Answer: 3.5, 4.5, 9.5, 3.5
Step-by-step explanation:
Look at the image below to see where A, B, C, and D are.
A + B = 8
B + D = 8
A + C = 13
C - D = 6
we can see that A + B = 8 and D + B = 8, so A = D
substitute this into A + C = 13 to get D + C = 13
from D + C = 13 we can get D = 13 - C
plug this into C - D = 6 to get C - (13 - C) = 6
2C - 13 = 6
2C = 19
C = 9.5
Now we can find D = 13 - C = 13 - 9.5 = 3.5
D = 3.5
Now we can find A = D = 3.5
A = 3.5
Now we can find B from A + B = 8
B = 8 - A = 8 - 4.5 = 4.5
B = 4.5
8800 yards is equal to five miles
The answer is A) 3:6 = 7:14
Answer with Step-by-step explanation:
We are given that

Let g(x,y)=
We have to find the extreme values of the given function


Using Lagrange multipliers



Possible value x=0 or 
If x=0 then substitute the value in g(x,y)
Then, we get 


If
and substitute in the equation
Then , we get possible value of y=0
When y=0 substitute in g(x,y) then we get

Hence, function has possible extreme values at points (0,1),(0,-1), (1,0) and (-1,0).




Therefore, the maximum value of f on the circle
is
and minimum value of 