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3 years ago

(810^7)(210^6) simplify the expression and write in scientific notation

Mathematics

1 answer:

lesya [120]3 years ago

8 0

Grouping similar factors together makes this easier:

(8)(2)(10^7)(10^6) = 16(10^13) = 1.6(10^14)

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What is the solution to this system of equations?

antoniya [11.8K]

Answer:

Step-by-step explanation:

Since both equations express y in terms of x we can equate the right sides

3x + 5 = - x + 3 ( add x to both sides )

4x + 5 = 3 ( subtract 5 from both sides )

4x = - 2 ( divide both sides by 4 )

x = - 0.5

substitute x = - 0.5 in either of the 2 equations

using y = - x + 3 then y = 0.5 + 3 = 3.5

solution is (- 0.5, 3.5 ) → c

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3 years ago

2 times the square root of 25 plus the square root of negative 16

PSYCHO15rus [73]

Answer:

10 + 4 i

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3 years ago

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For a craft , each student will need 5 rubber bands . There are 8 students . Rubber bands comes in bags of 9 . How many bags wil

bixtya [17]

You would need 5 bags

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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

Mary ran 1/5 of 1 km each day. How many meters did she run from Monday to Wednesday?

lord [1]

Answer:

3/5 or 0.6 in decimal form

Step-by-step explanation:

All you have to do is multiply 1/5 x 3 for the three days Monday, Tuesday and Wednesday and that gets you 3/5 or as 1/5 of a km is 0.2 you have to multiply 0.2 x 3 and you get 0.6 which is equivalent to 3/5

5 0

3 years ago

(8*10^7)(2*10^6) simplify the expression and write in scientific notation

(810^7)(210^6) simplify the expression and write in scientific notation