using only the information derived in question 2, 3, and 4, which theorem
D. AAS
 
        
                    
             
        
        
        
Answer:

Step-by-step explanation:
 
        
                    
             
        
        
        
Answer:
It will take  <u><em>80 days</em></u>  for the bull calf to reach a weight of 500 kilograms.
Step-by-step explanation:
Given:
The weight of a bull calf is 388 kilograms.
Now, to find the weight of bull calf of how long it will take to reach a weight of 500 kilograms, if it’s weight increases at a rate of 1 2/5 kilograms per day.
Required weight which to be increased = 500 - 388 = 112 kilograms.
Rate of weight increase = 
                                         =  
 
Thus, the time required = 
                                         =  
 
                                         =  
 
<em>The time required   =    80 days</em>.
Therefore, it will take 80 days for the bull calf to reach a weight of 500 kilograms.
 
        
             
        
        
        
Answer:
a = 9
Step-by-step explanation:
Simplifying
8.1 = 0.9a
Solving
8.1 = 0.9a
Solving for variable 'a'.
Move all terms containing a to the left, all other terms to the right.
Add '-0.9a' to each side of the equation.
8.1 + -0.9a = 0.9a + -0.9a
Combine like terms: 0.9a + -0.9a = 0.0
8.1 + -0.9a = 0.0
Add '-8.1' to each side of the equation.
8.1 + -8.1 + -0.9a = 0.0 + -8.1
Combine like terms: 8.1 + -8.1 = 0.0
0.0 + -0.9a = 0.0 + -8.1
-0.9a = 0.0 + -8.1
Combine like terms: 0.0 + -8.1 = -8.1
-0.9a = -8.1
Divide each side by '-0.9'.
a = 9
Simplifying
a = 9
 
        
             
        
        
        
Answer:
x=2.125
y=0
C=19.125
Step-by-step explanation:
To solve this problem we can use a graphical method, we start first noticing the restrictions  and
 and   , which restricts the solution to be in the positive quadrant. Then we plot the first restriction
, which restricts the solution to be in the positive quadrant. Then we plot the first restriction  shown in purple, then we can plot the second one
 shown in purple, then we can plot the second one  shown in the second plot in green.
 shown in the second plot in green.
The intersection of all three restrictions is plotted in white on the third plot. The intersection points are also marked.
So restrictions intersect on (0,0), (0,1.7) and (2.215,0). Replacing these coordinates on the objective function we get C=0, C=11.9, and C=19.125 respectively. So The function is maximized at (2.215,0) with C=19.125.