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Marat540 [252]
3 years ago
13

The points A(2,1), B(3,6), C(5,-2) and D (1,-2) are the vertices of a parallelogram.fine |AC| and |BD|.are they equal in length?

Mathematics
1 answer:
iren2701 [21]3 years ago
7 0

Answer:

|AC| =√18   and  |BD| =√68.   They are not equal in length.

Step-by-step explanation:

To find |AC| and |BD|  of the parallelogram, we will simply use the distance formula.

Using the line distance formula;

D = √(y_{2}-y_{1})² + (x_{2}-x_{1})²

A(2,1)     C(5,-2)

x_{1} =2    y_{1}=1   x_{2} = 5   y_{2} =-2

|AC|  = √(y_{2}-y_{1})² + (x_{2}-x_{1})²

          =√(-2-1)² + (5-2)²

           =√(-3)² + (3)²

           =√9+9

            =√18

Distance  |AC| =√18

B(3,6)     D (1,-2)

x_{1} =3    y_{1}=6   x_{2} = 1   y_{2} =-2

|BD|  = √(y_{2}-y_{1})² + (x_{2}-x_{1})²

          =√(-2-6)² + (1-3)²

           =√(-8)² + (-2)²

           =√64+4

            =√68

Distance  |BD| =√68

|AC|   and   |BD|   are not equal in length

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Answer:

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Step-by-step explanation:

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The square of y is y^2

The sum of the squares is x^2 + y^2

ii) Product of x and y subtracted from 9

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Subtracting these from 9, we have;

9-xy

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Explanation:

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lys-0071 [83]

\displaystyle
(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k

<em>-------------------------------------------------------------</em>


\displaystyle&#10;(x^2+y)^n=\sum_{k=0}^n\binom{n}{k}x^{2n-2k}y^k\\&#10;n=5\\&#10;k=3\\\\\binom{5}{3}=\dfrac{5!}{3!2!}=\dfrac{4\cdor5}{2}=10

<u>It's 10.</u>

----------------------------------------------------

\displaystyle&#10;(3x^2+y)^n=\sum_{k=0}^n\binom{n}{k}(3x)^{2n-2k}y^k=\sum_{k=0}^n\binom{n}{k}\cdot 3^{2n-2k}\cdot x^{2n-2k}y^k\\\\&#10;n=5\\&#10;k=4\\\\&#10;\binom{5}{3}\cdot3^{2\cdot5-2\cdot4}=10\cdot3^{2}=10\cdot9=90

<u>It's 90</u>

6 0
3 years ago
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