-36.12 divided by -5 3/5

Please help! (question in attachment)

igor_vitrenko [27]

Ayy you got it right thank you very much your

5 0

3 years ago

What number is 80% of 90%?

storchak [24]

Answering so you can mark the other person brainliest

6 0

3 years ago

A newspaper used a chart resembling the one to the right to illustrate the rising amounts that a video rental company spends to

Dmitry [639]

This question is incomplete- no bar chart was included.

The complete question was gotten from google.

A newspaper used a chart resembling the one to the right to illustrate the rising amounts that a video rental company spends to provide streaming content online. Is this the bar chart of a categorical variable, or is it a timeplot that uses bars to show the data?

Open the attachment below to see the bar chart'

Answer:

A newspaper used a chart resembling the one to the right to illustrate the rising amounts that a video rental company spends to provide streaming content online. Is this the bar chart of a categorical variable, or is it a time plot that uses bars to show the data? It is a time plot that uses bars to show the data; because it graphs a series of data recorded over time; presenting the values in sequential order.

Step-by-step explanation:

The chart used in the newspaper to illustrate the rising amounts that a video rental company spends to provide streaming content online is a time plot that uses bars to show the data; because it graphs a series of data recorded over time; presenting the values in sequential order.

6 0

3 years ago

Ms. Lou rolls out 2 3/5 ft. of dough. If she cuts the dough into sections that are 1/4 of a ft. each, how many sections of dough

Natasha_Volkova [10]

Answer:

10

Step-by-step explanation:

We need to divide 2 3/5 ft by 1/4 ft.

2 3/5 / 1/4 =

= 2 3/5 * 4

= 13/5 * 4

= 52/5

= 10 4/10

Answer: 10

5 0

2 years ago

Consider the transpose of Your matrix A, that is, the matrix whose first column is the first row of A, the second column is the

Zarrin [17]

Answer:The system could have no solution or n number of solution where n is the number of unknown in the n linear equations.

Step-by-step explanation:

To determine if solution exist or not, you test the equation for consistency.

A system is said to be consistent if the rank of a matrix (say B ) is equal to the rank of the matrix formed by adding the constant terms(in this case the zeros) as a third column to the matrix B.

Consider the following scenarios:

(1) For example:Given the matrix A= $\left[\begin{array}{ccc}1&2\\3&4\end{array}\right]$ , to transpose A, exchange rows with columns i.e take first column as first row and second column as second row as follows:

Let A transpose be B.

∵B= $\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]$

the system Bx=0 can be represented in matrix form as:

$\left[\begin{array}{ccc}1&3\\2&4\end{array}\right]$ $\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]$ = $\left[\begin{array}{ccc}0\\0\end{array}\right]$ ................................eq(1)

Now, to determine the rank of B, we work the determinant of the maximum sub-square matrix of B. In this case, B is a 2 x 2 matrix, therefore, the maximum sub-square matrix of B is itself B. Hence,

|B|=(1*4)-(3*2)= 4-6 = -2 i.e, B is a non-singular matrix with rank of order (-2).

Again, adding the constant terms of equation 1(in this case zeros) as a third column to B, we have $B_{0}$ :

$B_{0}$ = $\left[\begin{array}{ccc}1&3&0\\4&2&0\end{array}\right]$ . The rank of $B_{0}$ can be found by using the second column and third column pair as follows:

| $B_{0}$ |=(3*0)-(0*2)=0 i.e, $B_{0}$ is a singular matrix with rank of order 1.

Note: a matrix is singular if its determinant is = 0 and non-singular if it is $\neq$ 0.

Comparing the rank of both B and $B_{0}$ , it is obvious that

Rank of B $\neq$ Rank of $B_{0}$ since (-2)<1.

Therefore, we can conclude that equation(1) is inconsistent and thus has no solution.

(2) If B= $\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]$ is the transpose of matrix A= $\left[\begin{array}{ccc}-4&-8\\5&10\end{array}\right]$ , then

Then the equation Bx=0 is represented as:

$\left[\begin{array}{ccc}-4&5\\-8&10&\end{array}\right]$ $\left[\begin{array}{ccc}x_{1} \\x_{2} \end{array}\right]$ = $\left[\begin{array}{ccc}0\\0\end{array}\right]$ ..................................eq(2)

|B|= (-4*10)-(5*(-8))= -40+40 = 0 i.e B has a rank of order 1.

$B_{0}$ = $\left[\begin{array}{ccc}-4&5&0\\-8&10&0\end{array}\right]$ ,

| $B_{0}$ |=(5*0)-(0*10)=0-0=0 i.e $B_{0}$ has a rank of order 1.

we can therefor conclude that since

rank B=rank $B_{0}$ =1, equation(2) is consistent and has 2 solutions for the 2 unknown ( $X_{1}$ and $X_{2}$ ).

Summary:

Given an equation Bx=0, transform the set of linear equations into matrix form as shown in equations(1 and 2).
Determine the rank of both the coefficients matrix B and $B_{0}$ which is formed by adding a column with the constant elements of the equation to the coefficient matrix.
If the rank of both matrix is same, then the equation is consistent and there exists n number of solutions(n is based on the number of unknown) but if they are not equal, then the equation is not consistent and there is no number of solution.

5 0

3 years ago