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:
Step-by-step explanation:
Subtract 4 by 2.5 which is 1.5
Divide 4 by 2.5 which is 1.6
Im not sure if you have to divide or subtract so...
Answer:
13.12
Step-by-step explanation:
Answer:
The simplified form of -6.3x+14 and 1.5x-6 is -4.8x+8
Step-by-step explanation:
We have to simplify the following
-6.3x+14 and 1.5x-6
it can be written as:
=(-6.3x+14) + (1.5x-6)
Adding the like terms
=(-6.3x+1.5x)+(14-6)
= (-4.8x)+(8)
= -4.8x+8
So, the simplified form of -6.3x+14 and 1.5x-6 is -4.8x+8
<h2>Answer:</h2><h2>2(x^2 + 2x + 3) or
.</h2><h2>~</h2><h3>To find the answer, simply factor 2 out of 2x^2+4x+6 to get your final answer, </h3><h3>2(x^2 + 2x + 3)</h3><h2>~</h2><h3>Thank you for taking the time to read this!</h3><h3><u>And I hope you have a good day, Loves!</u></h3><h3><em><u>Also if I'm right can you let me know, please?</u></em></h3>
<em><u>Also Also if I am right can I please have Brainliest??</u></em>
<u>You don't need to, I'm just glad I could help. ^^</u>
(If I did help-)
And don't forget to eat a full meal, and have at least one cup of water today!
