Ab+bc+ca2

SolutionShow Solution Recall the formula ` (abc)^2 = a^b^2 c^2 2 (ab bc ca)` Given that `a^2 b^2 c^2 = ` ` (abc) = 0` Then we have ` (abc)^2 = a^2 b^2 c^2 2 (ab bc ca)` ` (0)^2 = 2 (ab bc ca)`.

Ab+bc+ca2.  Given in question (ab bc ca)= 9/2 and a 2 b 2 c 2 = 280 we know that (a b c) 2 = a 2 b 2 c 2 2ab 2ba 2ac =a 2 b 2 c 2 2 (ab bc ca) put the given value = 280 2 * 9/2 = 280 9 (a b c) 2 = 2 and ( a b c) = √2 = 17 so (a b c) 3 = (a b c) 2 * ( a b c) = 2 * 17 = 4913 answer is 4913. Using identity a2 b2 2ab = (a b) 2 We get, = (a b) 2 2bc 2ca = (a b) 2 2c (b a) or (a b) 2 2c (a b) Taking (a b) common = (a b) (a b 2c ) ∴ a 2 b 2 2 (ab bc ca) = (a b) (a b 2c) Concept Algebraic Expressions.  ∆ABC is such that AB=3 cm, BC= 2cm, CA= 25 cm If ∆ABC ~ ∆DEF and EF = 4cm, then perimeter of ∆DEF is (a) 75 cm (b) 15 cm (c) 225 cm (d) 30 cm.

Cyclic inequalities in three variables ab bc ca does not exceed aa bb cc A Cyclic Inequality in Three Variables \left (\displaystyle\frac {a^3} {b^2 (5a2b)}\frac {b^3} {c^2 (5b2c)}\frac {c^3} {a^2 (5c2a)}\ge\frac {3} {7}\right) A Cyclic Inequality in Three Variables II \left (\displaystyle\frac {10a^3} {3a^27bc}\frac {10b^3. Complete stepbystep answer We are given a triangle ABC and AM is median on the side BC We can prove ABBCCA> 2AM by starting with the triangles So, in the triangle ABC We have sub triangles ABM and AMC So, in triangle ABM Using the inequality of the triangle that the sum of any two sides is always greater than or equal to the third side. 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.

Factor abbcb^2ac ab bc b2 ac a b b c b 2 a c Reorder terms b2 abbcac b 2 a b b c a c Factor out the greatest common factor from each group Tap for more steps Group the first two terms and the last two terms ( b 2 a b) b c a c ( b 2 a b) b c a c Factor out the greatest common factor ( GCF) from each.  If x = a2 – bc y = b2 – ca and z = c2 – ab find the value of ax by cz asked in Class VI Maths by navnit40 Expert ( 405k points) substitution (including use of brackets as grouping symbols). Solution for abbcca=abc equation 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 right Add '1abc' to each side of the equation ab ac 1abc bc = abc 1abc Reorder the terms ab 1abc ac bc = abc 1abc Combine like terms abc 1abc = 0.

Precalculus Solve for a v=2 (abbcca) v = 2(ab bc ca) v = 2 ( a b b c c a) Rewrite the equation as 2(abbcca) = v 2 ( a b b c c a) = v 2(abbc ca) = v 2 ( a b b c c a) = v Divide each term by 2 2 and simplify Tap for more steps.  Given Two triangles ABC and DEF in which AB/DF =BC/FE=CA/ED To find The relationship between the given triangles ABC and DEF Solution As we know that according to the basic proportionality theorem (BPT), two triangles ABC and PQR are said to be similar if all the corresponding sides of the two triangles are in the same proportion. $abbcca=0$ $a(bc)= bc$ $bc= bc/a$ Adding both side $a$ then $bca=abc/a$ $a(abc)= a^2bc$ $1/(a^2bc)= 1/a(abc)$ Similarly $1/(b^2ca) = 1/b(abc)$ and $1/(c^2ab) = 1/c(abc)$ By adding all eqution we get $1/a(abc) 1/b(abc) 1/c(abc)$.

Therefore, the given equation is false _\square If a a a, b, b, b, and c c c all equal 1, then this equation is true But there are many other examples The only two values that satisfy the equation 1 c = c \frac1c = c c 1 = c are c = 1 c=1 c = 1 and c = − 1 c=1 c = − 1. BC Gov News Provincial state of emergency extended With recovery efforts still underway in communities affected by severe flooding and highways damaged by flooding and mudslides, the Province is extending the provincial state of emergency. Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!.

We know, (abc) 2=a 2b 2c 22ab2bc2ca ⇒(abc) 2=a 2b 2c 22(abbcca) Putting the values here, we get, (9) 2=a 2b 2c 22(26) ⇒81=a 2b 2c 252 ⇒a 2b 2c 2=81−52=29 Hence, option D is correct. Get an answer for 'If x = logabc y = logbca z = logcab ptove that xyz2 = xyz' and find homework help for other Math questions at eNotes Search this site Go iconquestion. #rbclasses9email rbclassesmtr@gmailcomwatsapp number R B Gautam R B Classes9th class Number Systemshttps//wwwyoutubecom/watch?v=PUUrm0qDQ&l.

Alberta, with an area of 661,848 km 2 (255,500 sq mi), is the fourthlargest province after Quebec, Ontario and British Columbia Alberta's southern border is the 49th parallel north, which separates it from the US state of MontanaThe 60th parallel north divides Alberta from the Northwest TerritoriesThe 110th meridian west separates it from the province of Saskatchewan;. Với abc là các số thực dương thỏa mãn ab bc ca = 3abc Tìm giá trị nhỏ nhất của P = Câu hỏi và hướng dẫn giải Nhận biết Với a,b,c là các số thực dương thỏa mãn ab bc ca = 3abc Tìm giá trị nhỏ nhất của P = A P min = 3/2 B. Revenue Cycle Applications Analyst jobs at Randstad Technologies in , CA job summary We have a 10 Month contract opportunity for a Revenue Cycle Applications Analyst This role is 100% remote.

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.  After trying a lot, I've got a simple solution Suppose that, $2≤a≤b≤c$ Note that, $$abbcca2. (i) ab bc, bc ca, ca ab (ii) a b ab, b c bc , c a ac (iii) 2 p 2 q 2 3pq 4, 5 7pq 3 p 2 q 2 (iv) l 2 m 2, m 2 n 2, n 2 l 2, 2lm 2mn 2nl Solution The given algebraic expressions are written by rearranging and combining like terms followed by performing the addition operation as follows (i) (ab – bc) (bc – ca) (ca ab).

A triangle ABC can be constructed in which ∠B = 60°, ∠C = 45° and AB BC CA = 12 cm Write true or false and give reason for your answer asked in Class IX Maths by aditya23 Expert (737k points) construction 0 votes 1 answer asked in Class IX Maths by saurav24 Expert (90k points).  (abc) 2 = a 2 b 2 c 2 2(abbcca) () 2 =902(abbcca) 400=902(abbcca) =2(abbcca) 310=2(abbcca) Therfore, abbcca=155 8. 2 ABBC=AC (Segment Addition Postulate) 3 BCBC=AC (Substitution Property of Equality) 4 2BC=AC (Distributive Property) 5 AC=2BC (Identity) Identify the hypothesis, Argument and Conclusion Answer by MathLover1() (Show Source) You can.

If b) is Options A 1 B 0 C 1 D 3 Answer D 3 Justification bc b3/abc c3/abc = 3 or a2/bc b2/ac c2/ab = 3. NCERT Solutions for Class 7 Maths Chapter 6 Exercise 64 Question 3 Summary AM is a median of a triangle ABC The given expression AB BC CA > 2 AM is true.  If a>0, b>0, c>0 and a b > c, b c > a, c a > b then ( a b c )^2.

Solve for b,bc=acab \square!.  Using properties of determinants, prove the following (a 2 ,a 2 (bc) 2, bc) (b 2 ,b 2 (ca) 2 ,ca) (c 2 ,c 2 (ab) 2 ,ab) = (ab) (bc) (ca) (abc) (a 2 b 2 c 2) determinant class12 cbse Please log in or register to add a comment. Formula,(abc) 2=a 2b 2c 22ab2bc2ca=a 2b 2c 22(abbcca)Given,⇒9 2=a 2b 2c 22(23)⇒81−46=a 2b 2c 2∴a 2b 2c 2=35.

= (AB)^2 (BC)^2 (CA)^22ABC (ABC) (or) we know that , (abc)^2=a^2b^2c^22ab2bc2ca we also solve this by substituting AB in place of ‘a’ , ‘BC’in place of ‘b’ and ‘CA’ in place of ‘c’ in above equation 16K views View upvotes.  a^3b^3c^33abc = (abc)* (a^2b^2c^2 (abbcca) ) a^3b^3c^33abc = 6* (14 11) a^3b^3c^33abc = 18 heart outlined Thanks 0 star outlined star outlined star outlined star outlined. Ab bc ca.

 If O be a inner point of a triangle ABC Then prove that OAOBOC < ABBCCA ?. To prove this equation nonnegative, you will have to convert the equation in terms of perfect square form containing a,b and c Now, a²b²c²abbcca = ½ • ( 2a²2b²2c²2ab2bc 2ca ) = ½ • ( Can A^2B^2C^22AB2AC2BC be a perfect square https//mathstackexchangecom/questions/547/cana2b2c22ab2ac2bcbeaperfect. `bc^2 ac^2 abc = c(bc ac ab)` Note that the expressions in parenthesis are the same, so the three groups of terms have a common trinomial factor!.

 You can see that if you expand (abc)^2, simplify, multiply by 2, and use the trivial inequality Instead of doing AMGM, I managed to solve it using CauchySchwartz Inequality There is likely 1/b1/c) ≥(abc) 2 Now, the LHS = (a 3 b. If a^2b^2c^2abbcca=0 then prove that a=b=c a² b² c² = ab bc ca On 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 = 0 a² b² – 2ab b² c² – 2bc c² a² – 2ca = 0. 3 a) truth table b) sop y0 = (a’b’c’d)(a’b’cd’)(a’bc’d’)(a’bcd)(ab’c’d’)(ab’cd)(abc’d)(a bcd’) y1= (a’b’cd)(a’bc’d.

23 hours ago Stack Exchange network consists of 178 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers Visit Stack Exchange.  Since 16 1 a 1 b 1 c 1 abc = 16 bcacab1 abc and ab bc ac = 1, a b c abc ≥ 4√abc ≥ 8abc From 4√abc ≥ 8abc, we can conclude that abc ≤ 1 4 Then, a b c abc ≥ 2 Answer link Cesareo R.  Transcript Ex 91, 3 Add the following (i) ab − bc, bc − ca, ca − ab Expressions are ab − bc bc − ca ca − ab We put like terms below like terms So, required sum = 0.

 Ex 42, 7 By using properties of determinants, show that 8(−a2&ab&ac@ba&−b2&bc@ca&cb&−c2) = 4a2b2c2 Solving LHS 8(−a2&ab&ac@ba&−b2&bc@ca&cb&−c2) Taking a common from R1, b common from R2 , c common from R3 = abc 8(−a&b&c@a&−b&c@a&b&−c) Taking a common from C1, b common.