The ordered pair (2,-1) is a solution the inequality.
Given the inequality is 3x-4y<12.
An inequality is a statement that compares two quantities and claims that one is greater or less than the other (the equation may also be included). These inequalities are linear if the variables are only linear.
By substituting each of the given option in the given inequality we will check which ordered pair satisfy the inequality.
1. (4,0)
Here x=4 and y=0
Substitute this in the inequality 3x-4y<12, we get
3(4)-4(0)<12
12<12
This is not true
2. (0,-3)
Here x=3 and y=-3
Substitute this in the inequality 3x-4y<12, we get
3(0)-4(-3)<12
12<12
This is not true.
3. (2,-1)
Here x=2 and y=-1
Substitute this in the inequality 3x-4y<12, we get
3(2)-4(-1)<12
6+4<12
10<12
This is true.
4. (4,-3)
Here x=4 and y=-3
Substitute this in the inequality 3x-4y<12, we get
3(4)-4(-3)<12
12+12<12
24<12
This is not true.
Hence, the ordered pair that satisfy the given inequality 3x-4y<12 is (2,-1).
Learn more about inequality from here brainly.com/question/21819032.
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Answer:
where is the box ?
Step-by-step explanation:
Answer:
<u>Option b. (x = 3, y = 20, z = -14)</u>
Step-by-step explanation:
Given:
2x + 2y + 3z = 4
5x + 3y + 5z = 5
3x + 4y + 6z = 5
Solve using Cramer’s rule
∴ ![\left[\begin{array}{ccc}2&2&3\\5&3&5\\3&4&6\end{array}\right] =\left[\begin{array}{ccc}4\\5\\5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%263%5C%5C5%263%265%5C%5C3%264%266%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%5C%5C5%5C%5C5%5Cend%7Barray%7D%5Cright%5D)
∴A = ![\left[\begin{array}{ccc}2&2&3\\5&3&5\\3&4&6\end{array}\right] = -1](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%263%5C%5C5%263%265%5C%5C3%264%266%5Cend%7Barray%7D%5Cright%5D%20%3D%20-1)
Ax = ![\left[\begin{array}{ccc}4&2&3\\5&3&5\\5&4&6\end{array}\right] = -3](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%262%263%5C%5C5%263%265%5C%5C5%264%266%5Cend%7Barray%7D%5Cright%5D%20%3D%20-3)
Ay = ![\left[\begin{array}{ccc}2&4&3\\5&5&5\\3&5&6\end{array}\right] =-20\\](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%264%263%5C%5C5%265%265%5C%5C3%265%266%5Cend%7Barray%7D%5Cright%5D%20%3D-20%5C%5C)
Az = ![\left[\begin{array}{ccc}2&2&4\\5&3&5\\3&4&5\end{array}\right] = 14](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%262%264%5C%5C5%263%265%5C%5C3%264%265%5Cend%7Barray%7D%5Cright%5D%20%3D%2014)
∴ x = Ax/A = -3/-1 = 3
y = Ay/A = -20/-1 = 20
z = Az/A = 14/-1 = -14
<u>So, the answer is option b. (x = 3, y = 20, z = -14)</u>
Answer:
A. Square
Explanation:
The square with sides: x+1
Has an area of: (x+1)^2
The rectangle with sides: x+2 and x
Has an area of: x(x+2)
So we simplify the square: (x+1).(x+1)= x^2+2x+1
Simplifying the rectangle: x^2+2x
Therefore the square area is larger by one unit.
Hope you get it!
