the 2nd power is (3*2). Hence(3*2)2 is (3*2)(3*2) = (6)(6) = 36 d. Expand 3*22 and simplify the answer. The base for the 2nd power is 2. Hence 3*22 means 3*2*2 = 12 Order of Operations b. Expand – 32 The base of the 2nd power is 3. Hence – 32 means – (3*3) = – 9 Example B. (Exponential Notation) a. Expand (–3)2 and simplify the answer. The base is (–3). Hence (–3)2 is (–3)(–3) = 9
  • 67. Order of Operations e. Expand (–3y)3 and simplify the answer.
  • 68. Order of Operations e. Expand (–3y)3 and simplify the answer. (–3y)3 = (–3y)(–3y)(–3y)
  • 69. Order of Operations e. Expand (–3y)3 and simplify the answer. (–3y)3 = (–3y)(–3y)(–3y) (the product of three negatives number is negative)
  • 70. Order of Operations e. Expand (–3y)3 and simplify the answer. (–3y)3 = (–3y)(–3y)(–3y) (the product of three negatives number is negative) = –(3)(3)(3)(y)(y)(y)
  • 71. Order of Operations e. Expand (–3y)3 and simplify the answer. (–3y)3 = (–3y)(–3y)(–3y) (the product of three negatives number is negative) = –(3)(3)(3)(y)(y)(y) = –27y3
  • 72. Order of Operations e. Expand (–3y)3 and simplify the answer. (–3y)3 = (–3y)(–3y)(–3y) (the product of three negatives number is negative) = –(3)(3)(3)(y)(y)(y) = –27y3 From part b above, we see that the power is to be carried out before multiplication.
  • 73. Order of Operations e. Expand (–3y)3 and simplify the answer. (–3y)3 = (–3y)(–3y)(–3y) (the product of three negatives number is negative) = –(3)(3)(3)(y)(y)(y) = –27y3 From part b above, we see that the power is to be carried out before multiplication. Below is the complete rules of order of operations.
  • 74. Order of Operations e. Expand (–3y)3 and simplify the answer. (–3y)3 = (–3y)(–3y)(–3y) (the product of three negatives number is negative) = –(3)(3)(3)(y)(y)(y) = –27y3 Order of Operations (PEMDAS) From part b above, we see that the power is to be carried out before multiplication. Below is the complete rules of order of operations.
  • 75. Order of Operations e. Expand (–3y)3 and simplify the answer. (–3y)3 = (–3y)(–3y)(–3y) (the product of three negatives number is negative) = –(3)(3)(3)(y)(y)(y) = –27y3 Order of Operations (PEMDAS) 1st. (Parenthesis) Do the operations within grouping symbols, starting with the innermost one. From part b above, we see that the power is to be carried out before multiplication. Below is the complete rules of order of operations.
  • 76. Order of Operations e. Expand (–3y)3 and simplify the answer. (–3y)3 = (–3y)(–3y)(–3y) (the product of three negatives number is negative) = –(3)(3)(3)(y)(y)(y) = –27y3 Order of Operations (PEMDAS) 1st. (Parenthesis) Do the operations within grouping symbols, starting with the innermost one. 2nd. (Exponents) Do the exponentiation From part b above, we see that the power is to be carried out before multiplication. Below is the complete rules of order of operations.
  • 77. Order of Operations e. Expand (–3y)3 and simplify the answer. (–3y)3 = (–3y)(–3y)(–3y) (the product of three negatives number is negative) = –(3)(3)(3)(y)(y)(y) = –27y3 Order of Operations (PEMDAS) 1st. (Parenthesis) Do the operations within grouping symbols, starting with the innermost one. 2nd. (Exponents) Do the exponentiation 3rd. (Multiplication and Division) Do multiplications and divisions in order from left to right. From part b above, we see that the power is to be carried out before multiplication. Below is the complete rules of order of operations.
  • 78. Order of Operations e. Expand (–3y)3 and simplify the answer. (–3y)3 = (–3y)(–3y)(–3y) (the product of three negatives number is negative) = –(3)(3)(3)(y)(y)(y) = –27y3 Order of Operations (PEMDAS) 1st. (Parenthesis) Do the operations within grouping symbols, starting with the innermost one. 2nd. (Exponents) Do the exponentiation 3rd. (Multiplication and Division) Do multiplications and divisions in order from left to right. 4th. (Addition and Subtraction) Do additions and subtractions in order from left to right. From part b above, we see that the power is to be carried out before multiplication. Below is the complete rules of order of operations.
  • 79. Example C. Order of Operations a. 52 – 32 Order of Operations
  • 80. Example C. Order of Operations a. 52 – 32 = 25 – 9 Order of Operations
  • 81. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 Order of Operations
  • 82. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 Order of Operations
  • 83. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 = – (2)2 Order of Operations
  • 84. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 = – (2)2 = – 4 Order of Operations
  • 85. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 = – (2)2 = – 4 c. –2*32 + (2*3)2 Order of Operations
  • 86. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 = – (2)2 = – 4 c. –2*32 + (2*3)2 = –2*9 + (6)2 Order of Operations
  • 87. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 = – (2)2 = – 4 c. –2*32 + (2*3)2 = –2*9 + (6)2 = –18 + 36 Order of Operations
  • 88. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 = – (2)2 = – 4 c. –2*32 + (2*3)2 = –2*9 + (6)2 = –18 + 36 = 18 Order of Operations
  • 89. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 = – (2)2 = – 4 c. –2*32 + (2*3)2 = –2*9 + (6)2 = –18 + 36 = 18 d. –32 – 5(3 – 6)2 Order of Operations
  • 90. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 = – (2)2 = – 4 c. –2*32 + (2*3)2 = –2*9 + (6)2 = –18 + 36 = 18 d. –32 – 5(3 – 6)2 = –9 – 5(–3)2 Order of Operations
  • 91. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 = – (2)2 = – 4 c. –2*32 + (2*3)2 = –2*9 + (6)2 = –18 + 36 = 18 d. –32 – 5(3 – 6)2 = –9 – 5(–3)2 = –9 – 5(9) Order of Operations
  • 92. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 = – (2)2 = – 4 c. –2*32 + (2*3)2 = –2*9 + (6)2 = –18 + 36 = 18 d. –32 – 5(3 – 6)2 = –9 – 5(–3)2 = –9 – 5(9) = –9 – 45 Order of Operations
  • 93. Example C. Order of Operations a. 52 – 32 = 25 – 9 = 16 b. – (5 – 3)2 = – (2)2 = – 4 c. –2*32 + (2*3)2 = –2*9 + (6)2 = –18 + 36 = 18 d. –32 – 5(3 – 6)2 = –9 – 5(–3)2 = –9 – 5(9) = –9 – 45 = –54 Order of Operations
  • 94. Make sure that you interpret the operations correctly. Exercise A. Calculate the following expressions. Order of Operations 7. 1 + 2(3) 8. 4 – 5(6) 9. 7 – 8(–9) 1. 3(–3) 2. (3) – 3 3. 3 – 3(3) 4. 3(–3) + 3 5. +3(–3)(+3) 6. 3 + (–3)(+3) B.Make sure that you don’t do the ± too early. 10. 1 + 2(3 – 4) 11. 5 – 6(7 – 8) 12. (4 – 3)2 + 1 13. [1 – 2(3 – 4)] – 2 14. 6 + [5 + 6(7 – 8)](+5) 15. 1 + 2[1 – 2(3 + 4)] 16. 5 – 6[5 – 6(7 – 8)] 17. 1 – 2[1 – 2(3 – 4)] 18. 5 + 6[5 + 6(7 – 8)] 19. (1 + 2)[1 – 2(3 + 4)] 20. (5 – 6)[5 – 6(7 – 8)] C.Make sure that you apply the powers to the correct bases. 23. (–2)2 and –22 24 (–2)3 and –23 25. (–2)4 and –24 26. (–2)5 and –25 27. 2*32 28. (2*3)2 21. 1 – 2(–3)(–4) 22. (–5)(–6) – (–7)(–8)
  • 95. Order of Operations D.Make sure that you apply the powers to the correct bases. 29. (2)2 – 3(2) + 1 30. 3(–2)2 + 4(–2) – 1 31. –2(3)2 + 3(3) – 5 32. –3(–1)2 + 4(–1) – 4 33. 3(–2)3 – 4(–2)2 – 1 34. (2)3 – 3(2)2 + 4(2) – 1 35. 2(–1)3 – 3(–1)2 + 4(–1) – 1 36. –3(–2)3 – 4(–2)2 – 4(–2) – 3 37. (6 + 3)2 38. 62 + 32 39. (–4 + 2)3 40. (–4)3 + (2)3 E. Calculate. 41. 72 – 42 42. (7 + 4)(7 – 4 ) 43. (– 5)2 – 32 44. (–5 + 3)(–5 – 3 ) 45. 53 – 33 46. (5 – 3) (52 + 5*3 + 32) 47. 43 + 23 48. (4 + 2)(42 – 4*2 + 22) 7 – (–5) 5 – 353. 8 – 2 –6 – (–2) 54. 49. (3)2 – 4(2)(3) 50. (3)2 – 4(1)(– 4) 51. (–3)2 – 4(–2)(3) 52. (–2)2 – 4(–1)(– 4) (–4) – (–8) (–5) – 3 55. (–7) – (–2) (–3) – (–6) 56.