![\sqrt[3]{a} = {a}^{m} \\ = > { a}^{ \frac{1}{3} } = {a}^{m} \\ = > m = \frac{1}{3}](https://tex.z-dn.net/?f=%20%5Csqrt%5B3%5D%7Ba%7D%20%20%3D%20%20%7Ba%7D%5E%7Bm%7D%20%20%5C%5C%20%20%3D%20%20%3E%20%20%7B%20a%7D%5E%7B%20%5Cfrac%7B1%7D%7B3%7D%20%7D%20%20%3D%20%20%7Ba%7D%5E%7Bm%7D%20%20%5C%5C%20%20%3D%20%20%3E%20m%20%3D%20%20%5Cfrac%7B1%7D%7B3%7D%20)
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Answer:
A reflection followed by a translation!!!!!!!!!!!!!!!!!!!!!!!!!
Step-by-step explanation:
Answer:
61.71
Step-by-step explanation:
Answer: she has spent $60 dollars and she has 180 more to go
Step-by-step explanation:
Answer:
The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.
Step-by-step explanation:
The volume (
), in cubic centimeters, and surface area (
), in square centimeters, formulas for the candle are described below:
(1)
(2)
Where:
- Radius, in centimeters.
- Height, in centimeters.
By (1) we have an expression of the height in terms of the volume and the radius of the candle:

By substitution in (2) we get the following formula:


Then, we derive the formulas for the First and Second Derivative Tests:
First Derivative Test



![r = \sqrt[3]{\frac{V}{2\pi} }](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7BV%7D%7B2%5Cpi%7D%20%7D)
There is just one result, since volume is a positive variable.
Second Derivative Test

If
:

(which means that the critical value leads to a minimum)
If we know that
, then the dimensions for the minimum amount of plastic are:
![r = \sqrt[3]{\frac{V}{2\pi} }](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7BV%7D%7B2%5Cpi%7D%20%7D)
![r = \sqrt[3]{\frac{3217\,cm^{3}}{2\pi}}](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B3217%5C%2Ccm%5E%7B3%7D%7D%7B2%5Cpi%7D%7D)




And the amount of plastic needed to cover the outside of the candle for packaging is:



The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.