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.
Only the third one makes x=-1 true
24, just do 6x4 because there are 4 sides.
Answer: <span>A square inscribed in a circle.
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Justification:
Note that by making two perpendicular lines that intersect each other in the center of the circle, he obtains 4 equidistant points on the circumference.
So, joining each pair of neighbouring points, the image will reveal 4 congruent sides joining at right angles (90°). This is the image of a square with the four vertices on the circumference.
