Notice that 72 hours is 12 times the half-life of 6 hours. So, the starting amount of 100 mg will decay to half, 12 times. In other words
• after 1 hour, 100 mg decays to 50 mg
• after a total 2 hours, 100 mg decays to 25 mg
• after a total 3 hours, 100 mg decays to 12.5 mg
and so on, so that after a total of 72 hours, the hospital is left with
![\dfrac{100\,\rm mg}{2^{12}} = \boxed{0.0244140625 \, \mathrm{mg}}](https://tex.z-dn.net/?f=%5Cdfrac%7B100%5C%2C%5Crm%20mg%7D%7B2%5E%7B12%7D%7D%20%3D%20%5Cboxed%7B0.0244140625%20%5C%2C%20%5Cmathrm%7Bmg%7D%7D)
Answer:
5^16
Step-by-step explanation:
25^9 + 5^17
Replace 25 with 5^2
5^2^9 + 5^17
We know that a^b^c = a^(b*c)
5^(2*9)+ 5^17
5^18+ 5^17
We can factor out 5^17
5^17 (5+1)
5^17(6)
But we need 30 so I need one more 5 inside 5^17 = 5^16 *5
5^16 *5 *6
5^16(30)
Answer:
25
Step-by-step explanation:
1/x+x/(x+2)=1
take LCM
![(x+2)+x^2/(x(x+2)=1](https://tex.z-dn.net/?f=%28x%2B2%29%2Bx%5E2%2F%28x%28x%2B2%29%3D1)
x+2+x^2=x(x+2)
![x^2+x+2=x^2+2x](https://tex.z-dn.net/?f=x%5E2%2Bx%2B2%3Dx%5E2%2B2x)
x=2
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2)
![1/(x-3)+1/(x+5)=x+1/x-3](https://tex.z-dn.net/?f=1%2F%28x-3%29%2B1%2F%28x%2B5%29%3Dx%2B1%2Fx-3)
![x+5+x-3/(x+1)(x-3)=x+1/x-3](https://tex.z-dn.net/?f=x%2B5%2Bx-3%2F%28x%2B1%29%28x-3%29%3Dx%2B1%2Fx-3)
![2x+2=(x+1)(x+5)](https://tex.z-dn.net/?f=2x%2B2%3D%28x%2B1%29%28x%2B5%29%20)
![2x+2=x^2+5x+x+5](https://tex.z-dn.net/?f=2x%2B2%3Dx%5E2%2B5x%2Bx%2B5)
![x^2+4x+3=0](https://tex.z-dn.net/?f=x%5E2%2B4x%2B3%3D0)
![x^2+3x+x+3=0](https://tex.z-dn.net/?f=x%5E2%2B3x%2Bx%2B3%3D0)
(x+1)(x+3)=0
x= - 1, x= - 3
--------------------------------------------
3)
![3x/x^2+2x-8=1/x-2+x/x+4](https://tex.z-dn.net/?f=3x%2Fx%5E2%2B2x-8%3D1%2Fx-2%2Bx%2Fx%2B4)
after solving both side u will get
![3x=x+4+x-2](https://tex.z-dn.net/?f=3x%3Dx%2B4%2Bx-2)
3x=2x+2
x=2
--------------------------------
![1/x^2-1 =2/x^2+x-2](https://tex.z-dn.net/?f=1%2Fx%5E2-1%20%3D2%2Fx%5E2%2Bx-2)
1/(x-1)(x+1)=2/(x-1)(x+2)
u will get same will cross out each other
1/(x+1)=2/(x+2)
(x+2)=2(x+1)
x+2=2x+2
x=0
I believe it python idk any but due to my research i think it python