Answer:
Advance tickets=$25
Same-day tickets=$15
Step-by-step explanation:
Complete question below:
Suppose that there are two types of tickets to a show: advance and same-day. The combined cost of one advance ticket and one same-day ticket is $ 40. For one performance, 25 advance tickets and 30 same-day tickets were sold. The total amount paid for the tickets was $1075 . What was the price of each kind of ticket?
Let
advance tickets=x
Same-day tickets=y
Combined cost of advance and same-day tickets=$40
It means,
x+y=40 Equ (1)
25 advance tickets and 30 same-day tickets=$1075
It means,
25x+30y=1075 Equ(2)
From (1)
x+y=40
x=40-y
Substitute x=40-y into (2)
25x+30y=1075
25(40-y)+30y=1075
1000-25y+30y=1075
5y=1075-1000
5y=75
Divide both sides by 5
5y/5=75/5
y=15
Recall,
x+y=40
x+15=40
x=40-15
=25
x=25
Advance tickets=$25
Same-day tickets=$15
Check
25x+30y=1075
25(25)+30(15)=1075
625+450=1075
1075=1075
What’s something goes up but never comes down?
Answer: x = 560 miles/hr
Step-by-step explanation:
Let speed of first plane = x
Let speed of second plane = y
x-y =108 (first plane is faster)
It took first plane 2.25 hrs to be 1825 miles apart.
Distance moved after 2.25 hrs =2.25x
It took second plane 1.25 hrs to be 1825 miles apart
Distance moved after 1.25 hrs =1.25y
So 2.25x + 1.25y = 1825 --------1
Put x = 108+y in eqn 1
2.25(108+y) + 1.25y= 1825
243 + 3.5y = 1825
3.5y= 1582
y = 452
x = 108 + 452 = 560 miles/hr
Give more information, this gives us basically nothing.
