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2 years ago

What is the constant variation, k, of the direct variation, y=kx, through (-3,2)

Mathematics

2 answers:

podryga [215]2 years ago

7 0

Answer:

$k=-\frac{2}{3}$

Step-by-step explanation:

We are to find the constant of variation, $k$ , of the direct variation, $y = k x$ given the coordinates of the point $( - 3 , 2 )$ .

Direct variation is represented by:

$y = k x$

where $k$ is the constant of variation.

Substituting the coordinates of the given point to find the value of $k$ .

$2 = k (-3)$

$k=-\frac{2}{3}$

lukranit [14]2 years ago

5 0

For this case we have that by definition, two magnitudes are directly proportional when there is a constant such that:

$y = kx$

Where:

k: It is the constant of proportionality

We must find the value of "k" when $(x, y): (- 3,2)$

$k = \frac {y} {x} = \frac {2} {- 3} = - \frac {2} {3}$

Answer:

$k = - \frac {2} {3}$

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T - 6 1/3 = - 9 3/4 solve this !

Radda [10]

Answer: T=-3 5/12

Step-by-step explanation: 6 1/3-6 /13

6 1/3-9 3/4= -3 5/12

T= -3 5/12

3 0

3 years ago

Read 2 more answers

Suppose that the derivable functions x=x(t) and y=y(t) satisfy xcosy=2.

ololo11 [35]

Applying implicit differentiation, it is found that dy/dt when y=π/4 is of:

a-) -√2 / 2.

<h3>What is implicit differentiation?</h3>

Implicit differentiation is when we find the derivative of a function relative to a variable that is not in the definition of the function.

In this problem, the function is:

xcos(y) = 2.

The derivative is relative to t, applying the product rule, as follows:

$\cos{y}\frac{dx}{dt} - x\sin{y}\frac{dy}{dt} = 0$

$\frac{dy}{dt} = \frac{\cos{y}\frac{dx}{dt}}{x\sin{y}}$

Since dx/dt=−2, we have that:

$\frac{dy}{dt} = -2\frac{\cos{y}}{x\sin{y}}$

When y = π/4, x is given by:

xcos(y) = 2.

$x = \frac{2}{\cos{\frac{\pi}{4}}} = \frac{2}{\frac{\sqrt{2}}{2}} = \frac{4}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = 2\sqrt{2}$

Hence:

$\frac{dy}{dt} = -2\frac{\cos{y}}{x\sin{y}}$

$\frac{dy}{dt} = -\frac{1}{\sqrt{2}}\cot{y}$

Since cot(pi/4) = 1, we have that:

$\frac{dy}{dt} = -\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = -\frac{\sqrt{2}}{2}$

Which means that option a is correct.

More can be learned about implicit differentiation at brainly.com/question/25608353

#SPJ1

4 0

1 year 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

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