A square field has an area of 479 A2. What is the approximate length of a side of the field? Give your answer to the nearest foo
1 answer:
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There are two inequality equations to be graphed:
<span>x + y ≥ 4
2x + 3y < 12
For the first step, let's disregard the inequality symbols and take it like any conventional algebraic equation. This is to be able to graph the lines on a Cartesian planes first.
For the first equation, x+y=4. To find the x- and y-intercepts, let the other variable be 0. For example,
x-intercept:
x+0=4
x=4
y-intercept:
0+y=4
y=4
Therefore, you can graph the equation line by plotting the intercepts (4,0) and (0,4) and connecting them together. The same thing is done to the second equation:
x-intercept:
</span>2x + 0 = 12
x=12/2=6
y-intercept:
0 + 3y =12
y= 12/3 = 4
Therefore, you can graph the equation line by plotting the intercepts (6,0) and (0,4) and connecting them together. The graph is shown in the left side of the picture.
The next step would be testing the inequalities. Let's choose a point that does not coincide with the lines. That point could be (-5,-1).
x + y ≥ 4
-5 + -1 ≥4
-6 ≥ 4 --> this is not true. Thus, the solution of the graph must not include the area of this point. It includes everything to the right of the line denoted by the blue-shaded region.
2x + 3y < 12
2(-5) + 3(-1) <12
-13 < 12 ---> this is true. Thus, the solution would include this point. That includes all points to the left of the orange line denoted by the orange-shaded the region.
The region where blue and orange overlap is the solution of the system of equations, denoted by the green-shaded region.
<span>x + y ≥ 4
2x + 3y < 12
For the first step, let's disregard the inequality symbols and take it like any conventional algebraic equation. This is to be able to graph the lines on a Cartesian planes first.
For the first equation, x+y=4. To find the x- and y-intercepts, let the other variable be 0. For example,
x-intercept:
x+0=4
x=4
y-intercept:
0+y=4
y=4
Therefore, you can graph the equation line by plotting the intercepts (4,0) and (0,4) and connecting them together. The same thing is done to the second equation:
x-intercept:
</span>2x + 0 = 12
x=12/2=6
y-intercept:
0 + 3y =12
y= 12/3 = 4
Therefore, you can graph the equation line by plotting the intercepts (6,0) and (0,4) and connecting them together. The graph is shown in the left side of the picture.
The next step would be testing the inequalities. Let's choose a point that does not coincide with the lines. That point could be (-5,-1).
x + y ≥ 4
-5 + -1 ≥4
-6 ≥ 4 --> this is not true. Thus, the solution of the graph must not include the area of this point. It includes everything to the right of the line denoted by the blue-shaded region.
2x + 3y < 12
2(-5) + 3(-1) <12
-13 < 12 ---> this is true. Thus, the solution would include this point. That includes all points to the left of the orange line denoted by the orange-shaded the region.
The region where blue and orange overlap is the solution of the system of equations, denoted by the green-shaded region.