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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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5

Step-by-step explanation:

\triangle CDE \sim\triangle FGH...(given)

\therefore \frac{CD}{FG}=\frac{CE}{FH}...(csst)

\therefore \frac{4x}{28}=\frac{25}{35}

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Which represents the solution set of the inequality 2.9(x+8)<26.1
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The answer is x is less than 1.

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An object is launched from a launching pad 208 ft. above the ground at a velocity of 192ft/sec. what is the maximum height reach
Alexeev081 [22]

Answer:

h(x) = -16x² + 192x + 208

784ft

6 sec

13 sec

Step-by-step explanation:

a)

h(x) = -16x² +vx + h_{o}

here v represent velocity

         h_{o} represent initial height of launch

       

h(x) = -16x² + 192x + 208

b)

h(x) = -16x² + 192x + 208

here a = -16

        b = 192

        c = 208

x = -b/2a

  = -192/2(-16)

  = 6

plug this value in the equation

h(x) = -16(6)² + 192(6) + 208

      = 784ft

e)

Plug h(x)=0 in the equation

0 = -16x² + 192x + 208

divide equation by -16

x² - 12x - 13 = 0

Factors

1x * -13x = -13

1x - 13x = -12

Factorised form

x² - 12x - 13 = 0

x² + x - 13x - 13 = 0

x(x+1) -13(x+1) = 0

(x+1)(x-13) = 0

x = -1

x = 13

Since time can not be negative so we will reject x = -1  

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