If a = [1,2,3] and b = [1, 3, –2], find the angle, №18010395, 25.12.2020 13:51

Mathematics

If a = [1,2,3] and b = [1, 3, –2], find the angle between a and b.

Step-by-step answer

P Answered by Specialist

85.9°

Step-by-step explanation:

angle = $\arccos\frac{a\cdot b}{|a| |b|} = \arccos\frac{1+6-6}{{\sqrt{1+4+9}}\sqrt{1+9+4}}\\=\arccos\frac{1}{14}=85.9 degrees$

Mathematics

What is fc? f= [-2 0] c= [12 0 3/2] [0 8] [1 -6 7 ] [2 1] a. [-25 0 3] [8 -49 56] [25 -6 10] b. [-23 0 -3] [8 -49 56] [25 -6 9] c. [-25 0 -3] [8 -49 56] [25 -6 9] d. [-24 0 -3] [8 -48 56] [25 -6 10]

Step-by-step answer

P Answered by Master

6

The answer is D. $\left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]$

Step-by-step explanation:

We have:

$F=\left[\begin{array}{cc}-2&0\\0&8\\2&1\end{array}\right]$

$C=\left[\begin{array}{ccc}12&0&\frac{3}{2} \\1&-6&7\end{array}\right]$

And we need to find FC, this means that we have to find the product between this two matrices.

This product will be possible only if the number of columns of the first matrix is equal to the number of rows of the second matrix.

F is a 3x2 matrix, this means F has 3 rows and 2 columns.

C is a 2x3 matrix this means it has 2 rows and 3 columns.

Then the product is possible because the number of columns of F is equal to the number of rows of C.

And the resulting matrix will be a 3x3.

By definition the product between two matrices is:

Let's suppose both matrices are 2x2,

$A=\left[\begin{array}{cc}a_{11} &a_{12} \\a_{21} &a_{22}\end{array}\right]$

$B=\left[\begin{array}{cc}b_{11} &b_{12} \\b_{21} &b_{22}\end{array}\right]$

The product between A and B is:

$AB=\left[\begin{array}{cc}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{array}\right]$

IMPORTANT: It's not the same AB then BA the results of both products are differents.

Now, doing the product between F and C:

$FC=\left[\begin{array}{cc}-2&0\\0&8\\2&1\end{array}\right]X\left[\begin{array}{ccc}12&0&\frac{3}{2} \\1&-6&7\end{array}\right]$

$=\left[\begin{array}{ccc}(-2).12+0.1&(-2).0+0.(-6)&(-2).\frac{3}{2} +0.7\\0.12+8.1&0.0+8.(-6)&0.\frac{3}{2} +8.7\\2.12+1.1&2.0+1.(-6)&2.\frac{3}{2} +1.7\end{array}\right] \\ =\left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]$

And then the answer is option D.

Mathematics

What is fc? f= [-2 0] c= [12 0 3/2] [0 8] [1 -6 7 ] [2 1] a. [-25 0 3] [8 -49 56] [25 -6 10] b. [-23 0 -3] [8 -49 56] [25 -6 9] c. [-25 0 -3] [8 -49 56] [25 -6 9] d. [-24 0 -3] [8 -48 56] [25 -6 10]

Step-by-step answer

P Answered by Master

6

The answer is D. $\left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]$

Step-by-step explanation:

We have:

$F=\left[\begin{array}{cc}-2&0\\0&8\\2&1\end{array}\right]$

$C=\left[\begin{array}{ccc}12&0&\frac{3}{2} \\1&-6&7\end{array}\right]$

And we need to find FC, this means that we have to find the product between this two matrices.

This product will be possible only if the number of columns of the first matrix is equal to the number of rows of the second matrix.

F is a 3x2 matrix, this means F has 3 rows and 2 columns.

C is a 2x3 matrix this means it has 2 rows and 3 columns.

Then the product is possible because the number of columns of F is equal to the number of rows of C.

And the resulting matrix will be a 3x3.

By definition the product between two matrices is:

Let's suppose both matrices are 2x2,

$A=\left[\begin{array}{cc}a_{11} &a_{12} \\a_{21} &a_{22}\end{array}\right]$

$B=\left[\begin{array}{cc}b_{11} &b_{12} \\b_{21} &b_{22}\end{array}\right]$

The product between A and B is:

$AB=\left[\begin{array}{cc}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{array}\right]$

IMPORTANT: It's not the same AB then BA the results of both products are differents.

Now, doing the product between F and C:

$FC=\left[\begin{array}{cc}-2&0\\0&8\\2&1\end{array}\right]X\left[\begin{array}{ccc}12&0&\frac{3}{2} \\1&-6&7\end{array}\right]$

$=\left[\begin{array}{ccc}(-2).12+0.1&(-2).0+0.(-6)&(-2).\frac{3}{2} +0.7\\0.12+8.1&0.0+8.(-6)&0.\frac{3}{2} +8.7\\2.12+1.1&2.0+1.(-6)&2.\frac{3}{2} +1.7\end{array}\right] \\ =\left[\begin{array}{ccc}-24&0&-3\\8&-48&56\\25&-6&10\end{array}\right]$

And then the answer is option D.

Mathematics

!1. select the best answer for the question. multiply matrix d = [5,-2,1 ] by matrix e= [12] a. [4 8] b. 2 c. [5 -2 1 ,10 4 2] d. [5 10, -2 -4, 1 2] 2. subtract [-6 -4,6 0, 6 4] - [-5 5, -4 -4,6 -4] 3. add [1 -4,3 5] + [-2 6,-2 4] 4. find the determinant of a= [ 4 -7, 3 -2] a. 13 b. −2 c. 2 d. 29 5. what is matrix a + matrix b? a=[3 0, 2 -1] b= [ 2 8, .6 3] 6. what is 4c? c= [12 03/2, 1 -6 7] 7. evaluate the determinant for the following matrix: [-5 1, 5 4] 8. complete the multiplication: 2ea e= [12] a= [3 0, 2 -1] 9. multiply [5 0 , 3 -5] x [2

Step-by-step answer

P Answered by PhD

9

Question 1:

$\left[\begin{array}{ccc}5\\-2\\1\end{array}\right] \left[\begin{array}{ccc}12\end{array}\right] = \left[\begin{array}{ccc}(5)(12)+(-2)(12)+(1)(1)\end{array}\right]$
= $\left[\begin{array}{ccc}60-24+12\end{array}\right] = \left[\begin{array}{ccc}48\end{array}\right]$
A
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Question 2
$\left[\begin{array}{ccc}-6&-4\\6&0\\6&4\end{array}\right] - \left[\begin{array}{ccc}-5&5\\-4&-4\\6&-4\end{array}\right]$
$\left[\begin{array}{ccc}(-6--5)&(-4-5)\\(6--4)&(0--4)\\(6-6)&(4--4)\end{array}\right] = \left[\begin{array}{ccc}-1&-9\\10&4\\0&8\end{array}\right]$
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Question 3

$\left[\begin{array}{ccc}1&-4\\3&5\end{array}\right] + \left[\begin{array}{ccc}-2&6\\-2&4\end{array}\right]$
= $\left[\begin{array}{ccc}(1+-2)&(-4+6)\\(3+-2)&(5+4)\end{array}\right]$
= $\left[\begin{array}{ccc}-1&2\\1&9\end{array}\right]$
----------------------------------------------------------------------------------------------------------------

Question 4

$A= \left[\begin{array}{ccc}4&-7\\3&-2\end{array}\right]$
det A=(4)(-2)-(-7)(3)=-8+21=13

A
--------------------------------------------------------------------------------------------------------------

Question 5
$\left[\begin{array}{ccc}3&0\\2&-1\end{array}\right] + \left[\begin{array}{ccc}2&8\\6&3\end{array}\right]$
= $\left[\begin{array}{ccc}3+2&0+8\\2+6&(-1)+3\end{array}\right]$
= $\left[\begin{array}{ccc}5&8\\8&2\end{array}\right]$
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Question 6
$4 \left[\begin{array}{ccc}12&0& \frac{3}{2} \\1&-6&7\end{array}\right] = \left[\begin{array}{ccc}48&0&6\\4&-24&28\end{array}\right]$

________________________________________________________

Question 7

$C = \left[\begin{array}{ccc}-5&1\\5&4\end{array}\right]$
det C = (-5)(4) - (1)(5) = -20-5 = -25

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Question 8
$2 \left[\begin{array}{ccc}12\end{array}\right] \left[\begin{array}{ccc}3&0\\2&-1\end{array}\right]$
$\left[\begin{array}{ccc}24\end{array}\right] \left[\begin{array}{ccc}3&6\\2&-1\end{array}\right]$
$\left[\begin{array}{ccc}72&0\\48&-24\end{array}\right]$
-------------------------------------------------------------------------------------------------------------

Question 9
$\left[\begin{array}{ccc}5&0\\3&-5\end{array}\right] \left[\begin{array}{ccc}2&-1\\2&-2\end{array}\right]$
$\left[\begin{array}{ccc}(5)(2)+(0)(2)&(5)(-1)+(0)(-2)\\(3)(2)+(-5)(2)&(3)(-1)+(-5)(-2)\end{array}\right]$
$\left[\begin{array}{ccc}10+0&-5+0\\6-10&-3+10\end{array}\right] = \left[\begin{array}{ccc}10&-5\\-4&7\end{array}\right]$

Mathematics

!1. select the best answer for the question. multiply matrix d = [5,-2,1 ] by matrix e= [12] a. [4 8] b. 2 c. [5 -2 1 ,10 4 2] d. [5 10, -2 -4, 1 2] 2. subtract [-6 -4,6 0, 6 4] - [-5 5, -4 -4,6 -4] 3. add [1 -4,3 5] + [-2 6,-2 4] 4. find the determinant of a= [ 4 -7, 3 -2] a. 13 b. −2 c. 2 d. 29 5. what is matrix a + matrix b? a=[3 0, 2 -1] b= [ 2 8, .6 3] 6. what is 4c? c= [12 03/2, 1 -6 7] 7. evaluate the determinant for the following matrix: [-5 1, 5 4] 8. complete the multiplication: 2ea e= [12] a= [3 0, 2 -1] 9. multiply [5 0 , 3 -5] x [2

Step-by-step answer

P Answered by PhD

9

Question 1:

$\left[\begin{array}{ccc}5\\-2\\1\end{array}\right] \left[\begin{array}{ccc}12\end{array}\right] = \left[\begin{array}{ccc}(5)(12)+(-2)(12)+(1)(1)\end{array}\right]$
= $\left[\begin{array}{ccc}60-24+12\end{array}\right] = \left[\begin{array}{ccc}48\end{array}\right]$
A
------------------------------------------------------------------------------------------------------------

Question 2
$\left[\begin{array}{ccc}-6&-4\\6&0\\6&4\end{array}\right] - \left[\begin{array}{ccc}-5&5\\-4&-4\\6&-4\end{array}\right]$
$\left[\begin{array}{ccc}(-6--5)&(-4-5)\\(6--4)&(0--4)\\(6-6)&(4--4)\end{array}\right] = \left[\begin{array}{ccc}-1&-9\\10&4\\0&8\end{array}\right]$
---------------------------------------------------------------------------------------------------------------

Question 3

$\left[\begin{array}{ccc}1&-4\\3&5\end{array}\right] + \left[\begin{array}{ccc}-2&6\\-2&4\end{array}\right]$
= $\left[\begin{array}{ccc}(1+-2)&(-4+6)\\(3+-2)&(5+4)\end{array}\right]$
= $\left[\begin{array}{ccc}-1&2\\1&9\end{array}\right]$
----------------------------------------------------------------------------------------------------------------

Question 4

$A= \left[\begin{array}{ccc}4&-7\\3&-2\end{array}\right]$
det A=(4)(-2)-(-7)(3)=-8+21=13

A
--------------------------------------------------------------------------------------------------------------

Question 5
$\left[\begin{array}{ccc}3&0\\2&-1\end{array}\right] + \left[\begin{array}{ccc}2&8\\6&3\end{array}\right]$
= $\left[\begin{array}{ccc}3+2&0+8\\2+6&(-1)+3\end{array}\right]$
= $\left[\begin{array}{ccc}5&8\\8&2\end{array}\right]$
---------------------------------------------------------------------------------------------------------

Question 6
$4 \left[\begin{array}{ccc}12&0& \frac{3}{2} \\1&-6&7\end{array}\right] = \left[\begin{array}{ccc}48&0&6\\4&-24&28\end{array}\right]$

________________________________________________________

Question 7

$C = \left[\begin{array}{ccc}-5&1\\5&4\end{array}\right]$
det C = (-5)(4) - (1)(5) = -20-5 = -25

--------------------------------------------------------------------------------------------------------------

Question 8
$2 \left[\begin{array}{ccc}12\end{array}\right] \left[\begin{array}{ccc}3&0\\2&-1\end{array}\right]$
$\left[\begin{array}{ccc}24\end{array}\right] \left[\begin{array}{ccc}3&6\\2&-1\end{array}\right]$
$\left[\begin{array}{ccc}72&0\\48&-24\end{array}\right]$
-------------------------------------------------------------------------------------------------------------

Question 9
$\left[\begin{array}{ccc}5&0\\3&-5\end{array}\right] \left[\begin{array}{ccc}2&-1\\2&-2\end{array}\right]$
$\left[\begin{array}{ccc}(5)(2)+(0)(2)&(5)(-1)+(0)(-2)\\(3)(2)+(-5)(2)&(3)(-1)+(-5)(-2)\end{array}\right]$
$\left[\begin{array}{ccc}10+0&-5+0\\6-10&-3+10\end{array}\right] = \left[\begin{array}{ccc}10&-5\\-4&7\end{array}\right]$

Mathematics

Linear algebra is a fundamental component of continuum mechanics (and most engineering disciplines). Solve the following matrix and vector problems showing all steps for full credit. As it is always a good idea to be able to find and correct your own mistakes, you are encouraged to use MATLAB, Maple, Mathematica, etc. to check your answers, but you still need to show all step. a. Determine if the vectors [u] = [7 3 -2] and [v] = [2 -4 -1] are orthogonal to each other. b. A plane is described by the vectors [u] = [3 -9 2] and [v] = [11 2 -6]. Find

Step-by-step answer

P Answered by Master

A ) Not orthogonal to each other

B) 50i + 40j + 105k

C) The tensor product is attached below

D ) The value of X = F.X is attached below

Step-by-step explanation:

attached below is the detailed solution of the above problem

A) for the vectors ( u ) and ( v ) to be orthogonal to each other [ U.V has to be = 0 ] but in this scenario U.V = 4 hence they are not orthogonal to each other

b) The vector normal to plane is gotten by : U x V

= 50i + 40j + 105k

Mathematics

Linear algebra is a fundamental component of continuum mechanics (and most engineering disciplines). Solve the following matrix and vector problems showing all steps for full credit. As it is always a good idea to be able to find and correct your own mistakes, you are encouraged to use MATLAB, Maple, Mathematica, etc. to check your answers, but you still need to show all step. a. Determine if the vectors [u] = [7 3 -2] and [v] = [2 -4 -1] are orthogonal to each other. b. A plane is described by the vectors [u] = [3 -9 2] and [v] = [11 2 -6]. Find

Step-by-step answer

P Answered by Specialist

A ) Not orthogonal to each other

B) 50i + 40j + 105k

C) The tensor product is attached below

D ) The value of X = F.X is attached below

Step-by-step explanation:

attached below is the detailed solution of the above problem

A) for the vectors ( u ) and ( v ) to be orthogonal to each other [ U.V has to be = 0 ] but in this scenario U.V = 4 hence they are not orthogonal to each other

b) The vector normal to plane is gotten by : U x V

= 50i + 40j + 105k

Computers and Technology

You are given an array of integers a. A new array b is generated by rearranging the elements of a in the following way: b[0] is equal to a[0]; b[1] is equal to the last element of a; b[2] is equal to a[1]; b[3] is equal to the second-last element of a; and so on. Your task is to determine whether the new array b is sorted in strictly ascending order or not. Example For a = [1, 3, 5, 6, 4, 2], the output should be alternatingSort(a) = true. The new array b will look like [1, 2, 3, 4, 5, 6], which is in strictly ascending order, so the answer is

Step-by-step answer

P Answered by PhD

Explanation:

Take note to avoid Indentation mistake which would lead to error message.

So let us begin by;

import java.util.*; using Java as our programming language

class Mutation {

public static boolean alternatingSort(int n , int[] a)

{

int []b = new int[n];

b[0]=a[0];

int j=1,k=0;

for(int i=1;i<n;i++)

{

if(i%2==1)

{

b[i]=a[n-j];

j++;

}

else

b[i]=a[++k];

}

for(int i=0;i<n;i++)

{

System.out.print(b[i]+" ");

}

System.out.print("\n");

for(int i=1;i<n;i++)

{

if(b[i]<=b[i-1])

return false ;

}

return true;

}

public static void main (String[] args) {

Scanner sc = new Scanner(System.in);

int n= sc.nextInt();

int []a = new int [n];

for(int i=0;i<n;i++)

a[i]=sc.nextInt();

System.out.println(alternatingSort(n,a));

}

This code should give you the desired result (output).

Computers and Technology

You are given an array of integers a. A new array b is generated by rearranging the elements of a in the following way: b[0] is equal to a[0]; b[1] is equal to the last element of a; b[2] is equal to a[1]; b[3] is equal to the second-last element of a; and so on. Your task is to determine whether the new array b is sorted in strictly ascending order or not. Example For a = [1, 3, 5, 6, 4, 2], the output should be alternatingSort(a) = true. The new array b will look like [1, 2, 3, 4, 5, 6], which is in strictly ascending order, so the answer is

Step-by-step answer

P Answered by PhD

Explanation:

Take note to avoid Indentation mistake which would lead to error message.

So let us begin by;

import java.util.*; using Java as our programming language

class Mutation {

public static boolean alternatingSort(int n , int[] a)

{

int []b = new int[n];

b[0]=a[0];

int j=1,k=0;

for(int i=1;i<n;i++)

{

if(i%2==1)

{

b[i]=a[n-j];

j++;

}

else

b[i]=a[++k];

}

for(int i=0;i<n;i++)

{

System.out.print(b[i]+" ");

}

System.out.print("\n");

for(int i=1;i<n;i++)

{

if(b[i]<=b[i-1])

return false ;

}

return true;

}

public static void main (String[] args) {

Scanner sc = new Scanner(System.in);

int n= sc.nextInt();

int []a = new int [n];

for(int i=0;i<n;i++)

a[i]=sc.nextInt();

System.out.println(alternatingSort(n,a));

}

This code should give you the desired result (output).

Mathematics

1. henry s current age is 10 times justin s current age. in 6 years, henry s age will be 4 times justin s age. which system of equations can be used to calculate h, henry s current age and j, justin s current age? a: j=10h j+6=4 [h+6] b: j+6=h j+4=h+10 c: h=10j h+6=4 [j+6] d: h=10j h+4=6 [j+4] 2. kevin plotted a linear equation on a graph with 18 as the y-intercept. which of the following could be the equation plotted by kevin? a: 3x=6-1/3y b: 7x+2y=38 c: 4y+18y=21 d: 1/2y=9x+5 3. janie was given [-2,3] as the end point of a line segment and [1,0]

Step-by-step answer

P Answered by PhD

7

1. C
2. A
3. D
4. B
5. C
6. B
7. B
8. B

If a = [1,2,3] and b = [1, 3, –2], find the angle between a and b.

Step-by-step answer

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