You are playing an online game with your friend and they are on level 4. You are on level 6 and can beat, on average, 4 levels p
1 answer:
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Answer:
a) 40 feet
b) 54 ft/min
c) 4 mins
Step-by-step explanation:
Solution:-
- Kesha models the height ( h ) of the baton from the ground level but thrown from a platform of height hi.
- The function h ( t ) is modeled to follow a quadratic - parabolic path mathematically expressed as:
h ( t ) = −16t² + 54t + 40
Which gives the height of the baton from ground at time t mins.
- The initial point is of the height of the platform which is at a height of ( hi ) from the ground level.
- So the initial condition is expressed by time = 0 mins, the height of the baton h ( t ) would be:
h ( 0 ) = hi = -16*(0)^2 + 54*0 + 40
h ( 0 ) = hi = 0 + 0 + 40 = 40 feet
Answer: The height of the platform hi is 40 feet.
- The speed ( v ) during the parabolic path of the baton also varies with time t.
- The function of speed ( v ) with respect to time ( t ) can be determined by taking the derivative of displacement of baton from ground with respect to time t mins.
v ( t ) = dh / dt
v ( t )= d ( −16t² + 54t + 40 ) / dt
v ( t )= -2*(16)*t + 54
v ( t )= -32t + 54
- The velocity with which Kesha threw the baton is represented by tim t = 0 mins.
Hence,
v ( 0 ) = vi = -32*( 0 ) + 54
v ( 0 ) = vi = 54 ft / min
Answer: Kesha threw te baton with an initial speed of vo = 54 ft/min
- The baton reaches is maximum height h_max and comes down when all the kinetic energy is converted to potential energy. The baton starts to come down and cross the platform height hi = 40 feet and hits the ground.
- The height of the ball at ground is zero. Hence,
h ( t ) = 0
0 = −16t² + 54t + 40
0 = -8t^2 + 27t + 20
- Use the quadratic formula to solve the quadratic equation:
Answer: The time taken for the baton to hit the ground is t = 4 mins
Answer:
Unfortunately, your answer is not right.
Step-by-step explanation:
The functions whose graphs do not have asymptotes are the power and the root.
The power function has no asymptote, its domain and rank are all the real.
To verify that the power function does not have an asymptote, let us make the following analysis:
The function , when x approaches infinity, where does y tend? Of course it tends to infinity as well, therefore it has no horizontal asymptotes (and neither vertical nor oblique)
With respect to the function we can verify that if it has asymptote horizontal in y = 0. Since when x approaches infinity the function is closer to the value 0.
For example: 1/2 = 0.5; 1/1000 = 0.001; 1/100000 = 0.00001 and so on. As "x" grows "y" approaches zero
Also, when x approaches 0, the function approaches infinity, in other words, when x tends to 0 y tends to infinity. For example: 1 / 0.5 = 2; 1 / 0.1 = 10; 1 / 0.01 = 100 and so on. This means that the function also has an asymptote at x = 0