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3 years ago

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

2 answers:

8 0

To solve for this, use your points to plug them into the x and y values knowing that the x value in the point is -3 and the y value is 2. When you plug in the points, you should get this:

2=k(-3)

Then, because the question asks to solve for the constant of variation, or k, divide both sides by -3 to solve for k. This would get you to your answer,

k=-2/3

Svet_ta [14]3 years ago

7 0

Y = kx
so
k = y/x
k = 2 / -3 = -2/3

answer
-2/3

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Work out the shaded area

Dennis_Churaev [7]

Answer:

Step-by-step explanation:

Shaded area:

Area of big rectangle - Area of small rectangle

$= 10 \times 12 - (10 - 5) \times (12 - 2)$

Calculate the difference

$= 10 \times 12 - 5 \times 10$

Calculate the product

$= 120 - 50$

Calculate the difference

$= 70 \: {cm}^{2}$

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7 0

3 years ago

Hey how to do this and what the answer

frez [133]

Answer:

5x + y - 4 = 0

Step-by-step explanation:

When we are given two points and are asked to find the equation of the line we use the two - point form.

Two - point form: $\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}$ .

where: $(x_1,y_1)$ and $(x_2,y_2)$ are the two points on the line.

Given the two points are: (5,21) and (-5,-29).

Substituting in the two point form:

⇒ $\frac{y - 21}{-29 - 21} = \frac{x - 5}{-5 -5}$

$\implies \frac{y - 21}{-50} = \frac{x - 5}{-10}$

$\implies y - 21 = 5x - 25$

⇒ 5x + y - 4 = 0

8 0

3 years ago

How to determine if the columns of a matrix are linearly independent?

astra-53 [7]

Let's consider an arbitrary 2x2 matrix as an example,

$\mathbf A=\begin{bmatrix}\mathbf x&\mathbf y\end{bmatrix}=\begin{bmatrix}x_1&y_1\\x_2&y_2\end{bmatrix}$

The columns of $\mathbf A$ are linearly independent if and only if the column vectors $\mathbf x,\mathbf y$ are linearly independent.

This is the case if the only way we can make a linear combination of $\mathbf x,\mathbf y$ reduce to the zero vector is to multiply the vectors by 0; that is,

$c_1\mathbf x+c_2\mathbf y=\mathbf 0$

only by letting $c_1=c_2=0$ .

A more concrete example: suppose

$\mathbf A=\begin{bmatrix}1&2\\4&8\end{bmatrix}$

Here, $\mathbf x=\begin{bmatrix}1\\4\end{bmatrix}$ and $\amthbf y=\begin{bmatrix}2\\8\end{bmatrix}$ . Notice that we can get the zero vector by taking $c_1=-2$ and $c_2=1$ :

$-2\begin{bmatrix}1\\4\end{bmatrix}+\begin{bmatrix}2\\8\end{bmatrix}=\begin{bmatrix}-2+2\\-8+8\end{bmatrix}=\mathbf 0$

so the columns of $\mathbf A$ are not linearly independent, or linearly dependent.

8 0

3 years ago

Two angles are supplementary. The larger angle is 33 degrees more than 6 times the smaller angle. Find the measure of each angle

noname [10]

Sum of two angles that are supplementary = 180°

Let the smaller angle be = x

$Let \: the \: smaller \: angle \: be \: = \: x$