Answer:
1 C) 23
2 D) x + y + 2
Step-by-step explanation:
1.
a = 1 , b = -23, c = 130
b²-4ac = (-23)²- 4×1×130 = 529 -520 = 9>0
if solutions (A=x₁, B=x₂) exist (b²-4ac≥0) then based on vieta's formula:
[tex]A+B=x_1+x_2=-\dfrac{b}a=-\dfrac{-23}1=23[/tex]
2.
sum of two even integers is na even integer
sum of two odd integers is na even integer
only sum of an even integer and an odd integer is an odd integer
x is even and y is odd so 2x is even and ±2y is even and x+y is odd
1 is odd and 2 is even which means:
2x±2y is even,
x+y + 1 is even
x-2y is even
only x+y+2 is odd (as sum of odd x+y and even 2)
