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FinnZ [79.3K]
3 years ago
11

Choose the correct simplification of (7x^3y^3)^2. (5 points) 100 POINTS GIVING BRAINLIEST FOR GOOD EXPLANATION

Mathematics
2 answers:
In-s [12.5K]3 years ago
4 0

Answer:

Option C

Step-by-step explanation:

(7x^3y^3)^2

= (7)^2 * (x^3)^2 * (y^3)^2

= 49 * x^(3*2) * y^(3*2)

= 49x^6y^6

You have to distribute the terms in "7x^3 * y^3" each to the power of 2

(7)^2 * (x^3)^2 * (y^3)^2

Now you can apply the rule "(x^a)^b = x^a*b" and further simplify the expression

qaws [65]3 years ago
3 0

Answer:

it is c

Step-by-step explanation:

2x7=49 so its not a or d. 3x2=6 so both would be six so it is 49x^6y^6

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A village was founded four hundred years ago by a group of 20 people. In this village, the population triples every one hundred
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Answer:

Step-by-step explanation:

Treat this like compound interest:  Use A = P(1 + r)^t.

Here, P is the initial population and A is 3 times that, or 3P.  Since P = 20 people, 3P = 60 people,

and this population is reached after 100 years.

We need to determine r, substitute its value into the formula A = P(1 + r)^t, and then determine the population of the village after 400 years.

60 = 20(1 + r)^100

Simplifying, 3 = (1 + r)^100.

Taking the natural log of both sides,

ln 3 = 100 ln (1 + r), or

                   ln 3

ln (1 + r) = ---------------

                    100

              = 1.0986 / 100 = 0.01986

We must solve this for r.  Raising e to the power ln (1 + r), on the left side of an equation, and raising e to the power 0. 01986 on the right side, we get:

1 + r = 3, so r must = 2.

Now find the pop of the village today.  Use the same equation:  A = P (1+r)^t.

A = 20(1 +2)^4 (hundreds),

or

A = 20(3)^4, or

A = 81

The population after 400 years is 81.

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What is the least common multiple of 5, 10, and 12? What is the least common multiple of 6, 10, and 12? What is the least common
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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,

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

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