Answer:
You might want to put that in a picture, because that makes very little sense.
When a quadratic equation ax^2+bx+c has a double root, the discriminant,
D=b^2-4ac=0
Here
a=2,
b=b,
c=18
and
D=b^2-4ac=b^2-4*2*18=0
solve for b
b^2-144=0
=> b= ± sqrt(144)= ± 12
So in order that the given equation has double roots, the possible values of b are ± 12.
Answer:
Step-by-step explanation:
Do substitution by first writing an equation 15c + 29a=175 when you get your answer either substitute it in your other equation which would be 21c+27a=177, if you found c(Children) in your first equation substitute it in 21 by multipying if you found a do the same thing.
Answer:
I think it's 5 or7
Step-by-step explanation:
hope it helps you
Answer:
So both equations are true
Step-by-step explanation:
#4 was asking you to substitute x = 4 into both equations to see if it's the solution of the equations.
Substitute x = 4
Ritz equation
y = 30x + 20
y = 30(4) + 20
y = 120 + 20
y = 140
Smits equation:
y = 15x + 80
y = 15(4) + 80
y = 60 + 80
y = 140
Both companies charged same price $140 for 4 days
So both equations are true
