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
Hello!
If a you want to find an equation of the line that is parallel to the given line, we must know that if two lines are parallel, then their slope will <em>always</em> have to be equal.
This means that the new equation must have a slope of -1/3.
Also, you would need to substitute the given point (6, 8) into the equation with the slope: -1/3 as well. Following these steps allows you to find the equation of line.
y = -1/3x + b (6, 5)
5 = -1/3(6) + b
5 = -2 + b (add 2 to both sides)
7 = b
Therefore, the equation of the line is y = -1/3x + 7.
90
55.8 /? = 62/100
55.8 x 100= 62 x ?
5580=62 x ?
5580/62 = ?
?=90
Answer:
A is true but I dont know the rest I am very sorry .-.
Step-by-step explanation:
I Just added 18 + 12 = 30 + 10 = 40 and that is why A is true