For this question, we can use the identity (abc) 2 = a 2 b 2 c 2 2ab abc 2ca (abc) 2 = 250 2(ab bc ca) (abc) 2 = 250 2(3) (abc) 2 = 250 6 (abc) 2 = 256 abc = square root of 256 abc = 16 There you go Hope it helps! Transcript Misc 13 Using properties of determinants, prove that 3a ab ac ba 3b bc ca cb 3c = 3 ( a b c) (ab bc ac) Taking LH S 3a ab ac ba 3b bc ca cb 3c Applying C1 C1 C2 C3 = 3a ab ac ab ac ba 3b bc 3b bc ca cb 3c cb 3c = ab ac 3b bc cb 3c Taking (a b c) common from C1 = ( ) 1 abIn algebra, a quadratic equation (from the Latin quadratus for "square") is any equation that can be rearranged in standard form as = where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0If a = 0, then the equation is linear, not quadratic, as there is no term The numbers a, b, and c are the coefficients of the equation and may be distinguished by calling
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(ab+bc+ca)^3 formula-$$(a b c)^3 = (a^3 b^3 c^3) 3a(ab ac bc) 3b(ab bc ac) 3c(ac bc ab) 3abc$$ $$(a b c)^3 = (a^3 b^3 c^3) 3(a b c)(ab ac bc) 3abc$$ $$(a b c)^3 = (a^3 b^3 c^3) 3(a b c)(ab ac bc) abc$$ It doesn't look like I made careless mistakes, so I'm wondering if the statement asked is correct at allThe 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 whole
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Math formula is very important for solving math problems Here you can find All Mathematical formulas which you need in Competitive exams 21 Learn these basic mathematical formula 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 = 3abc;Without loss of generality, we may suppose that AD is the minimum side (1) When AB=AD, we have BC=CD In this case, letting O be the intersection point of AC and the bisector of \angle B,(4) If in the figure below AB = 15 cm, BC= cm and CA = 7 cm, find the area of the rectangle BDCE (5) The area of a trapezium is 98 cm 2 and the height is
Hence we have the other factor = (a2 b2 c2) k (ab bc ca) ;A3 −b3 =(a−b)33ab(a−b) 6 a2 −b2 =(ab)(a−b) 7 a3 −b3 =(a−b)(a2 ab b2) 8 a3 b3 =(ab)(a2 −ab b2) 9 a n−bn=(a−b)(an−1 a −2b an−3b2 Example 11 If a b c = 9 and ab bc ca = 40, find a 2 b 2 c 2 Solution We know that Example 12 If a 2 b 2 c 2 = 250 and ab bc ca = 3, find a b c Solution We know that Example 13 Write each of the following in expanded form (i) (2x 3y) 3 (ii) (3x – 2y) 3 Solution Example 14 If x y = 12 and xy = 27, find the value of x 3 y 3
= sin sin sinAB C 000 31 3 1 sin15 sin75 sin90 1 22 22 − == =− () 31 3 122() 7 Prove tht 2 cos c cos cos{bcA a Bab Cab =} 22 2c LHS 2 cos 2 cos 2 cosbc A ca B ab c From cosine rule 2 22 2 2 2 cos cos 22 bc a a c b AB bc ac − − ==The actual formula is (abc)² = a² b² c² 2 (ab bc ca) You can get this simple formula by multiplying (ab c) with (abc) (ab c)² = (ab c)* (ab c)RD 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
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Section Formula With Examples Type I On finding the section point when the section ratio is given Example 1 Find the coordinates of the point which divides the line segment joining the points (6, 3) and (– 4, 5) in the ratio 3 2 internally Sol Let P (x, y) be the required point Given #v= 2(ab bc ca)#, how do you solve for a?b2 =(ab)2−2ab 2 (a−b)2 = a 2−2ab b;
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1 (a b)2 = a2 2ab b2;If P, Q and R are the midpoints of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic asked in Class IX Maths by aman28 (872 points) circles Categories AllHence the other factor, (a2 b2 c2 ab bc ca)
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Justify whether it is true to say that 1,3/2,2,5/2forms an AP as a2 a1 = a3 a2 asked in Class X Maths by akansha Expert ( 22k points) arithmetic progressionsIf a^2b^2c^2abbcca=0 then prove that a=b=c Get the answer to this question along with unlimited Maths questions and prepare better for JEE exam3 views asked 1 hour ago in 3D Coordinate Geometry by Harshal01 ( 110 points) The coordinates of the midpoints of sides AB, BC and CA of ΔABC are D(1, 2, 3), E(3, 0, 1) and F(1, 1,
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Exercise 91 Question 1 Identify the terms, their coefficients for each of the following expressions Answer Term x x Coefficient = 1 = 1 and Term x2 x 2 Coefficient = 1 Answer 1 2 1 2 is the coefficient for x and y and for xy it is 1 Answer 03 is the coefficient for a, for b there are two coefficients 06 and 05 a 3 − b 3 = (a − b) (a 2 b 2 ab) a 3 b 3 = (a b) (a 2 b 2 − ab) (a b c) 3 = a 3 b 3 c 3 3 (a b) (b c) (c a) a 3 b 3 c 3 − 3abc = (a b c) (a 2 b 2 c 2 − ab − bc − ac) If (a b c) = 0, a 3 b 3 c 3 = 3abcA^2 b^2 = (a – b)^2 2ab (a – b)^2 = a^2 – 2ab b^2 (a b c)^2 = a^2 b^2 c^2 2ab 2ac 2bc (a – b – c)^2 = a^2 b^2 c^2 – 2ab – 2ac 2bc (a b)^3 = a^3 3a^2b 3ab^2 b3 (a b)^3 = a^3 b^3 3ab (a b) (a – b)^3 = a^3 – 3a^2b 3ab^2 –
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b2 =(a−b)22ab 3 (a b c)2 = a2 b2 c2 2(ab bc ca) 4 (a b) 3= a3 b3 3ab(a b);Click here👆to get an answer to your question ️ If a^2 b^2 c^2 = 250 and ab bc ca = 3 , then find a b c Get the list of basic algebra formulas in Maths at BYJU'S Stay tuned with BYJU'S to get all the important formulas in various chapters like trigonometry, probability and so on
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AP CALCULUS AB & BC FORMULA LIST Definition of e 1 lim 1 n n e of n §¨¸ ©¹ _____ Absolute value 0 0 x if x x x if x t ® ¯ _____ Definition of the derivative 0 ( ) lim h f x h f x fx o h c lim xa f x f a fa o xa c (Alternative form) _____ Definition of continuity f isExample 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 Multiply both sides with 2, we get 2 ( a2 b2 c2 – ab – bc – ca) = 0 ⇒ 2a2 2b2 2c2 – 2ab – 2bc – 2ca = 0 ⇒ (a2 – 2ab b2) (b2 – 2bc c2) (c2 – 2ca a2) = 0 ⇒ (a –b)2 (b – c)2 (c – a)2 = 0 Since the sum of square is zero then each term should be zero ⇒ (a –b)2 = 0, (b – c)2 = 0, (c – a)2 = 0
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#vinodmaths a³b³c³3abc=(abc)(a²b²c²abbcca) formula based questions This video is useful for all competitive exams specially ssc and delhi SI HereAP CALCULUS AB and BC Final Notes Trigonometric Formulas 1 sin θcos 2 θ=1 2 1tan θ=sec 2 θ 3 1cot θ=csc 2 θ 4 θ sin(−θ) =−sinθ 5If a b c = 9 and ab bc ca = 26, find the value of a 3 b 3 c 3 − 3abc Advertisement Remove all ads Solution Show Solution In the given problem, we have to find value of `a^3 b^3 c^3 3abc` Given `abc = 9, ab bc ca = 26` We shall use the identity
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(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)Example Solve 8a 3 27b 3 125c 3 – 90abc Solution This proceeds as Given polynomial (8a 3 27b 3 125c 3 – 90abc) can be written as (2a) 3 (3b) 3 (5c) 3 – 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(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 2
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Solution 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 equation a^3 b^3 c^3 a 3 b 3 c 3 = (a b c) (a 2 b 2 c 2 – ab – bc – ca) 3abc s Algebra, cube, sum, sum of cubes This entry was posted on at 554 pm and is filed under Algebra You can follow any responses to this entry through the RSS feed You can leave a response, or trackback from your own site=V384∠124 ca= 2°are applied to an inductionmachine The average value of the magnitudes will be (576 480 384)/3 = 480 V and the maximum deviation from average value is (576 480) = 96 V Therefore, the NEMA definition of % voltage unbalancewill be 100(96/480) = % The positive sequence voltage isV
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A3 b3 =(ab)−3ab(a b) 5 (a−b)3 = a3 −b3 −3ab(a−b);Recall the formula ` (abc)^2 = a^2 b^2 c^2 2 (ab bc ca)` Given that `a^2 b^2 c^2 = 250 , ab bc ca = 3 ` Then we have ` (abc)^2 = a^2 b^2 c^2 2 (ab bcca)` ` (abc)^2 = 250 2 (3)` ` (abc)^2 = 256` ` (abc) =± 16`If abc=8 and abbcca= Find the value of a3b3c3−3abcHow to do it null null null null abc=8abbcca=a3b3c3=abca2b2c2−ab−bc−ca3abc⇒a3b3c3−
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Algebra Linear Equations Formulas for Problem Solving 1 Answer P dilip_k What is the perimeter of the rectangle if the area of a rectangle is given by the formula How do you find the value of y that makes (3,y) a solution to the equation #3xy=4#?Simplify a b c = 25 and ab bc ca = 59 Find the value of a 2 b 2 c 2 Solution According to the question, a b c = 25 Squaring both the sides, we get (a b c) 2 = (25) 2 a 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 = 625Click here👆to get an answer to your question ️ If a^2 b^2 c^2 ab bc ca = 0 , prove that a = b = c
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Abc (1/a 1/b 1/c) = (3) (−2) bc ca ab = −6 Using the a 2 b 2 c 2 formula, a 2 b 2 c 2 = (a b c) 2 2 (ab bc ca) a 2 b 2 c 2 = (3) 2 2 (6) = 9 12 = 21 Answer a 2 b 2 c 2 = 21 Example 3 Find the value of a 2 b 2 c 2 if a b c = and ab bc ca = 100Where k is any integer (since net coefficients are integers) Now ((a2 b2 c2) k (ab bc ca) ) (abc) = a3b3c3−3abc The value of can be easily found out to be 1 (even by simply multiplying and comparing);Volume = (1/3)Π r²h L ² = r ² h² sphere Curved surface area = 4Π r² Volume = (4/3)Π r³ Hemisphere Curved surface area = 2Π r² Total Curved surface area=3Π r² Volume = (2/3)Π r³ Cuboid Curved surface area=4h(lb) Total surface area = 2(lbbhh l) Volume = l x b x h Cube Curved surface area=4a ² Total surface area = 6a² Volume = a ³
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The Formula is given below (a b c)³ = a³ b³ c³ 3 (a b) (b c) (a c) \(=> (abc)^3 = (a^3ab^2ac^2 2a^2b 2abc 2ca^2)\\ (a^2bb^3bc^2 2ab^2 2b^2c 2abc)\\ (ca^2 cb^2 c^3 2abc 2bc^2 2c^2a) \) Arrage value according power and similear \(=> (abc)^3 = a^3 b^3 c^3 \\ 6abc 3a^2b 3ab^2 \\ 3ac^2 3bc^2 3b^2c 3a^2c \)16 ejercicios resueltos productos notables nivel preuniversitarioTeoría https//wwwyoutubecom/watch?v=Qjes17MQXac
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