![The value of the determinant | b^2 - ab b - c bc - ac | ab - b^2 a - b b^2 - ab | bc - ac c - a ab - b^2 The value of the determinant | b^2 - ab b - c bc - ac | ab - b^2 a - b b^2 - ab | bc - ac c - a ab - b^2](https://d1hj4to4g9ba46.cloudfront.net/questions/2022827_1148955_ans_b6064ae55b644c44a4e9eb584afd2090.jpeg)
The value of the determinant | b^2 - ab b - c bc - ac | ab - b^2 a - b b^2 - ab | bc - ac c - a ab - b^2
If a, b, c are real, then f(x) = |((x + a^2), ab, ac), (ab, x + b^2, bc), ( ac, bc, x + c^2)| is decreasing in - Sarthaks eConnect | Largest Online Education Community
Prove the following identities – |(b^2+c^2,ab,ac)(ba,c^2+a^2,bc)(ca,cb,a^2 +b^2)| = 4a^2b^2c^2 - Sarthaks eConnect | Largest Online Education Community
![The determinant |{:(b^2-ab, b-c, bc-ac), (a b-a^2, a-b, b^2-ab) ,(b c-c a, c-a, a b-a^2):}| equals (a)a b c\ (b-c)(c-a)(a-b) (b) (b-c)(c-a)(a-b) (c) (a+b+c)(b-c)(c-a)(a-b) (d) none of these The determinant |{:(b^2-ab, b-c, bc-ac), (a b-a^2, a-b, b^2-ab) ,(b c-c a, c-a, a b-a^2):}| equals (a)a b c\ (b-c)(c-a)(a-b) (b) (b-c)(c-a)(a-b) (c) (a+b+c)(b-c)(c-a)(a-b) (d) none of these](https://d10lpgp6xz60nq.cloudfront.net/web-thumb/642579426_web.png)
The determinant |{:(b^2-ab, b-c, bc-ac), (a b-a^2, a-b, b^2-ab) ,(b c-c a, c-a, a b-a^2):}| equals (a)a b c\ (b-c)(c-a)(a-b) (b) (b-c)(c-a)(a-b) (c) (a+b+c)(b-c)(c-a)(a-b) (d) none of these
![SOLVED: P and Q are the points on the side BC of triangle ABC and AP=AQ prove that :AC+AB+BC is greater than 2AP +PQ SOLVED: P and Q are the points on the side BC of triangle ABC and AP=AQ prove that :AC+AB+BC is greater than 2AP +PQ](https://cdn.numerade.com/ask_previews/85447aea-613e-4130-85b6-88231d8d83ff_large.jpg)
SOLVED: P and Q are the points on the side BC of triangle ABC and AP=AQ prove that :AC+AB+BC is greater than 2AP +PQ
![In any triangle ABC, prove that `AB^2 + AC^2 = 2(AD^2 + BD^2)` , where D is the midpoint of BC. - YouTube In any triangle ABC, prove that `AB^2 + AC^2 = 2(AD^2 + BD^2)` , where D is the midpoint of BC. - YouTube](https://i.ytimg.com/vi/wu1SuC41ez4/maxresdefault.jpg)
In any triangle ABC, prove that `AB^2 + AC^2 = 2(AD^2 + BD^2)` , where D is the midpoint of BC. - YouTube
![PROVE STATEMENTS ABOUT SEGMENTS & ANGLES. EXAMPLE 1 Write a two-column proof Write a two-column proof for the situation in Example 4 on page 107. GIVEN: - ppt download PROVE STATEMENTS ABOUT SEGMENTS & ANGLES. EXAMPLE 1 Write a two-column proof Write a two-column proof for the situation in Example 4 on page 107. GIVEN: - ppt download](https://images.slideplayer.com/25/7927734/slides/slide_4.jpg)
PROVE STATEMENTS ABOUT SEGMENTS & ANGLES. EXAMPLE 1 Write a two-column proof Write a two-column proof for the situation in Example 4 on page 107. GIVEN: - ppt download
![Suppose A, B, C are defined as A = a^(2)b + ab^(2) - a^(2)c - ac^(2), B = b^(2)c + bc^(2) - a^(2)b - ab^(2), and C = a^(2)c + ac^(2) - Suppose A, B, C are defined as A = a^(2)b + ab^(2) - a^(2)c - ac^(2), B = b^(2)c + bc^(2) - a^(2)b - ab^(2), and C = a^(2)c + ac^(2) -](https://d10lpgp6xz60nq.cloudfront.net/web-thumb/644180564_web.png)
Suppose A, B, C are defined as A = a^(2)b + ab^(2) - a^(2)c - ac^(2), B = b^(2)c + bc^(2) - a^(2)b - ab^(2), and C = a^(2)c + ac^(2) -
![The determinant |{:(b^2-ab, b-c, bc-ac), (a b-a^2, a-b, b^2-ab) ,(b c-c a, c-a, a b-a^2):}| equals (a)a b c\ (b-c)(c-a)(a-b) (b) (b-c)(c-a)(a-b) (c) (a+b+c)(b-c)(c-a)(a-b) (d) none of these The determinant |{:(b^2-ab, b-c, bc-ac), (a b-a^2, a-b, b^2-ab) ,(b c-c a, c-a, a b-a^2):}| equals (a)a b c\ (b-c)(c-a)(a-b) (b) (b-c)(c-a)(a-b) (c) (a+b+c)(b-c)(c-a)(a-b) (d) none of these](https://d10lpgp6xz60nq.cloudfront.net/ss/web/731558.jpg)