we have a maximum at t = 0, where the maximum is y = 30.
We have a minimum at t = -1 and t = 1, where the minimum is y = 20.
<h3>
How to find the maximums and minimums?</h3>
These are given by the zeros of the first derivation.
In this case, the function is:
w(t) = 10t^4 - 20t^2 + 30.
The first derivation is:
w'(t) = 4*10t^3 - 2*20t
w'(t) = 40t^3 - 40t
The zeros are:
0 = 40t^3 - 40t
We can rewrite this as:
0 = t*(40t^2 - 40)
So one zero is at t = 0, the other two are given by:
0 = 40t^2 - 40
40/40 = t^2
±√1 = ±1 = t
So we have 3 roots:
t = -1, 0, 1
We can just evaluate the function in these 3 values to see which ones are maximums and minimums.
w(-1) = 10*(-1)^4 - 20*(-1)^2 + 30 = 10 - 20 + 30 = 20
w(0) = 10*0^4 - 20*0^2 + 30 = 30
w(1) = 10*(1)^4 - 20*(1)^2 + 30 = 20
So we have a maximum at x = 0, where the maximum is y = 30.
We have a minimum at x = -1 and x = 1, where the minimum is y = 20.
If you want to learn more about maximization, you can read:
brainly.com/question/19819849
Answer:
b = -6.5
Step-by-step explanation:
I am 100% sure thts the right answer
Step-by-step explanation:
6/10
divided by 2
3/10
divided by 1/2
3/5
Question:
A sample of cans of peaches was taken from a warehouse, and the content of each can mearsed for weight. the sample means was 486g with stand deviation 6g. state the weight percentage of cans with weight:
Draw normal curve to help - split into 8 section ( i can't draw it here)
a) 34.13% of cans will be between 480g and 486g.
b) 13.59 + 2.15 + 0.13= 15.87% of cans greater than 492g.
Look at the Stand Dev Graph where the give you the number
Step-by-step explanation:
Let me give you a different question and i will answer it, which you help you answer your question.
918x3=2754 the answer is 2757 wecome
