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

which expression can be used to convert 80 us dollars (usd) to australian dollars (aud)?

1 usd = 1.0343 aud

1 aud = 0.9668 usd

step-by-step explanation:

sorry just need the points

step-by-step explanation:

i found the koficent of the linear function and i foun c acordind to the strict formula y=kx+c and then i took 2 as an x and replacet it on the formula i found and then foun the y