-6/5 is the slope. you start at point (0,0) and then move down 6 units and to the right 5 units and then create one point. from that point, go down 6 units and to the right 5 units and repeat. when you reach the bottom of the graph stop. go back to (0,0) and then move up 6 units and to the left 5 units. connect the points with a ruler and add arrows to each side. label your line with "y=-6/5x" on the line
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
N*3.25=325.00
N=325.00/3.25
N=100
Well x=19/3
Decimal form is 0.6333333....
Answer:
2673
Step-by-step explanation: