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aev [14]
3 years ago
10

1. A super-deadly strain of bacteria is causing the zombie population to double every two days. currently, there are 25 zombies.

After how many days will there be 25600 zombies?
2. Use the function from the previous question. If the current population is 5,000,000,000 people, after how many days will there be no humans left?
Mathematics
1 answer:
Dvinal [7]3 years ago
4 0
1. Is 124 days

2. ia 200,000,000

Ur welcome
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If u answer pls explain how u did it so i can understand … ty!
Nataliya [291]

Answer:

hope this helps :)

Step-by-step explanation:

7 0
3 years ago
AS ALWAYS, DUE TODAY! Pls and thanks. NOTE- Legit answers only! Legit answers= Brainliest, 5.0 stars, and a Thanks. NOTE 2- Dist
olganol [36]

Answer:

5

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
A scuba diver starts 8 feet below sea level. She then dives down another 13 feet. Which expression gives her total number of fee
scoray [572]

I see this as a basic equation.

If she starts at point x, 8 feet, then goes down another 13, she is adding to the number of feet she traveled down.

So your answer would be 8+13=21

If this isn't what you needed, because I am not sure what math you are taking, let me know if I can help you more!


3 0
3 years ago
Read 2 more answers
A savings account earns 5% simple interest per year. The principal is $1200. What is the balance after 4 years?
stiv31 [10]
The formula of the Simple Interest is:
I=PRT
P for Principle Amount     ($1200)
R for Rate                        (5%=\frac{5}{100} = 0.05)
T for Time in years          (4 years)
I = 1200 × 0.05 × 4
  = $240

Add the interest to the principle amount to check the balance
$1200 + $240 = $1440

7 0
3 years ago
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3y''-6y'+6y=e*x sexcx
Simora [160]
From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

which has roots at r=1\pm i. This admits the two fundamental solutions

y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
7 0
3 years ago
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