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d1i1m1o1n [39]
3 years ago
13

85 orders in 17 days = ? orders in 12 days

Mathematics
1 answer:
tamaranim1 [39]3 years ago
4 0

Answer:

60

Step-by-step explanation:

Let the order be x

Solution

85/x= 17/12

1020=17x

1020/17=x

x=60

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A customer buys 15.5 cheese pizzas which are $10.50 each. What is the total? Round it to the nearest hundredth.
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5 0
3 years ago
​
pashok25 [27]

Answer:

Let's call:

f = price of 1 cup of dried fruit

a = price of 1 cup of almonds

In order to build the linear system, you need to consider that the total price of a bag is given by the sum of the price of cups times the number of cups in each bag, therefore:

Solve for a in first equation:

a = (6 - 3f) / 4

Then substitute in the second equation:

41/2 f + 6 · (6 - 3f) / 4 = 9

41/2 f + 9 - 9/2 f = 9

16 f = 0

f = 0

Now, substitute this value in the formula found for a:

a = (6 - 3·0) / 4

  = 3/2 = 1.5

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Step-by-step explanation:

8 0
2 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}.

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\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:

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

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

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\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.

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