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raketka [301]
3 years ago
5

The following set of coordinates represents which figure? (0, 4), (0, 7), (1, 5), (1, 2)

Mathematics
2 answers:
BigorU [14]3 years ago
7 0

Answer with explanation:

The Coordinates of four points in two dimensional plane is  (0, 4), (0, 7), (1, 5), (1, 2).

Point (0,4) and point (0,7) lies on y axis.

And point (1,5) and point (1,2) lies in First Quadrant.

If you will join these four points , we will get a Quadrilateral, as it consist of four distinct line segments ,ends being joined with one another.

We can also check this by plotting the points in Two dimensional coordinate plane.

Marking, the points as  A(0, 4), B(0, 7), C(1, 5), D(1, 2).

We will check which kind of Quadrilateral is this

1. Parallelogram

2. Rhombus

3. Rectangle

4. Square

5. Kite

6. Trapezium

We will use distance formula to find the length of sides of Quadrilateral.

Distance formula

=\sqrt{({x_{2}-x_{1})^2+(y_{2}-y_{1})^2}}\\\\ AB=\sqrt{(0-0)^2+(7-4)^2} =3\\\\ BC=\sqrt{(1-0)^2+(5-7)^2}=\sqrt{1+4}\\\\ BC=\sqrt{5}\\\\ CD=\sqrt{(1-1)^2+(5-2)^2}=3\\\\ DA=\sqrt{(1-0)^2+(4-2)^2}\\\\ DA=\sqrt{5}

Length of Opposite sides are equal. So,it can be parallelogram or Rectangle.

Now, we will find the length of Diagonal.

AC=\sqrt{(1-0)^2+(5-4)^2}\\\\ AC=\sqrt{2} \\\\ BD=\sqrt{(1-0)^2+(7-2)^2}\\\\ BD=\sqrt{1+25}\\\\ BD=\sqrt{26}

Length of Diagonal is not equal, but opposite sides are equal.

So, the given Quadrilateral is Parallelogram.

Romashka [77]3 years ago
6 0

Answer:

The figure represent a parallelogram

Step-by-step explanation:

we have

(0, 4), (0, 7), (1, 5), (1, 2)

using a graphing tool

Plot the points

see the attached figure  

The figure has opposite sides parallel and equal in length

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Given that;

1.Let W₁ be the set of all polynomials of the form p(t) = at², where a is in R

2.Let W₂ be the set of all polynomials of the form p(t) = t² + a, where a is in R

3.Let W₃ be the set of all polynomials of the form p(t) = at² + at, where a is in R

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let W₁ = { at² ║ a∈ R }

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let c₁, c₂ be two scalars

c₁∝ + c₂β = c₁(a₁t²) + c₂(a₂t²)

= c₁a₁t² + c²a₂t²

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2)

let W₂ = { t² + a ║ a∈ R }

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let W₃ = { at² + a ║ a∈ R }

let ∝ = a₁t² +a₁t  and β = a₂t² + a₂t ∈ W₃

let c₁, c₂ be two scalars

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