Review of basic MathsReview of basic Maths(ctd) a) 0. ഥThis means x= 0.66 so we have one repeating...
Transcript of Review of basic MathsReview of basic Maths(ctd) a) 0. ഥThis means x= 0.66 so we have one repeating...
Review of basic Maths
1.Natural Numbers:
N = {1, 2, 3…..}
Closed for addition and multiplication
2.Whole Numbers:
N0 = {0, 1, 2, 3…..}
Closed for addition, multiplication
3.Integers:
Z = {…..-3, -2, -1, 0, 1, 2, 3…..}
Closed for addition, multiplication and subtraction but not division4.
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Review of basic Maths(ctd) 4.Rational Numbers:
• Q = {a/b : a, b, є Z; b ≠ 0}• e.g. a = 3 b = 4 a/b = ¾• closed for all four operations
have finite decimals or repeating decimals example: 2.6 , 2. 121212
Express each of the following decimals in the form of a/ba) 0.6b) 0. 35c) 0.585
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Review of basic Maths(ctd)
a) 0.ഥ𝟔 This means x= 0.66 so we have one repeating digit
i) x= 0.66ii)10x=6.6
10x-x= 6.6-0.6=9x=6X= 6/9 = 2/3
b) 0.𝟑𝟓 this means x=0.3535 so we have two repeating numbers
i) x=0.3535ii) 100x=35.3535
100x-1x=35.3535-0.353599x=35X=35/99
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Review of basic Maths(ctd) c) 0.𝟓𝟖𝟓 this means x=0.585585585 so we have two repeating numbers
i) x=0.585585585ii) 1000x=585.585585585
999x=585X=585/999x= 61/111
5.Irrational Numbers:
• Lie between the rational numbers• Have an infinite number of decimals• e.g. √2 = 1.4142135……..• e.g. π =3.1415………….[ NOTE: If the decimal place is finite or recurring the number is rational 3/4 = 0.7 ; 61/2 = 6.21
3= 0.3333= 0.ത3
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11=7.4545=7..45
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Review of basic Maths(ctd) 6.Real Numbers:
R = {Rational numbers} + {Irrational Numbers}Some properties of Real Numbers
1. The Transitive PropertyLet a,b and c be real numbers If a = b and b = c then a = c
2. Commutative Propertya+b = b+a ↔ ab = bacan add or multiply two real numbers in any order
3. Associative Property
a + (b+c) = (a+b) + c a(bc) = (ab)c
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Review of basic Maths(ctd)
4.Inverse Property
i) Additive Inverse
a + (-a) = 0
-6 + 6 = 0ii) Multiplicative Inverse
𝑎 ∗ 𝑎−1 = 𝑎 ∗1
𝑎= 1
3 ∗ 3−1 = 3 ∗1
3= 1
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Review of basic Maths(ctd)
Operations with signed numbers
2 – 7 = 2 + (-7) = -52 – (-7) = 2 +7 = 96 (7-2) = 6 (7) – 6(2) = 42 – 12 = 30-(7+2) = -7 - 2 = -9-(2-7) = -2 + 7 = 52(0) = 0
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Review of basic Maths(ctd)
Fundamental Principle of Fractions
𝑎
𝑏=𝑎
𝑏∗𝑐
𝑐= 𝑎𝑐
𝑏𝑐
7
1/8=7∗81
8∗8
= 56
1=56
The order of mathematical operationsI) BracketsII) ExponentsIII) Multiplication and division (equal status but perform operations from left to right)IV) Addition and subtraction (equal status but perform operations from left to right)
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Review of basic Maths(ctd)
e.g. 2 + 3 * (4/2 ÷ 8/16) – 10Treat item in brackets first:
4/2 ÷ 8/16 = 2 ÷ 1/2 = 2 * 2/1 = 4Next treat the multiplication:
2 + 3 x 4 – 10 = 2 + 12 – 10 = 4Note: 5 * 4 – 10 = 10 is WRONG
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Review of basic Maths(ctd) I) Exponent𝑎4 where 4 is the exponent and 𝑎 is the base 23= 2 x 2 x 2 = 8Special exponents:𝑎1 = a41 = 4𝑞0 = 140 = 1
II)Dot for multiplication:a . b ↔ a x b see also a*b
III) Coefficient:
3X where 3 is the coefficient and χ the variable2ab where 2 is the coefficient and ab the variables
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Review of basic Maths(ctd)
IV)Terms of an expression:
Terms are formed by positive (+) or negative (-)e.g. 2a + 3 is an expression with two terms
2.a.3 one term6χ4 – 5y + 8 three terms
5(3χ + 4) + 4(2 – 3y) two terms
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Review of basic Maths(ctd) V) Variables and constants:
Variable: -a symbol e.g. (x;y;p)-can be assigned any numerical valueConstants: -numbers with fixed value-e.g. 18; 2; 341
Note: χy = yχ but 25 ≠ 52
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Review of basic Maths(ctd) VI) Addition and subtraction of variables
Variables with the same powers can be added or subtractede.g. 7χ – 3χ = 4χ ; 7χ2 – 4χ2 = 3χ27𝑋2 – 3X cannot be simplified 6mn + 9mn = 15mn9a + 2ab cannot be simplified
Rules of Exponents
1. Multiplication Rule
𝑘𝑚* 𝑘𝑛 = 𝑘𝑚+𝑛
e.g. 24* 25 = 24+5
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Review of basic Maths(ctd)
2. Division Rule
𝑥𝑚 ÷ 𝑥𝑛 = 𝑥𝑚/ 𝑥𝑛 = 𝑥𝑚−𝑛
e.g. 57 ÷ 54 = 57/ 54 = 57−4
3. Involution Rule (Raising to a power)
(𝑘𝑚)𝑛 = 𝑘𝑚∗𝑛
e.g. (𝑥4)3 = 𝑥4∗3
Negative Exponents
𝑥−5= 1
𝑥5alternatively
1
𝑥−5=𝑥5
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Review of basic Maths(ctd) Fractions as exponents (also referred as roots and radicals)
𝑥1/2 = 𝑥Note: 𝑥 * 𝑥= 𝑥1/2*𝑥1/2
e.g. 𝑥1/4=4 𝑥
e.g. χ3/4 =𝑥3/4=4𝑥4
LOGARITHMS
Intuitive approach using examples16 = 42 where 2 is the log and 4 is the base , 2 is the log of 16 to the base of 416 = 24 4 is the log of 16 to the base of 2Note: → exponential form 16 = 42 → log form 𝑙𝑜𝑔416 = 2
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Review of basic Maths(ctd)
i) The natural number 10-the log key on your calculator
ii) The irrational number e = 2,7182818
-the ln key on your calculator some examples:e.g. 𝑙𝑜𝑔10100 = 2 ↔ 102 = 100
𝑙𝑜𝑔101000 = 3 ↔ 103 = 1000
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Review of basic Maths(ctd)
THE LAWS OF LOGARITHMSI)Log of a Productln a.b = ln a + ln be.g. ln 3χ2 = ln 3 + ln χ2
II)Log of a Quotientln (a/b) = ln a – ln be.g. ln (6/5χ) = ln 6 – ln 5χ = ln 6 – (ln 5 + lnχ)
iii) Log of a Powerln aχ = χ ln ae.g. ln χ4 = 4 ln χe.g. ln (3X)2 = 2 ln 3X = 2 (ln3 + lnX)
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Review of basic Maths(ctd)
Further Notes on Logarithms
•The log of 0 or a negative number (-) is undefined•The log of 1 to any base = 0
100 = 1 20 = 1 𝑒0 = 1
𝑙𝑜𝑔𝑚1 = 0 ↔ 𝑚0 = 11 cannot be a base of a log
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Review of basic Maths(ctd)
Some application examplesa) 𝑙𝑜𝑔3𝑥 = 4 solve for x34 = x81= x
c) 𝑙𝑜𝑔𝑥49 = 2𝑥2 = 4x= 7
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Review of basic Maths(ctd) Review of basic algebra Operations that guarantee equivalence
i) Adding (subtracting) the same polynomial to both sides of an equation
e.g 3χ = 5 – 6χ3χ + 6χ = 59χ = 59χ/9 = 5/9χ = 5/9
ii) Multiplying (dividing) both sides of an equation by the same constant except 0e.g 10χ = 510χ/10 = 5/10χ = 5/10
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Review of basic Maths(ctd)
(iii) Replacing either side of an equation by an equal expression
e. g χ3 + 4 = y and y = a2 + b + 4 then χ3 + 4 = a2 + b + 4
Linear Equations
An equation of degree 1ax + b = 0 a and b are constants
a) e.g. 5χ – 6 = 3χ 5χ – 3χ = 6
2χ = 6
2χ/2 = 6/2
χ = 3
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Review of basic Maths(ctd)
b) I =Prt solve for rI/Pt = Prt/Pt (divide through by Pt) r = I/Pt
c) S = P + Prt solve for PS = P (1 + rt) (factor out P)
S / (1 + rt) = P (1 + rt) / (1 + rt) (divide through by (1+ rt) P = S/(1+rt)
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• Applications
Application 1
A certain municipality has 1,000,000 population. Of this population, 20%are women of child bearing age. A woman of child bearing age in thismunicipality has on average 3 children. One of these 3 children attendhigh school and the rest attend primary school. The fees per yea forprimary school education is R3000 and the fee per year for secondaryschool education is R10,000
a) How many women of child bearing age are in the municipality
b) How many children attending primary school and how many attending high school in the municipality
c) What is the total fees of primary school and total fees of secondary per year in the municipality
d) What the total amount of school fees in the municipality23
Solutions
a) The number of women of child bearing age is
1,000,000*0.20=200,000 women
b) Because each woman has on average 3 children, one of whom
attending high school
The municipality has 2*200,000=400,000 children attending primary
school and 1*200,000=200,000 children attending high school
c) The total fees for primary school per year is obtained by multiplying
fees by the number of children in primary school
R3000*400,000=R1200,000
d) The total fees for high school per year in the municipality is obtained
by multiplying the school fees per year with the number of high school
children R10,000*200,000=
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FractionsObjectives pursued consists of having participants
Section 1
• Distinguish between various types of fractions
• Converting improper fractions to whole or mixed
numbers
• Converting mixed numbers to improper fractions
• Reducing fractions to lowest terms
• Raising fractions to higher terms
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Fractions(contd)
Section 2
• Adding fractions and mixed numbers
• Subtracting fractions and mixed numbers
• Multiplying fractions and mixed numbers
• Dividing fractions and mixed numbers
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Fraction (contd)
A mathematical way of expressing a part of a whole thing. ¼ is a
fraction expressing one part out of a total of four parts.
numerator
The number on top of the division line of a fraction. It
represents the dividend in the division. In the fraction ¼,
1 is the numerator.
denominator
• The number on the bottom of the division line of a
fraction. It represents the divisor in the division. In the
fraction ¼, 4 is the denominator
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Fraction (contd)
Various types of fractions
common or proper fraction
• A fraction in which the numerator is less than the
denominator. It represents less than a whole unit. The fraction
¼ is a common or proper fraction.
improper fraction
• A fraction in which the denominator is equal to or less than the
numerator. It represents one whole unit or more. The fraction 4/1
is an improper fraction.
mixed number
• A number that combines a whole number with a proper
fraction. The fraction 10¼ is a mixed number.
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Fraction (contd)
• To convert improper fractions to whole or mixed number
• STEP 1 Divide the numerator of the improper fraction by
the denominator.
• STEP 2a If there is no remainder, the improper fraction
becomes a whole number.
• STEP 2b If there is a remainder, write the whole number
and then write the fraction as
𝑤ℎ𝑜𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑟𝑒𝑚𝑖𝑛𝑎𝑑𝑛𝑒𝑟
𝑑𝑖𝑣𝑖𝑠𝑜𝑟
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Fraction (contd)
Converting improper fraction to whole and mixed numbers examples25
5=5
4
3=1
1
3
66
10=6
6
10
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Fraction (contd)
• Converting mixed numbers to improper fractions
STEP 1 Multiply the denominator by the whole number.
STEP 2 Add the numerator to the product from Step 1.
STEP 3 Place the total from Step 2 as the “new” numerator.
STEP 4 Place the original denominator as the “new”
Denominator
Examples:
3 2
5= 5∗3+2
5=
17
5
11
2=2∗1+1
2=
3
2
226
7=22∗7+6
7=
160
7
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Fraction (contd)
Reducing Fractions to Lowest Terms
reduce to lowest terms
The process of dividing whole numbers, known as common
divisors or common factors, into both the numerator and
the denominator of a fraction.
• Raising Fractions to Higher Terms
raise to higher terms
The process of multiplying the numerator and denominator
of a fraction by a common multiple.
Needed to have fractions with differing denominators to
have the same denominator
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Fraction (contd)
Examples
2
5=
?
10one does as follows
10÷5 =2, then 2*2 =4 so ?=4
The answer becomes:
2
5=4
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Adding up fractions:
Like fractions: these are fractions with the same
denominator but different numerators
Example: 2
7+5
7+
3
7+
1
7=11
7= 1
4
7
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Fraction (contd)
When adding like fractions you add up the numerators and
the denominator remain the same as for individual fractions
The same rule applies when subtracting fractions
unlike fractions: these are fractions with the different
denominators
Example
2
5+3
6+
1
4=
48+60+30
120=138
120=1
18
120=1
9
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Adding unlike fractions requires fining a common
denominator. The easy way to get it is multiply all the
denominators. The numerator is found by dividing the
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Fraction (contd)
common denominator by each individual denominators and
multiplying the result with the numerator of individual
fractions. Adding mixed numbers
STEP 1 Add the fractional parts. If the sum is an improper
fraction, convert it to a mixed number.
STEP 2 Add the whole numbers.
STEP 3 Add the fraction from Step 1 to the whole number
from Step 2.
STEP 4 Reduce the answer to lowest terms if necessary.
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Fraction (contd)
1
2*3
4=3
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To multiply mixed number, convert each mixed number into
an improper fraction before multiplying
Example:
21
2*3
3
4=
5
2*15
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Dividing fractions: multiply the first fraction with the inverse
of the second fractions
Example: 3
7÷*
4
9=
3
7*9
4
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Fraction (contd)
Fractions: applications
A certain ward received money from the government to be
spent on activities aimed at uplifting the community . The
municipality decided to spend 1/3 of the money on
assisting the disabled. It then spent 1/9 of the money on
the youth empowerment by informing them on how to get
jobs
a) What fraction of the money received did the municipality
use
b) If the municipality received R360,000 what amount was
spent on community uplifiment
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Fraction (contd)
c) How much was spend on the disabled
d) How much was spend on the youth empowerment
see solutions at the end of the notes
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Fraction (contd)
solutions
a)1
3+1
9=3
9+1
9=
4
9the fraction of the money is 4/9
b) The money spent on community upliftment was
R360,000* 4/9=160,000
c) The money spent on the disabled is
1/3*R360000=120,000
d) Money spend on the youth empowerment was R
1/9*360000= R40,000
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