y - 3
g(y) = ------------------
y^2 - 3y + 9
To find the c. v., we must differentiate this function g(y) and set the derivative equal to zero:
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3)
g '(y) = --------------------------------------------
(y^2 - 3y + 9)^2
Note carefully: The denom. has no real roots, so division by zero is not going to be an issue here.
Simplifying the denominator of the derivative,
(y^2 - 3y + 9)(1) - (y - 3)(2y - 3) => y^2 - 3y + 9 - [2y^2 - 3y - 6y + 9], or
-y^2 + 6y
Setting this result = to 0 produces the equation y(-y + 6) = 0, so
y = 0 and y = 6. These are your critical values. You may or may not have max or min at one or the other.
Answer:
36°
Step-by-step explanation:
3+5+8°+10+6= 36°
<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em> </em>
In money if you have £6.25 it is easier to write it in decimals
If you are measuring something e.g. 10.3cm long no one writes it 10 3/10 cm
Answer:
1 / 18
Step-by-step explanation:
In a roll of two dice :
Number of faces on a dice = 6
Total sample space for 2 6-sided dice = (number of faces)^Number of dice = 6^2 = 36
Total possible outcomes = 36
Required outcome = sum of 11
11 = {(5,6) ; (6, 5)}) = 2 possibilities
Probability = required outcome / Total possible outcomes
P(obtaining a sum of 11) = 2 / 36 = 1/18
5 and 6. square of 5 is 25 and square of 6 is 36. must be between 5 and 6