1. Perimeter
22a + 2(6a + 8) -4 = p
[I got 22a from 12a + 10a]
2. If p = 250, what is a?
22a + 2(6a + 8) - 4 = 250
Distributive property
22a + 12a + 16 - 4 = 250
Add like terms
34a + 12 = 250
Subtract 12 by both sides
34a = 238
Divide 34 by both sides
a = 7
Answer:
The probabily of P(not less than 6) is
P= the favorable outcomes/ the total outcomes
favorable out come is 1 case(7)
the total outcomes are 5 cases(2,3,5,6)
P=1/5=20%
Dividing two decimals means that we need to make the following math operation:
![\frac{p}{q} \\ \\ where \ p \ and \ q \ are \ two \ decimals](https://tex.z-dn.net/?f=%5Cfrac%7Bp%7D%7Bq%7D%20%5C%5C%20%5C%5C%20where%20%5C%20p%20%5C%20and%20%5C%20q%20%5C%20are%20%5C%20two%20%5C%20decimals)
According to the following definition:
![If \ p \ and \ d \ are \ two \ real \ numbers \ (two \ decimals) \ it \ is \ true \ that:\\ \\ p=d\times q+r \\ \\ where:\\ \\ p:Dividend \\ d:Divisor \\q:Quotient \\ r:Reminder](https://tex.z-dn.net/?f=%20If%20%5C%20p%20%5C%20and%20%5C%20d%20%5C%20are%20%5C%20two%20%5C%20real%20%5C%20numbers%20%5C%20%28two%20%5C%20decimals%29%20%5C%20it%20%5C%20is%20%5C%20true%20%5C%20that%3A%5C%5C%20%5C%5C%20%20p%3Dd%5Ctimes%20q%2Br%20%5C%5C%20%5C%5C%20where%3A%5C%5C%20%5C%5C%20p%3ADividend%20%5C%5C%20d%3ADivisor%20%5C%5Cq%3AQuotient%20%5C%5C%20r%3AReminder%20)
We can prove that our result is correct by multiplying the divisor by the quotient plus the reminder and verifying that this result matches the dividend. Also, we can use a calculator to prove that.
Answer:
1.![\frac{dy}{dt}=ky](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3Dky)
2.543.6
Step-by-step explanation:
We are given that
y(0)=200
Let y be the number of bacteria at any time
=Number of bacteria per unit time
![\frac{dy}{dt}\proportional y](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%5Cproportional%20y)
![\frac{dy}{dt}=ky](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3Dky)
Where k=Proportionality constant
2.
,y'(0)=100
Integrating on both sides then, we get
![lny=kt+C](https://tex.z-dn.net/?f=lny%3Dkt%2BC)
We have y(0)=200
Substitute the values then , we get
![ln 200=k(0)+C](https://tex.z-dn.net/?f=ln%20200%3Dk%280%29%2BC)
![C=ln 200](https://tex.z-dn.net/?f=C%3Dln%20200)
Substitute the value of C then we get
![ln y=kt+ln 200](https://tex.z-dn.net/?f=ln%20y%3Dkt%2Bln%20200)
![ln y-ln200=kt](https://tex.z-dn.net/?f=ln%20y-ln200%3Dkt)
![ln\frac{y}{200}=kt](https://tex.z-dn.net/?f=ln%5Cfrac%7By%7D%7B200%7D%3Dkt)
![\frac{y}{200}=e^{kt}](https://tex.z-dn.net/?f=%5Cfrac%7By%7D%7B200%7D%3De%5E%7Bkt%7D)
![y=200e^{kt}](https://tex.z-dn.net/?f=y%3D200e%5E%7Bkt%7D)
Differentiate w.r.t
![y'=200ke^{kt}](https://tex.z-dn.net/?f=y%27%3D200ke%5E%7Bkt%7D)
Substitute the given condition then, we get
![100=200ke^{0}=200 \;because \;e^0=1](https://tex.z-dn.net/?f=100%3D200ke%5E%7B0%7D%3D200%20%5C%3Bbecause%20%5C%3Be%5E0%3D1)
![k=\frac{100}{200}=\frac{1}{2}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B100%7D%7B200%7D%3D%5Cfrac%7B1%7D%7B2%7D)
![y=200e^{\frac{t}{2}}](https://tex.z-dn.net/?f=y%3D200e%5E%7B%5Cfrac%7Bt%7D%7B2%7D%7D)
Substitute t=2
Then, we get ![y=200e^{\frac{2}{2}}=200e](https://tex.z-dn.net/?f=y%3D200e%5E%7B%5Cfrac%7B2%7D%7B2%7D%7D%3D200e)
![y=200(2.718)=543.6=543.6](https://tex.z-dn.net/?f=y%3D200%282.718%29%3D543.6%3D543.6)
e=2.718
Hence, the number of bacteria after 2 hours=543.6