Answer:
(-2, 1)
(2, 5)
Please solve the third question and answer in the comments and I will tell you if you are correct.
Step-by-step explanation:
1)Find the solution of the following system of equations; express the answer as a coordinate pair (x,y):
y = x + 3 subtract from the bottom and solve for x
y = -2x - 3
y - y = x - (-2x) + 3 - (-3)
0 = 3x + 6
-6 = 3x
-2 = x
y = x + 3 y = -2 + 3 y = 1 (-2, 1)
2)Find the solution of the following system of equations; express the
answer as a coordinate pair (x,y):
y = 2x + 1 subtract from the bottom and solve for x
y = x/2 + 4
y-y = 2x - 1/2 x +1 - 4
0 = x(2-1/2) - 3
3 = 1.5x
2 = x y = 2x + 1 y = 2(2) + 1 y = 5 (2, 5)
3) Find the solution of the following system of equations; express the
answer as a coordinate pair (x,y):
y = -3x + 6 subtract from the bottom and solve for x
y= -1/2x+1
y-y = x(-3 - (-1/2)) + (6 - 1) follow the examples above
Answer:
a) ![P(t) = 660 e^{\frac{ln(6)}{6} t}](https://tex.z-dn.net/?f=%20P%28t%29%20%3D%20660%20e%5E%7B%5Cfrac%7Bln%286%29%7D%7B6%7D%20t%7D)
b) ![P(t=3) = 660 e^{\frac{ln(6)}{6}*3} = 1616.663 \approx 1617](https://tex.z-dn.net/?f=%20P%28t%3D3%29%20%3D%20660%20e%5E%7B%5Cfrac%7Bln%286%29%7D%7B6%7D%2A3%7D%20%3D%201616.663%20%5Capprox%201617)
Step-by-step explanation:
For this case we have the following info:
represent the initial amount of bacteria
represent the population of bacteria after 6 hours
Part a
For this case we use the proportional model given by:
![\frac{dP}{dt}= kP](https://tex.z-dn.net/?f=%20%5Cfrac%7BdP%7D%7Bdt%7D%3D%20kP)
Where P represent the population at time t and k is a constant of proportionality.
We can rewrite the model like this:
![\frac{dP}{P} = k dt](https://tex.z-dn.net/?f=%20%5Cfrac%7BdP%7D%7BP%7D%20%3D%20k%20dt)
And if we integrate both sides we got:
![ln |P|= kt + C](https://tex.z-dn.net/?f=%20ln%20%7CP%7C%3D%20kt%20%2B%20C)
Now if we apply exponential we got:
![P = e^{kt +c}= e^{kt} e^C](https://tex.z-dn.net/?f=%20P%20%3D%20e%5E%7Bkt%20%2Bc%7D%3D%20e%5E%7Bkt%7D%20e%5EC)
![P(t)= P_o e^{kt}](https://tex.z-dn.net/?f=%20P%28t%29%3D%20P_o%20e%5E%7Bkt%7D)
For this case using the initial condition:
![660 = P_o e^{0k}, P_o = 660](https://tex.z-dn.net/?f=%20660%20%3D%20P_o%20e%5E%7B0k%7D%2C%20P_o%20%3D%20660)
We have the following model:
![P(t) = 660 e^{kt}](https://tex.z-dn.net/?f=%20P%28t%29%20%3D%20660%20e%5E%7Bkt%7D)
Using the second condition
we have this:
![3960 = 660e^{6k}](https://tex.z-dn.net/?f=%203960%20%3D%20660e%5E%7B6k%7D)
We can divide both sides by 660 and we got:
![6= e^{6k}](https://tex.z-dn.net/?f=%206%3D%20e%5E%7B6k%7D)
Now we can apply natural log on both sides and we got:
![ln (6)= 6k](https://tex.z-dn.net/?f=%20ln%20%286%29%3D%206k)
![k = \frac{ln(6)}{6}=0.2986265782](https://tex.z-dn.net/?f=%20k%20%3D%20%5Cfrac%7Bln%286%29%7D%7B6%7D%3D0.2986265782)
So then the model would be given by:
![P(t) = 660 e^{\frac{ln(6)}{6} t}](https://tex.z-dn.net/?f=%20P%28t%29%20%3D%20660%20e%5E%7B%5Cfrac%7Bln%286%29%7D%7B6%7D%20t%7D)
Part b
For this case we just need to replace t=3 into the model and we got:
![P(t=3) = 660 e^{\frac{ln(6)}{6}*3} = 1616.663 \approx 1617](https://tex.z-dn.net/?f=%20P%28t%3D3%29%20%3D%20660%20e%5E%7B%5Cfrac%7Bln%286%29%7D%7B6%7D%2A3%7D%20%3D%201616.663%20%5Capprox%201617)
Answer:
160 is greater than 2 and a half minutes.
Step-by-step explanation:
You're just dividing in this problem
which if you divide both bottom and to portion by 3 it leads to
4/3
Answer:
2
Step-by-step explanation:
1+1=2