Answer: he purchased 16 ride tickets.
Step-by-step explanation:
Let x represent the number of ride tickets that he purchased.
Let y represent the number of game tickets that he purchased.
Levi purchased a total of 50 ride tickets and game tickets at the amusement park. It means that
x + y = 50
If ride tickets cost $.75 each and game tickets cost $.50 each and the total amount spent on the tickets is $29, it means that
0.75x + 0.5y = 29- - - - - - - - - - - -1
Substituting x = 50 - y into equation 1, it becomes
0.75(50 - y) + 0.5y = 29
37.5 - 0.75y + 0.5y = 29
- 0.75y + 0.5y = 29 - 37.5
- 0.25y = - 8.5
y = - 8.5/-0.25
y = 34
x = 50 - y = 50 - 34
x = 16
Answer:
Option D
Y=2.6x
Step-by-step explanation:
When y=10.4 and x=4, these two can be related by the equation
10.4=4x
Making x the subject of the formula then x=10.4/4=2.6
Relating x and y then we have the equation
Y=2.6x
For example, when x is 4 then y=2.6x=2.6(4)=10.4 as given in the question. Therefore, the right option is D
Y=2.6x
Assuming that the figures given are square such that the scale factor between them is equal to 28/8 which can be further simplified into 7/2. The ratio of the perimeter is also equal to this value, 7/2. However, the ratio of the areas is equal to the square of this value giving us an answer of 49/4.
<h2>
Hello!</h2>
The answer is:
The ball will hit the ground after 
<h2>
Why?</h2>
Since we are given a quadratic function, we can calculate the roots (zeroes) using the quadratic formula. We must take into consideration that we are talking about time, it means that we should only consider positive values.
So,

We are given the function:

Where,

Then, substituting it into the quadratic equation, we have:

Since negative time does not exists, the ball will hit the ground after:

Have a nice day!