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blagie [28]
4 years ago
15

1. The following is a parallelogram. Find x and y using the given information.

Mathematics
1 answer:
Tom [10]4 years ago
3 0
Given : ABCD is a parallelogram
            AB =5x-2y
<span>             DC = 4 </span>
           AD = y
<span>           BC =5-x

AB = DC ( opposite sides of parallelogram are </span>equal)
5x -2y = 4 (1)

AD = BC
y = 5-x

Putting the value of y in equation 1

5x-2y =4
5x-2(5-x) =4

5x -10+2x =4

7x = 14

x = 2

Putting the value of x in eqution 2

5-x = y
5-2 =y
3 =y
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132

Step-by-step explanation:

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evablogger [386]

Answer:

  • Solution of equation ( q ) = <u>1</u><u>6</u>

Step-by-step explanation:

In this question we have given an equation that is <u>3 </u><u>(</u><u> </u><u>q </u><u>-</u><u> </u><u>7</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>2</u><u>7</u><u> </u>and we have asked to solve this equation that means to find the value of <u> </u><u>q</u><u> </u><u>.</u>

<u>Solution : -</u>

\quad \: \longmapsto \:  3(q - 7) = 27

<u>Step </u><u>1</u><u> </u><u>:</u> Solving parenthesis :

\quad \: \longmapsto \:3q - 21 = 27

<u>Step </u><u>2</u><u> </u><u>:</u> Adding 21 on both sides :

\quad \: \longmapsto \:3q -  \cancel{ 21} +  \cancel{21} = 27  +  21

On further calculations we get :

\quad \: \longmapsto \:3q = 48

<u>Step </u><u>3 </u><u>:</u> Dividing by 3 from both sides :

\quad \: \longmapsto \: \frac{ \cancel{3}q}{ \cancel{3}}  =  \cancel {\frac{48}{3} }

On further calculations we get :

\quad \: \longmapsto \:   \pink{\boxed{\frak{q = 16}}}

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<u>Verifying</u><u> </u><u>:</u><u> </u><u>-</u>

Now we are very our answer by substituting value of q in the given equation . So ,

  • 3 ( q - 7 ) = 27

  • 3 ( 16 - 7 ) = 27

  • 3 ( 9 ) = 27

  • 27 = 27

  • L.H.S = R.H.S

  • Hence, Verified.

<u>Therefore</u><u>,</u><u> </u><u>our </u><u>solution</u><u> </u><u>is </u><u>correct</u><u> </u><u>.</u>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
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Answer:

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Step-by-step explanation:

given

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= 2 : 2/3

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Step-by-step explanation:

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