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Lelechka [254]
3 years ago
5

Important for my exam tomorrow ! Please help

Mathematics
1 answer:
Cerrena [4.2K]3 years ago
4 0

Answer:

Step-by-step explanation:

Adjacent angles of parallelogram are supplementary.

∠A + ∠D = 180

Divide both sides by 2

\frac{1}{2} ∠A + \frac{1}{2} ∠D = 90

∠PAD + ∠ADP = 90       --------------------(I)

IN ΔPAD,

∠PAD + ∠ADP + ∠APD = 180 {angle sum property of triangle}

90  + ∠APD = 180         {from (I)}

          ∠APD = 180 - 90

         ∠APD = 90

∠SPQ = ∠APD             {vertically opposite angles}

∠SPQ = 90°

Similarly, we can prove ∠PQR = 90° ; ∠QRS = 90° and ∠RSP = 90°

In a quadrilateral if each angle is 90°, then it is a rectangle.

PQRS is a rectangle.

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Answer:

<h2> </h2><h2>x = 5 -  \frac{4}{3} y - 2z</h2>

Step-by-step explanation:

<h3><u>Question</u><u>:</u><u>-</u></h3>
  • To solve for x

<h3><u>Equation</u><u>:</u><u>-</u></h3>
  • 3x + 4y + 6z = 15

<h3><u>Solution</u><u>:</u><u>-</u></h3>

=> 3x + 4y + 6z = 15

  • <em>[</em><em>On</em><em> </em><em>subtracting</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>with</em><em> </em><em>4y</em><em>]</em>

=> 3x + 4y + 6z - 4y = 15 - 4y

  • <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>

=> 3x + 6z = 15 - 4y

  • <em>[</em><em>On</em><em> </em><em>subtracting</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>with</em><em> </em><em>6z</em><em>]</em>

=> 3x + 6z - 6z = 15 - 4y - 6z

  • <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>

=> 3x = 15 - 4y - 6z

  • <em>[</em><em>On</em><em> </em><em>dividing</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>with</em><em> </em><em>3</em><em>]</em>

=  >  \frac{3x}{3}  =  \frac{15 - 4y - 6z}{3}

  • <em>[</em><em>On</em><em> </em><em>Simplification</em><em>]</em>

=  > x = 5 -  \frac{4}{3} y - 2z \: (ans)

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