If J = 40° and G = 60°, then HJ _____ GJ
2 answers:
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Height of wall = 8 feet
1 feet = 12 inches
⇒ Height of wall = 8 × 12 inches
⇒ Height of wall = 96 inches
We know that the 4 shelves are to be placed such that that the 5 spaces created decrease by 7 inches as we move up the wall
Let height of 1st space = h inches
⇒ Height of 2nd space = h - 7 inches
⇒ Height of 3rd space = h - 14 inches
⇒ Height of 4th space = h - 21 inches
⇒ Height of 5th space = h - 28 inches
Thickness of each shelf = 3/4 inch
⇒ Thickness of 4 shelves = × 4 inches = 3 inches
So, Height of Wall = Height of 1st space + Height of 2nd space + Height of 3rd space + Height of 4th space + Height of 5th space + Thickness of 4 shelves
⇒ 96 = h + h - 7 + h - 14 + h -21 + h - 28 + 3
⇒ 95 = 5h - 70 + 3
⇒ 95 = 5h - 67
⇒ 5h = 162
⇒ h = 32.4 inches
So the first shelf is placed 32.4 inches from the floor
First, we have to make sure that the number of columns in the first matrix is equal to the number of rows in the second matrix.
![\left[\begin{array}{cc}1&-3&2&0\\\end{array}\right] * \left[\begin{array}{ccc}2&3&4\\1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-3%262%260%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2A%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%264%5C%5C1%262%263%5Cend%7Barray%7D%5Cright%5D%20)
Since this is true, we can continue to solve the problem.
To multiply two matrices, multiply each row element in the first matrix by each column element in the second matrix. For example:
1*2 = 2
-3*1=-3
Then we add them to get our new matrix element.
-3+2=-1
Then we move to the next column of the second matrix.
1*3=3
-3*2=-6
-6+3=-3
Then the final column of the second matrix.
1*4=4
-3*3=-9
-9+4=-5
Our matrix so far:
![\left[\begin{array}{ccc}-1&-3&-5\\x&x&x\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-3%26-5%5C%5Cx%26x%26x%5Cend%7Barray%7D%5Cright%5D%20)
We do the same for the bottom row of the first matrix.
<em>First Column</em>
2*2=4
0*1=0
4+0=4
<em>Second Column
</em>2*3=6
0*2=0
6+0=6
<em>Third Column</em>
2*4=8
0*3=0
8+0=8
Our final matrix is:
![\left[\begin{array}{ccc}-1&-3&-5\\4&6&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%26-3%26-5%5C%5C4%266%268%5Cend%7Barray%7D%5Cright%5D)
:)
Since this is true, we can continue to solve the problem.
To multiply two matrices, multiply each row element in the first matrix by each column element in the second matrix. For example:
1*2 = 2
-3*1=-3
Then we add them to get our new matrix element.
-3+2=-1
Then we move to the next column of the second matrix.
1*3=3
-3*2=-6
-6+3=-3
Then the final column of the second matrix.
1*4=4
-3*3=-9
-9+4=-5
Our matrix so far:
We do the same for the bottom row of the first matrix.
<em>First Column</em>
2*2=4
0*1=0
4+0=4
<em>Second Column
</em>2*3=6
0*2=0
6+0=6
<em>Third Column</em>
2*4=8
0*3=0
8+0=8
Our final matrix is:
:)