So prime numbers are numbers which cant be divided by any number other than one and these are <span>2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, </span>61<span>, </span>67<span>, </span>71,73<span>, </span>79<span>, </span><span>83
83+2 is 85
so the product is the two numbers multiplied together which is 166</span>
Answer: ![11\frac{2}{15}](https://tex.z-dn.net/?f=11%5Cfrac%7B2%7D%7B15%7D)
Step-by-step explanation:
What you are trying to do is isolate the variable x. For you to do that you have to first simplify the left side of the equation and then you solve for x.
Answer:
You:If it’s was 6 cups of citric I’d be to 32 water
Your friend: if it was 4 cups it’d be 12 water
Step-by-step explanation:
I believe your friends is more acidic because your Is being diluted more
Answer:
As a statistical tool, a frequency distribution provides a visual representation for the distribution of observations within a particular test. Analysts often use frequency distribution to visualize or illustrate the data collected in a sample.
Answer:
The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.
Step-by-step explanation:
The volume of the sphere (
), measured in cubic feet, is represented by the following formula:
![V = \frac{4\pi}{3}\cdot r^{3}](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7B4%5Cpi%7D%7B3%7D%5Ccdot%20r%5E%7B3%7D)
Where
is the radius of the sphere, measured in feet.
The change in volume is obtained by means of definition of total difference:
![\Delta V = \frac{\partial V}{\partial r}\Delta r](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%5Cfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20r%7D%5CDelta%20r)
The derivative of the volume as a function of radius is:
![\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpartial%20V%7D%7B%5Cpartial%20r%7D%20%3D%204%5Cpi%20%5Ccdot%20r%5E%7B2%7D)
Then, the change in volume is expanded:
![\Delta V = 4\pi \cdot r^{2}\cdot \Delta r](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%204%5Cpi%20%5Ccdot%20r%5E%7B2%7D%5Ccdot%20%5CDelta%20r)
If
and
, the change in the volume of the sphere is approximately:
![\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)](https://tex.z-dn.net/?f=%5CDelta%20V%20%5Capprox%204%5Cpi%5Ccdot%20%2840%5C%2Cft%29%5E%7B2%7D%5Ccdot%20%280.05%5C%2Cft%29)
![\Delta V \approx 1005.310\,ft^{3}](https://tex.z-dn.net/?f=%5CDelta%20V%20%5Capprox%201005.310%5C%2Cft%5E%7B3%7D)
The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.