Question

A+b+c=6 a²+b²+c²=16 ab+bc+ac=?​

Answer 1

$a+b+c=6\\\\a^2+b^2+c^2=16\\\\\\(a+b+c)^2=36\\\\(a+b+c)(a+b+c)=36\\\\a^2+ab+ac+ab+b^2+bc+ac+bc+c^2=36\\\\\underbrace {a^2+b^2+c^2}_{16}+2(ab+ac+bc)=36\\\\2(ab+ac+bc)=36-16\\\\2(ab+ac+bc)=20\\\\\boxed {ab+bc+ac=10}$

Answer 2

Ответ:ab+bc+ac=10

Ответ:10

Объяснение:a+b+c=6

Возводим в квадрат

a²+b²+c²+2ab+2ac+2bc=6²

Так как

A²+b²+c²=16,то

2аb+2ac+2bc=36-16

Значит,что

ab+bc+ac=10

Answer 3

To find the value of ab+bc+ac, we can use the given equations to solve for a, b, and c. Let's solve the equations step by step.

From the equation a + b + c = 6, we can express one variable in terms of the other two. Let's solve for c:

c = 6 - a - b

Now, substitute this value of c into the second equation:

a² + b² + c² = 16 a² + b² + (6 - a - b)² = 16 a² + b² + 36 - 12a - 12b + a² + b² = 16 2a² + 2b² - 12a - 12b + 20 = 0 a² + b² - 6a - 6b + 10 = 0

Now we have a quadratic equation in terms of a and b. To simplify it further, we can complete the square:

(a² - 6a) + (b² - 6b) + 10 = 0 (a² - 6a + 9) + (b² - 6b + 9) + 10 = 9 + 9 (a - 3)² + (b - 3)² = 10

This equation represents a circle in the (a, b) coordinate plane centered at (3, 3) with a radius of √10.

Now, let's solve for the value of ab+bc+ac. We can use the identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ac):

(a + b + c)² = a² + b² + c² + 2(ab + bc + ac) 6² = 16 + 2(ab + bc + ac) 36 = 16 + 2(ab + bc + ac) ab + bc + ac = (36 - 16) / 2 ab + bc + ac = 20 / 2 ab + bc + ac = 10

Therefore, the value of ab + bc + ac is 10.