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Unit 1: Algebra foundations
Unit 2: solving equations & inequalities, unit 3: working with units, unit 4: linear equations & graphs, unit 5: forms of linear equations, unit 6: systems of equations, unit 7: inequalities (systems & graphs), unit 8: functions, unit 9: sequences, unit 10: absolute value & piecewise functions, unit 11: exponents & radicals, unit 12: exponential growth & decay, unit 13: quadratics: multiplying & factoring, unit 14: quadratic functions & equations, unit 15: irrational numbers, unit 16: creativity in algebra.
Curriculum / Math / 9th Grade / Unit 1: Functions, Graphs and Features / Lesson 2
Functions, Graphs and Features
Lesson 2 of 11
Criteria for Success
Tips for teachers, anchor problems, problem set, target task, additional practice.
Define functions and evaluate points in tables, graphs, and contextual situations using function notation.
Common Core Standards
Core standards.
The core standards covered in this lesson
Interpreting Functions
F.IF.A.1 — Understand that a function from one set (called the domain) to another set (called the range) assigns to each element of the domain exactly one element of the range. If f is a function and x is an element of its domain, then f(x) denotes the output of f corresponding to the input x. The graph of f is the graph of the equation y = f(x).
F.IF.A.2 — Use function notation, evaluate functions for inputs in their domains, and interpret statements that use function notation in terms of a context.
Foundational Standards
The foundational standards covered in this lesson
8.F.A.1 — Understand that a function is a rule that assigns to each input exactly one output. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output. Function notation is not required in Grade 8.
The essential concepts students need to demonstrate or understand to achieve the lesson objective
- Define functions as relationships where each input has only one output in contextual and non-contextual situations represented in tables and graphs.
- Use function notation to describe a function of a situation by denoting the function f. Describe the input as $$x$$ , the output as f( $$x$$ ), and a coordinate point in function notation (e.g., f(3) = 5 is the coordinate point (3, 5)).
- Evaluate a function represented graphically given in function notation.
- Use appropriate language to represent functions ("f of $$x$$ "), and describe a situation in terms of independent variables (i.e., "Time is a function of distance." Represent this as function notation where t is the time, and f(t) represents the distance.)
Suggestions for teachers to help them teach this lesson
- Students may need to review the definition of functions, as represented in the standard 8.F.1, specifically graphically and in a table, before they can fully access this lesson.
- A misconception students may have is to treat function notation as multiplication. Use the oral language of “f of x” and name this misconception to students to help alleviate confusion.
- This is the first lesson that introduces function notation to students. We are specifically focusing on naming functions and identifying/writing points, inputs, and outputs in function notation. Using function notation to define equations that represent a function will come later.
- The following resource may be helpful to see all the ways function notation can be represented.
Unlock features to optimize your prep time, plan engaging lessons, and monitor student progress.
Problems designed to teach key points of the lesson and guiding questions to help draw out student understanding
Below is an example of a relation that is a function and an example of a relation that is not a function .
What are the defining characteristics of a function?
NOT A FUNCTION:
Guiding Questions
Use the graph below to answer the questions that follow:
Function notation describes a particular relationship and implies the dependency of variables.
The general form of function notation is $${f(x) = y}$$ . $$y$$ is dependent on $${x }$$ .
Part A: The function $$f$$ above has a point $$f(-2) = 1$$ . Mark this point on the graph above.
Part B: Find the $$x$$ or $$y$$ values from the graph noted by the function notation below.
$$f(x)=-2$$
$${g(0)=}$$
John was running a race, and his coach was timing him. John’s race can be described as the distance in meters, $$d$$ , as a function of time in seconds, $$t$$ . Partway through the race, John’s position could be described as $$d(6)=50$$ . Describe John’s time and distance at this point.
A set of suggested resources or problem types that teachers can turn into a problem set
Give your students more opportunities to practice the skills in this lesson with a downloadable problem set aligned to the daily objective.
A task that represents the peak thinking of the lesson - mastery will indicate whether or not objective was achieved
Suppose $$f$$ is a function.
- If $$10 = f(-4)$$ , give the coordinates of a point on the graph of $$f$$ .
- If 6 is a solution to the equation $$f(w) = 1$$ , give a point on the graph of $$f$$ .
Points on a Graph , accessed on June 22, 2017, 3:51 p.m., is licensed by Illustrative Mathematics under either the CC BY 4.0 or CC BY-NC-SA 4.0 . For further information, contact Illustrative Mathematics .
Let $${f(t)}$$ be the number of people, in millions, who own cell phones $$t$$ years after 1990. Explain the meaning of the following statements.
- $${f(10)=100.3}$$
- $${f(a)=20}$$
- $${f(20)=b}$$
- $$n= {f(t)}$$
Cell Phones , accessed on June 22, 2017, 3:53 p.m., is licensed by Illustrative Mathematics under either the CC BY 4.0 or CC BY-NC-SA 4.0 . For further information, contact Illustrative Mathematics .
The following resources include problems and activities aligned to the objective of the lesson that can be used for additional practice or to create your own problem set.
- Reason and Wonder, “The Date Game,” blog entry by Michael Fenton.
- Sample problem: When does $${{f(x)} = {g(x)}}$$ ? What is $${f(x)}$$ at this point? What is $${g(x)}$$ at this point? What is $${f(-6)}$$ ? What is the solution of $${g(x)} = 1$$ ?
- Illustrative Mathematics Yam in the Oven — Include this as a follow-up to this problem: Write this statement in function notation: “The temperature of the yam after being in the oven for 30 minutes is 120 degrees.
- Illustrative Mathematics The Parking Lot
- Mathematics Vision Project: Secondary Mathematics One Module 5: Systems of Equations and Inequalities — Lesson 5.4
- Mathematics Vision Project: Secondary Mathematics One Module 3: Features of Functions — Lesson 3.7: To Function or Not to Function - A Practice Understanding Task
Topic A: Features of Functions
Model a contextual situation graphically using appropriate scales and features.
F.IF.B.4 F.IF.B.6 N.Q.A.1 N.Q.A.2 N.Q.A.3
F.IF.A.1 F.IF.A.2
Identify features of functions, including x-intercept and y-intercept, in context. Evaluate function notation in context.
F.IF.A.2 F.IF.C.9
Identify the domain and range through contextual situations, and explore domain and range on a graph. Represent domain and range with inequalities.
F.IF.A.1 F.IF.B.5
Calculate and interpret the rate of change from two points on a graph, in a situation, or in function notation.
F.IF.A.2 F.IF.B.4 F.IF.B.6
Describe and sketch functions using the features of domain and range, intercepts, function behavior, and the value of the function.
Analyze the key features of a contextual situation and model these graphically.
F.IF.B.4 F.IF.B.5 F.IF.B.6 N.Q.A.2
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Topic B: Nonlinear Functions
Draw quadratic functions represented contextually. Identify key features of the graph and relate to context.
A.CED.A.2 F.IF.B.5 F.IF.C.7.A F.LE.A.3
Sketch an exponential function that represents a situation. Identify key features of the graph and relate to context.
A.CED.A.2 F.IF.B.5 F.IF.C.7.E F.LE.A.3
Draw a graph to represent a system of functions. Identify the solution to a system represented graphically and in context.
A.REI.D.11 F.IF.A.2 F.IF.B.5
Analyze functions and identify parent functions of graphs. Identify variables of a situation and the scale of the associated graph. Represent a situation in a graph, table, and description.
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Mathematics Part II (Solutions) Solutions for Class 9 Math
- Textbook Solutions
- class 9 math
Mathematics Part II (Solutions) Solutions are considered an extremely helpful resource for exam preparation. Meritnation.com gives its users access to a profuse supply of Mathematics Part II (Solutions) questions and their solutions. MAHARASHTRA Board Class 9 math Mathematics Part II (Solutions) Solutions are created by experts of the subject, hence, sure to prepare students to score well. The questions provided in Mathematics Part II (Solutions) Books are prepared in accordance with MAHARASHTRA Board, thus holding higher chances of appearing on MAHARASHTRA Board question papers. Not only do these Mathematics Part II (Solutions) Solutions for Class 9 math strengthen students’ foundation in the subject, but also give them the ability to tackle different types of questions easily.
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- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 1 - Basic concepts in Geometry
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 2 - Parallel Lines
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 3 - Triangles
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 4 - Constructions of Triangles
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- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 8 - Trigonometry
- Mathematics Part - II (Solutions) Solutions for Class 9 Math Chapter 9 - Surface Area and Volume
- NCERT Solutions
- NCERT Class 9
- NCERT 9 Maths
- Chapter 2: Polynomial
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials
Ncert solutions class 9 maths chapter 2 – polynomials free pdf download.
NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. BYJU’S expert faculty create these NCERT Solutions to help students in preparation for their exams. BYJU’S provides NCERT Solutions for Class 9 Maths which will help students to solve problems easily. They give a detailed and stepwise explanation of each answer to the problems given in the exercises in the NCERT textbook for Class 9.
Download Exclusively Curated Chapter Notes for Class 9 Maths Chapter – 2 Polynomials
Download most important questions for class 9 maths chapter – 2 polynomials.
In NCERT Solutions for Class 9, students are introduced to many important topics that will be helpful for those who wish to pursue Mathematics as a subject in higher studies. NCERT Solutions help students to prepare for their upcoming exams by covering the updated CBSE syllabus for 2023-24 and its guidelines.
- Chapter 1: Number System
- Chapter 2: Polynomials
- Chapter 3: Coordinate Geometry
- Chapter 4: Linear Equations in Two Variables
- Chapter 5: Introduction to Euclid’s Geometry
- Chapter 6: Lines and Angles
- Chapter 7: Triangles
- Chapter 8: Quadrilaterals
- Chapter 9: Areas of Parallelograms and Triangles
- Chapter 10: Circles
- Chapter 11: Constructions
- Chapter 12: Heron’s Formula
- Chapter 13: Surface Areas and Volumes
- Chapter 14: Statistics
NCERT Class 9 Maths Chapter 2 Polynomials Topics
As this is one of the important Chapters in Class 9 Maths, it comes under the unit – Algebra and has a weightage of 12 marks in the Class 9 Maths CBSE examination. This chapter talks about:
- Polynomials in One Variable
- Zeroes of a Polynomial
- Remainder Theorem
- Factorisation of Polynomials
- Algebraic Identities
Students can refer to the NCERT Solutions for Class 9 while solving exercise problems and preparing for their Class 9 Maths exams.
NCERT Class 9 Maths Chapter 2 – Polynomials Summary
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials is the second chapter of Class 9 Maths. Polynomials are introduced and discussed in detail here. The chapter discusses Polynomials and their applications. The introduction of the chapter includes whole numbers, integers, and rational numbers.
The chapter starts with the introduction of Polynomials in section 2.1, followed by two very important topics in sections 2.2 and 2.3
- Polynomials in one Variable – Discussion of Linear, Quadratic and Cubic Polynomial.
- Zeroes of a Polynomial – A zero of a polynomial need not be zero and can have more than one zero.
- Real Numbers and their Decimal Expansions – Here, you study the decimal expansions of real numbers and see whether they can help in distinguishing between rational and irrational numbers.
Next, it discusses the following topics:
- Representing Real Numbers on the Number Line – In this, the solutions for 2 problems in Exercise 2.4.
- Operations on Real Numbers – Here, you explore some of the operations like addition, subtraction, multiplication, and division on irrational numbers.
- Laws of Exponents for Real Numbers – Use these laws of exponents to solve the questions.
Key Advantages of NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials
- These NCERT Solutions for Class 9 Maths help you solve and revise the updated CBSE syllabus of Class 9 for 2023-24.
- After going through the stepwise solutions given by our subject expert teachers, you will be able to score more marks.
- It follows NCERT guidelines which help in preparing the students accordingly.
- It contains all the important questions from the examination point of view.
- It helps in scoring well in Class 10 CBSE Maths exams.
To learn the NCERT solutions for Class 9 Maths Chapter 2 Polynomials offline, click on the below link:
NCERT Solutions for Class 9 Maths Chapter 2 – Polynomials
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List of Exercises in Class 9 Maths Chapter 2 Polynomials
Class 9 Maths Chapter 2 Polynomials contains 5 exercises. Based on the concept of polynomials, each exercise provides a number of questions. Click on the below links to access the exercise-wise NCERT solutions for Class 9 Maths Chapter 2 polynomials.
Exercise 2.1 Solutions 5 Questions
Exercise 2.2 Solutions 4 Questions
Exercise 2.3 Solutions 3 Questions
Exercise 2.4 Solutions 5 Questions
Exercise 2.5 Solutions 16 Questions
Access Answers of NCERT Class 9 Maths Chapter 2 – Polynomials
Exercise 2.1 page: 32.
1. Which of the following expressions are polynomials in one variable, and which are not? State reasons for your answer.
(i) 4x 2 –3x+7
The equation 4x 2 –3x+7 can be written as 4x 2 –3x 1 +7x 0
Since x is the only variable in the given equation and the powers of x (i.e. 2, 1 and 0) are whole numbers, we can say that the expression 4x 2 –3x+7 is a polynomial in one variable.
(ii) y 2 +√2
The equation y 2 + √2 can be written as y 2 + √ 2y 0
Since y is the only variable in the given equation and the powers of y (i.e., 2 and 0) are whole numbers, we can say that the expression y 2 + √ 2 is a polynomial in one variable.
(iii) 3√t+t√2
The equation 3√t+t√2 can be written as 3t 1/2 +√2t
Though t is the only variable in the given equation, the power of t (i.e., 1/2) is not a whole number. Hence, we can say that the expression 3√t+t√2 is not a polynomial in one variable.
The equation y+2/y can be written as y+2y -1
Though y is the only variable in the given equation, the power of y (i.e., -1) is not a whole number. Hence, we can say that the expression y+2/y is not a polynomial in one variable.
(v) x 10 +y 3 +t 50
Here, in the equation x 10 +y 3 +t 50
Though the powers, 10, 3, 50, are whole numbers, there are 3 variables used in the expression
x 10 +y 3 +t 50 . Hence, it is not a polynomial in one variable.
2. Write the coefficients of x 2 in each of the following:
(i) 2+x 2 +x
The equation 2+x 2 +x can be written as 2+(1)x 2 +x
We know that the coefficient is the number which multiplies the variable.
Here, the number that multiplies the variable x 2 is 1
Hence, the coefficient of x 2 in 2+x 2 +x is 1.
(ii) 2–x 2 +x 3
The equation 2–x 2 +x 3 can be written as 2+(–1)x 2 +x 3
We know that the coefficient is the number (along with its sign, i.e. – or +) which multiplies the variable.
Here, the number that multiplies the variable x 2 is -1
Hence, the coefficient of x 2 in 2–x 2 +x 3 is -1.
(iii) ( π /2)x 2 +x
The equation (π/2)x 2 +x can be written as (π/2)x 2 + x
Here, the number that multiplies the variable x 2 is π/2.
Hence, the coefficient of x 2 in (π/2)x 2 +x is π/2.
The equation √2x-1 can be written as 0x 2 +√2x-1 [Since 0x 2 is 0]
Here, the number that multiplies the variable x 2 is 0
Hence, the coefficient of x 2 in √2x-1 is 0.
3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Binomial of degree 35: A polynomial having two terms and the highest degree 35 is called a binomial of degree 35.
For example, 3x 35 +5
Monomial of degree 100: A polynomial having one term and the highest degree 100 is called a monomial of degree 100.
For example, 4x 100
4. Write the degree of each of the following polynomials:
(i) 5x 3 +4x 2 +7x
The highest power of the variable in a polynomial is the degree of the polynomial.
Here, 5x 3 +4x 2 +7x = 5x 3 +4x 2 +7x 1
The powers of the variable x are: 3, 2, 1
The degree of 5x 3 +4x 2 +7x is 3, as 3 is the highest power of x in the equation.
Here, in 4–y 2 ,
The power of the variable y is 2
The degree of 4–y 2 is 2, as 2 is the highest power of y in the equation.
(iii) 5t–√7
Here, in 5t –√7
The power of the variable t is: 1
The degree of 5t –√7 is 1, as 1 is the highest power of y in the equation.
Here, 3 = 3×1 = 3× x 0
The power of the variable here is: 0
Hence, the degree of 3 is 0.
5. Classify the following as linear, quadratic and cubic polynomials:
We know that,
Linear polynomial: A polynomial of degree one is called a linear polynomial.
Quadratic polynomial: A polynomial of degree two is called a quadratic polynomial.
Cubic polynomial: A polynomial of degree three is called a cubic polynomial.
The highest power of x 2 +x is 2
The degree is 2
Hence, x 2 +x is a quadratic polynomial
The highest power of x–x 3 is 3
The degree is 3
Hence, x–x 3 is a cubic polynomial
(iii) y+y 2 +4
The highest power of y+y 2 +4 is 2
Hence, y+y 2 +4 is a quadratic polynomial
The highest power of 1+x is 1
The degree is 1
Hence, 1+x is a linear polynomial.
The highest power of 3t is 1
Hence, 3t is a linear polynomial.
The highest power of r 2 is 2
Hence, r 2 is a quadratic polynomial.
The highest power of 7x 3 is 3
Hence, 7x 3 is a cubic polynomial.
Exercise 2.2 Page: 34
1. Find the value of the polynomial (x)=5x−4x 2 +3.
(ii) x = – 1
(iii) x = 2
Let f(x) = 5x−4x 2 +3
(i) When x = 0
f(0) = 5(0)-4(0) 2 +3
(ii) When x = -1
f(x) = 5x−4x 2 +3
f(−1) = 5(−1)−4(−1) 2 +3
(iii) When x = 2
f(2) = 5(2)−4(2) 2 +3
2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y)=y 2 −y+1
p(y) = y 2 –y+1
∴ p(0) = (0) 2 −(0)+1 = 1
p(1) = (1) 2 –(1)+1 = 1
p(2) = (2) 2 –(2)+1 = 3
(ii) p(t)=2+t+2t 2 −t 3
p(t) = 2+t+2t 2 −t 3
∴ p(0) = 2+0+2(0) 2 –(0) 3 = 2
p(1) = 2+1+2(1) 2 –(1) 3 =2+1+2–1 = 4
p(2) = 2+2+2(2) 2 –(2) 3 =2+2+8–8 = 4
(iii) p(x)=x 3
∴ p(0) = (0) 3 = 0
p(1) = (1) 3 = 1
p(2) = (2) 3 = 8
(iv) P(x) = (x−1)(x+1)
p(x) = (x–1)(x+1)
∴ p(0) = (0–1)(0+1) = (−1)(1) = –1
p(1) = (1–1)(1+1) = 0(2) = 0
p(2) = (2–1)(2+1) = 1(3) = 3
3. Verify whether the following are zeroes of the polynomial indicated against them.
(i) p(x)=3x+1, x = −1/3
For, x = -1/3, p(x) = 3x+1
∴ p(−1/3) = 3(-1/3)+1 = −1+1 = 0
∴ -1/3 is a zero of p(x).
(ii) p(x) = 5x–π, x = 4/5
For, x = 4/5, p(x) = 5x–π
∴ p(4/5) = 5(4/5)- π = 4-π
∴ 4/5 is not a zero of p(x).
(iii) p(x) = x 2 −1, x = 1, −1
For, x = 1, −1;
p(x) = x 2 −1
∴ p(1)=1 2 −1=1−1 = 0
p(−1)=(-1) 2 −1 = 1−1 = 0
∴ 1, −1 are zeros of p(x).
(iv) p(x) = (x+1)(x–2), x =−1, 2
For, x = −1,2;
p(x) = (x+1)(x–2)
∴ p(−1) = (−1+1)(−1–2)
= (0)(−3) = 0
p(2) = (2+1)(2–2) = (3)(0) = 0
∴ −1, 2 are zeros of p(x).
(v) p(x) = x 2 , x = 0
For, x = 0 p(x) = x 2
p(0) = 0 2 = 0
∴ 0 is a zero of p(x).
(vi) p(x) = lx +m, x = −m/ l
For, x = -m/ l ; p(x) = l x+m
∴ p(-m/ l) = l (-m/ l )+m = −m+m = 0
∴ -m/ l is a zero of p(x).
(vii) p(x) = 3x 2 −1, x = -1/√3 , 2/√3
For, x = -1/√3 , 2/√3 ; p(x) = 3x 2 −1
∴ p(-1/√3) = 3(-1/√3) 2 -1 = 3(1/3)-1 = 1-1 = 0
∴ p(2/√3 ) = 3(2/√3) 2 -1 = 3(4/3)-1 = 4−1 = 3 ≠ 0
∴ -1/√3 is a zero of p(x), but 2/√3 is not a zero of p(x).
(viii) p(x) =2x+1, x = 1/2
For, x = 1/2 p(x) = 2x+1
∴ p(1/2) = 2(1/2)+1 = 1+1 = 2≠0
∴ 1/2 is not a zero of p(x).
4. Find the zero of the polynomials in each of the following cases:
(i) p(x) = x+5
∴ -5 is a zero polynomial of the polynomial p(x).
(ii) p(x) = x–5
∴ 5 is a zero polynomial of the polynomial p(x).
(iii) p(x) = 2x+5
p(x) = 2x+5
∴x = -5/2 is a zero polynomial of the polynomial p(x).
(iv) p(x) = 3x–2
p(x) = 3x–2
∴ x = 2/3 is a zero polynomial of the polynomial p(x).
(v) p(x) = 3x
∴ 0 is a zero polynomial of the polynomial p(x).
(vi) p(x) = ax, a≠0
∴ x = 0 is a zero polynomial of the polynomial p(x).
(vii) p(x) = cx+d, c ≠ 0, c, d are real numbers.
p(x) = cx + d
∴ x = -d/c is a zero polynomial of the polynomial p(x).
Exercise 2.3 Page: 40
1. Find the remainder when x 3 +3x 2 +3x+1 is divided by
∴ Remainder:
p(−1) = (−1) 3 +3(−1) 2 +3(−1)+1
p(1/2) = (1/2) 3 +3(1/2) 2 +3(1/2)+1
= (1/8)+(3/4)+(3/2)+1
p(0) = (0) 3 +3(0) 2 +3(0)+1
p(0) = (−π) 3 +3(−π) 2 +3(−π)+1
= −π 3 +3π 2 −3π+1
(-5/2) 3 +3(-5/2) 2 +3(-5/2)+1 = (-125/8)+(75/4)-(15/2)+1
2. Find the remainder when x 3 −ax 2 +6x−a is divided by x-a.
Let p(x) = x 3 −ax 2 +6x−a
p(a) = (a) 3 −a(a 2 )+6(a)−a
= a 3 −a 3 +6a−a = 5a
3. Check whether 7+3x is a factor of 3x 3 +7x.
3(-7/3) 3 +7(-7/3) = -(343/9)+(-49/3)
= (-343-(49)3)/9
= (-343-147)/9
= -490/9 ≠ 0
∴ 7+3x is not a factor of 3x 3 +7x
Exercise 2.4 Page: 43
1. Determine which of the following polynomials has (x + 1) a factor:
(i) x 3 +x 2 +x+1
Let p(x) = x 3 +x 2 +x+1
The zero of x+1 is -1. [x+1 = 0 means x = -1]
p(−1) = (−1) 3 +(−1) 2 +(−1)+1
∴ By factor theorem, x+1 is a factor of x 3 +x 2 +x+1
(ii) x 4 +x 3 +x 2 +x+1
Let p(x)= x 4 +x 3 +x 2 +x+1
The zero of x+1 is -1. [x+1= 0 means x = -1]
p(−1) = (−1) 4 +(−1) 3 +(−1) 2 +(−1)+1
= 1−1+1−1+1
∴ By factor theorem, x+1 is not a factor of x 4 + x 3 + x 2 + x + 1
(iii) x 4 +3x 3 +3x 2 +x+1
Let p(x)= x 4 +3x 3 +3x 2 +x+1
The zero of x+1 is -1.
p(−1)=(−1) 4 +3(−1) 3 +3(−1) 2 +(−1)+1
∴ By factor theorem, x+1 is not a factor of x 4 +3x 3 +3x 2 +x+1
(iv) x 3 – x 2 – (2+√2)x +√2
Let p(x) = x 3 –x 2 –(2+√2)x +√2
p(−1) = (-1) 3 –(-1) 2 –(2+√2)(-1) + √2 = −1−1+2+√2+√2
∴ By factor theorem, x+1 is not a factor of x 3 –x 2 –(2+√2)x +√2
2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x 3 +x 2 –2x–1, g(x) = x+1
p(x) = 2x 3 +x 2 –2x–1, g(x) = x+1
∴ Zero of g(x) is -1.
p(−1) = 2(−1) 3 +(−1) 2 –2(−1)–1
∴ By factor theorem, g(x) is a factor of p(x).
(ii) p(x)=x 3 +3x 2 +3x+1, g(x) = x+2
p(x) = x 3 +3x 2 +3x+1, g(x) = x+2
∴ Zero of g(x) is -2.
p(−2) = (−2) 3 +3(−2) 2 +3(−2)+1
= −8+12−6+1
∴ By factor theorem, g(x) is not a factor of p(x).
(iii) p(x)=x 3 –4x 2 +x+6, g(x) = x–3
p(x) = x 3 –4x 2 +x+6, g(x) = x -3
∴ Zero of g(x) is 3.
p(3) = (3) 3 −4(3) 2 +(3)+6
= 27−36+3+6
3. Find the value of k, if x–1 is a factor of p(x) in each of the following cases:
(i) p(x) = x 2 +x+k
If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem
⇒ (1) 2 +(1)+k = 0
⇒ 1+1+k = 0
(ii) p(x) = 2x 2 +kx+ √2
⇒ 2(1) 2 +k(1)+√2 = 0
⇒ 2+k+√2 = 0
⇒ k = −(2+√2)
(iii) p(x) = kx 2 – √ 2x+1
If x-1 is a factor of p(x), then p(1)=0
⇒ k(1) 2 -√2(1)+1=0
(iv) p(x)=kx 2 –3x+k
⇒ k(1) 2 –3(1)+k = 0
⇒ k−3+k = 0
4. Factorise:
(i) 12x 2 –7x+1
Using the splitting the middle term method,
We have to find a number whose sum = -7 and product =1×12 = 12
We get -3 and -4 as the numbers [-3+-4=-7 and -3×-4 = 12]
12x 2 –7x+1= 12x 2 -4x-3x+1
= 4x(3x-1)-1(3x-1)
= (4x-1)(3x-1)
(ii) 2x 2 +7x+3
We have to find a number whose sum = 7 and product = 2×3 = 6
We get 6 and 1 as the numbers [6+1 = 7 and 6×1 = 6]
2x 2 +7x+3 = 2x 2 +6x+1x+3
= 2x (x+3)+1(x+3)
= (2x+1)(x+3)
(iii) 6x 2 +5x-6
We have to find a number whose sum = 5 and product = 6×-6 = -36
We get -4 and 9 as the numbers [-4+9 = 5 and -4×9 = -36]
6x 2 +5x-6 = 6x 2 +9x–4x–6
= 3x(2x+3)–2(2x+3)
= (2x+3)(3x–2)
(iv) 3x 2 –x–4
We have to find a number whose sum = -1 and product = 3×-4 = -12
We get -4 and 3 as the numbers [-4+3 = -1 and -4×3 = -12]
3x 2 –x–4 = 3x 2 –4x+3x–4
= x(3x–4)+1(3x–4)
= (3x–4)(x+1)
5. Factorise:
(i) x 3 –2x 2 –x+2
Let p(x) = x 3 –2x 2 –x+2
Factors of 2 are ±1 and ± 2
p(x) = x 3 –2x 2 –x+2
p(−1) = (−1) 3 –2(−1) 2 –(−1)+2
Therefore, (x+1) is the factor of p(x)
Now, Dividend = Divisor × Quotient + Remainder
(x+1)(x 2 –3x+2) = (x+1)(x 2 –x–2x+2)
= (x+1)(x(x−1)−2(x−1))
= (x+1)(x−1)(x-2)
(ii) x 3 –3x 2 –9x–5
Let p(x) = x 3 –3x 2 –9x–5
Factors of 5 are ±1 and ±5
By the trial method, we find that
So, (x-5) is factor of p(x)
p(x) = x 3 –3x 2 –9x–5
p(5) = (5) 3 –3(5) 2 –9(5)–5
= 125−75−45−5
Therefore, (x-5) is the factor of p(x)
(x−5)(x 2 +2x+1) = (x−5)(x 2 +x+x+1)
= (x−5)(x(x+1)+1(x+1))
= (x−5)(x+1)(x+1)
(iii) x 3 +13x 2 +32x+20
Let p(x) = x 3 +13x 2 +32x+20
Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
So, (x+1) is factor of p(x)
p(x)= x 3 +13x 2 +32x+20
p(-1) = (−1) 3 +13(−1) 2 +32(−1)+20
= −1+13−32+20
Now, Dividend = Divisor × Quotient +Remainder
(x+1)(x 2 +12x+20) = (x+1)(x 2 +2x+10x+20)
= (x+1)x(x+2)+10(x+2)
= (x+1)(x+2)(x+10)
(iv) 2y 3 +y 2 –2y–1
Let p(y) = 2y 3 +y 2 –2y–1
Factors = 2×(−1)= -2 are ±1 and ±2
So, (y-1) is factor of p(y)
p(y) = 2y 3 +y 2 –2y–1
p(1) = 2(1) 3 +(1) 2 –2(1)–1
Therefore, (y-1) is the factor of p(y)
Now, Dividend = Divisor × Quotient + Remainder
(y−1)(2y 2 +3y+1) = (y−1)(2y 2 +2y+y+1)
= (y−1)(2y(y+1)+1(y+1))
= (y−1)(2y+1)(y+1)
Exercise 2.5 Page: 48
1. Use suitable identities to find the following products:
(i) (x+4)(x +10)
Using the identity, (x+a)(x+b) = x 2 +(a+b)x+ab
(x+4)(x+10) = x 2 +(4+10)x+(4×10)
= x 2 +14x+40
(ii) (x+8)(x –10)
(x+8)(x−10) = x 2 +(8+(−10))x+(8×(−10))
= x 2 +(8−10)x–80
= x 2 −2x−80
(iii) (3x+4)(3x–5)
(3x+4)(3x−5) = (3x) 2 +[4+(−5)]3x+4×(−5)
= 9x 2 +3x(4–5)–20
= 9x 2 –3x–20
(iv) (y 2 +3/2)(y 2 -3/2)
Using the identity, (x+y)(x–y) = x 2 –y 2
(y 2 +3/2)(y 2 –3/2) = (y 2 ) 2 –(3/2) 2
2. Evaluate the following products without multiplying directly:
(i) 103×107
103×107= (100+3)×(100+7)
Using identity, [(x+a)(x+b) = x 2 +(a+b)x+ab
Here, x = 100
We get, 103×107 = (100+3)×(100+7)
= (100) 2 +(3+7)100+(3×7)
= 10000+1000+21
(ii) 95×96
95×96 = (100-5)×(100-4)
Using identity, [(x-a)(x-b) = x 2 -(a+b)x+ab
We get, 95×96 = (100-5)×(100-4)
= (100) 2 +100(-5+(-4))+(-5×-4)
= 10000-900+20
(iii) 104×96
104×96 = (100+4)×(100–4)
Using identity, [(a+b)(a-b)= a 2 -b 2 ]
Here, a = 100
We get, 104×96 = (100+4)×(100–4)
= (100) 2 –(4) 2
3. Factorise the following using appropriate identities:
(i) 9x 2 +6xy+y 2
9x 2 +6xy+y 2 = (3x) 2 +(2×3x×y)+y 2
Using identity, x 2 +2xy+y 2 = (x+y) 2
Here, x = 3x
= (3x+y)(3x+y)
(ii) 4y 2 −4y+1
4y 2 −4y+1 = (2y) 2 –(2×2y×1)+1
Using identity, x 2 – 2xy + y 2 = (x – y) 2
Here, x = 2y
4y 2 −4y+1 = (2y) 2 –(2×2y×1)+1 2
= (2y–1)(2y–1)
(iii) x 2 –y 2 /100
x 2 –y 2 /100 = x 2 –(y/10) 2
Using identity, x 2 -y 2 = (x-y)(x+y)
Here, x = x
= (x–y/10)(x+y/10)
4. Expand each of the following using suitable identities:
(i) (x+2y+4z) 2
(ii) (2x−y+z) 2
(iii) (−2x+3y+2z) 2
(iv) (3a –7b–c) 2
(v) (–2x+5y–3z) 2
(vi) ((1/4)a-(1/2)b +1) 2
Using identity, (x+y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx
(x+2y+4z) 2 = x 2 +(2y) 2 +(4z) 2 +(2×x×2y)+(2×2y×4z)+(2×4z×x)
= x 2 +4y 2 +16z 2 +4xy+16yz+8xz
(ii) (2x−y+z) 2
Here, x = 2x
(2x−y+z) 2 = (2x) 2 +(−y) 2 +z 2 +(2×2x×−y)+(2×−y×z)+(2×z×2x)
= 4x 2 +y 2 +z 2 –4xy–2yz+4xz
Here, x = −2x
(−2x+3y+2z) 2 = (−2x) 2 +(3y) 2 +(2z) 2 +(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x)
= 4x 2 +9y 2 +4z 2 –12xy+12yz–8xz
Using identity (x+y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx
Here, x = 3a
(3a –7b– c) 2 = (3a) 2 +(– 7b) 2 +(– c) 2 +(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a)
= 9a 2 + 49b 2 + c 2 – 42ab+14bc–6ca
Here, x = –2x
(–2x+5y–3z) 2 = (–2x) 2 +(5y) 2 +(–3z) 2 +(2×–2x × 5y)+(2× 5y×– 3z)+(2×–3z ×–2x)
= 4x 2 +25y 2 +9z 2 – 20xy–30yz+12zx
(vi) ((1/4)a-(1/2)b+1) 2
Here, x = (1/4)a
y = (-1/2)b
(i) 4x 2 +9y 2 +16z 2 +12xy–24yz–16xz
(ii) 2x 2 +y 2 +8z 2 –2√2xy+4√2yz–8xz
We can say that, x 2 +y 2 +z 2 +2xy+2yz+2zx = (x+y+z) 2
4x 2 +9y 2 +16z 2 +12xy–24yz–16xz = (2x) 2 +(3y) 2 +(−4z) 2 +(2×2x×3y)+(2×3y×−4z)+(2×−4z×2x)
= (2x+3y–4z) 2
= (2x+3y–4z)(2x+3y–4z)
Using identity, (x +y+z) 2 = x 2 +y 2 +z 2 +2xy+2yz+2zx
2x 2 +y 2 +8z 2 –2√2xy+4√2yz–8xz
= (-√2x) 2 +(y) 2 +(2√2z) 2 +(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)
= (−√2x+y+2√2z) 2
= (−√2x+y+2√2z)(−√2x+y+2√2z)
6. Write the following cubes in expanded form:
(i) (2x+1) 3
(ii) (2a−3b) 3
(iii) ((3/2)x+1) 3
(iv) (x−(2/3)y) 3
Using identity,(x+y) 3 = x 3 +y 3 +3xy(x+y)
(2x+1) 3 = (2x) 3 +1 3 +(3×2x×1)(2x+1)
= 8x 3 +1+6x(2x+1)
= 8x 3 +12x 2 +6x+1
Using identity,(x–y) 3 = x 3 –y 3 –3xy(x–y)
(2a−3b) 3 = (2a) 3 −(3b) 3 –(3×2a×3b)(2a–3b)
= 8a 3 –27b 3 –18ab(2a–3b)
= 8a 3 –27b 3 –36a 2 b+54ab 2
((3/2)x+1) 3 =((3/2)x) 3 +1 3 +(3×(3/2)x×1)((3/2)x +1)
(iv) (x−(2/3)y) 3
Using identity, (x –y) 3 = x 3 –y 3 –3xy(x–y)
7. Evaluate the following using suitable identities:
(ii) (102) 3
(iii) (998) 3
We can write 99 as 100–1
(99) 3 = (100–1) 3
= (100) 3 –1 3 –(3×100×1)(100–1)
= 1000000 –1–300(100 – 1)
= 1000000–1–30000+300
We can write 102 as 100+2
(100+2) 3 =(100) 3 +2 3 +(3×100×2)(100+2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
We can write 99 as 1000–2
(998) 3 =(1000–2) 3
=(1000) 3 –2 3 –(3×1000×2)(1000–2)
= 1000000000–8–6000(1000– 2)
= 1000000000–8- 6000000+12000
= 994011992
8. Factorise each of the following:
(i) 8a 3 +b 3 +12a 2 b+6ab 2
(ii) 8a 3 –b 3 –12a 2 b+6ab 2
(iii) 27–125a 3 –135a +225a 2
(iv) 64a 3 –27b 3 –144a 2 b+108ab 2
(v) 27p 3 –(1/216)−(9/2) p 2 +(1/4)p
The expression, 8a 3 +b 3 +12a 2 b+6ab 2 can be written as (2a) 3 +b 3 +3(2a) 2 b+3(2a)(b) 2
8a 3 +b 3 +12a 2 b+6ab 2 = (2a) 3 +b 3 +3(2a) 2 b+3(2a)(b) 2
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y) 3 = x 3 +y 3 +3xy(x+y) is used.
The expression, 8a 3 –b 3 −12a 2 b+6ab 2 can be written as (2a) 3 –b 3 –3(2a) 2 b+3(2a)(b) 2
8a 3 –b 3 −12a 2 b+6ab 2 = (2a) 3 –b 3 –3(2a) 2 b+3(2a)(b) 2
= (2a–b)(2a–b)(2a–b)
Here, the identity,(x–y) 3 = x 3 –y 3 –3xy(x–y) is used.
(iii) 27–125a 3 –135a+225a 2
The expression, 27–125a 3 –135a +225a 2 can be written as 3 3 –(5a) 3 –3(3) 2 (5a)+3(3)(5a) 2
27–125a 3 –135a+225a 2 = 3 3 –(5a) 3 –3(3) 2 (5a)+3(3)(5a) 2
= (3–5a)(3–5a)(3–5a)
Here, the identity, (x–y) 3 = x 3 –y 3 -3xy(x–y) is used.
The expression, 64a 3 –27b 3 –144a 2 b+108ab 2 can be written as (4a) 3 –(3b) 3 –3(4a) 2 (3b)+3(4a)(3b) 2
64a 3 –27b 3 –144a 2 b+108ab 2 = (4a) 3 –(3b) 3 –3(4a) 2 (3b)+3(4a)(3b) 2
=(4a–3b)(4a–3b)(4a–3b)
Here, the identity, (x – y) 3 = x 3 – y 3 – 3xy(x – y) is used.
(v) 27p 3 – (1/216)−(9/2) p 2 +(1/4)p
The expression, 27p 3 –(1/216)−(9/2) p 2 +(1/4)p can be written as
(3p) 3 –(1/6) 3 −(9/2) p 2 +(1/4)p = (3p) 3 –(1/6) 3 −3(3p)(1/6)(3p – 1/6)
Using (x – y) 3 = x 3 – y 3 – 3xy (x – y)
27p 3 –(1/216)−(9/2) p 2 +(1/4)p = (3p) 3 –(1/6) 3 −3(3p)(1/6)(3p – 1/6)
Taking x = 3p and y = 1/6
= (3p–1/6) 3
= (3p–1/6)(3p–1/6)(3p–1/6)
(i) x 3 +y 3 = (x+y)(x 2 –xy+y 2 )
(ii) x 3 –y 3 = (x–y)(x 2 +xy+y 2 )
We know that, (x+y) 3 = x 3 +y 3 +3xy(x+y)
⇒ x 3 +y 3 = (x+y) 3 –3xy(x+y)
⇒ x 3 +y 3 = (x+y)[(x+y) 2 –3xy]
Taking (x+y) common ⇒ x 3 +y 3 = (x+y)[(x 2 +y 2 +2xy)–3xy]
⇒ x 3 +y 3 = (x+y)(x 2 +y 2 –xy)
(ii) x 3 –y 3 = (x–y)(x 2 +xy+y 2 )
We know that, (x–y) 3 = x 3 –y 3 –3xy(x–y)
⇒ x 3 −y 3 = (x–y) 3 +3xy(x–y)
⇒ x 3 −y 3 = (x–y)[(x–y) 2 +3xy]
Taking (x+y) common ⇒ x 3 −y 3 = (x–y)[(x 2 +y 2 –2xy)+3xy]
⇒ x 3 +y 3 = (x–y)(x 2 +y 2 +xy)
10. Factorise each of the following:
(i) 27y 3 +125z 3
(ii) 64m 3 –343n 3
The expression, 27y 3 +125z 3 can be written as (3y) 3 +(5z) 3
27y 3 +125z 3 = (3y) 3 +(5z) 3
We know that, x 3 +y 3 = (x+y)(x 2 –xy+y 2 )
= (3y+5z)[(3y) 2 –(3y)(5z)+(5z) 2 ]
= (3y+5z)(9y 2 –15yz+25z 2 )
The expression, 64m 3 –343n 3 can be written as (4m) 3 –(7n) 3
64m 3 –343n 3 = (4m) 3 –(7n) 3
We know that, x 3 –y 3 = (x–y)(x 2 +xy+y 2 )
= (4m-7n)[(4m) 2 +(4m)(7n)+(7n) 2 ]
= (4m-7n)(16m 2 +28mn+49n 2 )
11. Factorise: 27x 3 +y 3 +z 3 –9xyz.
The expression 27x 3 +y 3 +z 3 –9xyz can be written as (3x) 3 +y 3 +z 3 –3(3x)(y)(z)
27x 3 +y 3 +z 3 –9xyz = (3x) 3 +y 3 +z 3 –3(3x)(y)(z)
We know that, x 3 +y 3 +z 3 –3xyz = (x+y+z)(x 2 +y 2 +z 2 –xy –yz–zx)
= (3x+y+z)[(3x) 2 +y 2 +z 2 –3xy–yz–3xz]
= (3x+y+z)(9x 2 +y 2 +z 2 –3xy–yz–3xz)
12. Verify that:
x 3 +y 3 +z 3 –3xyz = (1/2) (x+y+z)[(x–y) 2 +(y–z) 2 +(z–x) 2 ]
x 3 +y 3 +z 3 −3xyz = (x+y+z)(x 2 +y 2 +z 2 –xy–yz–xz)
⇒ x 3 +y 3 +z 3 –3xyz = (1/2)(x+y+z)[2(x 2 +y 2 +z 2 –xy–yz–xz)]
= (1/2)(x+y+z)(2x 2 +2y 2 +2z 2 –2xy–2yz–2xz)
= (1/2)(x+y+z)[(x 2 +y 2 −2xy)+(y 2 +z 2 –2yz)+(x 2 +z 2 –2xz)]
= (1/2)(x+y+z)[(x–y) 2 +(y–z) 2 +(z–x) 2 ]
13. If x+y+z = 0, show that x 3 +y 3 +z 3 = 3xyz.
x 3 +y 3 +z 3 -3xyz = (x +y+z)(x 2 +y 2 +z 2 –xy–yz–xz)
Now, according to the question, let (x+y+z) = 0,
Then, x 3 +y 3 +z 3 -3xyz = (0)(x 2 +y 2 +z 2 –xy–yz–xz)
⇒ x 3 +y 3 +z 3 –3xyz = 0
⇒ x 3 +y 3 +z 3 = 3xyz
Hence Proved
14. Without actually calculating the cubes, find the value of each of the following:
(i) (−12) 3 +(7) 3 +(5) 3
(ii) (28) 3 +(−15) 3 +(−13) 3
Let a = −12
We know that if x+y+z = 0, then x 3 +y 3 +z 3 =3xyz.
Here, −12+7+5=0
(−12) 3 +(7) 3 +(5) 3 = 3xyz
= 3×-12×7×5
(28) 3 +(−15) 3 +(−13) 3
We know that if x+y+z = 0, then x 3 +y 3 +z 3 = 3xyz.
Here, x+y+z = 28–15–13 = 0
(28) 3 +(−15) 3 +(−13) 3 = 3xyz
= 0+3(28)(−15)(−13)
15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area: 25a 2 –35a+12
(ii) Area: 35y 2 +13y–12
We have to find a number whose sum = -35 and product =25×12 = 300
We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20 = 300]
25a 2 –35a+12 = 25a 2 –15a−20a+12
= 5a(5a–3)–4(5a–3)
= (5a–4)(5a–3)
Possible expression for length = 5a–4
Possible expression for breadth = 5a –3
We have to find a number whose sum = 13 and product = 35×-12 = 420
We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=420]
35y 2 +13y–12 = 35y 2 –15y+28y–12
= 5y(7y–3)+4(7y–3)
= (5y+4)(7y–3)
Possible expression for length = (5y+4)
Possible expression for breadth = (7y–3)
16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume: 3x 2 –12x
(ii) Volume: 12ky 2 +8ky–20k
3x 2 –12x can be written as 3x(x–4) by taking 3x out of both the terms.
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (x–4)
12ky 2 +8ky–20k can be written as 4k(3y 2 +2y–5) by taking 4k out of both the terms.
12ky 2 +8ky–20k = 4k(3y 2 +2y–5)
= 4k(3y 2 +5y–3y–5)
= 4k[y(3y+5)–1(3y+5)]
= 4k(3y+5)(y–1)
Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y -1)
Disclaimer:
Dropped Topics – 2.4 Remainder theorem.
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Grade 9 Math - 3 - Unit 2: Powers and Exponent Laws Example: Identify the base and the exponent for each. (a) 53 base is 5, exponent is 3 (b) (-3)4 base is -3, exponent is 4 (c) ( )2 base is and exponent is 2 Example: Write each power as repeated multiplication and then evaluate (write the answer in standard form). (a) 5 2 = 5 5 = 25
Free lessons, worksheets, and video tutorials for students and teachers. Topics in this unit include: solving degree 1 (linear) equations, balance method, inverse operations, solving simple and multi-step equations, solving equations involving fractions, and rearranging formulas. This follows chapter 4 of the principles of math grade 9 McGraw ...
Grade 9 Mathematics Unit 2: Powers and Exponent Rules Sec 2 What is a Power. 2 5. 2 is the BASE 5 is the EXPONENT The entire 2 5 is called a POWER. 25 = 2 × 2 × 2 × 2× 2 written as repeated multiplication. 25 = 32 written in standard form. 25 = 2 × 2 × 2 × 2× 2 = 32 Power Repeated Standard Multiplication Form
The worksheets cover a variety of math topics that are typically covered in any 9th grade math curriculum, including fractions and decimals, exponents, working with radicals, data and central tendency, solving and graphing linear functions, solving and graphing linear inequalities, geometry, algebra, word problems, and much more!
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163 likes • 161,703 views. Paolo Dagaojes. Grade 9: Mathematics Unit 2 Quadratic Functions. Education. 1 of 71. Download Now. Download to read offline. Grade 9: Mathematics Unit 2 Quadratic Functions.
The Algebra 1 course, often taught in the 9th grade, covers Linear equations, inequalities, functions, and graphs; Systems of equations and inequalities; Extension of the concept of a function; Exponential models; and Quadratic equations, functions, and graphs. Khan Academy's Algebra 1 course is built to deliver a comprehensive, illuminating, engaging, and Common Core aligned experience!
Math 9 Ch 2: Exponent Laws: Assignments • Unit 2.1What are Exponents? ... Math 9 Ch 11: Analysing Data : Assignments • 11.1 Introduction to Non Linear Functions • 11.2 Introduction to Function Notations • 11.3 Introduction to Domain and Range • 11.4 Non Permissible Values ` SOL
Symmetric Property. when you reflect something=6=y y=6. Transitive Property. if a=b and b=c the a=c. Additive Property. adding the same thing to both sides=x-3=x+4. Subtractive property of equality. subtracting the same thing to both sides. Study with Quizlet and memorize flashcards containing terms like plane, coplanar, colinear and more.
10-Problem set 2 11 Nov 25/26. Solving equations - variable on both sides. 11-Variable on both sides notes 11-Practice problems with solutions Nov 2 7/28 Equations Assignment Dec 1-9 December exam review 12-13 Solving equations involving fractions 12- Solving equations with fractions note 13-Fractions Assignment: 14. Jan 9/12
NCERT Solutions for Class 9 Maths Chapter 2. Class 9 Maths Polynomials Exercise 2.1; Class 9 Maths Polynomials Exercise 2.2; Class 9 Maths Polynomials Exercise 2.3; ... CBSE Class 9 Maths Unit-wise Weightage: UNIT NO: UNIT Name: Scoring Marks: UNIT I: Number Systems: 8: UNIT II: Algebra: 17: UNIT III: Coordinate Geometry: 4: UNIT IV: Geometry ...
Problem 2. Use the graph below to answer the questions that follow: Function notation describes a particular relationship and implies the dependency of variables. The general form of function notation is f ( x) = y. y is dependent on x. Part A: The function f above has a point f ( − 2) = 1. Mark this point on the graph above.
9th Maths Assignment 2 Answers | Unit - 2 Real Numbers10th Maths Video Classes Full Pdf English Mediumhttps://drive.google.com/file/d/1ApdFjnJZRf3gf-bmrcXIiU...
Class 9 Maths Notes (Unit # 2) for FBISE (Federal Board) Islamabad (Matric, SSC Part-I, 9th): Very Easy, Updated & Comprehensive Notes for all types of students i.e. Average / Extra Ordinary. All academic material which includes notes, key books, guides, handouts, assignments, solutions, solved exercises, review questions, important questions ...
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These textbook solutions help students in exams as well as their daily homework routine. The solutions included are easy to understand, and each step in the solution is described to match the students' understanding. Mathematics Part 2 Solutions Textbook Solutions for Class 9 MATH. Homework Help with Chapter-wise solutions and Video explanations.
Topics Covered in Class 9 Maths Chapter 2 Polynomials. Definition of a polynomial in one variable, its coefficients, with examples and counter examples, its terms, zero polynomial. Degree of a polynomial. Constant, linear, quadratic, cubic polynomials; monomials, binomials, trinomials. Factors and multiples. Zeros/roots of a polynomial/equation.
maths notes assignment class subject: maths unit real numbers part i.one mark questions. 82 18 an bn (ab)n scientific notation of 102345 is. Skip to document. University; High School; Books; ... Class : 9 Subject: Maths Unit 2- REAL NUMBERS. Part - A I Mark Questions. 2 5 3 2 a) 6 10 b)5 7 c)6 7 d)6 10;
NCERT Solutions for Class 9 Maths. Chapter 1 Number systems. Chapter 2 Polynomials. Chapter 3 Coordinate Geometry. Chapter 4 Linear Equations in Two Variables. Chapter 5 Introduction to Euclid Geometry. Chapter 6 Lines and Angles. Chapter 7 Triangles. Chapter 8 Quadrilaterals.
The powers of the variable x are: 3, 2, 1. The degree of 5x 3 +4x 2 +7x is 3, as 3 is the highest power of x in the equation. (ii) 4-y2. Solution: The highest power of the variable in a polynomial is the degree of the polynomial. Here, in 4-y 2, The power of the variable y is 2.
Math 1201 DF unit 2 - Math 1201 DF unit 2; Math 1201 Unit1 Written Assignment; Written Assignment Unit 1; MATH 1201 College Algebra Written Assignment Unit 1; Related documents. ... Carnival Game Project; Preview text. Written Assignment Unit 2: Graphs, Linear & Quadratic Equations. University of the People. MATH 1201, College Algebra. Man-Fung ...
Mathematics document from University of the People, 4 pages, PROGRAMMING ASSIGNMENT 1. The circumference of a circle is calculated by 2πr, where π = 3.14159 (rounded to five decimal places). Write a function called print_circum that takes an argument for the circle's radius and prints the circle's circumference. Ca
Running head: Written Assignment Unit 2 1 Written Assignment Unit 2 University of the People Math 1211 Calculus Ernest Ampadu, Instructor September 15, 2021 Written Assignment Unit 2 1. Find the limit: 𝐥𝐢𝐦𝒙→𝟎(𝟏 + 𝒙) 𝟏. 𝒙. 𝑥→0lim(1 + 𝑥) 1 𝑥= 𝑒. 2. Find the limit: 𝐥𝐢𝐦𝒙→ ...