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Chapter 1: Rational Numbers In the rational numbers chapter, we learnt how to compare and order rational numbers, to solve problems involving rational numbers, to determine the square root of a perfect square rational number, and to determine the approximate square root of a non-perfect 8 Terminology I learnt… Rational number: A number that can be expressed as the quotient of two integers, where the divisor is not zero. Example; 0.75, ¾, and -2 are all rational numbers. Key Words Non-perfect square: A rational number that cannot be expressed as the product of two equal rational factors. For example, you cannot multiply any rational number by itself and get an answer of 3, 5, 1.5, or 7/8. The square root of a non-perfect square is a non-repeating, non-terminating decimal. Perfect Square: A number that is the product of two identical factors. 2 X 2 = 4, so 4 is a perfect square 6 X 6 = 36, so 36 is a perfect square. Compare Rational Numbers Which fraction is bigger 3/4 or 2/3? 1. Find a common denominator: 3/4 and 2/3 share 12 as a common denominator 2. Whatever you do to the top, you do to the bottom: We multiplied 4 by 3 so now we have 12 and we multiply 3 by 3 to get nine. New fraction equals 9/12. We multiplied 3 by 4 to get 12 so we multiply 2 by 4 to get 8. New fractions equals 8/12. 9/12 is larger than 8/12. Option 2; use decimals 1. Convert fractions into decimals and compare. 3/4= 0.75 2/3= 0.6’ 0.75 > 0.6’ ¾ is the greater fraction. *Now that you can compare fractions in different ways you can order them from least to greatest.* Problem Solving with Rational Numbers, as decimals and fractions. a) 2.93+ (-3.87)= ? b) 6.42 - 7.83 =? C) 5.32 X 2.14=? Estimate and Calculate a)3+(-4)= (-1) estimation 2.93+(-3.87)= (-0.94) calculation b) 6.42 – 7.83 *add the opposite* 6.42 + (-7.83) = (-1.41) c) 5 X 2 = 10 5.32 X 2.14= 11.38 Adding and Subtracting Fractions 1. Find a common denominator 2. Keep the denominator the same 3. Add or subtract the numerators 2/3 + 5/6 4/6 + 5/6= 9/6 reduced to 3/2 or 1 ½ Multiplying and Dividing Fractions Multiplying 1. Multiply the numerators and multiply the denominators ¾ x ¼ = 3/12 = ¼ Dividing 1. Multiply by the reciprocal *Change only second fraction to its reciprocal* 2/3 3/5 = 2/3 X 5/3 = 10/9 = 1 1/9 Determining Square Roots of Rational Numbers: Determine whether each of the following numbers is a perfect square; A) 25/49 B) 0.4 A) Both the denominator and numerator are a perfect square because it is expressed as the product of two equal rational factors. 25= 5X5 , 49= 7X7 B) 0.4 can be expressed as 4/10 , the numerator is a perfect square (2X2) but the numerator is not… so 0.4 is not a perfect square Area= 0.64 units squared, is it perfect square? 0.64= 16/25 The square root of 16 is 4 and the square root of 25 is 5, therefore a *use square roots to determine perfect squares and non-perfect squares. Chapter 2: Exponents In the exponent unit we learnt how to represent repeated multiplication exponents, how to describe how powers represent repeated multiplication, demonstrate the difference between the exponent and the base by building models of a given power. We demonstrated the difference between two given powers in which the exponent and the base are interchanged by using repeated multiplication. We also evaluated powers with integral bases and whole exponents. We learnt that if the base is negative, use brackets (-3)7 means -3 multiplied 7 times. But -37 means -1 time 37. We also learnt how to model addition and subtraction of polynomial expressions concretely, pictorially, and record the process symbolically. Also, how to solve problems using addition and subtraction of polynomials and how to simplify the expressions. Terminology I learnt… Key words Power; An expression Base; the number you Exponent: the number Exponential form; a made up of a base and exponent multiply by of times you multiply the base in a power shorter way to write repeated multiplication Examples of learning: How do you write 4x4x4x4x4x4x4x4x4x4x4 in exponential form? = 411 What is the difference between 43 and 34? The difference is, for 43 you would multiply 4 three times; 4x4x4 but for 34 you would multiply 3 four times; 3x3x3x3. Use a model to represent 52, 25 5 5 The Exponent Laws A) When you multiply the same bases, add exponents; 74 x 73 = 77 B) When you divide the same bases subtract exponents; 69 x 64 = 65 C) When you raise a power to a power, multiply exponents; (45)3= 415 D) Anything to the power of 0 is 1; 80= 1 *Last tip*- When using BEDMAS, Exponents are done second* Chapter 3: Polynomials In this chapter I learnt how to write expressions represented by algebra tiles, how to write expressions for a given situation and solve them. How to find the perimeter of shapes containing polynomials, how to add and subtract polynomials and simplify the expressions. Terminology I learnt… Degree of a term: The sum of exponents in the variable in a single term (monomial) Degree of a polynomial: The degree of the highestdegree in a polynomial. Number of variables Coefficients: The number with a variable attached to the end. Eg; 7x and 5y are coefficients but 3-x is not a coefficient. Constant: The number with no variable attached to the end. Like terms: Terms that differ only by their numerical coefficients. Eg; 3x2+7x2 Zero Pairs: Algebra tiles that have a combined value of zero. x2 -x2 =0 What polynomial expression is represented by the set of algebra tiles = -X2+2x+1 Write an expression for this statement Twice the sum of 5 times a number and 8= 2(5x+8) Seven and a number are added together= 7+x The product of a number and 8 is decreased by 3= 8x-3 A triangles base and height are equal. Find an expression for the area. A= X2 , 2 Tickets for the basketball game are $7 for adults and $4 for students. Write an expression that shows the total income. I=7a+4s Write an expression represented by the algebra tiles after you’ve gotten rid of zero pairs. = X2 –X Adding and subtracting polynomials. Adding; *To add, just find zero pairs and simplify (2x2 – 2x +1) + (-x2+ 2x+2) = (Find zero pairs) X2-x+3 Subtraction; *To subtract a polynomial, you add the opposite terms and then simplify Chapter 4:Multiplying and Dividing Polynomials. In this short chapter, I learnt how to model multiplication of a monomial by monomial and a polynomial by a monomial and record the process algebraically. How to model division of a monomial by a monomial and a polynomial by a monomial and record the process and then to simplify polynomial expressions. Legend: = -x2 =x2 = x =-x 1 1 =y = -y =xy2 =xy2 Examples of learning: = 2x * -2= -4x =2y * (-2x+2) = -4xy+4y Or, -4x or, -4xy+4y 2x= -2 2y= -2x+2 You can also do these questions without the algebra tiles, this is a more efficient and faster way to do it. a) 2xy(3xy-4)= 6xy2 – 8xy a) 3x2 – 6x , 3x = x-2 b) 3x(-x+5y-3) = -3x2 + 15 xy- 9x b) 2x2+3x -x = -2x - 3 Using this process in certain situations. 4y 3x A=24x2 4x 3x * 2 *4y = volume 2 24x2 4x= 6x y-2 y-2 – y-2 =y2 +4 5y * 5y=25y2-y2+4 5y 25y2+ y2 -4 27y2 - 4 Chapter 5: Surface Area In this chapter, we learnt how to the find surface area of 1 dimensional to 3 dimensional shapes. We learnt how to find the area of a circle, trapezoid, triangle, rectangle, parallelogram, and a square. We learnt that symmetry can help us find surface area of a 3 dimensional shape because there are 2 of every side so we only need to find half of the dimensions and then multiply by 2 Vocab and property’s I learnt throughout this chapter Area=the number of square units contained in a two-dimensional region. Circumference=The boundary of or distance around a circle. This is a linear measurement. Diameter=the distance across a circle through its centre. Perimeter=the distance around the outside of a two dimensional shape Polygon = a two dimensional closed figure made of three or more line segments, triangles, squares, octagon, etc. Radius= a line segment joining the centre of a circle to the outside edge. Half the circumference. Line symmetry=A line running through the centre of an object or design such that the halves on each side are mirror images. Surface Area=The number of square units needed to cover a 3-D object. The sum of all the faces of an object. Area Area of a circle = pi * radius2 Area of a rectangle = base * height Area of a parallelogram= base * height Area of a trapezoid= ½ (b1 +b2) Area of a triangle= ½ base * height Area of a square= s2 How to apply this to finding the surface area of shapes Top= 15X5=75cm2 (so bottom equals 75cm2)=150cm2 10cm Front= 15X10=150cm2(so back equals 150cm2)=300cm2 Sides= 5X10= 50cm2 (other side equals 50cm2)=100cm2 15cm 5cm Add all sides together, 150+300+100=550cm2 Find the surface area of the cylinder (Height=18cm, radius=11cm) 18x22=396cm2 18cm C=2pi r, C=22cm Area of a circle= pi x r2 3.141592653589793… X 112= 69.1cm2 Both circles equal 69.1, so add them together (138.2) Add this number and the rectangle side together, 132.2 + 396 = 534.2cm2 Chapter 6: Linear Relations In this chapter I learnt how to interpret graphs and apply it to real life circumstances, how to find patterns in figures and/or graphs and write an equation for it, read and make tables relating to linear relations and make an equation about it, and how to graph the linear relations. Some terminology I learnt is… Key Words Linear patterns: a sequence of numbers in which the pattern inly involves addition or subtraction Common difference: the difference between any two consecutive numbers in a linear pattern Dependent variable: graphed on the vertical axis e.g. cost Independent variable Graphed on the horizontal axis e.g. time Interpolate: Estimate a value between two given values Extrapolate: estimate a value beyond a given set of values Slope: describes the steepness of a line y-intercept: Where the graph crosses the y-axis. The x coordinate is zero. Finding relations in patterns Fig.1=3 fig.2=5 fig.3=7 fig.4=9 The relation is, you add 2 squares each time, The equation is, the difference multiplied by the fig. number and then add or subtract to get the right number of squares needed. 2xf+1 How many squares might the next three figures contain? Figure 5, 2x5+1= 11 Figure 6, 2x6+1= 13 Figure 7, 2x7+1= 15 Putting the numbers into a graph… Figure # #of squares 1 3 2 5 3 7 4 9 5 11 6 13 And now you can look at the table and figure out an equation that applies to this series of numbers. (2xf+1=) (difference x figure # plus or minus a number.) Interpreting graphs and relating them to real life. 40 30 20 10 0 2 4 6 8 This graph represents how much money Briana gets in her bank each day. She gets exactly $10 every day. put the information in a table. Days Money 1 $5 2 $15 3 $25 4 $35 5 $45 6 $55 7 $65 8 $75 We are able to extrapolate this graph because she makes exactly $10 every day so we know what she will make in 17 days. An equation we can use is, 10d-5=p (d representing the day and p representing profit. We can use this equation to find out how much money she will make in 15 days, 10x15-5= $145 *If you can extrapolate on the graph, you put an arrow through the dots. Because it will always go up or down the same amount.* Graphing linear relations. I learnt, if you are given an equation you can graph it. Y=-3x-4 X Y 1 -7 2 -10 3 -13 4 -16 And now you can graph it, 0 -1 -3 -5 -7 -9 -11 -13 -15 -17 1 2 3 4 Chapter 7: Solving Linear Equations In this chapter I learnt how solve equations algebraically using inverse/opposite operations. We reviewed that addition and subtraction and opposite and multiplying and dividing are opposite operations. I learnt how to check my answer by putting the answer I got into the equation. I found this chapter super useful and know I will continue to use it throughout my high-school career. To solve linear equations we use inverse operations, for example… 3x+4=-5 3x=-9 (we subtract 4 from each side.) X=-3 (we divide 3 by each side because 3 is being multiplied by x) *Another property we learnt was distributive property* 2(-2+x)=18 -4+2x=18 (distributive property, multiply -2 and x by 2) 2x=22 (add 4 to each side to isolate the variable.) X=11 *GOAL! TO ISOLATE THE VARIABLE!!* “Balance the scale” To check you answer, just put the answer you got into the equation. 2(-2+11)=18 2x9=18 18=18 *Yup it works!* Chapter 8: Linear Inequalities I learnt how to solve a linear inequality algebraically and explain how to do it. How to graph the answer of linear inequality on a number line. I learnt how to translate an equation into a single variable linear inequality using the symbols <, >, <, >, or =. I learnt the difference between solving a linear equation and linear inequality, which is if you multiply or divide by a negative while solving a linear inequality you must switch the direction of the sign to make it true. And lastly how to check if your answer is correct. Inequality: a mathematical statement comparing expressions that may not be equal Key Words Boundary Point: Open circle boundary separates the values point: the boundary less than and great point is not included in than a specified value. the solution. Might be the solution <,> Closed circle boundary point: the boundary point is included in the solution. <,> 3<5, 6>4, 8>4, 19<22, -10<-3, 12<13, 5>1, -5>-17, 15>-18, 17<18, 3>1, 1<6, 9=9, 1<2 Translating verbal statements into inequalities on a number line X is larger than or equal to 2, x>2 -2 -1 0 1 2 Up above is the three ways you can express inequalities. The first way is verbally, describing the inequality in words. The second way is algebraically, using <,>, <,>. The last way is graphically, using visuals and in this case a number line. Solving Inequalities. Solving inequalities is the same as solving equations in the previous chapters. The differences though, are if you multiply or divide by a negative when working with inequalities you must switch the direction of the sign to be true. The other difference is equations we use = sign and inequalities use <,>, <,>. Example -3x -10 < 5x+38 -3x<5x +48 (added 10 to each side.) -8x< 48 (subtracted 5 to each side.) X>-6 (divided each side by -8 and then switched the sign.) To check if the answer is correct put any number equal or greater than -6 into the equation. -3*-4 < 5*-4 +38 12< -20+38 12< 18 It is correct. 12 is smaller than 18. Chapter 9: Scale Factor and Similarity In this chapter, I learnt that scale factor is when you take an image and divide by the actual object length or the scale factor and that will equal the actual diagram measurements. I learnt how to represent an enlargement and reduction by drawing them. I also learnt how to determine if a diagram is proportional or not. When a polygon is enlarged or reduced its corresponding angle measurements will stay the same but the corresponding side lengths are enlarged or reduced. When a triangle is enlarged or reduced its corresponding angle measurements will remain the same. And then the corresponding side lengths are proportional. 11 8 4 5.5 4.5 The triangles are similar because the side lengths are proportional. That means the angles are similar as well because it is a triangle. 9 The black triangle is the original one and the blue one is the image. The scale factor is… 4/8= 0.5. The scale factor is 0.5. Since it’s less than 1 it is a reduction. 26cm 16 17 9 Although the angles are the same, the side lengths are not proportional therefore this polygon is not similar. Key Words Enlargement: An increase in the dimensions of an object by a scale factor Reduction: A decrease in the dimensions of an object by a scale factor Scale factor: The constant amount by which all dimensions of an object are multiplying to draw an enlargement or reduction Corresponding angles/sides: Have the same relative position in a geometric figure Similar figures: Have the same shape but different size Scale: A comparison between the actual size of an object and the size of its diagram. Expressed as a ratio, fraction, percent, words, or a diagram Scale Diagram: A drawing similar to the actual figure or object. Might be smaller or larger but is the same proportion. Proportion: Is a relationship that shows two ratios are equal. Written as a fraction or ratio Scale Factor=Image length/object length Scale= Diagram measurement/actual measurement. Fraction Ratios Percent 1 /4 1:4 25% 2/5 2:5 20% 6/7 6:7 85% 3 /4 3:4 75% Chapter 10: Circle Geometry This chapter was completely new to me and I found it really interesting. I learnt how to find angles within circles, the different properties to use. I learnt how to solve problems involving using these circle properties. Key words and properties I learnt. Key Words Inscribed Angle: An angle formed by two chords that share a common endpoint. Central Angle: An angle formed by two radii of a circle Chord: A line between two points on a circle. Arc: A section of the circumference of the circle. Tangent: A line outside the circle touching the circle at one point Important Properties, to find angles within circles. An inscribed angle subtended to the same arc of a central angle is half the angle of the central angle. (Central angle is double the inscribed.) When two or more inscribed angles are connected to the same arc they feature the same angles. Complementary <’s add to 90o Supplementary <’s add to 180o Angles on a line add to 1800 Right angles are 90o Straight angles are 180o Angles at a point add to 360o Vertically opposite angles are equal I learnt how to use Pythagorean theorem to find the side lengths of RIGHT TRIANGLES in a circle. The formula is, a2+b2+=c2 Example, 52+62=c2 25+36=c2 61=c2 7.8cm X=7.8cm Perpendicular Bisector of a Chord This was an important property I learnt. You use this when you want to find the center of a circle. It is a line that passes through the centre of a circle and is perpendicular to a chord bisects the chord. And lastly, we learnt about tangents. At the top of this chapter it explains that I tangent is a line outside of a circle that touches the circle and precisely one point. We can use these tangents to find side lengths, angles, and bisectors just like how we chords inside the circle. If a tangent is perpendicular to the point of tangency, that angle made is 90o.