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Oduvanchick [21]
3 years ago
11

Determine what type of quadrilateral ABCD is, given the points. A(2,0) B(3,5) C(5,0) D(6,5) Rhombus Parallelogram Rectangle I kn

ow you find the distance formula, but I’m actually confuse in how to KNOW which quadrilateral it is.

Mathematics
2 answers:
NikAS [45]3 years ago
7 0

Answer:

Parallelogram

Step-by-step explanation:

A quadrilateral is a shape with only 4 lines.

Such as: Square, Kite, Rhombus etc

Angelina_Jolie [31]3 years ago
5 0

Answer:

Brainliest!!!

Step-by-step explanation:

quadrilateral

is

DescriptionIn Euclidean plane geometry, a quadrilateral is a polygon with four edges and four vertices or corners.

-google

In plane Euclidean geometry, a rhombus is a quadrilateral whose four sides all have the same length. Another name is equilateral quadrilateral, since equilateral means that all of its sides are equal in length.

-google

In Euclidean geometry, a parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.

-Google

In Euclidean plane geometry, a rectangle is a quadrilateral with four right angles. It can also be defined as an equiangular quadrilateral, since equiangular means that all of its angles are equal. It can also be defined as a parallelogram containing a right angle

-google

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Step-by-step explanation:

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Answer:

\large\boxed{x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi,\ k\in\mathbb{Z}}

Step-by-step explanation:

\sin x-(3\sin x-1)=0\\\\\sin x-3\sin x+1=0\qquad\text{subtract 1 from both sides}\\\\-2\sin x=-1\qquad\text{divide both sides by 2}\\\\\sin x=\dfrac{1}{2}\Rightarrow x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi,\ k\in\mathbb{Z}

\text{Equation:}\\\\\sin x=a\\\\\text{has solutions}\\\\x=\theta+2k\pi\ \vee\ x=(\pi-\theta)+2k\pi\\\\\text{Why}\ 2k\pi?\\\text{Because the sine function has a period of}\ 2\pi.

\text{look at the table}\\\\\sin x=\dfrac{1}{2}\to x=\dfrac{\pi}{6}\ \vee\ x=\pi-\dfrac{\pi}{6}=\dfrac{5\pi}{6}

\text{Other solution:}\\\\\sin x=\dfrac{1}{2}\Rightarrow x=\sin^{-1}\dfrac{1}{2}\\\\x=\dfrac{\pi}{6}+2k\pi\ \vee\ x=\dfrac{5\pi}{6}+2k\pi

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