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3 years ago

PLEASE PLEASE PLEASE PLEASE HURRY I ONLY HAVE 59 MINUTES LEFT REEEEEE

Mathematics

1 answer:

KengaRu [80]3 years ago

3 0

Answer:

The second and fourth one

Step-by-step explanation:

The open circle means more/less than or equal to and the closed circle is only more than or less than and it is greater so it goes right

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What is the y-intercept? y-3=1\2(x+2)

Margaret [11]

Y-3=1/2(x+2)

To find the y-intercept, we must first get the equation in slope intercept form. Start off by distributing 1/2 into the parentheses.

y-3= 1/2x+1
Add 3 to both sides

y= 1/2x+4

The slope is whatever is in front of x, in this case, 1/2. The y-intercept is the number after that which would be 4

6 0

3 years ago

Select the equivalent expression that shows the fewest possible terms.

LenaWriter [7]

Answer:

Step-by-step explanation:

7a-10c-2a-0.5c+5

5a-10.5c+5

it is your first choice

7 0

3 years ago

Read 2 more answers

Ill give you brainliest pleaseeeeeeee help me

ruslelena [56]

Line D and C because they are not parallel but distorted or unsymmetrical

4 0

3 years ago

6. Craig and James both live 4 miles up a hill. Craig hikes uphill at a rate of 1 mile every two hours.

Iteru [2.4K]

Answer:

James

Step-by-step explanation:

James rate:

2x + 2 = 4

2x = 2

x = 1 hour

Rate = distance/time = 4/1 = 4 miles per hour

Craig's rate = 1/2 = 0.5 miles per hour

James is faster

6 0

3 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