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3 years ago

Multiply and simplify. 3xy4z2⋅5x2yx

Mathematics

2 answers:

Olenka [21]3 years ago

4 0

I think the answer is 15x^4y^5z^2

vazorg [7]3 years ago

4 0

Answer:

Step-by-step explanation:

15x2y4z2

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Jakes has $20 to spend on notebooks and pencils the notebooks and pencils. the notebooks cost 3.25 and the pencils cost $.50 wha

Leokris [45]

Hey there. So basically, find out how much the pencils and notebooks cost first.
The notebooks cost = $3.25
The pencils cost = $0.50

Then, think about what you need to figure out in this problem.
Jake has $20. You need to find how many notebooks Jake can buy in maximum after buying 8 pencils.

If Jake buys 8 pencils that costs $0.50 each, he spends $4 on the pencils.

So now, to find out how many notebooks he can buy, do 20 minus 4.
Jake's got $16 left.

If the notebooks cost $3.25 each, we need to find out how many notebooks he can buy by dividing them. So, 16 divided by 3.25 equals 4.923... and so on.

That means, Jake can buy 4 notebooks with his remaining money.

3 0

3 years ago

Kelly has 4 times as many songs on her music player as Lou

jeka57 [31]

Answer:4 times x

4(x)

Step-by-step explanation:

4 times as many songs on Lou's music player

Lou's music player is x

7 0

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:

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

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

Is the point (4,2) the solution to the system? Why or why not? Explain your reasoning.

VMariaS [17]

Solve the top equation to get on variable on one side
2x-y=6
-y=-2x+6
y=2x-6
then plug in the coordinate for x and y for both equations. if both equations are true then it’s a solution
2=2(2)+6
2=10

4+2=6

so it’s not a solution

4 0

3 years ago

Select the correct answer from each drop-down menu.If /(=) = 0.522 - 2 and g(a) = 853 + 2, find the value of the following funct

Alik [6]

In general,

$(f\cdot g)(x)=f(x)\cdot g(x)$

Therefore, in our case,

$(f\cdot g)(x)=(0.5x{}^2-2)(8x^3+2)=4x^5-16x^3+x^2-4$ <h2>The answer is 4x^5-16x^3+x^2-4. Select 4, 16, 1, 4 (left to right)</h2>

5 0

11 months ago