You could use the information from part A to get B. I'm not sure if you want A or not, so I'll do it as well.
A
Eo = 10^4.4 Joules
E = 2 * 10^15
Formula
M = (2/3) log (E/Eo)
M = 2/3 * log (2 * 10^15/10^(4.4) )
M = 2/3 * log( 7.9621* 10^10)
M = 2/3 * 10.901
M = 7.26735 on the Richter scale. That is a huge amount of energy.
Part B
Suppose that you use Eo and your base. Eo is 10^4.4
Now the new earthquake is E = 10000 * Eo
So what you get now is M = (2/3)* Log(10000 * Eo / Eo )
The Eo's cancel out.
M = 2/3 * log(10000)
M = 2/3 of 4
M = 8/3
M = 2.6667 difference in the Richter Scale Reading. It is still an awful lot of energy.
What this tells you is that if the original reading was (say) 6 then the 10000 times bigger reading would 8.266667
Answer: M = 2.6667
Answer:
An obtuse triangle?
Step-by-step explanation:
Answer:
![Equation~=~\Large\boxed{2x+3=x+10}](https://tex.z-dn.net/?f=Equation~%3D~%5CLarge%5Cboxed%7B2x%2B3%3Dx%2B10%7D)
![x~=~\Large\boxed{7}](https://tex.z-dn.net/?f=x~%3D~%5CLarge%5Cboxed%7B7%7D)
![RS~=~\Large\boxed{6}](https://tex.z-dn.net/?f=RS~%3D~%5CLarge%5Cboxed%7B6%7D)
![RT~=~\Large\boxed{17}](https://tex.z-dn.net/?f=RT~%3D~%5CLarge%5Cboxed%7B17%7D)
Step-by-step explanation:
<u>Given information</u>
![RS=2x-8](https://tex.z-dn.net/?f=RS%3D2x-8)
![ST=11](https://tex.z-dn.net/?f=ST%3D11)
![RT=x+10](https://tex.z-dn.net/?f=RT%3Dx%2B10)
<u>Derived expression from the given information</u>
<em>Presumably, I think this is a combination of segments</em>
![RS+ST=RT](https://tex.z-dn.net/?f=RS%2BST%3DRT)
<u>Substitute values into the given expression</u>
![(2x-8)+(11)=(x+10)](https://tex.z-dn.net/?f=%282x-8%29%2B%2811%29%3D%28x%2B10%29)
<u>Combine like terms</u>
<em>The following is the expression</em>
![\Large\boxed{2x+3=x+10}](https://tex.z-dn.net/?f=%5CLarge%5Cboxed%7B2x%2B3%3Dx%2B10%7D)
<u>Subtract 3 on both sides</u>
![2x+3-3=x+10-3](https://tex.z-dn.net/?f=2x%2B3-3%3Dx%2B10-3)
![2x=x+7](https://tex.z-dn.net/?f=2x%3Dx%2B7)
<u>Subtract x on both sides</u>
![2x-x=x+7-x](https://tex.z-dn.net/?f=2x-x%3Dx%2B7-x)
![\Large\boxed{x=7}](https://tex.z-dn.net/?f=%5CLarge%5Cboxed%7Bx%3D7%7D)
<u>Substitute the x value into corresponding expressions to determine the final value</u>
![RS=2x-8=2(7)-8=14-8=\Large\boxed{6}](https://tex.z-dn.net/?f=RS%3D2x-8%3D2%287%29-8%3D14-8%3D%5CLarge%5Cboxed%7B6%7D)
![RT=x+10=(7)+10=\Large\boxed{17}](https://tex.z-dn.net/?f=RT%3Dx%2B10%3D%287%29%2B10%3D%5CLarge%5Cboxed%7B17%7D)
Hope this helps!! :)
Please let me know if you have any questions
Ok so lets start with the 5 dozen there are 12 eggs in one dozen so we do 12*5 which gives us 60 and then we figure out what 3/4 of 12 is that sounds confusing the way i typed it. an example is make four piles each evenly numbered so four piles of three so take three of those piles which gives you the number nine so 60 + 9 gives you 69 so Jesse has 69 eggs.