we have
Group terms that contain the same variable, and move the constant to the opposite side of the equation
Factor the leading coefficient
Complete the square. Remember to balance the equation by adding the same constants to each side
Rewrite as perfect squares
-----> equation in vertex form
therefore
the answer is the option C
Answer:
A ( t ) = -4t + 51
B ( t ) = -2t + 29
t < 11 hours ... [ 0 , 11 ]
Step-by-step explanation:
Given:-
- The height of snowman A, Ao = 51 in
- The height of snowman B, Bo = 29 in
Solution:-
- The day Mason made two snowmans ( A and B ) with their respective heights ( A(t) and B(t) ) will be considered as the initial value of the following ordinary differential equation.
- To construct two first order Linear ODEs we will consider the rate of change in heights of each snowman from the following day.
- The rate of change of snowman A's height ( A ) is:
- The rate of change of snowman B's height ( B ) is:
Where,
t: The time in hours from the start of melting process.
- We will separate the variables and integrate both of the ODEs as follows:
- Evaluate the constant of integration ( c ) for each solution to ODE using the initial values given: A ( 0 ) = Ao = 51 in and B ( 0 ) = Bo = 29 in:
- The solution to the differential equations are as follows:
A ( t ) = -4t + 51
B ( t ) = -2t + 29
- To determine the time domain over which the snowman A height A ( t ) is greater than snowman B height B ( t ). We will set up an inequality as follows:
A ( t ) > B ( t )
-4t + 51 > -2t + 29
2t < 22
t < 11 hours
- The time domain over which snowman A' height is greater than snowman B' height is given by the following notation:
Answer: [ 0 , 11 ]
