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victus00 [196]
2 years ago
12

Please help!!

Mathematics
1 answer:
Natasha2012 [34]2 years ago
8 0

Answer: area = 30 square units

The diagram is below. Ignore point G and the red dashed line when drawing the final hexagon needed.

======================================================

Explanation:

On the grid your teacher has given you, plot the 6 points. Then connect them to form hexagon ABCDE. All of this is shown in blue in the diagram below.

Let's add point G as shown in red. I'll connect G to point C with a red dashed line. These will be temporary and erased later. Though if your teacher doesn't mind you leaving it in (so you can show your steps), then its probably not a bad idea to keep it. It's best to ask your teacher.

Anyways, the red segment GC splits the original hexagon into a trapezoid up top and a rectangle down below.

------------

The trapezoid has the parallel bases of b1 = GC = 4 and b2 = AB = 2. The height is h = GA = 3

This leads to,

area = h*(b1+b2)/2

area = 3*(4+2)/2

area = 3(6)/2

area = 18/2

area = 9

Trapezoid ABCG has area of 9 square units.

-----------

The rectangle has length EF = 7 and width ED = 3

area = length*width

area = 7*3

area = 21

Rectangle DEFG has area of 21 square units.

------------

Add up those two area results to get the area of the overall hexagon.

trapezoid + rectangle = 9 + 21 = 30 square units.

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Answer: \left \{ {{y\leq -\frac{5}{3}x + \frac{11}{3}} \atop {y < 8}} \right

<u>Step-by-step explanation:</u>

(1, 2) and (4, -3)

First, find the slope (m): \frac{y_{2}-y_{1}}{x_{2}-x_{1}}

m = \frac{-3-2}{4-1}

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Next, choose ONE of the points and input the point and slope into the Point-Slope formula: y - y₁ = m(x - x₁)

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Then, determine which inequality symbol will result in (-2, 8) being False (since it is not a solution):

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so ≤ makes the statement False

⇒ y ≤ -\frac{5}{3}x + \frac{11}{3}

****************************************************************

If you need a "system" of inequalities, then you need another equation.

There are infinite possibilities. Graph the points to see what works.  

y < 8   or  x > -2 are two examples.


---> I just realized that the system can be simpler than what I did above:

\left \{ {{x>-2} \atop {y < 8}} \right



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