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3 years ago

How do you convert a decimal into a fraction?

Mathematics

1 answer:

Maurinko [17]3 years ago

7 0

Answer:

I got you

Step-by-step explanation:

Step 1: Write down the decimal divided by 1, like this: decimal 1.

Step 2: Multiply both top and bottom by 10 for every number after the decimal point. (For example, if there are two numbers after the decimal point, then use 100, if there are three then use 1000, etc.)

Step 3: Simplify (or reduce) the

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1. calculate the cost of 10m cloth if the cost of per metre of cloth is Rs.119.75<br><br>

vova2212 [387]

Answer:11.95

Step-by-step explanation:

119.5 --- 1m (100cm)

119.5/100

1cm --- 1.195

10cm --- 11.95

4 0

3 years ago

Pl help it’s for a grade

uysha [10]

Answer:

23 miles south I'm pretty sure.

Step-by-step explanation:

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3 years ago

Matt is a software engineer writing a script involving 6 tasks. Each must be done one after the other. Let ti be the time for th

Masteriza [31]

Answer:

Let $t_i$ be the time for the $i$ th task.

We know these times have a certain structure:

Any 3 adjacent tasks will take half as long as the next two tasks.

In the form of an equations we have

$t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5 \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6$

The second task takes 1 second $t_2=1$

The fourth task takes 10 seconds $t_4=10$

So, we have the following system of equations:

$t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0 \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10$

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

Here is the augmented matrix for this system.

$\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]$

b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.

Subtract row 2 from row 1 $\left(R_1=R_1-R_2\right)$ .
Subtract row 2 from row 3 $\left(R_3=R_3-R_2\right)$ .
Add row 3 to row 2 $\left(R_2=R_2+R_3\right)$ .
Multiply row 3 by −1 $\left({R}_{{3}}=-{1}\cdot{R}_{{3}}\right)$ .
Add row 4 multiplied by $\frac{3}{2}$ to row 1 $\left(R_1=R_1+\left(\frac{3}{2}\right)R_4\right)$ .
Subtract row 4 from row 3 $\left(R_3=R_3-R_4\right)$ .

Here is the reduced echelon form for the augmented matrix.

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]$

c) The additional rows are

$\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right$

and the augmented matrix is

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right]$

d) To solve the system you must use these row operations.

Subtract row 1 from row 6 $\left(R_6=R_6-R_1\right)$ .
Subtract row 2 from row 6 $\left(R_6=R_6-R_2\right)$ .
Subtract row 3 from row 6 $\left(R_6=R_6-R_3\right)$ .
Swap rows 5 and 6.
Add row 5 to row 3 $\left(R_3=R_3+R_5\right)$ .
Multiply row 5 by 2 $\left(R_5=\left(2\right)R_5\right)$ .
Subtract row 6 multiplied by 1/2 from row 1 $\left(R_1=R_1-\left(\frac{1}{2}\right)R_6\right)$ .
Add row 6 multiplied by 1/2 to row 3 $\left(R_3=R_3+\left(\frac{1}{2}\right)R_6\right)$ .

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right]$