Let x,y be two different numbers
suppose x^2=y^2
then x^2-y^2=0
which yields (x+y)(x-y)=0
so either x=y or x=-y
In any case, x and y must be the same value
also when a vairable is squared like y=x^2
we must note that there are 2 possible solutions
x=(+/-)sqrt(y)

7 0

3 years ago

Help Please!!!!!!

Vanyuwa [196]

Answer:

centre = (2, - 1), radius = 6

Step-by-step explanation:

Rearrange the equation by placing the x and y terms together and adding 31 to both sides

Given

x² + y² - 4x + 2y - 31 = 0, then

x² - 4x + y² + 2y = 31

Use the method of completing the square

add ( half the coefficient of the x/y term )² to both sides

x² + 2(- 2)x + 4 + y² + 2(1)y + 1 = 31 + 4 + 1

(x - 2)² + (y + 1)² = 36

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r the radius

compare to (x - 2)² + (y + 1)² = 36, then

centre = (2, - 1) and r = $\sqrt{36}$ = 6

3 0

3 years ago

**Write them into an inequality!**

Alex73 [517]

1. 12in (1ft)+X=28in
2. 6in.+X=42
3. 170.5sq.ft./15.5=x
4. 120ft/12ft=x10sec.

5 0

2 years ago