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:
$22.68
Step-by-step explanation:
Meal Price =$18
Sales tax =6%
Tip = 20%
Total Price of the Meal =$18 + (6% of 18)+ (20% of 18)
=$18 + (0.06 X 18)+ (0.2 X 18)
=18+1.08+3.6
=$22.68
The total price of Alyssa's meal is $22.68.
Answer:
x = 45 when y = 20 and z = 6
Step-by-step explanation:
The general formula here is y = kx/z²
If x = 24 and y = 6, z = 8, so:
y = kx/z² becomes 6 = k(24)/8², and:
6 1
k = ---------------- = -------------- = 16
24/64 4/64
Then y = 16x/z², and so, if y = 20 and z = 6 (check this!), then:
20 = 16x/36, or 16x = 720, or x = 45
x = 45 when y = 20 and z = 6
Answer: ![\frac{\sqrt[4]{10xy^3}}{2y}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D)
where y is positive.
The 2y in the denominator is not inside the fourth root
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Work Shown:
![\sqrt[4]{\frac{5x}{8y}}\\\\\\\sqrt[4]{\frac{5x*2y^3}{8y*2y^3}}\ \ \text{.... multiply top and bottom by } 2y^3\\\\\\\sqrt[4]{\frac{10xy^3}{16y^4}}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{16y^4}} \ \ \text{ ... break up the fourth root}\\\\\\\frac{\sqrt[4]{10xy^3}}{\sqrt[4]{(2y)^4}} \ \ \text{ ... rewrite } 16y^4 \text{ as } (2y)^4\\\\\\\frac{\sqrt[4]{10xy^3}}{2y} \ \ \text{... where y is positive}\\\\\\](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%7D%7B8y%7D%7D%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B5x%2A2y%5E3%7D%7B8y%2A2y%5E3%7D%7D%5C%20%5C%20%5Ctext%7B....%20multiply%20top%20and%20bottom%20by%20%7D%202y%5E3%5C%5C%5C%5C%5C%5C%5Csqrt%5B4%5D%7B%5Cfrac%7B10xy%5E3%7D%7B16y%5E4%7D%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B16y%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20break%20up%20the%20fourth%20root%7D%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%7D%20%5C%20%5C%20%5Ctext%7B%20...%20rewrite%20%7D%2016y%5E4%20%5Ctext%7B%20as%20%7D%20%282y%29%5E4%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5Csqrt%5B4%5D%7B10xy%5E3%7D%7D%7B2y%7D%20%5C%20%5C%20%5Ctext%7B...%20where%20y%20is%20positive%7D%5C%5C%5C%5C%5C%5C)
The idea is to get something of the form 
in the denominator. In this case, 
To be able to reach the 
, your teacher gave the hint to multiply top and bottom by 
For more examples, search out "rationalizing the denominator".
Keep in mind that ![\sqrt[4]{(2y)^4} = 2y](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%20%3D%202y)
only works if y isn't negative.
If y could be negative, then we'd have to say ![\sqrt[4]{(2y)^4} = |2y|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%282y%29%5E4%7D%20%3D%20%7C2y%7C)
. The absolute value bars ensure the result is never negative.
Furthermore, to avoid dividing by zero, we can't have y = 0. So all of this works as long as y > 0.
