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Evaluate by using suitable identity 104*105

WebEvaluate 102×98 using suitable standard identity. Medium Solution Verified by Toppr Using the identity (x+a)(x+b) = x 2+(a+b)x+ab Writing 102 as 100+2 and 98 as 100−2 Hence (100+2)(100−2) = 100 2+[2+(−2)]100+(2)(−2) = 10000+(0)100−4 = 9996 The answer is 9996 Was this answer helpful? 0 0 Similar questions Evaluate: 53×55 Medium View … WebMar 28, 2024 · Transcript Ex 2.5, 7 Evaluate the following using suitable identities: (iii) (998)3 We write 998 = 1000 – 2 (998)3 = (1000 − 2)3 Using (a – b)3 = a3 – b3 – 3ab (a – b) Where a = 1000 & b = 2 = (1000)3 − (2)3 − 3 (1000) (2) (1000 − 2) = 1000000000 − 8 − 6000 (998) = 1000000000 − 8 − 5988000 = 1000000000 − 5988008 = 994011992

Ex 2.5, 2 (iii) - Evaluate 104 × 96 without multiplying directly

WebMar 23, 2024 · Transcript Ex 2.2, 4 Find the product using suitable properties. (a) 738 × 103 738 × 103 We write 103 as (100 + 3) = 738 × (100 + 3) = 738 × 100 + 738 x 3 = 73800 + 2214 = 76014 Next: Ex 2.2, 4 (b) Important → Ask a doubt Chapter 2 Class 6 Whole Numbers Serial order wise Ex 2.2 WebUsing suitable identities, evaluate the following: 98×103 Solution We have, 98×103=(100−2)(100+3) = (100)2+(−2+3)100+(−2)×3 = 10000+100−6 = 10094 [using the identity, (x+a)(x+b)= x2+(a+b)x+ab] Suggest Corrections 31 Video Solution EXEMP - Grade 08 - Mathematics - Algebraic Expressions, Identities and Factorisation - Q86_12 … spider-man always home the movie https://mjengr.com

Evaluate the following products without multiplying directly: - Toppr

WebUsing algebraic Identities, (x + a) (x + b) = x 2 + (a + b)x + ab. (a + b) (a - b) = a 2 - b 2. (i) 103 × 107. Identity: (x + a) (x + b) = x 2 + (a + b)x + ab. 103 × 107 = (100 + 3) (100 + 7) … WebThis video explains the concept of Expansion using Algebraic Identities. All Questions Ask a Doubt Answered Unanswered My Questions My Answers Filter Questions CBSE 8 - Maths Draw and show (A+B) ² =A²+2 A B + B². Asked by muskaanmenuk 10 Jun, 2024, 08:59: PM ANSWERED BY EXPERT CBSE 8 - Maths WebEvaluate (102) 2 using suitable identity. Q. Using suitable identity, evaluate 136 ... spiderman and batman crossover

Ex 2.5, 7 - Evaluate the following using suitable identities: (99)^3

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Evaluate by using suitable identity 104*105

Evaluate the following by using identities: 103 × 97 - Toppr

WebEvaluate the following products without multiplying directly: 104 × 96. Class 8. >> Maths. >> Squares and Square Roots. >> Finding Square of a Number. >> Evaluate the following … WebDec 30, 2024 · answered Solve by using suitable identity 103×105 Advertisement Loved by our community 18 people found it helpful identity (x+a) (x+b)=x²+xb+ax+ab =x²+ (a+b)x+ab Mark any one as brainliest Advertisement Loved by our community 16 people found it helpful Brainly User 103×105 (100+3) (100+5) Identity (x+a) (x+b)=x^2+ …

Evaluate by using suitable identity 104*105

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WebMar 28, 2024 · Transcript Ex 2.5, 7 Evaluate the following using suitable identities: (i) (99)3 We write 99 = 100 – 1 (99)3 = (100 − 1)3 Using (a – b)3 = a3 – b3 – 3ab (a – b) Where a = 100 & b = 1 = (100)3 − (1)3 − 3 (100) … WebSep 15, 2024 · Using suitable identity, evaluate: a) 104 X105 See answer Advertisement Advertisement zaid1233 zaid1233 Answer: answer is 10920. Step-by-step explanation: …

WebSolution: We will be using the algebraic Identity (x + a) (x + b) = x 2 + (a + b)x + ab (i) 103 × 104 = (100 + 3) (100 + 4) = (100) 2 + (3 + 4) (100) + (3) (4) = 10000 + 700 + 12 = 10712 (ii) 5.1 × 5.2 = (5 + 0.1) (5 + 0.2) = (5) 2 + (0.1 + 0.2) (5) + (0.1) (0.2) = 25 + 1.5 + 0.02 = 26.52 (iii) 103 × 98 = (100 + 3) (100 - 2) WebJul 13, 2024 · Evaluate The Following Using Suitable Identities (i) (99)3 (ii) (102)3 (iii) (998)3 AK MtCourse 59K views 2 years ago Factoring Quadratics in 5 seconds! Trick for factorising easily …

WebApr 3, 2024 · The number inside the bracket is 105. We can break the number 105 as 105 = 100 + 5. Substitute the value of 105 = 100 + 5 in equation (1) ⇒ (105)3 = (100 + 5)3. … WebMar 28, 2024 · Ex 2.5, 7 Evaluate the following using suitable identities: (i) (99)3 We write 99 = 100 – 1 (99)3 = (100 − 1)3 Using (a – b)3 = a3 – b3 – 3ab (a – b) Where a = 100 & …

WebEvaluate the following by using identities: 103×97 Medium Solution Verified by Toppr Correct option is A) 103×97=(100+3)(100−3) Using, a 2−b 2=(a+b)(a−b) =(100) 2−(3) 2 …

WebMar 22, 2024 · Example 17 Evaluate 105 × 106 without multiplying directly. 105 × 106 = (100 + 5) × (100 + 6) Using Identity (x + a) (x + b) = x2 + (a + b)x + ab, where x = 100 , … spider-man and black widowWebJul 15, 2024 · We have to evaluate the given expressions using suitable identities. Solution: We know that, (a−b)2 = a2−2ab+b2 ( a − b) 2 = a 2 − 2 a b + b 2 (a−b)2 = a2−2ab+b2 ( a − b) 2 = a 2 − 2 a b + b 2 a2 −b2 = (a+b)(a−b) a 2 − b 2 = ( a + b) ( a − b) Therefore, (a) (102)2 =(100+2)2 ( 102) 2 = ( 100 + 2) 2 spider man and black catWebApr 29, 2024 · 105 × 95 = 9975. Given : 105 × 95. To find : The value using proper algebraic identity. Formula: a² - b² = ( a + b ) ( a - b ) Solution : Step 1 of 2 : Write down … spiderman and black cat kissingWebMar 31, 2024 · Use suitable identity to find (2y + 5) (2y + 5) - Algebra Class 8 Chapter 9 Class 8 Algebraic Expressions and Identities Serial order wise Ex 9.5 Ex 9.5, 1 (ii) - Chapter 9 Class 8 Algebraic Expressions and Identities Last updated at March 22, 2024 by Teachoo Get live Maths 1-on-1 Classs - Class 6 to 12 Book 30 minute class for ₹ 499 ₹ 299 spiderman a long way home streamingWebAug 2, 2024 · a) 104 ×105 (100+4)×(100+5) using identity (x+a) (x+b) = x square ( a+b ) x + ab , where x =100 , a =4 , b=5. (100 ) square +(4+5)(100)+(4×5) 10000 + (9) (100) + … spiderman and black cat lemon fanfictionWebGiven expression can be written as104 x 96 = ( 100 + 4) x ( 100 – 4)We know that( a + b) ( a – b) = a 2 – b 2Herea = 100, b = 4⇒ ( 100 + 4) x ( 100 – 4) = 100 2 - 4 2 = 10000 - 16 = … spider man and black cat ps4 fanficWebTo evaluate volume of rectangular boxes, multiply all the monomials. (i) 5a x 3a2 x 7a4 = (5 × 3 × 7) (a × a2 × a4 ) = 105a7 (ii) 2p x 4q x 8r = (2 × 4 × 8 ) (p × q × r ) = 64pqr (iii) y × 2x2 y × 2xy2 = (1 × 2 × 2 ) ( x × x2 × x × y × y × y2 ) = 4x4 y 4 (iv) a x 2b x 3c = (1 × 2 × 3 ) (a × b × c) = 6abc 5. Obtain the product of (i) xy, yz, zx spiderman and black cat partners in crime