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Class 9 Tuition > CBSE - Class 9 > Learn Class 9 Mathematics

Class 9 Mathematics

Class 9 Mathematics relates to CBSE - Class 9

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Deepashree
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Basaveshwara Nagar, Bangalore
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I have working experience of 2 years teaching for 8th and 9th

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Vigneshwaran
Sholinganallur, Chennai
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With 6 years of teaching experience in both home and online tutoring, I have developed a deep understanding of how to adapt my teaching methods to...

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Riya A.
Rehabari, Gmc
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With 4+ years of experience, I specialize in teaching Class 9 students Science, English, and Social Studies in a way that makes tough topics simple...

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Akshat Soni
Daulah, Sohna
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I have experience teaching class 9 students, focusing on building a strong academic foundation in subjects like Mathematics, Science, and English....

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Karthik K
Gosha Mahal, Hyderabad
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I am a engineer, i have completed my graduation in 2020, I have one years of experience in teaching, i teach mathematics, science, social, biology,...

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Hemant Kumar
Nangloi, Delhi
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I have taught hundreds of students of class 9th so far from various boards such as CBSE, ICSE and state boards. I am also running my own coaching...

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Nihal O.
Uttam Nagar, Delhi
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I'm Nihal, an experienced tutor with over 7 years of teaching expertise. I have done my bachelors in Chemistry, scoring 8.3 CGPA from University of...

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Lavanya S.
Urmelalagian, Urmelazhagian
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Iam a engineering student.iam giving online tuition since mathamatics, chemistry, home science. Iam completed in schooling in amali girls hr sec school...

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Abhishek Lodha
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Jayanagar RBI Extension, Bangalore
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I am passionate about helping those who struggle with understanding Maths and Science. I firmly believe in conceptual learning and strive for my students...

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Aditya Kumar Singh
Old Katra, Prayagraj
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I am a teacher. I am giving tuition since 2021. I have a degree in engineering. I am an experienced tuition teacher specializing in mathematics, science. Skilled...

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Class 9 Mathematics Questions

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Answered on 09 Aug Learn CBSE - Class 9/Class 9 Mathematics/Chapter 2 Polynomials/EXERCISE - 2.2

Rakhi Yadav

more than 5 year experience tutor

Okay, let's find the values of the polynomials at the specified points. (i) p(y) = y2 – y + 1 p(0) = 02 – 0 + 1 = 1 p(1) = 12 – 1 + 1 = 1 p(2) = 22 – 2 + 1 = 3 (ii) p(t) = 2 + t + 2t2 – t3 p(0) = 2 + 0 + 2(0)2 – 03 = 2 p(1) = 2 + 1 + 2(1)2 – 13 = 4 p(2) =... read more

Okay, let's find the values of the polynomials at the specified points.

(i) p(y) = y2 – y + 1

p(0) = 02 – 0 + 1 = 1

p(1) = 12 – 1 + 1 = 1

p(2) = 22 – 2 + 1 = 3

(ii) p(t) = 2 + t + 2t2 – t3

p(0) = 2 + 0 + 2(0)2 – 03 = 2

p(1) = 2 + 1 + 2(1)2 – 13 = 4

p(2) = 2 + 2 + 2(2)2 – 23 = 4

(iii) p(x) = x3

p(0) = 03 = 0

p(1) = 13 = 1

p(2) = 23 = 8

(iv) p(x) = (x – 1) (x + 1)

p(0) = (0 – 1) (0 + 1) = -1

p(1) = (1 – 1) (1 + 1) = 0

p(2) = (2 – 1) (2 + 1) = 3

Here are the final answers:

(i) p(0) = 1, p(1) = 1, p(2) = 3

(ii) p(0) = 2, p(1) = 4, p(2) = 4

(iii) p(0) = 0, p(1) = 1, p(2) = 8

(iv)

p(0) = -1, p(1) = 0, p(2) = 3

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Answered on 02 Sep Learn CBSE - Class 9/Class 9 Mathematics/Chapter 2 Polynomials/EXERCISE - 2.4

Evaluate the following products without multiplying directly: (i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96

Math Decode Institute

i) 103×107=(100+3)(100+7)=1002+100(3+7)+3⋅7=10000+1000+21=11021. (ii) 95×96=(100−5)(100−4)=1002−100(5+4)+5⋅4=10000−900+20=9120. (iii) 104×96=(100+4)(100−4)=1002−42=10000−16=9984. read more

i) 103×107=(100+3)(100+7)=1002+100(3+7)+3⋅7=10000+1000+21=11021.

(ii) 95×96=(100−5)(100−4)=1002−100(5+4)+5⋅4=10000−900+20=9120.

(iii) 104×96=(100+4)(100−4)=1002−42=10000−16=9984.

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Answered on 23 Aug Learn CBSE - Class 9/Class 9 Mathematics/Chapter 2 Polynomials/EXERCISE - 2.4

Rakhi Yadav

more than 5 year experience tutor

Using (A+B+C)2=A2+B2+C2+2AB+2AC+2BC(A+B+C)^2=A^2+B^2+C^2+2AB+2AC+2BC: (i) (x+2y+4z)2=x2+4y2+16z2+4xy+8xz+16yz(x+2y+4z)^2=x^2+4y^2+16z^2+4xy+8xz+16yz (ii) (2x−y+z)2=4x2+y2+z2−4xy+4xz−2yz(2x-y+z)^2=4x^2+y^2+z^2-4xy+4xz-2yz (iii) (−2x+3y+2z)2=4x2+9y2+4z2−12xy−8xz+12yz(-2x+3y+2z)^2=4x^2+9y^2+4z^2-12xy-8xz+12yz (iv)... read more

Using (A+B+C)2=A2+B2+C2+2AB+2AC+2BC(A+B+C)^2=A^2+B^2+C^2+2AB+2AC+2BC:

(i) (x+2y+4z)2=x2+4y2+16z2+4xy+8xz+16yz(x+2y+4z)^2=x^2+4y^2+16z^2+4xy+8xz+16yz

(ii) (2x−y+z)2=4x2+y2+z2−4xy+4xz−2yz(2x-y+z)^2=4x^2+y^2+z^2-4xy+4xz-2yz

(iii) (−2x+3y+2z)2=4x2+9y2+4z2−12xy−8xz+12yz(-2x+3y+2z)^2=4x^2+9y^2+4z^2-12xy-8xz+12yz

(iv) (3a−7b−c)2=9a2+49b2+c2−42ab−6ac+14bc(3a-7b-c)^2=9a^2+49b^2+c^2-42ab-6ac+14bc

(v) (−2x+5y−3z)2=4x2+25y2+9z2−20xy+12xz−30yz(-2x+5y-3z)^2=4x^2+25y^2+9z^2-20xy+12xz-30yz

(vi) (14a−12b+1)2=116a2+14b2+1−14ab+12a−b\left(\tfrac14 a-\tfrac12 b+1\right)^2=\tfrac1{16}a^2+\tfrac14 b^2+1-\tfrac14 ab+\tfrac12 a-b

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Answered on 22 Aug Learn CBSE - Class 9/Class 9 Mathematics/Chapter 1 Number Systems/EXERCISE - 1.5

Shivam Yadav

i) 8²×½ =8 ii) 2⁵×⅕=2 iii) 5³×⅓=5
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Answered on 15 Aug Learn CBSE - Class 9/Class 9 Mathematics/Chapter 1 Number Systems/EXERCISE - 1.3

Radhika V

Physics tutor with 5 years experience

1/7= 0.142857 2/7=2*1/7=2*0.142857=0.285714 3/7=3*1/7=3*0.142857=0.428571
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