Find the critical points of f(y):Compute the critical points of -5 y^2
To find all critical points, first compute f'(y):( d)/( dy)(-5 y^2) = -10 y:f'(y) = -10 y
Solving -10 y = 0 yields y = 0:y = 0
f'(y) exists everywhere:-10 y exists everywhere
The only critical point of -5 y^2 is at y = 0:y = 0
The domain of -5 y^2 is R:The endpoints of R are y = -∞ and ∞
Evaluate -5 y^2 at y = -∞, 0 and ∞:The open endpoints of the domain are marked in grayy | f(y)-∞ | -∞0 | 0∞ | -∞
The largest value corresponds to a global maximum, and the smallest value corresponds to a global minimum:The open endpoints of the domain are marked in grayy | f(y) | extrema type-∞ | -∞ | global min0 | 0 | global max∞ | -∞ | global min
Remove the points y = -∞ and ∞ from the tableThese cannot be global extrema, as the value of f(y) here is never achieved:y | f(y) | extrema type0 | 0 | global max
f(y) = -5 y^2 has one global maximum:Answer: f(y) has a global maximum at y = 0
Answer:
T,F,T,F
Step-by-step explanation:
I got x is less than or equal to -3/2
Answer:
Step-by-step explanation:
given that the Chocolate House specializes in hand-dipped chocolates for special occasions. Three employees do all of the product packaging
Clerk I II III total
Pack 0.33 0.23 0.44 1
Defective 0.02 0.025 0.015
Pack&def 0.0066 0.00575 0.0066 0.01895
a) probability that a randomly selected box of chocolates was packed by Clerk 2 and does not contain any defective chocolate
= P(II clerk) -P(II clerk and defective) = 
b) the probability that a randomly selected box contains defective chocolate=P(I and def)+P(ii and def)+P(iiiand def)
=0.01895
c) Suppose a randomly selected box of chocolates is defective. The probability that it was packaged by Clerk 3
=P(clerk 3 and def)/P(defective)
=
In an arithmetic equation, there is no variable in the 'meat' of the equation(example: 5-5=0). In an algebraic equation, there is a variable in the meat of the equation(example: 5-x=0).
Hope this helps and please give brainliest!
