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Rasek [7]
3 years ago
15

The vertices of a quadrilateral are (–1, 2), (5, 2), (–1, –3), and (5, –3). What type of quadrilateral is it? A. rectangle B. sq

uare C. rhombus D. trapezoid
Mathematics
1 answer:
Mandarinka [93]3 years ago
6 0

Answer: The answer is B: Square

Step-by-step explanation:

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8v=8
v=1. answer.......
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Help I need done ASAP
lora16 [44]

Answer:

3/15 5/25 9/45

Step-by-step explanation:

If you look at 12/60 and simplify it you will get 1/5. Now you just have to apply this to the rest.

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3 years ago
The angle bisectors of AXYZ are XG, YG, and ZG. They meet at a single point G.
Harrizon [31]

Answer:

m∠FXD = 36°

EG = 10

m∠FZG = 24°

Step-by-step explanation:

In triangle XYZ, G is the incenter of the triangle.

Since, m∠FXG = 18°

And m∠FXD = 2(m∠FXG)

                     = 2 × 18°

                     = 36°

Since, point G is equidistant from all sides (Property of incenter of a triangle)

Therefore, DG = EG = GF = 10

Since, m∠X + m∠Y + m∠Z = 180°

2(m∠FXG) + m∠DYE + 2(m∠FZG) = 180°

2(18)° + 96° + 2(m∠FZG) = 180°

2(m∠FZG) = 180° - 132°

m∠FZG = 24°

3 0
3 years ago
how do I solve this quadratic simultaneous equation 2y+2x=7 and x^2-4y^2=8 it has two sets of solutions and I'm solving by both
Ghella [55]
These are all the possible answers i know from experience.

2x + 2y = 7
x^2 - 4y^2 = 8

2x = -2y + 7
x = -y + 7/2

(-y + 7/2)^2 - 4y^2 = 8

y^2 - 7y + 49/4 - 4y^2 = 8

0 = 3y^2 + 7y + 8 - 49/4

0 = 12y^2 + 28y + 32 - 49

0 = 12y^2 + 28y - 17

y = (-b ± √(b^2 - 4ac))/(2a)

y = (-28 ± √(28^2 - 4(12)(-17)))/(2(12))

Dividing by 2 is like dividing by √(4):

y = (-14 ± √(7*28 + (12)(17)))/(2*6)

y = (-7 ± √(7*7 + (3)(17)))/6

y = (-7 ± 10)/6 = 1/2 or -17/6

x = -y + 7/2 = 3 or 19/3

<span>Solutions are (3,1/2) and (19/3,-17/6)</span>
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jesse kicks a soccer ball off the ground and in the air, with an initial velocity of 29 feet per second. using the formula h(t)
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