<span>68=-16t2+64t+8
16t2-64t+60=0
4t2-16t+15=0
(2t-5)(2t-3)=0
t=5/2 or 3/2
The object is at 68 feet after 3/2 secs, and returns to 68 feet after another second..</span>
Set x as adult tickets.
Set y as children's tickets.
x + y = 15
30x + 20y = 270
Solve for x in the first equation.
x + y = 15
x = 15 - y
Plug this into the second equation.
30x + 20y = 270
30(15 - y) + 20y = 270
450 - 30y + 20y = 270
450 - 10y = 270
-10y = -180
y = 18
If there is 18 childrens tickets, there should be -3 adult tickets.
This is impossible, and this impossible answer occured because the question is written wrong.
There are a total of 15 tickets
The smallest costing ticket is the childrens ticket, which costs 20$.
If he only bought children tickets, this would be 20x15 which is 300$.
300$ is over 270$, which makes the question impossible.
-10y^2 + (-3y^2) - 4y^2 - (-6y^2) =
-10y^2 - 3y^2 - 4y^2 + 6y^2 =
-17y^2 + 6y^2 =
- 11y^2
Answer:
Umm….. I really don’t know……..
Step-by-step explanation:
Sorry…..