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Quantitative Aptitude Arithmetic Ability Questions & Answers for Bank Exams If a b c = 8 and ab bc ca = , then a^3 b^3 c^3 3abc is equal toSolution for ABBC=AC equation Simplifying AB BC = AC Solving AB BC = AC Solving for variable 'A' Move all terms containing A to the left, all other terms to the right Add '1AC' to each side of the equation AB 1AC BC = AC 1AC Combine like terms AC 1AC = 0 AB 1AC BC = 0 Add '1BC' to each side of the equationView the profiles of people named AB BC Join Facebook to connect with AB BC and others you may know Facebook gives people the power to share and makes
The following map contains all ABB locations in CanadaUpdate 234 COVID19 pandemic in Alberta (June 29) Update 233 COVID19 pandemic in Alberta (June 22) Update 232 COVID19 pandemic in Alberta (June 15) Update 231 COVID19 pandemic in Alberta (June 10) Update 230 COVID19 pandemic in Alberta (June 8) Update 229 COVID19 pandemic in Alberta (June 3) Update 228 COVID19 pandemic in AlbertaHowever, there is a problem when trying to prove (abc)^2 ≤ 3 (abbcca), because, in fact, the opposite is true (abc)^2 ≥ 3 (abbcca) You can see that if you expand (abc)^2, simplify, multiply by 2, and use the trivial inequality
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Your approach is intuitive and that was also the first thing I thought;Without loss of generality, we may suppose that AD is the minimum side (1) When AB = AD, we have BC = C D In this case, letting O be the intersection point of AC and the bisector of ∠B, Equivalence classes and rational numbersA 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 3abc = (a b c)(a 2 b 2 c 2 ab bc ca) and we get = (2a 3b 5c) { (2a) 2 (3b) 2 (5c) 2 – (2a)(3b) – (3b)(5c) – (5c)(2a) } Expand the
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We know, the third side of the triangle would be greater than the difference of the rest two sides and lesser than the sum of the rest two sides AB – BC < CA < AB BC 8 – 3 < CA < 8 3 5(abbcca) ( bc ca ab ) Get the answers you need, now!In ΔABC, AB, BC, CA are produce such that AB = AE, BC = BD, AC = CF then find ∠α ∠β ∠γ = ?
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= BCACAB Share Cite Follow answered Mar 14 '17 at 607 Shenal Madhushan Silva Shenal Madhushan Silva 11 1 1 silver badge 3 3 bronze badges $\endgroup$ 1 1 $\begingroup$ Welcome to mathstackexchange You can type mathematical formulas in a beautiful way using MathJaxFigure 7 Number and rate of COVID19 cases in Alberta by age group 0 5000 000 Male Under 1 year 14 years 59 years 1019 years 29 years 3039 years 4049 years 5059 years 6069 years 7079 years 80 years 0 5000 000 Female Count Age group Figure 8 COVID19 cases in Alberta by age group and gender Table 1SOLUTION a b c = 11, ab bc ca = 3 and abc = 135 जैसा कि हम जानते हैं कि, a 3 b 3 c 3 – 3abc = (a b c) (a b c) 2 – 3(ab bc ca) ⇒ a 3 b 3 c 3 – 3 × (135) = 11 11 2 – 3 × 3 ⇒ a 3 b 3 c 3 405 = 11 × 121 – 9 = 11 × 112 ⇒ a 3 b 3 c 3 = 1232 – 405 = 7
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Simplifying ab bc ca = abc Reorder the terms ab ac bc = abc Solving ab ac bc = abc Solving for variable 'a' Move all terms containing a to the left, all other terms to the rightWe 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 have (12) 2 = 50 2 (ab bc ca)F = ab bc ca?
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