Create a equation for this exponential function

Sistemas de ecuaciones

kolezko [41]

Answer:

El Dr. Potter aplicó 32 vacunas de polio y 28 de sarampión.

Step-by-step explanation:

- First, the data must be taken into account to create the equations:

X + Y = 60; (Equation 1)

- Where:

X = Number of polio vaccines and Y = Number of measles vaccines.

4X + 2Y = 184; (Equation 2)

Where for each polio vaccine there are 4 doses and for each measles vaccine there are 2 doses.

- We clear the variable that we want to find in each equation:

Y = 60 - X; (And we call it equation 3)

Y = (184-4X) / 2; (And we call it equation 4)

- Now both equations are equalized in terms of the same variable:

60 - X = (184 - 4X) / 2

- Then X is cleared to find its value:

120 - 2X = 184 - 4X

4X - 2X = 184 - 120

2X = 64

X = 32

- Now we demonstrate the value of X obtained in equation 1 or 2 to give the values of Y

(32) + Y = 60

Y = 60 - 32

Y = 28

- Finally, both values are used to see if equation 2 gives us the correct data:

4 (32) + 2 (28) = 184

128 + 56 = 184

- The answer is given:

Dr. Potter gave 32 polio and 28 measles vaccines.

6 0

3 years ago

Number 3 plz I don't get when they say dimensions

STatiana [176]

Well 1 cm = 2ft, so it might be asking the scale?

7 0

3 years ago

Read 2 more answers

Eleven plus forty-one is divided by a number. If the result is 13, what's the number?

sergij07 [2.7K]

Answer:

4

Step-by-step explanation:

Let the number be x.

(11+41)/x = 13

52/x = 13

13x = 52

x = 52/13

x = 4

4 0

3 years ago

Pls help extra points and mark brainlist easy reading

galben [10]

C. Explaining the reasons for the change in Elsa

5 0

3 years ago

The population of mosquitoes in a certain area increases at a rate proportional to the current pop-ulation, and in the absence o

ollegr [7]

Answer:

The population of mosquitoes in the area at any time t is:

$P(t)=201977.31-1977.31\times 2^{t}$

Step-by-step explanation:

The rate of growth of mosquitoes can be expressed as:

$\frac{dP}{dt}=kP$

$\frac{dP}{P}=k\ dt$

Integrate the above expression as follows:

$\int {\frac{dP}{P}} \, =\int {k\ dt} \, \\\ln|P|=kt+c\\e^{\ln|P|}=e^{kt+c}\\P=Ce^{kt}$

$\Rightarrow P=P_{0}e^{kt}$

It is provided that the population doubles every day.

Compute the value of k as follows:

$2=1\times e^{k\times1}\\2=e^{k}\\k=\ln (2)$

It is also provided that every day 20,000 mosquitoes are eaten.

The rate of growth per week can be expressed as:

$\frac{dP}{dt}=\ln(2)P-14000\\\frac{dP}{dt}-\ln(2)P=14000$

The integrating factor for this is:

$e^{\int {\ln(2)dP}}=e^{\ln(2)\int {dt}}=e^{\ln(2)t}$

Then,

$P(t)\ e^{-\ln(2)t}=\int {e^{-\ln(2)t}}-14000\, dt\\=-14000\int {e^{-\ln(2)t}}\, dt\\=-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C\\P(t)=(e^{-\ln(2)t})\times [-14000\times \frac{e^{-\ln(2)t}}{-\ln(2)}+C]\\=\frac{14000}{\ln(2)}+Ce^{-\ln(2)t}$

The initial population is 200,000.

Compute the value of C as follows:

$P(t)=\frac{140000}{\ln(2)}+Ce^{-\ln(2)t}\\200000=\frac{14000}{\ln(2)}+Ce^{-\ln(2)(0)}\\C=200000-\frac{140000}{\ln(2)}\\C=-1977.31$

Now substitute C in P (t),

$P(t)=\frac{140000}{\ln(2)}+Ce^{\ln(2)t}\\P(t)=201977.31-1977.31\times 2^{t}$

6 0

3 years ago