Byjus solution class 8 maths ch 3
WebAccess Answers to NCERT Class 8 Maths Chapter 1 Rational Numbers Exercise 1.1 Page: 14 1. Using appropriate properties, find: (i) -2/3 × 3/5 + 5/2 – 3/5 × 1/6 Solution: -2/3 × 3/5 + 5/2 – 3/5 × 1/6 = -2/3 × 3/5– 3/5 × 1/6+ 5/2 (by commutativity) = 3/5 (-2/3 – 1/6)+ 5/2 = 3/5 ( (- 4 – 1)/6)+ 5/2 = 3/5 ( (–5)/6)+ 5/2 (by distributivity) WebIn this page, you will complete detailed NCERT Solutions for Class 10 of all subjects …
Byjus solution class 8 maths ch 3
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WebExercise 8.3 of Class 10 Maths recalls the concept of complementary angles. In a right-angle triangle two angles are said to be complementary when their sum equals 90°. In this chapter, the complementary angles of six trigonometric functions, sin, cos, tan, cosec, sec and cot, are formulated with respect to the angles that lie between 0° and 90°. WebAccess answers to Maharashtra Board Solutions For Class 8 Maths Part 1 Chapter 3- Indices and Cube Root. Practice set 3.1 PAGE NO: 15. 1. Express the following numbers in index form. (1) Fifth root of 13. (2) Sixth root of 9. (3) Square root of 256. (4) Cube root of 17. (5) Eighth root of 100.
WebSolution: By resolving 46656 into prime factor, 46656 = 2×2×2×2×2×2×3×3×3×3×3×3 By grouping the factors in triplets of equal factors, 46656 = (2×2×2)× (2×2×2)× (3×3×3)× (3×3×3) Here, 46656 can be grouped into triplets of equal factors, ∴ 46656 = (2×2×3×3) = 36 Hence, 46656 is the cube of 36. 7 2. WebCBSE Class 8 Maths Chapter 14 Factorisation Exercise 14.3, is based on the topic of the division of algebraic expressions and subtopics like division of a monomial by another monomial, division of a polynomial by a monomial and division of algebraic expressions continued (Polynomial ÷ Polynomial).
WebNCERT Solutions for Class 10 Maths Chapter 3; NCERT Solutions for Class 10 Maths Chapter 4; NCERT Solutions for Class 10 Maths Chapter 5; NCERT Solutions for Class 10 Maths Chapter 6; ... Give the BNAT exam to get a 100% scholarship for BYJUS courses. C. b ... WebClass 8 Maths Chapter 3 Understanding Quadrilaterals MCQs These objective questions are designed for students, as per the CBSE syllabus (2024-2024) and NCERT guidelines. Solving the chapter-wise questions will help students understand each concept and help to score good marks in exams.
WebChapter 3 Squares and Square Roots of RD Sharma Class 8 Maths gives an account of the various topics like a square of a number and square numbers, properties and patterns of some square numbers, some short-cuts to finding squares (column method, visual method, diagonal method), square roots, square root of a perfect square by prime factorisation, …
WebSolution: (i) 3 -2 = (1/3) 2 = 1/9 (ii) (-4)-2 = (1/-4)2 = 1/16 (iii) (1/2)-5 = (2/1)5 = 2 5 = 32 2. Simplify and express the result in power notation with a positive exponent: (i) (-4)4 ÷ (-4)8 (ii) (1/23)2 (iii) - (3)4× (5/3)4 (iv) (3-7÷3-10)×3-5 (v) 2-3× (-7)-3 Solution: (i) = (-4) 5 / (-4) 8 = (-4) 5-8 = 1/ (-4) 3 (ii) (1/23)2 = 1 2 / (2 3) 2 hard dollar estimating toolWebSolution. The correct option is A The Rajputs had a standing army. Instructions: Causes for the failure of the Rajputs: Rajputs had slow-moving elephants as opposed to Turkish cavalry. Rajput states were feudal in nature. This divided the revenue between feudal lords and kings. Rajputs did not have a standing army. changa marathi.comchanga manga is located nearWebNCERT Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Solution:-By ASA congruence property: Two triangles are congruent if the two angles and the included side of one are respectively equal to the two angles and the included side of the other. ΔLMN ≅ ΔGFH (d) Given: EB = DB AE = BC ∠A = ∠C = 90 o So, ΔABE ≅ ΔACD Solution:- chang alvinWebNCERT Solutions for Class 10 Maths Chapter 8; NCERT Solutions for Class 10 Maths Chapter 9; NCERT Solutions for Class 10 Maths Chapter 10; ... Give the BNAT exam to get a 100% scholarship for BYJUS courses. B. 4.00. Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses. C. 4. hard donationWebSolution: 3x = 2x + 18 ⇒ 3x – 2x = 18 ⇒ x = 18 Putting the value of x in RHS and LHS we get, 3 × 18 = (2 × 18) +18 ⇒ 54 = 54 ⇒ LHS = RHS 2. 5t – 3 = 3t – 5 Solution: 5t – 3 = 3t – 5 ⇒ 5t – 3t = -5 + 3 ⇒ 2t = -2 ⇒ t = -1 Putting the value of t in RHS and LHS we get, 5× (-1) – 3 = 3× (-1) – 5 ⇒ -5 – 3 = -3 – 5 ⇒ -8 = -8 ⇒ LHS = RHS 3. 5x + 9 = 5 + 3x hard dollar softwareWebNCERT Solutions for Class 10 Maths Chapter 3; NCERT Solutions for Class 10 … chan game tren may tinh