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daser333 [38]
3 years ago
14

The coordinate plane below represents a city. Points A through F are schools in the city. graph of coordinate plane. Point A is

at (-3, -4). Point B is at (-4, 3). Point C is at (2, 2). Point D is at (1, -2). Point E is at (5, -4). Point F is at (3, 4). Part A: Using the graph above, create a system of inequalities that only contain points C and F in the overlapping shaded regions. Explain how the lines will be graphed and shaded on the coordinate grid above. Part B: Explain how to verify that the points C and F are solutions to the system of inequalities created in Part A. Part C: Natalie can only attend a school in her designated zone. Natalie's zone is defined by y < −2x + 2. Explain how you can identify the schools that Natalie is allowed to attend.
Mathematics
1 answer:
Firlakuza [10]3 years ago
7 0
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.
</span>
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Svet_ta [14]

Answer:

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Step-by-step explanation:

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2 years ago
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What are the coordinates of the centroid of a triangle with vertices A(−3, 1) , B(1, 6) , and C(5, 2) ? Enter your answer in the
kirza4 [7]

<u>ANSWER</u>: The centroid is (1,3)


<u>Explanation:</u>

The centroid is the intersection of the medians of the triangle.

So we need to find the equation of any two of the medians and solve simultaneously.


Since the median is the straight line from one vertex to the midpoint of the opposite side, we find the midpoint of any two sides.


We find the midpoint of AC using the formula;

N=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})



N=(\frac{-3+5}{2}, \frac{1+2}{2})


N=(\frac{2}{2}, \frac{3}{2})


N=(1, \frac{3}{2})


The equation of the median passes through B(1,6) and N=(1, \frac{3}{2}).


This line is parallel to the y-axis hence has equation

x=1-------first median.



We also find the midpoint M of BC.

M=(\frac{1+5}{2}, \frac{6+2}{2})


M=(\frac{6}{2}, \frac{8}{2})


M=(3, 4)


The slope of the median, AM is


Slope_{AM}=\frac{4-1}{3--3}


Slope_{AM}=\frac{3}{6}


Slope_{AM}=\frac{1}{2}


The equation of the median AM is given by;


y-y_1=m(x-x_1)


We use the point M and the slope of AM.


y-4=\frac{1}{2}(x-3)


2y-8=(x-3)


2y-x=-3+8


2y-x=5-------Second median

We now solve the equation of the two medians simultaneously by putting x=1 in to the equation of the second median.


2y-1=5


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-Dominant- [34]

Answer:

  7 square units

Step-by-step explanation:

As with many geometry problems, there are several ways you can work this.

Label the lower left  and lower right vertices of the rectangle points W and E, respectively. You can subtract the areas of triangles WSR and EQR from the area of trapezoid WSQE to find the area of triangle QRS.

The applicable formulas are ...

  area of a trapezoid: A = (1/2)(b1 +b2)h

  area of a triangle: A = (1/2)bh

So, our areas are ...

  AQRS = AWSQE - AWSR - AEQR

  = (1/2)(WS +EQ)WE -(1/2)(WS)(WR) -(1/2)(EQ)(ER)

Factoring out 1/2, we have ...

  = (1/2)((2+5)·4 -2·2 -5·2)

  = (1/2)(28 -4 -10) = 7 . . . . square units

4 0
3 years ago
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