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2 years ago

12 cm 3 cm 3 cm 3.14

Mathematics

2 answers:

gogolik [260]2 years ago

6 0

Answer:i need more information/quistion

Step-by-step explanation:

SVEN [57.7K]2 years ago

3 0

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Tim and Tom are trying to earn money to buy a new game system over a 3-month period. Tim saved $45.28 each month. If they need a

ki77a [65]

Answer:

$25.47

Step-by-step explanation:

Tim saved $45.28 each month. So, he will save (45.28 × 3) = $135.84 in three months.

If Tim and Tom need a total of $212.25 to buy the game system, then Tom has to save (212.25 - 135.84) = $76.41 in three months,

Therefore, each of the three months Tom has to earn in order to buy the game system, $\frac{76.41}{3} = 25.47$ dollars. (Answer)

5 0

2 years ago

Solve each of the following systems by Gauss-Jordan elimination. (b) X1-2x2+ x3- 4x4=1 X1+3x2 + 7x3 + 2x4=2 -12x2-11x3- 16x4 5 (

Len [333]

Answer:

a) The set of solutions is $\{(0,-3x_3,x_3): x_3\; \text{es un real}\}$ y b) the set of solutions is $\{(-6,\frac{-41}{17}-\frac{30}{17}x_4 , \frac{37}{17}+\frac{8}{17} x_4 ,x_4): x_4\;\text{es un real}\}$ .

Step-by-step explanation:

a) Let's first find the echelon form of the matrix $\left[\begin{array}{ccc}5&2&6\\-2&1&3\end{array}\right]$ .

We add $\frac{2}{5}$ from row 1 to row 2 and we obtain the matrix $\left[\begin{array}{ccc}5&2&6\\0&\frac{9}{5} &\frac{27}{5}\end{array}\right]$

From the previous matrix, we multiply row 1 by $\frac{1}{5}$ and the row 2 by $\frac{5}{9}$ and we obtain the matrix $\left[\begin{array}{ccc}1&\frac{2}{5} &\frac{6}{5} \\0&1&3\end{array}\right]$ . This matrix is the echelon form of the initial matrix.

The system has a free variable (x3).

x2+3x3=0, then x2=-3x3

0=x1+ $\frac{2}{5}$ x2+ $\frac{6}{5}$ x3=

x1+ $\frac{2}{5}$ (-3x3)+ $\frac{6}{5}$ x3=

x1- $\frac{6}{5}$ x3+ $\frac{6}{5}$ x3

then x1=0.

The system has infinite solutions of the form (x1,x2,x3)=(0,-3x3,x3), where x3 is a real number.

b) Let's first find the echelon form of the aumented matrix $\left[\begin{array}{ccccc}1&-2&1&-4&1\\1&3&7&2&2\\0&-12&-11&-16&5\end{array}\right]$ .

To row 2 we subtract row 1 and we obtain the matrix $\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&-12&-11&-16&5\end{array}\right]$

From the previous matrix, we add to row 3, $\frac{12}{5}$ of row 2 and we obtain the matrix $\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&5&6&6&1\\0&0&\frac{17}{5}&\frac{-8}{5}&\frac{37}{5} \end{array}\right]$ .

From the previous matrix, we multiply row 2 by $\frac{1}{5}$ and the row 3 by $\frac{5}{17}$ and we obtain the matrix $\left[\begin{array}{ccccc}1&-2&1&-4&1\\0&1&\frac{6}{5} &\frac{6}{5}&\frac{1}{5}\\0&0&1&\frac{-8}{17}&\frac{37}{17} \end{array}\right]$ . This matrix is the echelon form of the initial matrix.

The system has a free variable (x4).

x3- $\frac{8}{17}$ x4= $\frac{37}{17}$ , then x3= $\frac{37}{17}$ + $\frac{8}{17}x4.x2+[tex]\frac{6}{5}$ x3+ $\frac{6}{5}$ x4= $\frac{1}{5}$ , x2+ $\frac{6}{5}$ ( $\frac{37}{17}$ + $\frac{8}{17}x4)+[tex]\frac{6}{5}$ x4= $\frac{1}{5}$ , then

x2= $\frac{-41}{17}-\frac{30}{17}$ x4.

x1-2x2+x3-4x4=1, x1+ $\frac{82}{17}$ + $\frac{60}{17}$ x4+ $\frac{37}{17}$ + $\frac{8}{17}$ x4-4x4=1, then x1= $1-\frac{119}{17}=-6$

The system has infinite solutions of the form (x1,x2,x3,x4)=(-6, $\frac{-41}{17}-\frac{30}{17}$ x4, $\frac{37}{17}$ + $\frac{8}{17}$ x4,x4), where x4 is a real number.

6 0

3 years ago

Jennifer wrote 8 poems at school and Il poems at home. She

swat32

Answer:

Nell wrote 4 poems.

Step-by-step explanation:

8+2=10

10-6=4

4= Number of poems that Nell wrote

5 0

2 years ago