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MATH. Part 1. INTEGERS. ADDITION - Same sign : add the numbers and copy the sign - Opposite signs : subtract the numbers and copy the sign of the larger number MULTIPLICATION/ DIVISION - Same sign : positive (+) - Opposite signs : negative (-). Order of Operations. GEMDAS

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### MATH

Part 1

INTEGERS
• ADDITION

- Same sign: add the numbers and copy the sign

- Opposite signs: subtract the numbers and copy the sign of the larger number

• MULTIPLICATION/ DIVISION

- Same sign: positive (+)

- Opposite signs: negative (-)

Order of Operations

GEMDAS

Groups, Exponent, Multiplication, Division, Addition, Subtraction

Prime and Composite Numbers

PRIME NUMBERS

- Number greater than 1 whose only divisor is 1 and the number itself

- example: 7 : 1, 7

COMPOSITE NUMBERS

- Any integer which has other divisors aside from 1 and the number itself

- example: 12: 1, 2, 3, 4, 6, 12

Fractions
• Addition/Subtraction

- Similar fractions: Add or subtract the numerators and copy the denominators.

- Dissimilar fractions: Get the Least Common Denominator (LCD) , get the equivalent fractions, add or subtract the numerators.

• Multiplication

- Multiply the numerators. Multiply the Denominators.

• Division

- Multiply the dividend by the reciprocal of the divisor

Decimals
• Addition/Subtraction

- Align the decimal points. Add zeroes so they have the same length. Add or subtract them.

- example:

12.34 + 5.6789 = ?

12.3400

+ 5.6789

18.0189

• Multiplication

- Multiply the numbers normally. Count the total number of decimal points of both numbers. This will determine the number of decimal places of the product.

- Example:

4.2 x 1.2 = ?

42

x 12

504

*2 decimal points

5.04

Variation
• Direct Variation

- quantities change in the same manner

-solved using proportionality or by division

- or

- example:

How many hours would a 675 mile trip by train take if a 45 mile trip takes 30 minutes?

45 mile : 30 minutes = 675 mile : x hr

30 minutes = ½ hr

45x = ½ (675)

x = 7.5 hr

Variation
• Inverse Variation

- quantities change in the opposite manner

- solve by multiplying the first quantity by the second and setting an equation for the different cases

- or

- example:

Fifty workers can assemble one lot of an item in one day(24 hr). If 25 more workers were hired, how many more hours would it take to assemble one lot?

50 (24 hr) = (50 + 25)(x hr)

x = 16

Converting Percent/Decimal/Fractions
• Percent  Decimal

- move the decimal point two places to the left

- example: 42%  0.42

• Decimal  Percent

- move the decimal point two places to the right

- example: 0.1314  13.14%

Converting Percent/Decimal/Fractions
• Fraction  Decimal

- divide the numerator by the denominator

- example: = 0.67

• Decimal  Fraction

- divide the decimal digits by the last decimal place

- example: 4.5 = 4 = 4

Converting Percent/Decimal/Fractions
• Percent  Fraction

- divide the percent value by 100

- example : 78%  =

• Fraction  Percent

- divide the numerator by the denominator, the answer should be in decimal form to be changed to percent OR

- multiply the fraction by 100, simplify then add the % sign

- example:

= 0.375(100) = 37.5%

Percentage Problems
• “of”  multiply or “times”
• “is”  equals
• “What number/Find”  x or any variable

*Always change % to decimals or fractions before solving

Example :

What is 35% of 720?

x = (720)

x = 252

Discounts
• Discount or Savings

= %Discount (Original Price)

• Discounted Price

= Original Price – Discount

= Original Price – (1 - %Discount)

- Example:

Tom paid only P1020 for a pair of running shoes that was marked 15% OFF. What was the original price?

1020 = x(1 – 0.15)

1020 = 0.85x

X= 1200

Discounts
• % Discount

= Discount/ Original Price

- example:

What was the percent discount if a souvenir item originally marked P75 was bought for only P68.25?

% Discount = 75 – 68.25

75

% discount = 9 %

Relations and Functions
• Relation – a set of ordered pairs
• Domain – a set of all first elements of the ordered pairs
• Range – a set of all second elements of the ordered pairs
• Function – a set of ordered pairs in which no two ordered pairs have the same first element
• Vertical Line Test – given a graph, a relation can be identified as a function if there is only one intersection

Operations on Functions- substitute the variable in the function with the value inside the parenthesis, simplify and then perform the indicated operation

• Addition

- (f + g)x = f(x) + g(x)

- example:

f(x) = 2x3 – 2x2 + 5; g(x) = x2 + x - 6

(f+g)x = (2x3 – 2x2 + 5) + (x2 + x – 6)

(f+g)x = 2x3-x2 + x – 1

• Subtraction

- (f - g)x = f(x) - g(x)

- example:

f(x) = 4x - 2; g(x) = x + 8

(f - g)x = (4x - 2) - ( x + 8)

(f - g)x = 3x - 10

Operations on Functions
• Multiplication

- (f • g)x = f(x) • g(x)

- example:

f(x) = x - 1; g(x) = x

(f • g)x = (x - 1) • (x)

(f • g)x = x2 – x

• Division

- ; g(x) ≠ 0

- example:

f(x) = x -2; g(x) = 3x - 5

Operations on Functions
• Composite Functions

- function within a function

= (f  g)x = f(g(x))

- example

f(x) = 2x + 1; g(x) = x2 - 1

(f  g)x = 2(x2 – 1) + 1

(f  g)x = 2x2 - 1

Domain and Range of a Function
• Restrictions on variables

- DENOMINATOR of a function should not be equal to zero

- RADICAND with an even number as its index should be always be greater than or equal to zero.

- example:

Find the domain of the function:

4x -2 ≠ 0

4x ≠ 2

x ≠ ½

Linear Equations
• Equations dealing with variables whose maximum exponent is 1

Equations involving 1 variable

- Combine similar terms and solve algebraically, using PEMDAS.

Example:

3x – 9 = 2x – 6

3x – 2x = -6+9

x = 3

Equations involving 2 variables

- Use elimination or substitution

Example:

2x + 2y = -2

x - y = 1

ELIMINATION METHOD

2x + 2y = -2

2(x – y = 1)

2x + 2y = -2

2x – 2y = 2

4x = 0

x = 0

0 – y = 1

y = -1

SUBSTITUTION METHOD

x = y + 1

2(y + 1) + 2y = -2

2y + 2 + 2y = -2

4y = -4

y = -1

x = -1 + 1

x = 0

Equations with 1 Variable with Absolute Value
• Isolate the absolute value group of the equation
• Remove the absolute value signs by equating the absolute value group to both the (+) and (-) of the constant on the other side and then solve the 2 resulting equations

Example:

|2x – 4| = 6

2x – 4 = 6  2x = 10  x = 5

2x – 4 = -6  2x = -2  x = -1