Find all pairs $(x, y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 56$.

13.31 and - 3.31

Step-by-step explanation:

(x + y)^2 = x^2 + y^2 + 2xy ; x + y = 10; x^2 + y^2 = 56

This means ;

10^2 = 56 + xy;

100-56 = XY

44 = xy(3)

From x + y = 10

x = 10- y

Substitute that in eqn 3;

We have: 44 = (10-y) y

44 =10y - y^2

Y^2-10y -44 = 0;

From formula method of quadratic equation;

Y =[ - (-10) +_ √ (-10)^2 + 4 × 1 x (-44)] / 2 × 1

y = [ 10 +_√ (100 + 176) ]/ 2]

y = [ 10 +_ √ 276]/ 2

y = [ 10 + √ 276]/ 2 or [ 10 _ √ 276]/ 2

y = 13.31 or - 3.31

Similarly x = 10 - 13.31 or 10 - (-3.31)

x = - 3.31 or 13.31

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation: