All categories

3 years ago

Fill in the blank by performing the indicated elementary row operation(s).

Mathematics

1 answer:

Scilla [17]3 years ago

6 0

In this question, we are given a matrix, and we have to perform the given operation.

The matrix is:

$\left[\begin{array}{ccc}6&-1&|5\\1&-5&|0\end{array}\right]$

The following operation is given:

$R_1 \rightarrow -6R_2 + R_1$

In which $R_1$ is the element at the first line and $R_2$ is the element at the second line.

Updating the first line:

$R_{1,1} = -6*1 + 6 = 0$

$R_{1,2} = -6*-5 - 1 = 30 - 1 = 29$

$R_{1,3} = -6*0 + 5 = 5$

Thus, the filled matrix will be given by:

$\left[\begin{array}{ccc}0&29&|5\\1&-5&|0\end{array}\right]$

For another example where row operations are applied on a matrix, you can check brainly.com/question/18546657

You might be interested in

Given that P=x+y Find P when: x=11and y=6 what is p ?

frutty [35]

Answer:

P = 17

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Step-by-step explanation:

Step 1: Define

P = x + y

x = 11

y = 6

Step 2: Evaluate

Substitute: P = 11 + 6
Add: P = 17

And we have our final answer!

4 0

3 years ago

The ravens scored 1 touchdown and an extra point . how many total points did they score

ch4aika [34]

The ravens scored 7 points

8 0

3 years ago

Use set of notation to write the domain and range of the function y=-3x(x+1)^2+3 shown in graph

anygoal [31]

The domain of a function is the set of values of x in which the function can be calculated.

The range of a function is the associated values of the function respect to the domain.

You have the following function:

$y=-3(x+1)^2+3$

In order to determine the domain and range of the previous function, you consider that the given function is available for any value of x. Then, you have:

dom y: -∞ ≤ x ≤ ∞ : (-∞,∞)

The range of the function can be determine by noticing, on the graph, that the maximum value of the function is y=3 and also you can notice that there is no a limit for negatives values of the function. Then, you have:

ran y: -∞ ≤ y ≤ 3 : (-∞,3]

7 0

1 year ago

The value of k 4x^2 +20k+k

Wittaler [7]

Step-by-step explanation:

You want the to find the value of k such that the y coordinate of the vertex is 0.

0=x2−6x+k

The x coordinate, h, of the vertex is found, using the following equation:

h=−b2a

h=−−62(1)=3

Evaluate at x = 3:

0=32−6(3)+k

k=9

8 0

3 years ago

Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 4 in the manner described. (Ent

QveST [7]

Answer:

$x=2\cos(t)$ and $y=-2\sin(t)+1$

Step-by-step explanation:

$(x-h)^2+(y-k)^2=r^2$ has parametric equations:

$(x-h)=r\cos(t) \text{ and } (y-k)=r\sin(t)$ .

Let's solve these for x and y respectively.

$x-h=r\cos(t)$ can be solved for x by adding h on both sides:

$x=r\cos(t)+h$ .

$y-k=r \sin(t)$ can be solve for y by adding k on both sides:

$y=r\sin(t)+k$ .

We can verify this works by plugging these back in for x and y respectively.

Let's do that:

$(r\cos(t)+h-h)^2+(r\sin(t)+k-k)^2$

$(r\cos(t))^2+(r\sin(t))^2$

$r^2\cos^2(t)+r^2\sin^2(t)$

$r^2(\cos^2(t)+\sin^2(t))$

$r^2(1)$ By a Pythagorean Identity.

$r^2$ which is what we had on the right hand side.

We have confirmed our parametric equations are correct.

Now here your h=0 while your k=1 and r=2.

So we are going to play with these parametric equations:

$x=2\cos(t)$ and $y=2\sin(t)+1$

We want to travel clockwise so we need to put -t and instead of t.

If we were going counterclockwise it would be just the t.

$x=2\cos(-t)$ and $y=2\sin(-t)+1$

Now cosine is even function while sine is an odd function so you could simplify this and say:

$x=2\cos(t)$ and $y=-2\sin(t)+1$ .

We want to find $\theta$ such that

$2\cos(t-\theta_1)=2 \text{ while } -2\sin(t-\theta_2)+1=1$ when t=0.

Let's start with the first equation:

$2\cos(t-\theta_1)=2$

Divide both sides by 2:

$\cos(t-\theta_1)=1$

We wanted to find $\theta_1$ for when $t=0$

$\cos(-\theta_1)=1$

Cosine is an even function:

$\cos(\theta_1)=1$

This happens when $\theta_1=2n\pi$ where n is an integer.

Let's do the second equation:

$-2\sin(t-\theta_2)+1=1$

Subtract 2 on both sides:

$-2\sin(t-\theta_2)=0$

Divide both sides by -2:

$\sin(t-\theta_2)=0$

Recall we are trying to find what $\theta_2$ is when t=0:

$\sin(0-\theta_2)=0$

$\sin(-\theta_2)=0$

Recall sine is an odd function:

$-\sin(\theta_2)=0$

Divide both sides by -1:

$\sin(\theta_2)=0$

$\theta_2=n\pi$

So this means we don't have to shift the cosine parametric equation at all because we can choose n=0 which means $\theta_1=2n\pi=2(0)\pi=0$ .

We also don't have to shift the sine parametric equation either since at n=0, we have $\theta_2=n\pi=0(\pi)=0$ .

So let's see what our equations look like now:

$x=2\cos(t)$ and $y=-2\sin(t)+1$

Let's verify these still work in our original equation:

$x^2+(y-1)^2$

$(2\cos(t))^2+(-2\sin(t))^2$

$2^2\cos^2(t)+(-2)^2\sin^2(t)$

$4\cos^2(t)+4\sin^2(t)$

$4(\cos^2(t)+\sin^2(t))$

$4(1)$

$4$

It still works.

Now let's see if we are being moving around the circle once around for values of t between $0$ and $2\pi$ .

This first table will be the first half of the rotation.

t 0 pi/4 pi/2 3pi/4 pi

x 2 sqrt(2) 0 -sqrt(2) -2

y 1 -sqrt(2)+1 -1 -sqrt(2)+1 1

Ok this is the fist half of the rotation. Are we moving clockwise from (2,1)?

If we are moving clockwise around a circle with radius 2 and center (0,1) starting at (2,1) our x's should be decreasing and our y's should be decreasing at the beginning we should see a 4th of a circle from the point (x,y)=(2,1) and the point (x,y)=(0,-1).

Now after that 4th, the x's will still decrease until we make half a rotation but the y's will increase as you can see from point (x,y)=(0,-1) to (x,y)=(-2,1). We have now made half a rotation around the circle whose center is (0,1) and radius is 2.

Let's look at the other half of the circle:

t pi 5pi/4 3pi/2 7pi/4 2pi

x -2 -sqrt(2) 0 sqrt(2) 2

y 1 sqrt(2)+1 3 sqrt(2)+1 1

So now for the talk half going clockwise we should see the x's increase since we are moving right for them. The y's increase after the half rotation but decrease after the 3/4th rotation.

We also stopped where we ended at the point (2,1).

3 0

3 years ago