exists and is bounded for all 
. We're told that 
. Consider the interval [0, 3]. The mean value theorem says that there is some 
such that
Since 
, we have
so 19 is the largest possible value.
Step-by-step explanation:
From Pythagorean theorem, one of the sides can be determined as x^2 + y^2 =8^2
or y = (8^2 - x^2)^(1/2)
we can write the perimeter P as
P = 2x + 2y ---> 20 = 2x + 2(8^2 - x^2)^(1/2)
Dividing by 2, we get
10 = x + (8^2 - x^2)^(1/2)
Move the x to the other side,
10 - x = (8^2 - x^2)^(1/2)
Take the square of both sides to get rid of the radical sign:
(10 - x)^2 = 8^2 - x^2
Move everything to the left and expand the quantity inside the parenthesis:
x^2 + (100 - 20x + x^2) - 64 = 0
2x^2 - 20x + 64 = 0
or
x^2 - 10x + 32 = 0
Now we can see that a = -10 and b = 32
3/8 is greater than 1/8 because it is the same proportion and 3>1.
Answer:
C
Step-by-step explanation:
Using the law of cosines then cosΘ is the following quotient
cosΘ = 
where a = 5.8, b = 10.5 and c = 10.9
cosΘ = 
= 
= 
= 0.21 → C
