This is !
y=x^2
2y + 6 = 2(x+3)
if (x, y) is a solution of the system of equations above and x > 0 what is the value of xy ?
a) 1
b) 2
c) 3
d) 9

A

Step-by-step explanation:

Given the 2 equations

y = x² → (1)

2y + 6 = 2(x + 3) ← distribute

2y + 6 = 2x + 6 ( subtract 6 from both sides )

2y = 2x ( divide both sides by 2 )

y = x → (2)

Substitute y = x² into (2)

x² = x ( subtract x from both sides )

x² - x = 0 ← in standard form

x(x - 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x = 0 or x - 1 = 0 ⇒ x = 1

x > 0 ⇒ x = 1

From (2), x = 1 ⇒ y = 1

and xy = 1 × 1 = 1 → A

there would be 365 different 4 digit numbers

step-by-step explanation:

i belive that it is 365 because i asked google

c is the correct answer

step-by-step explanation: