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3 years ago

HELP PLS MATH WILL GIVE YOU BRANLYY

Mathematics

2 answers:

Tema [17]3 years ago

7 0

Answer: A and E

Step-by-step explanation:

Gelneren [198K]3 years ago

5 0

D and B are both the answers

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A belt and a wallet cost $42 while 7 belts and 4 wallets cost $213. Calculate the cost of each item.

Fittoniya [83]

Let, the cost of a belt = x
cost of a wallet = y

Then, system of equations would be:
x + y = 42
7x + 4y = 213

Multiply 1st equation by 4,
4x + 4y = 168
Substitute it from 2nd equation,
3x = 45
x = 15

Now, substitute it in 1st equation,
15 + y = 42
y = 42 - 15 = 27

In short, Belt costs $15 and wallet costs $27

Hope this helps!

3 0

3 years ago

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Answer My Question Please!!!<br>Its 20 Points!!

saul85 [17]

Yo yo yo so first u simplify everything as 4^2= 16 and 25 squared is 5 so order of operations make 5*2 and that equals 10 so finally

16-10=6 so the answer is 6

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3 years ago

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You purchase a painting for 9,000 that has an appreciation rate of 12%. What is the growth

Alenkinab [10]

Answer:

Step-by-step explanation:

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3 years ago

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The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population,

uranmaximum [27]

Answer:

Step-by-step explanation:

Assuming that the differential equation is

$\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right)$ .

We need to solve it and obtain an expression for $P(t)$ in order to complete the exercise.

First of all, this is an example of the logistic equation, which has the general form

$\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right)$ .

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that $k=0.04$ and $K=500$ and the initial condition $P(0)=100$ .

Notice that this equation is separable, then

$\frac{dP}{P(1-P/K)} = kdt$ .

Now, intagrating in both sides of the equation

$\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C$ .

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

$\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}$ .

So,

$\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|$ .

We have obtained that:

$-\ln\left| \frac{K-P}{P}\right| = kt +C$

which is equivalent to

$\ln\left| \frac{K-P}{P}\right|= -kt -C$

Taking exponentials in both hands:

$\left| \frac{K-P}{P}\right| = e^{-kt -C}$

Hence,

$\frac{K-P(t)}{P(t)} = Ae^{-kt}$ .

The next step is to substitute the given values in the statement of the problem:

$\frac{500-P(t)}{P(t)} = Ae^{-0.04t}$ .

We calculate the value of $A$ using the initial condition $P(0)=100$ , substituting $t=0$ :

$\frac{500-100}{100} = A}$ and $A=4$ .

So,

$\frac{500-P(t)}{P(t)} = 4e^{-0.04t}$ .

Finally, as we want the value of $t$ such that $P(t)=200$ , we substitute this last value into the above equation. Thus,

$\frac{500-200}{200} = 4e^{-0.04t}$ .

This is equivalent to $\frac{3}{8} = e^{-0.04t}$ . Taking logarithms we get $\ln\frac{3}{8} = -0.04t$ . Then,

$t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325$ .

So, the population of rats will be 200 after 25 months.

6 0

3 years ago

If you were to solve the following system by substitution, what would be the best variable to solve for and from what equation?2

aivan3 [116]

The second equation because they all have a common factor of 3. You can get x by itself. However, with the first one, you'll end up with fractions instead.

8 0

3 years ago