4-0-0
2-1-1
1-2-1
1-0-3
0-1-3
Those are just the ones I can think of so about 5 ways
The given system of equations in augmented matrix form is
![\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D3%262%26-4%262%26-23%5C%5C-6%261%262%264%26-12%5C%5C1%26-3%26-3%265%26-20%5C%5C-2%265%266%260%2612%5Cend%7Barray%7D%5Cright%5D)
If you need to solve this, first get the matrix in RREF:
- Add 2(row 1) to row 2, row 1 to -3(row 3), and 2(row 1) to 3(row 4):
![\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D3%262%26-4%262%26-23%5C%5C0%265%26-6%268%26-58%5C%5C0%2611%265%26-13%2637%5C%5C0%2619%2610%264%26-10%5Cend%7Barray%7D%5Cright%5D)
- Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):
![\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D3%262%26-4%262%26-23%5C%5C0%265%26-6%268%26-58%5C%5C0%260%26-91%26153%26-823%5C%5C0%260%26-164%26132%26-1052%5Cend%7Barray%7D%5Cright%5D)
- Add 164(row 3) to -91(row 4):
![\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D3%262%26-4%262%26-23%5C%5C0%265%26-6%268%26-58%5C%5C0%260%26-91%26153%26-823%5C%5C0%260%260%2613080%26-39240%5Cend%7Barray%7D%5Cright%5D)
- Multiply row 4 by 1/13080:
![\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D3%262%26-4%262%26-23%5C%5C0%265%26-6%268%26-58%5C%5C0%260%26-91%26153%26-823%5C%5C0%260%260%261%26-3%5Cend%7Barray%7D%5Cright%5D)
- Add -153(row 4) to row 3:
![\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D3%262%26-4%262%26-23%5C%5C0%265%26-6%268%26-58%5C%5C0%260%26-91%260%26-364%5C%5C0%260%260%261%26-3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D3%262%26-4%262%26-23%5C%5C0%265%26-6%268%26-58%5C%5C0%260%261%260%264%5C%5C0%260%260%261%26-3%5Cend%7Barray%7D%5Cright%5D)
- Add 6(row 3) and -8(row 4) to row 2:
![\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D3%262%26-4%262%26-23%5C%5C0%265%260%260%26-10%5C%5C0%260%261%260%264%5C%5C0%260%260%261%26-3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D3%262%26-4%262%26-23%5C%5C0%261%260%260%26-2%5C%5C0%260%261%260%264%5C%5C0%260%260%261%26-3%5Cend%7Barray%7D%5Cright%5D)
- Add -2(row 2), 4(row 3), and -2(row 4) to row 1:
![\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D3%260%260%260%263%5C%5C0%261%260%260%26-2%5C%5C0%260%261%260%264%5C%5C0%260%260%261%26-3%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7Cc%7D1%260%260%260%261%5C%5C0%261%260%260%26-2%5C%5C0%260%261%260%264%5C%5C0%260%260%261%26-3%5Cend%7Barray%7D%5Cright%5D)
So the solution to this system is
.
Answer:
600 m³
Step-by-step explanation:
Given:
- length of swimming pool = 30 m
- width of swimming pool = 10 m
- depth at deep end = 3 m
- depth at shallow end = 1 m
To find the volume of the water in the pool, find the area of a cross section (see attached image) and multiply it by the width of the pool.
<u>Area of cross section</u>
= area of rectangle - area of triangle
= (3 × 30) - [1/2 × 30 × (3 - 1)]
= 90 - 30
= 60 m²
<u>Volume of pool</u>
= area of cross section × width
= 60 × 10
= 600 m³
Answer:
66 and 62
Step-by-step explanation:
128 divided by 2 gives you 64, and by taking two away from one number and giving it to the other, you're final answers are 66 and 62
Answer:
40
Step-by-step explanation:
210-x=50×2
or, 210-x=100
or, x=210-100
or, x=110
so,
z = 360-210-110
or, z = 40