The answer to the question is the second option.
The first one is 4n+7=15, the second one is 1/4(n-15), the third one is 1/4n-15, the fourth one is n/7=4, and the last one is n4+7=15
Answer:
![\frac{4.1}{100} \times 714.73 = 29.30 \\ 714.73 - 29.30 = 685.43](https://tex.z-dn.net/?f=%20%5Cfrac%7B4.1%7D%7B100%7D%20%20%5Ctimes%20714.73%20%3D%2029.30%20%20%5C%5C%20714.73%20-%2029.30%20%3D%20685.43)
divide 4.1 with 100percent because per5 is always over 100 them multiply by the amount given
Step-by-step explanation:
subtract 29.30 from 714.30
subtract calculated amount from amount given
get yr final answer
The equation that is equivalent to the given logarithmic equation is 3² = x + 5. The correct option is the third option 3² = x + 5
<h3>Logarithmic equations</h3>
From the question, we are to determine the equation that is equivalent to the given logarithmic equation
The given logarithmic equation is
![log_{3}(x+5) = 2](https://tex.z-dn.net/?f=log_%7B3%7D%28x%2B5%29%20%3D%202)
From one of the laws of logarithm
If ![log_{x}(y) = z](https://tex.z-dn.net/?f=log_%7Bx%7D%28y%29%20%3D%20z)
Then,
![y = x^{z}](https://tex.z-dn.net/?f=y%20%3D%20x%5E%7Bz%7D)
Thus,
becomes
![x+5 = 3^{2}](https://tex.z-dn.net/?f=x%2B5%20%3D%203%5E%7B2%7D)
∴ ![3^{2} = x+5](https://tex.z-dn.net/?f=3%5E%7B2%7D%20%3D%20x%2B5)
Hence, the equation that is equivalent to the given logarithmic equation is 3²= x + 5. The correct option is the third option 3² = x + 5
Learn more on Logarithmic equations here: brainly.com/question/23268551
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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:
![h = \frac{V}{\pi\cdot r^{2}}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7BV%7D%7B%5Cpi%5Ccdot%20r%5E%7B2%7D%7D)
By substitution in (2) we get the following formula:
![A_{s} = 2\pi \cdot r^{2} + 2\pi\cdot r\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)](https://tex.z-dn.net/?f=A_%7Bs%7D%20%3D%202%5Cpi%20%5Ccdot%20r%5E%7B2%7D%20%2B%202%5Cpi%5Ccdot%20r%5Ccdot%20%5Cleft%28%5Cfrac%7BV%7D%7B%5Cpi%5Ccdot%20r%5E%7B2%7D%7D%20%5Cright%29)
![A_{s} = 2\pi \cdot r^{2} +\frac{2\cdot V}{r}](https://tex.z-dn.net/?f=A_%7Bs%7D%20%3D%202%5Cpi%20%5Ccdot%20r%5E%7B2%7D%20%2B%5Cfrac%7B2%5Ccdot%20V%7D%7Br%7D)
Then, we derive the formulas for the First and Second Derivative Tests:
First Derivative Test
![4\pi\cdot r -\frac{2\cdot V}{r^{2}} = 0](https://tex.z-dn.net/?f=4%5Cpi%5Ccdot%20r%20-%5Cfrac%7B2%5Ccdot%20V%7D%7Br%5E%7B2%7D%7D%20%3D%200)
![4\pi\cdot r^{3} - 2\cdot V = 0](https://tex.z-dn.net/?f=4%5Cpi%5Ccdot%20r%5E%7B3%7D%20-%202%5Ccdot%20V%20%3D%200)
![2\pi\cdot r^{3} = V](https://tex.z-dn.net/?f=2%5Cpi%5Ccdot%20r%5E%7B3%7D%20%3D%20V)
![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
![A_{s}'' = 4\pi + \frac{4\cdot V}{r^{3}}](https://tex.z-dn.net/?f=A_%7Bs%7D%27%27%20%3D%204%5Cpi%20%2B%20%5Cfrac%7B4%5Ccdot%20V%7D%7Br%5E%7B3%7D%7D)
If
:
![A_{s} = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }](https://tex.z-dn.net/?f=A_%7Bs%7D%20%3D%204%5Cpi%20%2B%20%5Cfrac%7B4%5Ccdot%20V%7D%7B%5Cfrac%7BV%7D%7B2%5Cpi%7D%20%7D)
(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)
![r = 8\,cm](https://tex.z-dn.net/?f=r%20%3D%208%5C%2Ccm)
![h = \frac{V}{\pi\cdot r^{2}}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7BV%7D%7B%5Cpi%5Ccdot%20r%5E%7B2%7D%7D)
![h = \frac{3217\,cm^{3}}{\pi\cdot (8\,cm)^{2}}](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B3217%5C%2Ccm%5E%7B3%7D%7D%7B%5Cpi%5Ccdot%20%288%5C%2Ccm%29%5E%7B2%7D%7D)
![h = 16\,cm](https://tex.z-dn.net/?f=h%20%3D%2016%5C%2Ccm)
And the amount of plastic needed to cover the outside of the candle for packaging is:
![A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h](https://tex.z-dn.net/?f=A_%7Bs%7D%20%3D%202%5Cpi%5Ccdot%20r%5E%7B2%7D%20%2B%202%5Cpi%5Ccdot%20r%20%5Ccdot%20h)
![A_{s} = 2\pi\cdot (8\,cm)^{2} + 2\pi\cdot (8\,cm)\cdot (16\,cm)](https://tex.z-dn.net/?f=A_%7Bs%7D%20%3D%202%5Cpi%5Ccdot%20%288%5C%2Ccm%29%5E%7B2%7D%20%2B%202%5Cpi%5Ccdot%20%288%5C%2Ccm%29%5Ccdot%20%2816%5C%2Ccm%29)
![A_{s} \approx 1206.372\,cm^{2}](https://tex.z-dn.net/?f=A_%7Bs%7D%20%5Capprox%201206.372%5C%2Ccm%5E%7B2%7D)
The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.