DSC1520 NOTES WITH SOLVED PAST QUESTIONS.

DSC1520 NOTES WITH SOLVED PAST QUESTIONS. Numbers: different type of numbers – Natural, Real, etc. Also called constants • Basic operations o + (add); 2 + 3 = 5 o – (subtract); 3 – 2 = 1 o x (multiply); also •; 3 x 2 = 3•2 = 6 o ÷ (division) also / or fraction ( 1 2 = 1 divide by 2); 6 ÷ 3 = 6/3 = 6 3 = 2 Remember: 1 x anything = anything 1 x 8 = 8 0 x anything = 0 0 x 4 = 0 1 + anything = one more than anything 1 + 345 = 346 0 + anything = anything 0 + 34 = 34 anything ÷ 0 = not allowed 12 ÷ 0 = not allowed 0 ÷ anything = 0 0 ÷ 7 = 0 Page 1 of 167 email: Tel: 9 • Brackets ( ) : group operations together (3 + 4) – 3 = 7 – 3 = 4 • Order of operation: BODMAS Brackets; Of; Divide; Multiply; Add; Subtract 40 – 4 x (5 + 8) + 20 = 40 – 4 x (13) + 20 = 40 – 52 + 20 = 8 • Variables: used for unknown or generalisation of things: place holder: use alphabetic characters for example X or A or Y. Can take on different values 3x + 2y +7g + x 3x is known as a term with coefficient 3 and variable x Remember : the last term x has a coefficient value of 1 in front of it namely 1x o Operations on variables or unknown:  + and – : only if same variable, then + or – coefficients and variable stays the same 3x + 4x + 3 = (3 + 4)x + 3 = 7x + 3 5x – x – 6 = (5 – 1)x – 6 = 4x –6 Page 2 of 167 email: Tel: 9  x and ÷ : only if same variable, then x and ÷ coefficient and unknowns 3a x 4a = (3x4) (a x a) = 12a 2x 2 2 2 x x ÷ x = (2÷1) = 2 xx x = 2x • Laws of operations o Commutative law : order  a + b = b + a 3 + 4 = 4 + 3 = 7  a x b = b x a 3 × 4 = 4 × 3 = 12  a – b ≠ b – a 4 – 3 = 1 ≠ 3 – 4 = –1  a ÷ b ≠ b ÷ a 4 ÷ 2 = 2 ≠ 2 ÷ 4 = 0,5 o Associative law: ( )  (a + b) + c = a + (b + c) (3 + 4) + 2 = 7 + 2 = 9 3 + (4 + 2) = 3 + 6 = 9  (a x b) x c = a x (b x c) (3 × 4) ×2 = 12 × 2 = 24 3 × (4 × 2) = 3 × 8 = 24  (a – b) – c ≠ a – (b – c) (3 – 4) – 2 = –1 – 2 = –3 3 – (4 – 2) = 3 – 2 = 1  (a ÷ b) ÷ c ≠ a ÷ (b ÷ c) (12 ÷ 2) ÷ 2 = 6 ÷ 2 = 3 12 ÷ (2 ÷ 2) = 12 ÷ 1 = 12 Page 3 of 167 email: Tel: 9 o Distributive law (addition):  a x ( b + c) = ab + ac 3 × (4 + 2) = 3 × 6 = 18 (3 × 4) + (3 × 2) = 12 + 6 = 18 • Exponent or Power: (something)power: short way of writing something multiplied over and over with itself. The bottom number: base The top number: exponent or power 3 x 3 = 32 15 x 15 x 15 x 15 x 15 = 15 base = 3 power = 2 5 Y 3 = Y x Y x Y o Rules of exponents Let A and B be any two bases and x and y any two powers then 1. Ax x Ay = Ax + y 22 x 23 = 22+3 = 2 2 x 2 x 2 x 2 x 2 = 2 5 2. A 5 x ÷ Ay = Ax– y 24 ÷ 23 = 24-3 = 2 3. (A x B) 1 x = Ax x Bx (2 x 3)3 = 23 3 4. (A/B) 3 x = Ax / Bx (2 / 3)3 = 23 / 3 5. 3 ( )x y xy A A × = (23 ) 3 = 23x3 = 29 Page 4 of 167 email: Tel: 9 Remember : If a is any number 1. (a)0 = 1 but 00 = 0 40 2. (a) = 1 1 3. = a 1 n a = a–n 2/x4 = 2x 4. -4 1 n n a a = 1 2 24=24 5. a2x+4 = a15 (base the same) then 2x + 4 = 15 • Roots : Is the reverse of the power statement. 25 - what number must I multiply 2 times with itself to get an answer of 25 = 5 because 52 3 8 = 25 - what number must I multiply 3 times with itself to get an answer of 8 = 2 because 23 = 8 etc. • Simplify: write it another way • Solve for x: Determine an answer for x • Remember : When multiplying positive and negative numbers  – × – = +  – × + = –  + × – = –  + × + = + Page 5 of 167 email: Tel: 9 2. Fractions • Fraction is a part of a whole : like a slice of a pizza number of slices fraction = number of slices inwhole pizza numerator = denominator (name of fraction) For example 1 4 is one slice of a pizza consisting of 4 pieces. • Can only add and subtract “same pizzas” if not convert to “same pizzas” – common denominator 1 2 + = 4 4 + = or 1 2 1+2 3 += = 44 4 4 10 15 10+15 25 5 + = = =25÷20=1 2 1 20 5 25 1 +=+== 2 or 21 4 1 51 +=+== 2 1 1 2 6 15 20 41 11 + + = + + = =1 5 2 3 30 30 30 30 30 Page 6 of 167 email: Tel: 9 • If add or subtract whole and slices of pizzas : change whole pizzas to slices : b b (a c) b a a cc c × + = = 1 1 1 3 2 12 14 6 +1 = + = + = =1 • Multiply : multiply the numbers across the top lines and multiply the numbers across the bottom lines 1 2 1×2 2 ×= = 4 3 4×3 12 • Divide by = multiply by inverse of fraction 12133 ÷=×= 43428 No 1 of discussion class × + Page 7 of 167 email: Tel: 9 Question 1 Simplify 15213 66334 −÷+× Solution 15213 66334 15313 66234 1 15 3 6 12 12 2 15 3 2 15 3 10 12 5 6 −÷+× = −×+× ÷ = × =− + − + −+= = − = − Multiply fractions = Common denominator Add and subtract fractions Simplify by dividing nominator and denominator by 2 a c ad bd bc Page 8 of 167 email: Tel: 9 3. Solve equations in 1 variable • Move values so that unknown is on its own on one side of equation by +, - x or ÷ both sides with same values 4x + 7 = 14 – 3x + 5 4x + 7 + 3x = 14 – 3x +3x + 5 4x + 3x + 7 – 7 = 14 + 5 – 7 7x = 12 7x/7 = 12/7 x = 12/7 4. Simple inequalities • Equation if something = something • Inequality something or or ≥or ≤ • Use number line to demonstrate 3x + 20 14 – 16x 3x + 16x + 20 14 – 16x + 16x 19x + 20 – 20 14 – 20 19x –6 x –6/19 • When solving for x remember and change if multiply or divide by (–) value. –2x 4x + 4 –2x – 4x 4x – 4x + 4 –6x 4 –6x/–6 4/–6 x –4/6 No 2 of discussion class Page 9 of 167 email: Tel: 9 Question 2 Solve for x in 5 1 2 24 1 62 3 4 x x x x   − + + ≥− − − −     Solution 5 1 2 24 1 62 3 4 5 5 2 24 62 34 5 4 20 2 2 62 3 4 5 4 20 2 25 5 62 3 4 4 2 2 2 3   − + + ≥− − − −       − + + ≥− − − −       − + + ≥− + +     − + + ≥− + + = − + +− Change fraction Multiply 4 into ( ) Remember x x x x x x x x x x x x x x x x x x x x 5 5 6 4 55 2 3 16 3 8 30 5 6 6 5 25 6 6 5 6 25 6 61 61 5 ≥ − − ≥− − − ≥ − ≥ − ×≥ × − Move all the same terms to one side Common denominator Multiply both sides by 6 x x x x x x 25 5 5 25 5 5 5 ≥ − − ≤ − − ≤ − Divide both sides by Inequality sign changes because we divide by a negative number x x x Page 10 of 167 email: Tel: 9 5. Calculating percentages • Pizza with 100 slices • % = fraction = For example 70% = 70 100 • % always of something : 25% of 75 : of means multiply 15% of students at a university are male. How many male students are there in a total of 500 students? 15% of 500 15 100 x 500 = 75 The university expects a 10% increase in the number of students for the next year. How many students do they expect in total? New total = previous number + 10% of previous number = 500 + (10% x 500) = 500 + ( 10 100 x 500) = 500 + (50) = 550 % increase or decrease : value + (% increase of value) value + (% x value) value (1 + %) = 500 (0,1 + 1) = 500 (1,1) = 550 Page 11 of 167 email: Tel: 9 Study Unit 2 : Linear functions Chapter 2 : Sections 2.1 – 2.4 and 2.6 1. Function • Humans = relationships • Function = mathematical form of a relationship Temperature and number of ice cream sold • Independent variable – if variable : x • Dependent variable – then variable : y • Function of x : f(x) = y f(x) = 2x +3 or y = 2x + 3 • Relationship of x and y : ordered pair (x;y) If temp is 20° then number of ice creams sold is 400 If temp is 30° then number of ice creams sold is 600 (x1 ; y1) = (20 ; 400) and (x2 ; y2 • To graph relationship use Cartesian plane ) = (30 ; 600) • 2 number lines : x-axis : horizontal y-axis : vertical intersection : origin 300 200 100 -3 -2 -1 1 2 3 –100 –200 y-axis x-axis Origin (0;0) (2;300) Page 12 of 167 email: Tel: 9 2. Linear function • Relationship between 2 variables is linear and graph is a straight line y = mx + c or y = ax +b with y and x variables and m and c values and (x1;y1) and (x2;y2) are 2 points on the line. o m = slope or how steep is line or how does y-values change if x values change  − − 2 1 2 1 m= x x y y o c = y-intercept : cut y-axis : where x = 0 y-axis x-axis y-intercept = c x = 0 (0;c) x-intercept y = 0 (x;0) y = mx + c y1 y2 x1 x2 m=slope − − 2 1 2 1 = x x y y m positive Page 13 of 167 email: Tel: 9 Other lines: • If two lines are parallel they have the same slope y-axis x-axis y-intercept = c x = 0 x-intercept y =0 y = –mx + c m = slope = negative x = 5 y = 2 Page 14 of 167 email: Tel: 9 • How to determine equation of line: o Need 2 points on line (x1;y1) and (x2;y2 1. Calculate ) − − 2 1 2 1 m = x x y y 2. Substitute m and any one of the 2 points into function y = mx + c to determine c No 3a of discussion class • How to draw a line: o Need two points on line or o Equation of line 1. Calculate any 2 points on line by choosing a x-value and calculate the y-value No 3b of discussion class Page 15 of 167 email: Tel: 9 Question 3a Find the equation of the line passing through the points (1; 20) and (5; 60). Solution y = mx + c. Let (x1 ; y1) = (1 ; 20) and (x2 ; y2 The slope m is ) = (5 ; 60) 2 1 2 1 51 4 y y m x x − − = = = = − − Therefore y = 10x +c. Substitute any one of the points into the equation of the line to determine c. Let’s choose the point (1 ; 20). Then 10 20 10 1 20 10 10 20 10 10 y xc c c c c c = + = ×+ = + −= − − =− = The equation of the line is y x = + 10 10 . Page 16 of 167 email: Tel: 9 Question 3 b Draw the graph of the line y = 10 x + 10. Solution Need two points to draw line : Choose any x or y value and calculate y or x: Choose x = 0 then y = 10 (0) + 10 y = 10 - point 1 = (0 ; 10) Choose y = 0 then 0 = 10x + 10 –10x = 10 x = 10/–10 x = –1 - point2 = (–1 ; 0) 30 20 10 -3 -2 -1 1 2 3 –10 –20 y-axis x-axis (–1;0) (0;10) y = 10x + 10 Page 17 of 167 email: Tel: 9 • How to determine a slope, y-intercept and x-intercept if given the equation of a line: for example 3x + 4y – 8 = 4 or y = 4x +20 1. Write it in the format y = mx + c 2. Compare with standard form = slope is m, y-intercept is c 3. To calculate the x-intercept make y = 0 and solve for x 1. Write in format y = mx +c 3x + 4y – 8 = 4 4y = 4 + 8 – 3x 4y = 12 – 3x y = 12/4 – 3/4x y = 3 – 3/4x 2. slope = m = –3/4 y-intercept = c = 3 x-intercept is where y = 0 but y = 3 – 3/4x 0 = 3 – 3/4x 3/4x = 3 x = 3 × 4/3 x = 12/3 x = 4 Page 18 of 167 email: Tel: 9 3. Application in economics Relationship between price P and quantity Q of a product • Demand function • If the price of a product ↑ then the demand ↓ • P = a – bQ with • a = y-intercept (c) • b = slope = negative • Supply function • If the price of a product ↑ then the supply ↑ • P = c + dQ with • c = y-intercept • d = slope = positive a Q P c P Q Page 19 of 167 email: Tel: 9 Cost function Fixed cost • Cost Variable cost dependent on quantity Q • TC = FC + VC x Q = y = c + mx A supermarket’s fixed cost is R5000 per month and the salary per employees is R2000 per month. What is the supermarket’s linear cost function if the number of employees is Q? Cost = 5000 + 2000Q • Revenue • What you earn • R = Price x Quantity • R = P x Q = px • Profit • Revenue – cost • Depreciation A R200 000 car depreciates linearly to R40 000 in 8 years’ time. Derive a linear equation for the value of the car after x years with 0 ≤ x ≤ 8. Page 20 of 167 email: Tel: 9 Let y = value and x = time or years y = mx + c Need two points on graph: Given (8; 40 000) and (0; 200 000) Now 2 1 2 1 80 8 − −− = = = = − − − y y m x x y = –20 000x + c Take any one of two points: Say point 2 200 000 = -20 000(0) + c c = 200 000 Depreciation : y = –20 000x + 200 000 Page 21 of 167 email: Tel: 9 • Elasticity • Important in economics • Think what happens with an elastic band: if you apply little pressure the band expand a little bit and if you apply a lot of pressure the band expand a lot. • How sensitive demand is for price change • If the price P and % change in price goes up or down what will happen to the % change in quantity • Ratio of % change • % change in demand ε = % change in price • Price elasticity of demand or supply o Point : At a point o Arc : Over an interval 1. Price elasticity of demand • Point (P0;Q0 Demand : P = a – bQ ) 0 0 1 P ε = • b Q d − • In terms of P 1P P ε = • bQ P a d − = − Discussion class 4a + 4b Page 22 of 167 email: Tel: 9 Question 4a If the demand function is P = 80 – 2Q, where P and Q are the price and quantity respectively, determine the expression for price elasticity of demand if the price P = 20. Solution Now ε = −  1 d P b Q Given P = 80 – 2Q and P = 20. Comparing P = 80 – 2Q with P = a – bQ – a = 80 and b = 2. To determine the value of Q we substitute P = 20 into the equation and solve for Q = = − = − = 20 80 – 2Q 20 – 80 – 2Q 60 2 30 Q Q Now Page 23 of 167 email: Tel: 9 ε = − = − = − = −   1 1 20 2 30 1 3 0,33 d P b Q At P = 20 a 1% increase (decrease) in price will cause a 0,33% decrease (increase) in the quantity demanded Question 4 b If the demand function is P = 80 – 2Q, where P and Q are the price and quantity respectively, determine the expression for price elasticity of demand in terms of P only. Solution Now demand in terms of P d P P a ε = − Given P = 80 – 2Q and P = 20. Comparing P = 80 – 2Q with P = a – bQ,– a = 80 and b = 2. Thus . 80 ε = − d P P Page 24 of 167 email: Tel: 9 • Arc Over an interval Use the average P and Q at beginning and end of interval. P1 → P2 and Q1 →Q2 1 2 1 2 1 P +P ε = • b Q +Q d − Discussion class 5 Question 5 Given the demand function P = 60 – 0,2Q where P and Q is the price and quantity respectively, calculate the arc price elasticity of demand when the price decreases from R50 to R40. Solution 1 2 1 2 1 Arc elasticity of demand P P bQ Q + =− × + Given function P = 60 – 0,2Q, with a = 60 and b = 0, 2 Given P1 = 50 and P2 Need to determine Q = 40. 1 and Q2 60 0,2 0,2 60 300 5 P Q Q P Q P = − = − = − . Page 25 of 167 email: Tel: 9 Determine Q1 and Q2 by substituting P1 = 50 and P2 1 1 2 2 If then If then P Q P Q = = −× = = = −× = = 40 into the equation. Thus Therefore 1 2 1 2 1 1 50 40 0,2 50 100 1 90 0,2 150 90 30 3 Elasticity of demand P P bQ Q + =− × + + =− × + = − × − = = − 2. Price elasticity of supply Demand : P = c + dQ 0 0 1 P ε = • d Q s Page 26 of 167 email: Tel: 9 Study Unit 3 : Linear algebra Chapter 3 : Sections 3.1, 3.2.1, 3.2.5, 3.3 Study guide C.2, C.3 and C.4 Chapter 9 : Section 9.1 1. Two equations in two unknowns • Algebraically Method 1: Elimination Step 1: Eliminate 1 variable • —, + one equation or multiple of equation from other equation • Indication of size of multiple - number in front of variable Step 2: Solve variable 1 Step 3: Substitute value of variable1 back into any one of equations and solve variable2 Discussion class example 6(a) Page 27 of 167 email: Tel: 9 Question 6a Solve the following set of linear equations by using the elimination method: y + 2x = 3 –eq(1) y – x = 2 –eq(2) Solution Step 1: Eliminate 1 variable – say y Subtract eq2 from eq1 and solve x 23 23 2 2 03 1 1 3 yx yx yx yx x x += += − = − + =− + = = ( – ) or Step 2: Substitute value of x back into any one of equations and solve y 2 3 1 2( ) 3 3 2 3 3 1 2 3 y x y y y + = + = = − = Page 28 of 167 email: Tel: 9 Method 2: Substitution Step 1: Change one of the equation so that any variables is the subject of the equation– eq3 Step 2: Substitute eq3 into the unchanged equation and solve first variable Step 3: Substitute answer step2 into any equation and solve the second variable Discussion class example 6(b) Question 6b Solve the following set of linear equations by using the substitution method: y + 2x = 3 –eq(1) y – x = 2 –eq(2