Answer:
(2x + y)(2x - y)
Step-by-step explanation:
1. Simplify the square 4x²:
(2x)² - y²
2. Apple the difference of two squares formula <em>x² - y² = (x + y)(x - y)</em>:
(2x + y)(2x - y)
hope this helps!
1. 3(6+4m) + 5m
2. Distribute the 3 to the inside of the brackets = 18 + 12m + 5m
3. 17m+18
Hope this helps!
17÷25=0.68
So it terminates.
Answer:
The answer to your question is:
x = 1
Square's perimeter = 12
Rectangle perimeter = 12
The value of x is one and the perimeter of the square and the rectangle is the same.
Step-by-step explanation:
Data
Side of Square = 4x - 1
Length of rectangle = 2x + 1
Width = x + 2
Perimeters are the same
Process
Square perimeter = 4(4x - 1)
Rectangle's perimeter = 2(2x + 1) + 2(x+ 2)
Then
4(4x - 1) = 2(2x + 1) + 2(x+ 2) Equation
16x - 4 = 4x + 2 + 2x + 4
16x - 4 = 6x + 6
16x - 6x = 6 + 4
10x = 10
x = 10 / 10
x = 1
Perimeter of the square = 4(4(1) - 1)
= 4(3)
= 12
Perimeter of the rectangle = 2(2(1) + 1) + 2((1)+ 2)
= 2(3) + 2(3)
= 6 + 6
= 12
Part A;
There are many system of inequalities that can be created such that only contain points D and E in the overlapping shaded regions.
Any system of inequalities which is satisfied by (-4, 2) and (-1, 5) but is not satisfied by (1, 3), (3, 1), (3, -3) and (-3, -3) can serve.
An example of such system of equation is
x < 0
y > 0
The system of equation above represent all the points in the second quadrant of the coordinate system.The area above the x-axis and to the left of the y-axis is shaded.
Part B:It can be verified that points D and E are solutions to the system of inequalities above by substituting the coordinates of points D and E into the system of equations and see whether they are true.
Substituting D(-4, 2) into the system
we have:
-4 < 0
2 > 0
as can be seen the two inequalities above are true, hence point D is a solution to the set of inequalities.
Also, substituting E(-1, 5) into the system we have:
-1 < 0
5 > 0
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.
Part C:Given that chicken can only be raised in the area defined by y > 3x - 4.
To identify the farms in which chicken can be raised, we substitute the coordinates of the points A to F into the inequality defining chicken's area.
For point A(1, 3): 3 > 3(1) - 4 ⇒ 3 > 3 - 4 ⇒ 3 > -1 which is true
For point B(3, 1): 1 > 3(3) - 4 ⇒ 1 > 9 - 4 ⇒ 1 > 5 which is false
For point C(3, -3): -3 > 3(3) - 4 ⇒ -3 > 9 - 4 ⇒ -3 > 5 which is false
For point D(-4, 2): 2 > 3(-4) - 4; 2 > -12 - 4 ⇒ 2 > -16 which is true
For point E(-1, 5): 5 > 3(-1) - 4 ⇒ 5 > -3 - 4 ⇒ 5 > -7 which is true
For point F(-3, -3): -3 > 3(-3) - 4 ⇒ -3 > -9 - 4 ⇒ -3 > -13 which is true
Therefore, the farms in which chicken can be raised are the farms at point A, D, E and F.
