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4 years ago

What is 1 2/3 multiplied by 1 1/4

2 answers:

8 0

To help you understand this better, here are my steps.
You convert both the numbers to improper fractions- like this
5/3 multiplied by 5/4
Now, you multiply across the top and bottom
(5*5)/(3*4)
25/12

Hope this helped!

FromTheMoon [43]4 years ago

6 0

1 2/3 multiplied by 1 1/4 is 25 over 4 (25/4) or 1 1/4

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tyreee bought a collection comic book for 49.62 last year.this year he would sold it for 52.10 find the percent of change.

wlad13 [49]

Answer:

The percentage increased by approximately 5%.

Step-by-step explanation:

52.10 - 49.62 = 2.48

2.48/49.62 x 100 = 4.998%

The percentage increase in the price is approximately 5%. Hope this helps! :)

3 0

3 years ago

You have a solid Rectangle of dough. Consider the cross sections that are formed when you slice into the cube as described. Draw

Phantasy [73]

B is the answer I think trying it tho

5 0

2 years ago

ANOTHER ONE AGAIN PLSSSSS

mina [271]

Answer:

Luka had 6 slices, Kira had 9 slices and Ali 12 slices

Step-by-step explanation:

Each had 2/3 of their slices hence

Luka = 3/2 x 4 slices = 6

Kira = 3/2 x 6 slices = 9

Ali = 3/2 x 8 slices = 12

6 0

3 years ago

Is (-3,4) a solution of the inequality y> - 2x – 3?

jeyben [28]

Answer:

(-3, 4) is a solution

Step-by-step explanation:

The point (-3, 4) is inside the shaded area of the graph, so is a solution.

You can check in the inequality

y > -2x -3

4 > -2(-3) -3 . . . . substitute for x and y

4 > 3 . . . . . . . true; the given point is a solution

8 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