The perimeter of the given rectangle will be ( 12x + 10 ).
<h3>What is the perimeter?</h3>
The perimeter is defined as the sum of all the sides of the given shape. For a triangle, the perimeter will be the sum of all the sides of the rectangle.
It is given that the two sides of the rectangle are (2x + 2) and (4x + 3). Then the perimeter of the rectangle will be calculated as below:-
Perimeter = Sum of all the sides of the rectangle
Perimeter = 2 ( 2x + 2 + 4x + 3 )
Perimeter = 4x + 4 + 8x + 6
Perimeter = 12x + 10
Therefore, the perimeter of the given rectangle will be ( 12x + 10 ).
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Answer:
no solution
Step-by-step explanation:
examples: 4^1 + 2 = 2^ 1 -1 and the answer is 6 = 1 which is not true. So its gonna be hard for them to have a solution.
Answer:
2.5 bottles or 3 bottles
Step-by-step explanation:
Given that :
Dimension of lot :
22.5 yds by 20.4 yds
Area of lot = 22.5 * 20.4 = 459 yd²
Volume per bottle of herbicide = 12 fl oz
1.1 fl oz combined with water treats an area of 155 ft²
Number of 1.1 fl oz from 12 fl oz that can be obtained : 12 / 1.1 = 10.90
1 yard² = 9 feet²
459 yard² = (459 * 9) = 4131 feet²
Number of 151 feet² that can be obtained from 4131 feet²
4131 feet² / 155 ft²
= 27.3576 times
= 27.3576 / 10.909
= 2.507
This means he'll have to buy a minimum of 3 bottles if a fraction of a bottle can't be purchased.
Part A;
There are many system of inequalities that can be created such that only contain points C and F in the overlapping shaded regions.
Any system of inequalities which is satisfied by (2, 2) and (3, 4) but is not stisfied by <span>(-3, -4), (-4, 3), (1, -2) and (5, -4) 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 first quadrant of the coordinate system.
The area above the x-axis and to the right of the y-axis is shaded.
Part 2:
It can be verified that points C and F are solutions to the system of inequalities above by substituting the coordinates of points C and F into the system of equations and see whether they are true.
Substituting C(2, 2) into the system we have:
2 > 0
2 > 0
as can be seen the two inequalities above are true, hence point C is a solution to the set of inequalities.
Part C:
Given that </span><span>Natalie
can only attend a school in her designated zone and that Natalie's zone is
defined by y < −2x + 2.
To identify the schools that
Natalie is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining Natalie's zone.
For point A(-3, -4): -4 < -2(-3) + 2; -4 < 6 + 2; -4 < 8 which is true
For point B(-4, 3): 3 < -2(-4) + 2; 3 < 8 + 2; 3 < 10 which is true
For point C(2, 2): 2 < -2(2) + 2; 2 < -4 + 2; 2 < -2 which is false
For point D(1, -2): -2 < -2(1) + 2; -2 < -2 + 2; -2 < 0 which is true
For point E(5, -4): -4 < -2(5) + 2; -4 < -10 + 2; -4 < -8 which is false
For point F(3, 4): 4 < -2(3) + 2; 4 < -6 + 2; 4 < -4 which is false
Therefore, the schools that Natalie is allowed to attend are the schools at point A, B and D.
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