Quadrilateral pest has vertices (-1, -5), (8, 2), (11, 13), and (2, 6), respectively. classify the quadrilateral as a square, rhombus, rectangle or parallelogram. will mark brainliest

The figure PEST is a rhombus

Step-by-step explanation:

* Lets talk about the difference between all these shapes

- At first to prove the shape is a parallelogram you must have one

 of these conditions

# Each two opposite sides are parallel OR

# Each two opposite sides are equal in length OR

# Its two diagonals bisect each other

- After that to prove the parallelogram is:

* A rectangle you must have one of these conditions

# Two adjacent sides are perpendicular to each other OR

# Its two diagonals are equal in length

* A rhombus you must have one of these conditions

# Two adjacent sides are equal in length OR

# Its two diagonals perpendicular to each other OR

# Its diagonals bisect its vertices angles

* A square you must have two of these conditions

# Its diagonals are equal and perpendicular OR

# Two adjacent sides are equal and perpendicular

* Now lets solve the problem

∵ The vertices of the quadrilateral PEST are

   P (-1 , -5) , E (8 , 2) , S (11 , 13) , T (2 , 6)

- Lets find the slope from each two points using this rule :

 m = (y2 - y1)/(x2 - x1), where m is the slope and (x1 , y1) , (x2 , y2)

 are two points on the line

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ m of PE = (2 - -5)/(8 - -1) = 7/9

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ m of ES = (13 - 2)/(11 - 8) = 11/3  

- Let (x1 , y1) is (11 , 13) and (x2 , y2) is (2 , 6)

∴ m of ST = (6 - 13)/(2 - 11) = -7/-9 = 7/9

- Let (x1 , y1) is (2 , 6) and (x2 , y2) is (-1 , -5)

∴ m of TP = (-5 - 6)/(-1 - 2) = -11/-3 = 11/3

∵ m PE = m ST = 7/9

∴ PE // ST ⇒ opposite sides

∵ m ES = m TP = 11/3

∴ ES // TP ⇒ opposite sides

- Each two opposite sides are parallel

∴ PEST is a parallelogram

- Lets check if the parallelogram can be rectangle or rhombus or

 square by one of the condition above

∵ If two line perpendicular , then the product of their slops = -1

- Lets check the slopes of two adjacent sides (PE an ES)

∵ m PE = 7/9

∵ m ES = 11/3

∵ m PE × m ES = 7/9 × 11/3 = 77/27 ≠ -1

∴ PE and ES are not perpendicular

∴ PEST not a rectangle or a square (the sides of the rectangle and

  the square are perpendicular to each other)

- Now lets check the length of two adjacent side by using the rule

 of distance between two points (x1 , y1) and (x2 , y2)

 d = √[(x2 - x1)² + (y2 - y1)²]

- Let (x1 , y1) is (-1 , -5) and (x2 , y2) is (8 , 2)

∴ PE = √[(8 - -1)² + (2 - -5)²] = √[9² + 7²] = √[81 + 49] = √130 units

- Let (x1 , y1) is (8 , 2) and (x2 , y2) is (11 , 13)

∴ ES = √[(11 - 8)² + (13 - 2)²] = √[3² + 11²] = √[9 + 121] = √130 units

∴ PE = ES ⇒ two adjacent sides in parallelogram

∴ The four sides are equal

* The figure PEST is a rhombus

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3