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charle [14.2K]
3 years ago
7

Write the quadratic equation whose roots are -3 and 3 and whose leading coefficient is 5. (Use the letter x to represent the var

iable)
Mathematics
1 answer:
Bumek [7]3 years ago
4 0
A(x - x₁)(x - x₂) = 0
a=5,  x₁=-3,  x₂ = 3

so:
5(x-(-3))(x-3)=0
5(x+3)(x-3)=0
5(x²-9) = 0
5x² - 45 = 0   ←  answer
You might be interested in
Combine like terms and solve equations<br> 5x+18-8x=33
Igoryamba
First thing you have to do is take the 5 + -8 which equals -3 then you have to 33-18 which equals 15 then u have to divide by the -3 which gives you equal your answer of -5
6 0
3 years ago
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Which properties are best used prove a quadrilateral is a square?
adell [148]

Answer:

a, b, c, & d

Step-by-step explanation:

8 0
3 years ago
The table shows the results of an experiment in which subjects taking either a drug or a placebo either reported fatigue or did
Lady bird [3.3K]

Answer:

Therefore, the correct options are;

-P(fatigue) = 0.44

P(fatigue \left |  \right drug) = 0.533

--P(drug and fatigue) = 0.32

P(drug)·P(fatigue) = 0.264

Step-by-step explanation:

Here we have that for dependent events,

P(A \, and \, B)= P(A)\times  P(B\left |  \right A )

From the options, we have;

P(fatigue \left |  \right drug) = 0.533

P(drug) = 0.6

P(drug and fatigue) = 0.32

Therefore

P(drug and fatigue) = P(drug)×P(fatigue \left |  \right drug)  

= 0.6 × 0.533 = 0.3198 ≈ 0.32 = P(drug and fatigue)

Therefore, the correct options are;

-P(fatigue) = 0.44

P(fatigue \left |  \right drug) = 0.533

--P(drug and fatigue) = 0.32

P(drug)·P(fatigue) = 0.264

Since P(fatigue) = 0.44 ∴ P(drug) = 0.264/0.44 = 0.6.

4 0
3 years ago
Can someone explain how to do this
Svetlanka [38]

Answer:

مرحبًا ، سأكون سعيدًا لمساعدتك في حبك في العمل ، وأريد المزيد من التفاصيل ، وسأبذل قصارى جهدي لإرضائك

Step-by-step explanation:

8 0
3 years ago
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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
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