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# using identities evaluate 103 × 98

Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM. Expansion of algebraic expressions using identities. 102 × 106 = (100 + 2) (100 + 6) This is in the form. 1 Answer +1 vote . Evaluate using identities (a) 103×97. (ii) Two terms like 4mn 2 are: 6mn 2 and –2n 2 m. (iii) Two terms like 21 are: 51 and -7b. Using formula (a + b) 2 = a 2 + b 2 + 2ab = 70 2 + 1 2 + 140 = 4900 + 140 +1 = 5041 (ii) 99². Question Bank Solutions 15191. Ex 9.5, 8 - Chapter 9 Class 8 Algebraic Expressions and Identities Last updated at Dec. 24, 2018 by Teachoo Subscribe to our Youtube Channel - https://you.tube/teachoo using the identities evaluate 1)94 * 94 - 6 * 6/ (94 -6) 2) 9 2 *8 8 3) 103 * 98 4) 103 (square) - Math - Algebraic Expressions and Identities 105 × 106 = (100 + 5) × (100 + 6) Using Identity (x + a) (x + b) = x2 + (a + b)x + ab, .. Expand using suitable identity (-2x + 5y - 3z) to the whole square. Using suitable identity, evaluate the following. Share 4 . ∵ 297 = 300 - 3 and 303 = 300 + 3 ∴ 297 x 303 = (300 - 3) (300 + 3) = (300)2 - (3)2 [Using (a + b)(a - b) = a2 -b2] = 90000 - 9 = 89991 0 votes. Evaluate each of the following using identities: (i) (399)^2 (ii) (0.98)^2 (iii) 991 x 1009 (iv) 117 x 83 asked Feb 11 in Algebraic Expressions by ShasiRaj ( 62.4k points) algebraic identities = a 2 – b 2 + b 2 – c 2 + c 2 – a 2 = 0 = RHS. Call our LearnNext Expert on 1800 419 1234 (tollfree) OR submit details below for a call back Solution: 99² = (100 -1) 2 Using formula (a – … 1000000+9+90000+2700. (103)^2, Evaluate using suitable identities. asked Sep 13, 2018 in Class IX Maths by muskan15 (-3,443 points) polynomials. Another method to verify the algebraic identity is the activity method. Try squaring 1013. Tamil Nadu Board of Secondary Education SSLC (English Medium) Class 9th. Trigonometric identities like sin²θ+cos²θ=1 can be used to rewrite expressions in a different, more convenient way. Use Identities to Evaluate : (97)2 . Evaluate the following using suitable identities (98)3 - Math - Polynomials. Polynomials. Expand using suitable identity (-2x + 5y - 3z) to the whole square. Important Solutions 3. Evaluate the following using suitable identities (98) 3. 98 x 103. algebraic expressions; factorisation; class-8; Share It On Facebook Twitter Email. Class-9 » Math. Ex 9.5, 8 - Chapter 9 Class 8 Algebraic Expressions and Identities Last updated at Dec. 24, 2018 by Teachoo Subscribe to our Youtube Channel - https://you.tube/teachoo Related Videos. Evaluate:(104)cube using a suitable identity. NCERT Solutions; Board Paper Solutions; Ask & Answer; School Talk; Login; GET APP; Login Create Account. (2) Evaluate the following by using identities: (i) 98 3 (ii) 1001 3 Solution (3) If (x + y + z) = 9 and (xy + yz + zx) = 26, then find the value of x 2 + y 2 + z 2 Solution (4) Find 27a 3 + 64b 3, if 3a + 4b = 10 and ab = 2. Evaluate using suitable identities. Learn and explore the use of algebraic identities or formulas to evaluate the numerical expressions with this series of printable worksheets. Question Bank Solutions 1231. Available for CBSE, ICSE and State Board syllabus. Solution: 71 2 = (70+1) 2. Related Videos. (98)^2, Evaluate using suitable identities. (i) 71². (i) 103 3 (ii) 101 x 102 (iii) 999 2 Firstly adjust the given number into two number such that one is a multiple of 10 and use the proper identity. ... 103 × 98 (iv) 9.7 × 9.8. ∵ 78 = 80 - 2 and 82 = 80 + 2 ∴ 78 x 82 = (80 - 2) (80 + 2) = (80)2 - (2)2 [Using (a + b)(a - b) = a2 -b2] = 6400 - 4 = 6396 (i) Two terms like 7xy are: –3xy and 8xy. asked Jan 30, 2018 in Class IX Maths by navnit40 (-4,939 points) polynomials +1 vote. Share 0 (98) 3 = (100 2) 3. (i) 712 (ii) 992 (iii) (v) 5.22 (vi) 297 × 303 (vii) (ix) 1.05 × 9.5 7. Ask questions, doubts, problems and we will help you. Example 17 Evaluate 105 × 106 without multiplying directly. Using suitable identity, evaluate the following. Ask questions, doubts, problems and we will help you. Using identity evaluate: 107 x 103. Evaluate using identities:- (a) 103 X 97 (b) (0 99)(0 99) (c) 105 X 105 X 105 - Math - Polynomials Advertisement Remove all ads. Textbook Solutions 19009. Using identities, evaluate. evaluate using suitable identities 1 . (99)^3 2. Syllabus. Concept Notes & Videos 257. 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) (1) (100 − 1) = 1000000 − 1 − 300(99) = 1000000 − 1 − 29700 = 970299 Ex 2.5, 7 Eva Share with your friends.  (103)^3 BSEB matric hall ticket 2021 has been released, know steps to download. Syllabus. 102 × 106. algebraic expressions; class-8; Share It On Facebook Twitter Email. Using Identities to Evaluate Once you've learned some trig identities , you can use them to evaluate angles exactly. Using identities, evaluate : . Paiye sabhi sawalon ka Video solution sirf photo … To ask Unlimited Maths doubts download Doubtnut from - https://goo.gl/9WZjCW Evaluate using suitable identity (ii) 97 xx 103 Yes, you can always get a decimal approximation with your calculator, but it's good to learn the logic of using the identities, so you can exercise your skills at pattern-finding and -matching. 0 votes . Using identities, evaluate. 101 x 103, Evaluate using suitable identities. Get a free home demo of LearnNext. Using suitable identity evaluate (103) 3. Concept Notes & Videos 257. Do you see that Identity (I) has given us a less tedious method than the direct method of squaring 103? answered Aug 4, 2020 by Dev01 (51.7k points) selected Aug 5, 2020 by Rani01 . All Questions Ask Doubt (2/3m+3/2n) ² If x cube + 2 square equal to a square x 6 plus x cube + 4, then find a. (i) 1033           (ii) 101 x 102       (iii) 9992 Firstly adjust the given number into two number such that one is a multiple of 10 and use the proper identity. We may also directly multiply 103 by 103 and get the answer. Using suitable identity, evaluate the following. 1092709. thumbs up if u understood!! Question Papers 10. 1 Answer. Get NCERT Solution class 8 Maths Chapter 9 Algebraic Expressions and Identities Exercise-9.5. (i) 71 2 (ii) 99 2 (iii) 102 2 (iv) 998 2 (v) 5.2 2 (vi) 297 x 303 (vii) 78 x 82 (viii) 8.9 2 (ix) 1.05 x 9.5 Solution: Ex 9.5 Class 8 Maths Question 7. using identity evaluate 107 - Mathematics - TopperLearning.com | o5uhf4ybb Using identity evaluate: 107 x 103. 12 ; View Full Answer (100+3)3. By using identity evaluate the following: 73 – 103 + 33 . 10.1 x 10.2, Evaluate using suitable identities. 100^3 + 3^3+ 3.100^2.3+3.100.3^2. Using factor theorem, factorise the polynomial x3 + x2 - 4x - 4. Textbook Solutions 1231. we have followed CBSE evaluation pattern where you can score on steps and not just on final answer. answered Mar 31, 2020 by Sunil01 (67.6k points) selected Apr 2, 2020 by Mohini01 . Using identities evaluate 99×103. If p (x) = x2 – 4x + 3, then evaluate p(2) – p (-1) + p (1/2). Using identities, evaluate : . Concept Notes & Videos 256. Best answer. The algebraic identities are verified using the substitution method. Advertisement Remove all ads. 105 × 106 = (100 + 5) × (100 + 6) Using Identity (x + a) (x + b) = x2 + (a + b)x + ab, .. Using suitable identity, evaluate the following. 10 ; oops the answer is 1092727, in third step 3^3 will be 27. By using identity evaluate the following: 73 – 103 + 33 . Evaluate using suitable identities. (i) 103 3 (ii) 101 x 102 (iii) 999 2 Firstly adjust the given number into two number such that one is a multiple of 10 and use the proper identity. (i) 71 2 (ii) 99 2 (iii) 102 2 (iv) 998 2 (v) 5.2 2 (vi) 297 x 303 (vii) 78 x 82 (viii) 8.9 2 (ix) 1.05 x 9.5 Solution: Ex 9.5 Class 8 Maths Question 7. Asked by Topperlearning User | 4th Jun, 2014, 01:23: PM. Best answer. (103) 2 = (100 + 3) 2 = 100 2 + 2 × 100 × 3 + 3 2 (Using Identity I) = 10000 + 600 + 9 = 10609. In this method, substitute the values for the variables and perform the arithmetic operation. Click hereto get an answer to your question ️ Using suitable identity, evaluate (i) $$102 \times 98 ($$ (ii) $$) ( 2 x - 3 )$$ (iii) $$( 4 y + 3 ) ( 4 y$$ Using suitable identity, evaluate the following. (103) 2. algebraic expressions; factorisation; class-8; Share It On Facebook Twitter Email. 1 Answer +1 vote . Evaluate each of the following by using identities: (i)103\ xx\ 97 (ii) 103\ xx\ 103 (iii) (97)^2 Evaluate each of the following by using identities: (i)103\ xx\ 97 (ii) 103\ xx\ 103 (iii) (97)^2 Doubtnut is better on App. 98 x 103, Evaluate using suitable identities. Evaluate using suitable identities. We have (0.98) 2 = [1 - 0.02] 2 = (1) 2 + (0.02) 2 - 2 x 1 x 0.02 = 1 + 0.0004 - 0.04 [∵ a = 1, b = 0.02] All Questions Ask Doubt (2/3m+3/2n) ² If x cube + 2 square equal to a square x 6 plus x cube + 4, then find a. Use Identities to Evaluate : (97)2 - Mathematics. Share with your friends. answered Aug 4, 2020 by Dev01 (51.7k points) selected Aug 5, 2020 by Rani01 . Using factor theorem,show that (x-y) is a factor of x(y(square) - z(square)) + y(z(square) - x(square)) + z(x(square) - y(square) ). Evaluate the following using identities: (0.98) 2 Solution Show Solution. All the solution has been made by experts in detail. Best answer. Evaluate 105 x 108 without multiplying directly. Example 17 Evaluate 105 × 106 without multiplying directly. Question 6. Evaluate each of the following using identities: (i) (399)^2 (ii) (0.98)^2 (iii) 991 x 1009 (iv) 117 x 83 asked Aug 4, 2018 in Mathematics by Sonu khan ( 35.6k points) algebraic identities Expansion of algebraic expressions using identities. Expert Answer: Answered by | 4th Jun, 2014, 03:23: PM. Bihar board class 10 admit card 2021 latest update. Expert Answer: Answered by | 4th Jun, 2014, 03:23: PM. Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. 9.8 x 10.2. Question Bank Solutions 1231. Textbook Solutions 1231. Solution (5) Find x 3 - y 3, if x - y = 5 and xy = 14 Solution Evaluate using suitable identities. Advertisement Remove all ads. 1 answer. For example, (1-sin²θ)(cos²θ) can be rewritten as (cos²θ)(cos²θ), and then as cos⁴θ. Tamil Nadu Board of Secondary Education SSLC (English Medium) Class 9th. 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