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user100 [1]
3 years ago
12

Roberta is lining up nine different coloured blocks. There are five green blocks, two white blocks and two orange blocks. In how

many ways can she arrange these blocks?
Mathematics
1 answer:
Amanda [17]3 years ago
6 0
If the blocks were 9 different colors, then there would be

     9 !(factorial) = 362,880 different ways to line them up.

But for each different line-up, there are 5! =120 ways to arrange
the green blocks and you can't tell these apart, 2!= 2 ways to arrange
the white blocks and you can't tell these apart, and 2!=2 ways to arrange
the orange blocks and you can't tell these apart.

So the number of distinct, recognizable ways to arrange all 9 blocks
is

         (9!) / (5! · 2! · 2!)  =  (362,880) / (120 · 2 · 2)  =  756 ways.
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During a 14-day period, there is the following activity on your bank account. You deposit $100, withdraw $75, deposit $85, and w
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6 0
3 years ago
Solve the system by finding the reduced row-echelon form of the augmented matrix.
Alex

Answer:

The reduced row-echelon form is

\left[ \begin{array}{cccc} 1 & 0 & 6 & -10 \\\\ 0 & 1 & 2 & -3 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

The solutions to the system of equations are:

y=-2z-3,\:x=-6z-10

Step-by-step explanation:

To solve this system of linear equations,

x-4y-2z=2\\2x-11y-10z=13\\-x+6y+6z=-8

you must:

Step 1: Transform the augmented matrix to the reduced row echelon form.

In an augmented matrix, each row represents one equation in the system and each column represents a variable or the constant terms.

This is the augmented matrix that represents the system.

\left[ \begin{array}{cccc} 1 & -4 & -2 & 2 \\\\ 2 & -11 & -10 & 13 \\\\ -1 & 6 & 6 & -8 \end{array} \right]

It can be transformed by a sequence of elementary row operations to the matrix.

There are three kinds of elementary matrix operations.

  1. Interchange two rows (or columns).
  2. Multiply each element in a row (or column) by a non-zero number.
  3. Multiply a row (or column) by a non-zero number and add the result to another row (or column).

Using elementary matrix operations, we get that

Row Operation 1: add -2 times the 1st row to the 2nd row

Row Operation 2: add 1 times the 1st row to the 3rd row

Row Operation 3: multiply the 2nd row by -1/3

Row Operation 4: add -2 times the 2nd row to the 3rd row

Row Operation 5: add 4 times the 2nd row to the 1st row

\left[ \begin{array}{cccc} 1 & 0 & 6 & -10 \\\\ 0 & 1 & 2 & -3 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

Step 2: Interpret the reduced row echelon form

The reduced row echelon form of the augmented matrix is

\left[ \begin{array}{cccc} 1 & 0 & 6 & -10 \\\\ 0 & 1 & 2 & -3 \\\\ 0 & 0 & 0 & 0 \end{array} \right]

which corresponds to the system

x+6z=-10\\y+2z=-3\\0=0

The system has infinitely many solutions.

y=-2z-3,\:x=-6z-10

4 0
4 years ago
A farm distributor packed 2,650 melons in boxes.each box can hold 8 melons. All the boxes are full except for one box.
laiz [17]

Answer:

there are 331 full boxes but the one box that isnt full has two melons

Step-by-step explanation:

2,650 divided by 8 = 331

331 multiplied by 8 = 2648

2650 - 2648 = 2

so there are 332 boxes total 331 full ones and one not full on with 2 melons in it.

Hope that helps!

7 0
3 years ago
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