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Aleks [24]
3 years ago
15

What did the Concordat of 1801 accomplish?

Mathematics
1 answer:
Elena L [17]3 years ago
3 0

Answer:

Concordat of 1801, agreement reached on July 15, 1801, between Napoleon Bonaparte and papal and clerical representatives in both Rome and Paris, defining the status of the Roman Catholic Church in France and ending the breach caused by the church reforms and confiscations enacted during the French Revolution .Step-by-step explanation:

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Simply 3/4 - (6 - 3/4s)
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Answer:

-21/4 + 3s/4

Step-by-step explanation:

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3 years ago
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Write out the sample space for the given experiment. Use the following letters to indicate each choice: M for mushrooms, A for a
babymother [125]

Answer:

MBI, MBV, ABV, ABI, MPI, MPV, APV, API

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3 years ago
Solve the following systems of equations using the matrix method: a. 3x1 + 2x2 + 4x3 = 5 2x1 + 5x2 + 3x3 = 17 7x1 + 2x2 + 2x3 =
lara [203]

Answer:

a. The solutions are

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

b. The solutions are

\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}

c. The solutions are

\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}

Step-by-step explanation:

Solving a system of linear equations using matrix method, we may define a system of equations with the same number of equations as variables as:

A\cdot X=B

where X is the matrix representing the variables of the system,  B is the matrix representing the constants, and A is the coefficient matrix.

Then the solution is this:

X=A^{-1}B

a. Given the system:

3x_1 + 2x_2 + 4x_3 = 5 \\2x_1 + 5x_2 + 3x_3 = 17 \\7x_1 + 2x_2 + 2x_3 = 11

The coefficient matrix is:

A=\left[\begin{array}{ccc}3&2&4\\2&5&3\\7&2&2\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}5&17&11\\\end{array}\right]

First, we need to find the inverse of the A matrix. To find the inverse matrix, augment it with the identity matrix and perform row operations trying to make the identity matrix to the left. Then to the right will be inverse matrix.

So, augment the matrix with identity matrix:

\left[ \begin{array}{ccc|ccc}3&2&4&1&0&0 \\\\ 2&5&3&0&1&0 \\\\ 7&2&2&0&0&1\end{array}\right]

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

\left[ \begin{array}{ccc|ccc}1&0&0&- \frac{2}{39}&- \frac{2}{39}&\frac{7}{39} \\\\ 0&1&0&- \frac{17}{78}&\frac{11}{39}&\frac{1}{78} \\\\ 0&0&1&\frac{31}{78}&- \frac{4}{39}&- \frac{11}{78}\end{array}\right]

And the inverse of the A matrix is

A^{-1}=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right]

Next, multiply A^ {-1} by B

X=A^{-1}\cdot B

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\left[ \begin{array}{ccc} - \frac{2}{39} & - \frac{2}{39} & \frac{7}{39} \\\\ - \frac{17}{78} & \frac{11}{39} & \frac{1}{78} \\\\ \frac{31}{78} & - \frac{4}{39} & - \frac{11}{78} \end{array} \right] \cdot \left[\begin{array}{c}5&17&11\end{array}\right]

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}-\frac{2}{39}&-\frac{2}{39}&\frac{7}{39}\\ -\frac{17}{78}&\frac{11}{39}&\frac{1}{78}\\ \frac{31}{78}&-\frac{4}{39}&-\frac{11}{78}\end{pmatrix}\begin{pmatrix}5\\ 17\\ 11\end{pmatrix}=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

The solutions are

\left[\begin{array}{c}x_1&x_2&x_3\\\end{array}\right]=\begin{pmatrix}\frac{11}{13}\\ \frac{50}{13}\\ -\frac{17}{13}\end{pmatrix}

b. To solve this system of equations

x -y - z = 0 \\30x + 40y = 12 \\30x + 50z = 12

The coefficient matrix is:

A=\left[\begin{array}{ccc}1&-1&-1\\30&40&0\\30&0&50\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x&y&z\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}0&12&12\\\end{array}\right]

The inverse of the A matrix is

A^{-1}=\left[ \begin{array}{ccc} \frac{20}{47} & \frac{1}{94} & \frac{2}{235} \\\\ - \frac{15}{47} & \frac{4}{235} & - \frac{3}{470} \\\\ - \frac{12}{47} & - \frac{3}{470} & \frac{7}{470} \end{array} \right]

The solutions are

\left[\begin{array}{c}x&y&z\\\end{array}\right]=\begin{pmatrix}\frac{54}{235}\\ \frac{6}{47}\\ \frac{24}{235}\end{pmatrix}

c. To solve this system of equations

4x_1 + 2x_2 + x_3 + 5x_4 = 0 \\3x_1 + x_2 + 4x_3 + 7x_4 = 1\\ 2x_1 + 3x_2 + x_3 + 6x_4 = 1 \\3x_1 + x_2 + x_3 + 3x_4 = 4\\

The coefficient matrix is:

A=\left[\begin{array}{cccc}4&2&1&5\\3&1&4&7\\2&3&1&6\\3&1&1&3\end{array}\right]

The variable matrix is:

X=\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]

The constant matrix is:

B=\left[\begin{array}{c}0&1&1&4\\\end{array}\right]

The inverse of the A matrix is

A^{-1}=\left[ \begin{array}{cccc} - \frac{1}{9} & - \frac{1}{9} & - \frac{1}{9} & \frac{2}{3} \\\\ - \frac{32}{9} & - \frac{5}{9} & \frac{13}{9} & \frac{13}{3} \\\\ - \frac{28}{9} & - \frac{1}{9} & \frac{8}{9} & \frac{11}{3} \\\\ \frac{7}{3} & \frac{1}{3} & - \frac{2}{3} & -3 \end{array} \right]

The solutions are

\left[\begin{array}{c}x_1&x_2&x_3&x_4\\\end{array}\right]=\begin{pmatrix}\frac{22}{9}\\ \frac{164}{9}\\ \frac{139}{9}\\ -\frac{37}{3}\end{pmatrix}

7 0
3 years ago
Read 2 more answers
Please help solve this, I’ve tried my best. My answers are a height of 6.7 using 30-60-90 as a last resort, and the glass volume
Neko [114]

Answer:

A) To ensure the correct and exact straight height

B) 15 inches

C) 24.87 in^{3}

Step-by-step explanation:

Hello! Lets start!

First, it asks us to explain why the two bases must be parallel. In order to measure the correct and exact height, we must be sure that the bases are parallel, or else there would be a range of heights; in order to ensure that you are measuring a straight distance, they must be parallel!

Second, it asks us to identify the height of cone. To do this, we must make a slope! We can see that, for every time we go down 5 inches, we lose 1 inch of our base! Our slope in this situation would be -5. So, lets continue to see how long before the cone reaches its tip, in which the base would be equal to 0. At 5 inches, our base is 2 inches. At 10 inches, our base is 1 inch. At 15 inches, we do not have a base, as it equals 0! So therefore, our cone height is 15 inches.

Third, it asks us to identify the volume of the cup, which is also called a <em>circular truncated cone. </em>To do so, we can use the formula:

₁V = \frac{1}{3} \pi (r_{1}^2 + r_{1}r_{2} + r_2^2)h

Lets plug in our information!

V = \frac{1}{3} \pi (1.5^{2} + (1.5)(1) + 1^{2}) (5)\\V = 24.87

Hope this helps! Remember, math is fun!

3 0
3 years ago
What is the equation that goes through (4,5) and is parallel to y=2x+1
forsale [732]

Answer:

The equation would be y = 2x - 3

Step-by-step explanation:

To start, we need to note that they will have the same slope, since parallel lines have the same slope. Then, we can use the point-slope form to find the equation.

y - y1 = m(x - x1)

y - 5 = 2(x - 4)

y - 5 = 2x - 8

y = 2x - 3

5 0
3 years ago
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