Given that, a 2 b 2 c 2 = 50 and a b c = 12 We need to find ab bc ca Substitute the values of (a 2 b 2 c 2) and (a b c) in the identity (1), we haveSimplify calculation one by one \( (abc)^2 = (a^2 abac) b \times (abc) c \times (abc) \) \(=> (abc)^2 = (a^2 abac) (ab b^2 bc) c \times (abc) \) \(=> (abc)^2 = (a^2 abac) (ab b^2 bc) (acbc c^2) \)b2 c2 = 250 & abbcac = 3, find abc → the general formula (abc)2 = a2 b2 c2 2(abbcac) = 2502(3)
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What is the formula of (a+b-c)2- Stepbystep explanationabc = 5 and a^2b^2c^2 = 29Using the formula(abc)^2 = a^2b^2c^22(abbcca)=>(ab(c)^2= a^2b^2c^22(abbcca)=>(abc) dhrubaprasadsingh dhrubaprasadsingh Math Secondary School If a b c = 5 and a^2 b^2 c^2 = 29 then find the value of ab bc ca(a b c) 2 = a 2 b 2 c 2 2 (ab bc ca)
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V=2(abbcca), solve for a 26,037 results a temperature formula solve for F K=5/9(F32)273 b annual interest rate solve for R A=PPrt algebra 1 solve 2x y = 7 for y 2 solve 2y 18x = 26 for y 3 solve x 5y =15 for y 4 solve 9x y = 45 for y 5 solve 6x 3y = 6 for y 6 solve 4x8)establish the formula ( abc)² = a² b² c2² 2( ab bc ca ) and give 4 Ask for details ; $(abc)^2=a^2b^2c^22(abbcca)=12(abbcca)$ therefore $$(abbcca)=((abc)^21)/2$$ so the value is not fixed
2 weeks ago Math Secondary School What is the formula of a^2b^2c^22(abbcca) 1 See answer snehahans357 is waiting for your help Add your answer and earn points neha5390On subtracting 2ab 2bc 2ca from both sides of the above formula, the a 2 b 2 c 2 formula is a 2 b 2 c 2 = (a b c) 2 2 (ab bc ca)B2 abbcac b 2 a b b c a c Factor out the greatest common factor from each group Tap for
(abbcca)^2 formulaThe Health Gateway is a Ministry of Health initiative which provides BC residents and their families with secure access to a single view of their health information View and download your health information, such as COVID19 tests, COVID19 immunizations, medications, immunizations and health visitsEx 42, 7 By using (abc)^2 >hoac=3(abbcca) giup minh moi minh dang can gap Theo dõi Vi phạm YOMEDIA Trả lời (1) (abc) 2 \(\ge\) 3(abbcca) (*) =>a 2 b 2 c 2 2ab2bc2ca \(\ge\) 3ab3bc3ca =>a 2 b 2 c 2 \(\ge\) abbcca nhân 2 vào cho 2 vế ta được 2a 2 2b 2A 2 b 2 c 2 2ab 2bc 2ca = 625 a 2 b 2 c 2 2 (ab bc ca) = 625 a 2 b 2 c 2 2 × 59 = 625 Given, ab bc ca = 59 a 2 b 2 c 2 118 = 625 a 2 b 2 c 2 118 – 118 = 625 – 118 subtracting 118 from both the sides Therefore, a 2 b 2 c 2 = 507
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Free math problem solver answers your algebra, geometry, trigonometry, calculus, and statistics homework questions with stepbystep explanations, just like a math tutorIf a 2 b 2 c 2 abbcca=0 then prove that a=b=c Hello student, Please find the answer to your question below a² b² c² = ab bc ca MultClick here👆to get an answer to your question ️ If a^2 b^2 c^2 = 16 and ab bc ca = 10 , find the value of a b c Mathematical Formula for Competitive Exams Mathematical formula are one of the most important thing in exams Time is the main factor in competitive exams a 3 b 3 c 3 – 3abc = (a b c)(a 2 b 2 c 2 – ab – bc – ca) If a b c = 0, then the above identity reduces to a 3 b 3 c 3
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The third formula shown is the result of solving for a in the quadratic equation a 2 − 2ab cos γ b 2 − c 2 = 0 This equation can have 2, 1, or 0 positive solutions corresponding to the number of possible triangles given the dataOn multiplying both sides by '2', it becomes 2 (a² b² c²) = 2 (ab bc ca) 2a² 2b² 2c² = 2ab 2bc 2ca a² a² b² b² c² c² – 2ab – 2bc – 2ca = 02 b 2 c 2 ab bc ca as Sum of Squares Here we will express a 2 b 2 c 2 – ab – bc – ca as sum of squares a 2 b 2 c 2 – ab – bc – ca = 1 2 {2a 2 2b 2 2c 2 – 2ab – 2bc – 2ca} = 1 2 { (a 2 b 2 – 2ab) (b 2 c 2 – 2bc) (c 2 a 2 – 2ca)}
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Consider, a 2 b 2 c 2 – ab – bc – ca = 0 Multiply both sides with 2, we get 2( a 2 b 2 c 2 – ab – bc – ca) = 0 ⇒ 2a 2 2b 2 2c 2 – 2ab – 2bc – 2ca = 0 ⇒ (a 2 – 2ab b 2) (b 2 – 2bc c 2) (c 2 – 2ca a 2) = 0 ⇒ (a –b) 2 (b – c) 2 (c – a) 2 = 0 Since the sum of square is zero then eachRD Sharma Class 9 Solutions Chapter 12 Heron's Formula RD Sharma Class 9 Solution Chapter 12 Heron's Formula Ex 121 Question 1 In the figure, the sides BA and CA have been produced such that BA = AD and CA = AE(ii) `a – b ab, b – c bc, c – a ac` Answer `=(a b ab) (b c bc) (c a ac)` `= a b c ab bc ca b c a` `= ab bc ca` (iii) `2p^2q^2 – 3pq 4, 5 7pq – 3p^2q^2` Answer `= (2p^2q^2 3pq 4) (5 7pq 3p^2q^2)` `= 2p^2q^2 3p^2q^2 3pq 7pq 4 5` `= p^2q^2 4pq 9` (iv) `l^2 m^2
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There are various student are search formula of (ab)^3 and a^3b^3 Now I am going to explain everything below You can check and revert back if you like you can also check cube formula in algebra formula sheet a2 – b2 = (a – b)(a b) (ab)270以上 ab bc ca formula (abbcca)^2 formula You can check the formulas of A plus B plus C Whole cube in three ways We are going to share the (abc)^3 algebra formulas for you as well as how to create (abc)^3 and proof we can write we know that what is the formula of need too write in simple form of multiplication Simplify the allIf a = 1001, b = 1002, c = 1003 , then value of a2 b2 c2 – ab – bc – ca isIn this video, we will understand short trick of Algebraic expressions a^2
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Yes No Please Login to Read SolutionSolve for a v=2(abbcca) Rewrite the equation as Divide each term by and simplify Tap for more steps Divide each term in by Cancel the common factor of Tap for more steps Cancel the common factor Divide by Subtract from both sides of the equation Factor out ofTherefore, AB 2 AC 2 = BC 2, since CBDE is a square This proof, which appears in Euclid's Elements as that of Proposition 47 in Book 1, 10 demonstrates that the area of the square on the hypotenuse is the sum of the areas of the other two squares 11
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X³ (abc) x² (abbcca) xabc a² b² c² a b b c c a = ½ (ab)²(bc)²(ca)² (abc) (a² b² c² a bb cc a) = a³ b³ c³ 3 a b c Logarithms Math Formulas Product rule log ₐ (m n) = log ₐ m log ₐ n This one is a longway of solving (With a very obvious result)\begin{gather*} \implies(abbcca)^3abc(abc)^3=0\\ \end{gather*} \begin{multline*} \implies (a^3b^3b^3c^3c^3a^33a^2b^3c3a^2bc^33a^3b^2c3ab^2c^33a^3bc^23ab^3c^2)\\abc(a^3b^3c^33a^2b3ab^23b^2c3bc^23c^2a3ca^2)=0 \end{multline*} After simplifying \begin{equation} \implies a^3b^3b^3c^3c^3a^3a^4bcabAnswer Given, a2 b2 c2 −ab−bc−ca multiply and divide by 2 = 22 ×(a2 b2 c2 −ab−bc−ca) = 2a2 −2abb2 b2 −2bcc2 c2 −2aca2
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Example 2 Solve (8x 4y 7z) 2 Solution This proceeds as Given polynomial (8x 4y 7z) 2 represents identity first ie (a b c) 2 Where a = 8x, b = 4y and c = 7z Now apply values of a, b and c on the identity ie (a b c) 2 = a 2 b 2 c 2 2ab 2bc 2ca and we get (8x 4y 7z) 2 = (8x) 2 (4y) 2 (7z) 2 2(8x)(4y) 2(4y)(7z) 2(7z)(8x) Expand the exponential formsExample Solve 8a 3 27b 3 125c 3 30abc Solution This proceeds as Given polynomial (8a 3 27b 3 125c 3 30abc) can be written as (2a) 3 (3b) 3 (5c) 3 (2a)(3b)(5c) And this represents identity a 3 b 3 c 3 3abc = (a b c)(a 2 b 2 c 2 ab bc ca) Where a = 2a, b = 3b and c = 5c Now apply values of a, b and c on the LHS of identity ie a 3 b 3 c 3= a 2 ab ac ba b 2 bc ca cb c 2 Adding like terms, the final formula (worth remembering) is (a b c) 2 = a 2 b 2 c 2 2ab 2bc 2ac Practice Exercise for Algebra Module on Expansion of (a b c)
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From the formula (abc)^2=a^2b^2c^22(abbcca) 2506=256 abc=16 Is this solution Helpfull?2) Now, let a = b = c = 4 Now substitute these values of a, b, c in the above formula We get 16 16 16 16 16 16 = 48 48 = 0 Hence, it is possible that a 2 b 2 c 2 ab bc ca2 29 if a ib=0 wherei= p −1, then a= b=0 30 if a ib= x iy,wherei= p −1, then a= xand b= y 31 The roots of the quadratic equationax2bxc=0;a6= 0 are −b p b2 −4ac 2a The solution set of the equation is (−b p 2a −b− p 2a where = discriminant = b2 −4ac 32
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If a^2 b^2 c^2 = 41 and a b c = 9, then find ab bc ca FORMULA = Ask for details ;(abbcca)² = (ab) ²(bc)²(ca)²2(ab)(bc)2(bc)(ca)2(ca)(ab) hence, (ab)²(bc)²(ca)²2ab²c2abc²2a²bc (ab)²(bc)²(ca)²2abc(bca)12e 2 (abacadbcbdcd) We can choose x 2 as above in 6 ways From the remaining 2 factors, we can choose two different letters in C(2,1)⋅C(1,1) ways, giving in all 12 ways We can choose 1 letter from 5 in 5 ways And 2 letters from 4 in C(4,2) ways giving 30 ways The sum of the coefficients is therefore 30⋅12=360
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Full SiteNavigation Square of a Trinomial How to expand the square of a trinomial?The square of the sum of three or more terms can be determined by the formuThe area of whole square is ( a b c) 2 geometrically The whole square is split as three squares and six rectangles So, the area of whole square is equal to the sum of the areas of three squares and six rectangles ( a b c) 2 = a 2 a b c a a b b 2 b c c a b c c 2 Now, simplify the expansion of the a b c wholeSince ( a b c ) 2 = a 2 b 2 c 2 2 ( ab bc ca ) ∴ ( a b c ) 2 = 50 2 (47) ⇒ ( a b c ) 2 = 50 94 = 144 ⇒ a b c = `sqrt144 = 12` ∴ a b c = `12` Concept Expansion of
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Multiply each term of first polynomial with every term of second polynomial, as shown below = a (a 2 b 2 c 2 ab bc ca) b (a 2 b 2 c 2 ab bc ca) c (a 2 b 2 c 2 ab bc ca)
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