WebPlease find below the solution to your problem. Given. a^2 + b^2 + c^2 = ab + bc + ca. 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 ... WebNov 18, 2024 · If a, b, c are real numbers and a2 + b2 + c2 = 2 (a - b - c) - 3, then the value of a + b + c is Q6. The value of ( 0.7) 3 − ( 0.43) 3 ( 0.7) 2 + 0.43 × 0.7 + ( 0.43) 2 is equal to: Q7. The value of ( 281 + 119) 2 − ( 281 − 119) 2 281 × 119 is equal to which of the following? Q8. If a + b 2 = 4 and ab = 7, then the value of (a - b) =
[Solved] If ab + bc + ca = 0, then the value of \(\frac{{\left( {{b^2
Web1 若a、b、c为三角形ABC的三边,且满足a2+b2-c2=ab+ac+bc,试判断三角形ABC的形状; 2 若a,b,c为三角形的三条边,且满足条件a2+b2+c2=ab+bc+ac,试判断该三角形的形状; 3 已 … WebSince c ≡ − c (mod 2), we have that a + b ≡ − c (mod2). This means that 4 ∣ (a + b − c)(a + b + c) = (a + b)2 − c2 = a2 + b2 − c2 + 2ab = 2ab. Since 4 ∣ 2ab, you have that 2 ∣ ab. So one of … notifications on ip
If a²+b²+c² = ab+bc+ca, Find (c+a)/b. [Solved] - Cuemath
WebThe correct option is A ∠ B = 60 ° Explanation for the correct Option: Determining the angle In ∆ A B C ⇒ ( a + b + c) ( a - b + c) = 3 a c [ Given] ⇒ ( ( a + c) + b) ( ( a + c) – b) = 3 a c ⇒ ( a + c) 2 – b 2 = 3 a c ⇒ a 2 + c 2 + 2 a c – b 2 – 3 a c = 0 ⇒ a 2 + c 2 – b 2 = a c … ( i) We know the cosine rule in ∆ A B C, having sides a, b, c WebThe Law of Cosines says: c2 = a2 + b2 − 2ab cos (C) Put in the values we know: c2 = 82 + 112 − 2 × 8 × 11 × cos (37º) Do some calculations: c2 = 64 + 121 − 176 × 0.798…. More … http://m.1010jiajiao.com/gzsx/shiti_id_78a79252fd3cc89065753c93b701ac39 notifications on cheap flights