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11Alexandr11 [23.1K]

3 years ago

15

Isabelle bought a new mountain bike for $1,149.79. She will finish paying for the bike after 6 months. Each month, she will pay

the same amount. About how much will she pay each month? A. $100 B. $200 C. $250 D. $300

2 answers:

vazorg [7]3 years ago

7 0

Answer:

$200

Step-by-step explanation:

1,149.79/6=*about* 191 which is rounded up to $200

So it's about $200

Good luck! :)

ankoles [38]3 years ago

7 0

Answer:

She will pay B. $200 each month.

6 * 200 = 1,200

Step-by-step explanation:

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Show all your work to receive full credit for this problem. Solve the system of equations using the elimination method. 2a + 3b

Sindrei [870]

Given equations are;<span>
2a + 3b = -1 ..................equation 1
3a + 5b = -2 ....................equation 2</span>

Now multiply equation 1 with (-3)

The equation will be;

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Now multiply equation 2 with (2)

The equation will be;

6a + 10b = -4 ……………..equation 4

Now add equation 3 and equation 4

-6a – 9b = 3

<span>6a + 10b = -4</span>

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0a + b = -1

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Now put the value of b in equation 1

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3 years ago

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:

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

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

Taking exponentials in both hands:

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

Hence,

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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 $t=0$ :

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So,

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

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Answer:

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