Ab Bc Ca

A 2 + b 2 + c 2 + 2 × 59 = 625 Given, ab + bc + ca = 59 a 2 + b 2 + c 2 + 118 = 625 a 2 + b 2 + c 2 + 118 – 118 = 625 – 118 subtracting 118 from both the sides Therefore, a 2 + b 2 + c 2 = 507 Thus, the formula of square of a trinomial will help us to expand. Sep 5 '16 at 22:08 Okay, yes then I understand. Ab + ac + bc = abc Solving ab + ac + bc = abc Solving for variable 'a'.
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Simplification Questions & Answers :.

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Ab bc ca. In ΔPQR measure of angle Q is 90°. From 4√abc ≥ 8abc, we can conclude that abc ≤ 1 4. What is the lewis structure for hcn?.

Ab + bc + ca does not exceed aa + bb + cc. Since 16 1 a + 1 b + 1 c + 1 abc = 16 bc+ac+ab+1 abc and ab + bc +ac = 1, a + b + c + abc ≥ 4√abc ≥ 8abc. To prove this equation non-negative, you will have to convert the equation in terms of perfect square form containing a,b and c.

1 Answer Jack H. Now, a²+b²+c²-ab-bc-ca = ½ • ( 2a²+2b²+2c²-2ab-2bc -2ca ) = ½ • ( a² -2ab +b² +b² -2bc +c² +c² -2ac +a² ) = ½ • { (. Then, a +b +c + abc ≥ 2.

How do you factor #(a+b+c)(ab+bc+ca)-abc#?. Oct 8, 17 #(a+b)(b+c)(c+a)# Explanation:. I thought you had negation on the variable to the left of your prime character.

If a + b + c = 13, a2+b2+c2=69 then find the ab + bc + ca :. If tanP = 4/3, and PQ = 1.5cm, then what is the length (in cm) of side PR?. Move all terms containing a to the left, all other terms to the right.

7th Grade Math Problems 8th Grade Math Practice From Square of a Trinomial to. The inequality below, though simple, is useful and often comes up in proofs of more involved inequalities. How do I determine the molecular shape of a molecule?.

Simplifying ab + bc + ca = abc Reorder the terms:. Just for clarification because my notation isn't that common the two equations are (AB) + (A* (¬C)) + (BC) and the other one is (A* (¬C)) + (B*C) – M. What is the lewis structure for co2?.

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$\\a)\ Solution:\\\\ \frac{(b+c)}{a},\ \frac{(c+a)}{b}\ and\ \frac{(a+b)}{c}\ are\ in \ AP\\\\\\ Considering\ the\ common\ difference\ of\ terms\\\\ \Rightarrow \frac{(c+a)}{b}-\frac{(b+c)}{a}=\frac{(a+b)}{c}-\frac{(c+a)}{b}\\\\\\ \Rightarrow \frac{a(c+a)-b(b+c)}{ab}=\frac{b(a+b)-c(c+a)}{bc}\\\\\\ \Rightarrow \frac{ac+a^2-b^2-bc}{ab}=\frac{ab+b^2-c^2-ac}{bc}\\\\\\ \Rightarrow \frac{c(a-b)+(a+b)(a-b)}{ab}=\frac{a(b-c)+(b+c)(b-c)}{bc}\\\\\\ \Rightarrow \frac{(a-b)(c+a+b)}{ab}=\frac{(b-c)(a+b+c))}{bc}\\\\\\ \Rightarrow \frac{a-b}{ab}=\frac{b-c}{bc}\\\\\\ \Rightarrow \frac{a}{ab}-\frac{b}{ab}=\frac{b}{bc}-\frac{c}{bc}\\\\\\ \Rightarrow \frac1b-\frac1a=\frac1c-\frac1b\\\\\\ \Rightarrow \frac1b+\frac1b=\frac1a+\frac1c$

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$\\Solution:\\\\ Given:\\\\ a^2+b^2+c^2-ab-bc-ca=0\\\\ \Rightarrow 2(a^2+b^2+c^2-ab-bc-ca)=0\\\\ \Rightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0\\\\ \Rightarrow (a^2+b^2-2ab)+(b^2+c^2-2bc)+(c^2+a^2-2ca)=0\\\\ \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2=0\\\\ Since\ sum\ of\ the\ squares\ are\ zero\ then\\\\ \Rightarrow (a-b)^2=0\Rightarrow a=b\\\\ \Rightarrow (b-c)^2=0\Rightarrow b=c\\\\ \Rightarrow (c-a)^2=0\Rightarrow c=a\\\\ \therefore a=b=c$

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$\\Solution:\\\\ Given:\ (a + b + c )= 7\ \ \ and\ \ \ (ab + bc + ca) = 20\\\\ \text{We have the identity:}\\\\ (a+b+c)^2= a^ 2 + b^ 2 + c^ 2+2(ab+bc+ca)\\\\ \Rightarrow ( a^ 2 + b^ 2 + c^ 2)=(a+b+c)^2-2(ab+bc+ca)\\\\ \Rightarrow ( a^ 2 + b^ 2 + c^ 2)=7^2-2\times20=49-40=9\\\\ \therefore ( a^ 2 + b^ 2 + c^ 2)=9$

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