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dezoksy [38]
3 years ago
7

What is 4 quarts converted to pints

Mathematics
2 answers:
Volgvan3 years ago
8 0

8 pints is 4 quarts

Lilit [14]3 years ago
5 0

4 liquid quarts = 8 liquid pints

You might be interested in
Solve the system of equations by row-reduction. At each step, show clearly the symbol of the linear combinations that allow you
adell [148]

Answer:

1) The solution of the system is

\left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right

2) The solution of the system is

\left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right

Step-by-step explanation:

1) To solve the system of equations

\left\begin{array}{ccccccc}&3x_2&-5x_3&=&89\\6x_1&&+x_3&=&17\\x_1&-x_2&+8x_3&=&-107\end{array}\right

using the row reduction method you must:

Step 1: Write the augmented matrix of the system

\left[ \begin{array}{ccc|c} 0 & 3 & -5 & 89 \\\\ 6 & 0 & 1 & 17 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 2: Swap rows 1 and 2

\left[ \begin{array}{ccc|c} 6 & 0 & 1 & 17 \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 3:  \left(R_1=\frac{R_1}{6}\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 1 & -1 & 8 & -107 \end{array} \right]

Step 4: \left(R_3=R_3-R_1\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 3 & -5 & 89 \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]

Step 5: \left(R_2=\frac{R_2}{3}\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & -1 & \frac{47}{6} & - \frac{659}{6} \end{array} \right]

Step 6: \left(R_3=R_3+R_2\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & \frac{37}{6} & - \frac{481}{6} \end{array} \right]

Step 7: \left(R_3=\left(\frac{6}{37}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{1}{6} & \frac{17}{6} \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 8: \left(R_1=R_1-\left(\frac{1}{6}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & - \frac{5}{3} & \frac{89}{3} \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 9: \left(R_2=R_2+\left(\frac{5}{3}\right)R_3\right)

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right]

Step 10: Rewrite the system using the row reduced matrix:

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 8 \\\\ 0 & 0 & 1 & -13 \end{array} \right] \rightarrow \left\begin{array}{ccc}x_1&=&5\\x_2&=&8\\x_3&=&-13\end{array}\right

2) To solve the system of equations

\left\begin{array}{ccccccc}4x_1&-x_2&+3x_3&=&12\\2x_1&&+9x_3&=&-5\\x_1&+4x_2&+6x_3&=&-32\end{array}\right

using the row reduction method you must:

Step 1:

\left[ \begin{array}{ccc|c} 4 & -1 & 3 & 12 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 2: \left(R_1=\frac{R_1}{4}\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 2 & 0 & 9 & -5 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 3: \left(R_2=R_2-\left(2\right)R_1\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 1 & 4 & 6 & -32 \end{array} \right]

Step 4: \left(R_3=R_3-R_1\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & \frac{1}{2} & \frac{15}{2} & -11 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 5: \left(R_2=\left(2\right)R_2\right)

\left[ \begin{array}{ccc|c} 1 & - \frac{1}{4} & \frac{3}{4} & 3 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 6: \left(R_1=R_1+\left(\frac{1}{4}\right)R_2\right)

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & \frac{17}{4} & \frac{21}{4} & -35 \end{array} \right]

Step 7: \left(R_3=R_3-\left(\frac{17}{4}\right)R_2\right)

\left[ \begin{array}{ccc|c} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & - \frac{117}{2} & \frac{117}{2} \end{array} \right]

Step 8: \left(R_3=\left(- \frac{2}{117}\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{2} & - \frac{5}{2} \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 9: \left(R_1=R_1-\left(\frac{9}{2}\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 15 & -22 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 10: \left(R_2=R_2-\left(15\right)R_3\right)

\left[ \begin{array}{cccc} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]

Step 11:

\left[ \begin{array}{ccc|c} 1 & 0 & 0 & 2 \\\\ 0 & 1 & 0 & -7 \\\\ 0 & 0 & 1 & -1 \end{array} \right]\rightarrow \left\begin{array}{ccc}x_1&=&2\\x_2&=&-7\\x_3&=&-1\end{array}\right

8 0
3 years ago
Which inequality is true if p = 3.4 ?
Anastaziya [24]

Answer:

the shape isnt angled thats what it means

Step-by-step explanation:

3 0
3 years ago
a farmer has 24 yards of rope that needs to be cut into 1/4 yards of pieces. How many pieces of the rope does the farmer collect
Mazyrski [523]

Answer:6

Step-by-step explanation:

7 0
3 years ago
2a - 4b +3ab -5a +2b
DaniilM [7]

Answer:

-3a-2b+3ab

Step-by-step explanation:

2a-5a

-4b+2b

6 0
3 years ago
Read 2 more answers
In a certain​ year, brand A of​ heart-rate watch cost ​$39.99 and brand B cost ​$59.99. A nonprofit community health organizatio
Oksi-84 [34.3K]

Answer:

Brand A = 21 heart-rate watches.

Brand B = 13 heart-rate watches.

Step-by-step explanation:

Let A be the number of Brand A heart watches and B be the number of Brand B heart watches.

We have been given that a nonprofit community health organization purchased 34 ​heart-rate watches for use at a wellness center.

We can represent this information in an equation as:

A+B=34...(1)

We have been given that in a certain​ year, brand A of​ heart-rate watch cost ​$39.99 and brand B cost ​$59.99 and the organization spent ​$1619.66 for the​ watches.

We can represent this information in an equation as:

39.99*A+59.99*B=1619.66...(2)

We will use substitution method to solve our system of equations.

From equation (1) we will get,

A=34-B

Substituting A=34-B in equation(2) we will get,

39.99*(34-B)+59.99*B=1619.66

1359.66-39.99*B+59.99*B=1619.66

-39.99*B+59.99*B=1619.66-1359.66

20*B=260

B=\frac{260}{20}

B=13

Therefore, the organisation purchased 13 heart-rate watches of brand B.

Now let us substitute B=13 in equation (1).

A+13=34

A+13-13=34-13

A=21

Therefore, the organisation purchased 21 heart-rate watches of brand A.

8 0
3 years ago
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