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8_murik_8 [283]
3 years ago
12

Katy's measured the growth of a plant seedling. The seedling grew 1/3 inch by the end of the first week. The seedling grew 5/6 i

nch by the end of the second week. About how much did the seedling grow in the fist 2 weeks?
Mathematics
2 answers:
Vikki [24]3 years ago
8 0
In the given problem it is already given that the seedling grew by 1/3 inch in the first week and 5/6 inch in the second week. Then it would not be a problem to find the total length of growth of the seedling.
The length up to which the seedling grew in 2 weeks = (1/3) + (5/6)
                                                                                     = (2 + 5)/6
                                                                                     = 7/6 inch
So the seedling grew by 7/6 inches in two weeks. I hope that this answer is what you were looking for. You can solve such problems in the future by following the method that is applied for solving this problem.
ddd [48]3 years ago
5 0
1/3 + 5/6 = 2/6 + 5/6 = 7/6
It grew by 7/6 or 1 1/6 inches by the first 2 weeks.
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Step-by-step explanation:

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What is the volume of the rectangular prism?<br><br> Type the answer in the boxes below.
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Step-by-step explanation:

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3 years ago
A small military base housing 1,000 troops, each of whom is susceptible to a certain virus infection. Assuming that during the c
slava [35]

Answer:

I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}

Step-by-step explanation:

The rate of infection is jointly proportional to the number of infected troopers and the number of non-infected ones. It can be expressed as follows:

\frac{dI}{dt}=a*I*(1000-I)

Rearranging and integrating

\frac{dI}{dt}=a*I*(1000-I)\\\\\frac{dI}{I*(1000-I)}=a*dt\\\\\int\frac{dI}{I*(1000-I)}=\int a*dt\\\\-\frac{ln(1000/I-1)}{1000}+C=a*t

At the initial breakout (t=0) there was one trooper infected (I=1)

-\frac{ln(1000/1-1)}{1000}+C=0\\\\-0,006906755+C=0\\\\C=0,006906755

In two days (t=2) there were 5 troopers infected

-\frac{ln(1000/5-1)}{1000}+0,006906755=a*2\\\\-0,005293305+0,006906755=2*a\\a = 0,00161345 / 2 = 0,000806725

Rearranging, we can model the number of infected troops (I) as

-\frac{ln(1000/I-1)}{1000}+0,006906755=0,000806725*t\\\\-\frac{ln(1000/I-1)}{1000}=0,000806725*t-0,006906755\\-ln(1000/I-1)=0,806725*t-0.6906755\\\\\frac{1000}{I}-1=exp^{0,806725*t-0.6906755}  \\\\\frac{1000}{I}=exp^{0,806725*t-0.6906755}+1\\\\I=\frac{1000}{exp^{0,806725*t-0.6906755}+1}

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