How to find a matrix that transform a column matrix from linear to circular quantum mechanic?

Find the solution to the system of equations 2x+3y+z=20 3y+5z=12 y-5z=4

podryga [215]

Answer:

x =4 ,y =4 and z =0 is the solution.

Step-by-step explanation:

We shall solve the equations using elimination method

In equation 2 and 3 we can see that coefficient of z in both equations are same with opposite sides , adding equations (2) and (3)

3y+5z = 12

y -5z = 4 adding the equations

___________

4y = 16

dividing both sides by 4

y = $\frac{16}{4}$

y = 4

Plugging the value of y in equation 2 or equation 3 ,we get

3(4) +5z = 12

12+5z = 12

5z = 12-12 or 5z =0 gives z =0

plugging y =4 and z =0 in equation 1

2x+3(4)+ 0 = 20

2x+12 = 20

2x =20-12

2x = 8

x = 8 divided by 2 gives

x =4

therefore solution of the system is given by

x=4 , y =4 and z =0

3 0

4 years ago

Read 2 more answers

use a graphing calculator or other technology to answer the question which quadratic regression equation best fits the data set

Greeley [361]

Answer:

Option C is the correct option.

In other words, the quadratic regression $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ is also attached.

Step-by-step explanation:

x y

3 470

4 416

5 403

Analyzing Option A:

Considering the equation

$y=32.86\:\left(x\right)^2+379.14\left(x\right)-1369.14\:$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2+379.14\left(3\right)-1369.14\:$

$y=64.02$

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2+379.14\left(4\right)-1369.14\:\:$

$y=673.18$

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2+379.14\left(5\right)-1369.14\:$

$y=1348.06$

Analyzing Option B:

$y=32.86\:\left(x\right)^2-379.14\left(x\right)$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2-379.14\left(3\right)$

$y=-841.68$

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2-379.14\left(4\right)$

$\:y=-990.8$

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2-379.14\left(5\right)$

$\:y=-1074.2$

Analyzing Option C:

Considering the equation

$y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$

From (3, 470), putting x = 3

$y=32.86\:\left(3\right)^2-379.14\left(3\right)+1369.14\:$

$y=527.46$

So, the approximately result is (3, 527)

From (4, 470), putting x = 4

$y=32.86\:\left(4\right)^2-379.14\left(4\right)+1369.14\:$

$y=378.34$

So, the approximately result is (4, 378)

From (5, 403), putting x = 5

$y=32.86\:\left(5\right)^2-379.14\left(5\right)+1369.14\:\:\:$

$y=294.94$

So, the approximately result is (5, 295)

Analyzing Option D:

Considering the equation

$y=-1369.14\:\left(x\right)^2-379.14\left(x\right)+32.86$

From (3, 470), putting x = 3

$y=-1369.14\:\left(3\right)^2-379.14\left(3\right)+32.86\:\:$

$y=-13426.82$

From (4, 470), putting x = 4

$y=-1369.14\:\left(4\right)^2-379.14\left(4\right)+32.86$

$y=-23389.94$

From (5, 403), putting x = 5

$y=-1369.14\:\left(5\right)^2-379.14\left(5\right)+32.86\:\:$

$y=-36091.34$

Therefore, from the above calculations and analysis, we conclude that Option C is the correct option.

In other words, the quadratic regression $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ best fits the data set, as it gets very much close to the data values given in the data table.

The graph of the equation $y=32.86\:\left(x\right)^2-379.14\left(x\right)+1369.14\:$ is also attached.

3 0

3 years ago

I don't understand this

igomit [66]

Looking at the picture attached, you will notice that on one side, we have a side of length 7+5=12. Next, we have a side length of 4+5+2=11. Hence, as length*width=area for a rectangle, this is the overall area of the rectangle. Next, in the picture, there are clearly two sides taken out - a 2x5 as well as a 4x5 rectangle, which are subtracted from the total area, resulting in 11*12-2*5-4*5
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