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.
Answer:
The horizontal distance from the plane to the person on the runway is 20408.16 ft.
Step-by-step explanation:
Consider the figure below,
Where AB represent altitude of the plane is 4000 ft above the ground , C represents the runner. The angle of elevation from the runway to the plane is 11.1°
BC is the horizontal distance from the plane to the person on the runway.
We have to find distance BC,
Using trigonometric ratio,
Here, ,Perpendicular AB = 4000
Solving for BC, we get,
(approx)
(approx)
Thus, the horizontal distance from the plane to the person on the runway is 20408.16 ft
Answer:
The value of is 17-18x and is -7-18x.
Step-by-step explanation:
It is given in the question functions f(x) as 3x+2 and g(x)=5-6x.
It is required to find and .
To find , substitute g(x) for x in f(x) and simplify the expression.
To find , substitute f(x) for x in g(x) and simplify the expression.
Step 1 of 2
Substitute g(x) for x in f(x) and simplify the expression.
Step 2 of 2
Substitute f(x) for x in g(x) and simplify the expression.
Answer: 115 miles
Step-by-step explanation:
just do 60+55 miles = 115
Answer:
The whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square is 6 ft
Step-by-step explanation:
Here we are required find the size of the sides of a dunk tank (cube with open top) such that the surface area is ≤ 160 ft²
For maximum volume, the side length, s of the cube must all be equal ;
Therefore area of one side = s²
Number of sides in a cube with top open = 5 sides
Area of surface = 5 × s² = 180
Therefore s² = 180/5 = 36
s² = 36
s = √36 = 6 ft
Therefore, the whole number dimension that would allow the student to maximize the volume while keeping the surface area at most 160 square = 6 ft.