Answer:
no
Step-by-step explanation:
yes
The coefficient matrix is build with its rows representing each equation, and its columns representing each variable.
So, you may write the matrix as
![\left[\begin{array}{cc}\text{x-coefficient, 1st equation}&\text{y-coefficient, 1st equation}\\\text{x-coefficient, 2nd equation}&\text{y-coefficient, 2nd equation} \end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Ctext%7Bx-coefficient%2C%201st%20equation%7D%26%5Ctext%7By-coefficient%2C%201st%20equation%7D%5C%5C%5Ctext%7Bx-coefficient%2C%202nd%20equation%7D%26%5Ctext%7By-coefficient%2C%202nd%20equation%7D%20%5Cend%7Barray%7D%5Cright%5D%20%20)
which means
![\left[\begin{array}{cc}4&-3\\8&-3\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-3%5C%5C8%26-3%5Cend%7Barray%7D%5Cright%5D%20%20)
The determinant is computed subtracting diagonals:
![\left | \left[ \begin{array}{cc}a&b\\c&d\end{array}\right]\right | = ad-bc](https://tex.z-dn.net/?f=%20%5Cleft%20%7C%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5Cright%20%7C%20%3D%20ad-bc%20)
So, we have
![\left | \left[\begin{array}{cc}4&-3\\8&-3\end{array}\right] \right | = 4(-3) - 8(-3) = -4(-3) = 12](https://tex.z-dn.net/?f=%20%5Cleft%20%7C%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-3%5C%5C8%26-3%5Cend%7Barray%7D%5Cright%5D%20%5Cright%20%7C%20%3D%204%28-3%29%20-%208%28-3%29%20%3D%20-4%28-3%29%20%3D%2012%20%20)
Answer:
128 boxes
Step-by-step explanation:
The dimensions of the trunk is 24 in. by 24 in by 48 in
The volume = 24(24)(48) = 27648 cu. in.
The cubed boxes each have a volume of 6(6)(6) = 216 cu. in.
27648/216 = 128 boxes
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).
