Answer:
(4,3,2)
Step-by-step explanation:
We can solve this via matrices, so the equations given can be written in matrix form as:
![\left[\begin{array}{cccc}3&2&1&20\\1&-4&-1&-10\\2&1&2&15\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D3%262%261%2620%5C%5C1%26-4%26-1%26-10%5C%5C2%261%262%2615%5Cend%7Barray%7D%5Cright%5D)
Now I will shift rows to make my pivot point (top left) a 1 and so:
![\left[\begin{array}{cccc}1&-4&-1&-10\\2&1&2&15\\3&2&1&20\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-4%26-1%26-10%5C%5C2%261%262%2615%5C%5C3%262%261%2620%5Cend%7Barray%7D%5Cright%5D)
Next I will come up with algorithms that can cancel out numbers where R1 means row 1, R2 means row 2 and R3 means row three therefore,
-2R1+R2=R2 , -3R1+R3=R3
![\left[\begin{array}{cccc}1&-4&-1&-10\\0&9&4&35\\0&14&4&50\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-4%26-1%26-10%5C%5C0%269%264%2635%5C%5C0%2614%264%2650%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&-4&-1&-10\\0&1&\frac{4}{9}&\frac{35}{9}\\0&14&4&50\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-4%26-1%26-10%5C%5C0%261%26%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B35%7D%7B9%7D%5C%5C0%2614%264%2650%5Cend%7Barray%7D%5Cright%5D)
4R2+R1=R1 , -14R2+R3=R3
![\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&-\frac{20}{9}&-\frac{40}{9}\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26%5Cfrac%7B7%7D%7B9%7D%26%5Cfrac%7B50%7D%7B9%7D%5C%5C0%261%26%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B35%7D%7B9%7D%5C%5C0%260%26-%5Cfrac%7B20%7D%7B9%7D%26-%5Cfrac%7B40%7D%7B9%7D%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&0&\frac{7}{9}&\frac{50}{9}\\0&1&\frac{4}{9}&\frac{35}{9}\\0&0&1&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%26%5Cfrac%7B7%7D%7B9%7D%26%5Cfrac%7B50%7D%7B9%7D%5C%5C0%261%26%5Cfrac%7B4%7D%7B9%7D%26%5Cfrac%7B35%7D%7B9%7D%5C%5C0%260%261%262%5Cend%7Barray%7D%5Cright%5D)
, 
![\left[\begin{array}{cccc}1&0&0&4\\0&1&0&3\\0&0&1&2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%260%260%264%5C%5C0%261%260%263%5C%5C0%260%261%262%5Cend%7Barray%7D%5Cright%5D)
Therefore the solution to the system of equations are (x,y,z) = (4,3,2)
Note: If answer choices are given, plug them in and see if you get what is "equal to". Meaning plug in 4 for x, 3 for y and 2 for z in the first equation and you should get 20, second equation -10 and third 15.
1. 5 in and 1/3 in: Area = 5/3 in^2
2. 5 in and 4/3 in: Area= 20/3 in^2
3. 5/2 in and 4/3 in: Area=10/3 in^2
4. 7/6 in and 6/7 in: Area = 1 in^2
Step-by-step explanation:
<u>1. 5 in and 1/3 in</u>
Here,
<u>2. 5 in and 4/3 in</u>
Here,
<u>3. 5/2 in and 4/3 in</u>
<u>4. 7/6 in and 6/7 in</u>
<u>Let</u>
<u>Hence,</u>
1. 5 in and 1/3 in: Area = 5/3 in^2
2. 5 in and 4/3 in: Area= 20/3 in^2
3. 5/2 in and 4/3 in: Area=10/3 in^2
4. 7/6 in and 6/7 in: Area = 1 in^2
Keywords: Rectangle, Area
Learn more about rectangles at:
#LearnwithBrainly
The answer is for your paper is 21
Answer: 14/25 = 56/100
11/20 = 55/100
14/25 > 11/20
Step-by-step explanation:
The least common multiple of 20 and 25 is 100.
(100 = 20*5, 25*4)
So we can take the common denominator 100.
14*4/25*4 =56/100. ∴14/25 = 56/100
11*5/20*5 =55/100. ∴11/20 = 55/100
56>55
∴56/100 > 55/100
∴14/25 > 11/20
Hope it's good;)
