Q#1 please help t o resolve

Mathematics

1 answer:

Veseljchak [2.6K]2 years ago

3 0

Answer: option. y=-1.33x; 2.67

Solution:

If y varies directly with x, the relation between y and x is:

y=kx

where k is a constant of proporcionality.

We know y=8 when x=-6. Replacing these values in the equation above:

8=k(-6)

8=-6k

Solving for k: Dividing both sides of the equation by -6:

8/(-6)=-6k/(-6)

-1.33=k

k=-1.33

Then the relation between y and x is:

y=kx→y=-1.33x

What is the value of y when x=-2?

Replacing x by -2 in the formula:

y=-1.33x→y=-1.33(-2)→y=2.66

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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

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