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3 years ago

I need to make a matrix in reduced row echelon form

Mathematics

2 answers:

a_sh-v [17]3 years ago

7 0

Let the three apartments be A, B & C

The rent of A is $x

The rent of B is $y

The rent of C is $z

So x + y + z = 1600 .......(1)

Now

Maintenance of A is 20% of x = 0.20x

Maintenance of B is 20% of y = 0.20y

Maintenance of C is 25% of z = 0.25z

0.20x + 0.20y + 0.25z = 345

Multiplying by 100 we get

20x + 20y + 25 z = 34500

Dividing by 5 we get

4x + 4y + 5z = 6900 .......(2)

Monthly fee of A is 10% of x = 0.10x

Monthly fee of B is 20% of y = 0.20 y

Monthly Fee of C is 10% of z = 0.10z

Now

0.10x + 0.20y + 0.10 z= 1820 - 1600

0.10x + 0.20y + 0.10z = 220

Multiplying by 100 we get

10x + 20y + 10z = 22000

Dividing by 10

x + 2y + z = 2200 ....(3)

Making a Matrix of equation (1), (2) & (3)

$\left[\begin{array}{ccc}1&1&1\\4&4&5\\1&2&1\end{array}\right] \left[\begin{array}{c}x\\y\\z\end{array}\right] =\left[\begin{array}{c}1600\\6900\\2200\end{array}\right]$

is the required matrix

gogolik [260]3 years ago

7 0

Answer: for plato I got the first three correct but not the last two so I can only help a little.

1 [ 1, 0.2 , 0.1]

2[ 1, 0.2, 0.2]

3 [ 1, 0.25, 0.1]

The last two answers I have in the pic are incorrect

Step-by-step explanation:

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Need help ...,,.....,,........

Hatshy [7]

Answer:

i’m pretty sure that it’s 50. i haven’t done this in a awhile though. but pretty positive it’s 50

Step-by-step explanation: :)

6 0

3 years ago

Which algebraic properties was applied to the expression a + (b + c) = (a + b) + c?

andrew11 [14]

Answer:

the Associative Property

Step-by-step explanation:

Notice that the ORDER doesn't change -- a, b, c on both sides. But the GROUPING changes.

Example:

3 + (4 + 5) = (3 + 4) + 5

3 + 9 = 7 + 5

12 = 12

Subtraction does NOT behave this way!

8 - (7 - 1) = 8 - 6 = 2

(8 - 7) - 1 = 1 - 1 = 0

It matters who you associate with! :-)

5 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:

$\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}$ .

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

For what value(s) of k is the linear system consistent? (Enter your answers as a comma-separated list.)

AleksAgata [21]

Answer:

Step-by-step explanation:

Given

$6x_1-9x_2=8$

$9x_1+kx_2=-1$

The given system is $AX=B$ can be represented by

$\begin{bmatrix}6 &-9 \\ 9 & k\end{bmatrix}\begin{bmatrix}x_1\\ x_2\end{bmatrix}=\begin{bmatrix}8\\ -1\end{bmatrix}$

The given system is consistent when determinant of A is not equal to zero

$|A|$

$|A|=6k-(-81)=6k+81$

$k\neq \frac{-27}{2}$

i.e. system is consistent for all value of k except $k=\frac{-27}{2}$

$R-\frac{-27}{2}$

4 0

3 years ago

Read 2 more answers

For which equations is 8 a solution? Select the four correct answer

maxonik [38]

Answer:

I <em>believe</em> the answers are= 2, 3, 6, 8

Step-by-step explanation:

For #2, you would subtract 2 from both sides. 10-2=8. x=8.

For #3, you would add 4 to both sides of the equation, making the final answer: x=8

For #6, you would divide both sides by 3. 24 divided by three is 8.

For #8, I multiplied the whole equation by 8, which is the denominator of x/8. 1 times 8=8. The final answer would be x=8.

Hope this helps!

8 0

2 years ago

Read 2 more answers