Answer:
All of them
Step-by-step explanation:
ツ
Answer:
Step-by-step explanation:
2x^2-6x+10=0
2(x^2-3x)+10=0
2(x-(3/2))^2-9/4)+10=0
2(x-(3/2))^2+10-9/2=0
2(x-(3/2))^2+(20-9)/2=0
2(x-(3/2))^2+11/2=0
2(x-(3/2))^2=-11/2
square is always positive so there is no solution
1/6 a cup. Since she was supposed to use 1/2 a cup for the original, just divide that by three and it’s 1/6
Answer:
9,000 ×(1+0.02)^6
Step-by-step explanation:
put that in yer calculator
Answer:
21 child tickets
Step-by-step explanation:
For this problem, you'd use a system of equations.
First, define your variables.
x = # of child tickets sold
y = # of adult tickets sold
There were 4 times as many adult tickets (y) as child tickets (x) sold, so:
4 x = y
4 x - y = 0
93 (4 x - y = 0)
372 x - 93 y = 0
The total revenue was $909.30, adult tickets were $9.30 each, child tickets were $6.10 each, so:
6.10 x + 9.30 y = 909.30
10 (6.10 x + 9.30 y = 909.30)
61 x + 93 y = 9093
372 x - 93 y = 0
433 x = 9093
433 x / 433 = 9093 / 433
x = 21
