Answer:
<em>It will occur zero times between midnight and one o'clock.</em>
Step-by-step explanation:
<u>Least Common Multiple (LCM)</u>
Three events keep James from sleeping: his clock ticking every 20 seconds, a tap dripping every 15 seconds, and his dog snoring every 27 seconds.
All three events happened together at midnight. They will happen together again the first time the numbers 20, 15, and 27 have a common multiple. This is the LCM.
List the prime factors of each number:
20: 2,2,5
15: 3,5
27: 3,3,3
Now multiply all the factors the maximum number of times they appear:
LCM=2*2*3*3*3*5=540
(a) All the events will happen together again after 540 minutes.
(b) Since 540 minutes = 9 hours, this event won't happen again until 9 am. Thus, it will occur zero times between midnight and one o'clock.
Answer:
Step-by-step explanation:
Applying the Laplace transform:
![\mathcal{L}[y'']+5\mathcal{L}[y']+4\mathcal{L}[y']=0](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%2B5%5Cmathcal%7BL%7D%5By%27%5D%2B4%5Cmathcal%7BL%7D%5By%27%5D%3D0)
With the formulas:
![\mathcal{L}[y'']=s^2\mathcal{L}[y]-y(0)s-y'(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%3Ds%5E2%5Cmathcal%7BL%7D%5By%5D-y%280%29s-y%27%280%29)
![\mathcal{L}[y']=s\mathcal{L}[y]-y(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%5D%3Ds%5Cmathcal%7BL%7D%5By%5D-y%280%29)
![\mathcal{L}[x]=L](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5Bx%5D%3DL)
Solving for 
Apply the inverse Laplace transform with this formula:
![\mathcal{L}^{-1}[\frac1{s-a}]=e^{at}](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs-a%7D%5D%3De%5E%7Bat%7D)
![y=3\mathcal{L}^{-1}[\frac1{s+4}]=3e^{-4t}](https://tex.z-dn.net/?f=y%3D3%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs%2B4%7D%5D%3D3e%5E%7B-4t%7D)
I see your last line is : c(x) = 0.9(x^2-10)^2 + 101.1
Let y = x^2, then c(y) = 0.9(y-10)^2 + 101.1
Apparently, c(y) is a parabola, min is 101.1 when y = 10, max is infinity
So let x^2 = 10 -> x = sqrt(10) or -sqrt(10), min is 101.1, max is infinity
Two fifths are 4 one fourth is 2 so there are 2 red guppies the fraction that they are red is 2 over 4 the fraction that are not red are 2 over 4 too
Answer:
The first one is 8/12.
Step-by-step explanation:Add numerator and keep denominator same.
