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  • 1 answers

Om Rajput 4 months, 1 week ago

Given:ABCD is parallelogram AC=BD(diagonal of parallelogram ) Proof:ABCD is rectangle In ∆ ABD and ∆ ABC AC=BD (given) AD=BC ( opposite side of parallelogram) AB=AB (Common) ∆ ABD and ∆ ABC (S.S.S) DAB=CAB ( CPCT ) DAB+CAB=180 DAB+DAB=180 2DAB=180 DAB=90 So, A=B=C=D ABCD is a rectangle
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Manish Desale 4 months, 1 week ago

2x+3x=5x ( ans) put vale of x=2 Therefore 2(2)+3(2)=5(2) 4+6=10 Hence proved LHS=RHS
  • 1 answers

Jigisha Mohanty 4 months, 1 week ago

4x³+3x²-4x+a=0 4(1)³+3(1)²-4(1)+a=0 4×1+3×1-4×1+a=0 4+3-4+a=0 3+a=0 a=-3
  • 1 answers

Aaditya Gupta 4 months, 1 week ago

Angle 1= Angle 2. { Law of Angle 3= Angle 4. Reflection} PQ//RS and BC is transversal line Angle 2= Angle 3. (Equation 1) Angle 1= Angle 2= Angle 3= Angle 4 Angle 1= Angle 4 (Equation 2) Adding eqn 1 and eqn 2 Angle 1+Angle 2=Angle 3+Angle 4 ABC=BCD
  • 3 answers

Mohd Gufran 4 months, 1 week ago

It is (a+b)^3

Jamil Ahmed 4 months, 1 week ago

Degree of the 4-y2

Shiwam Barnwal 4 months, 1 week ago

(a+b)(a+b)(a+b) =(a+b)³ =a³+b³+3ab(a+b) =a³+b³+3a²b+3ab²
  • 1 answers

Shiwam Barnwal 4 months, 1 week ago

Nice question
  • 1 answers

Shiwam Barnwal 4 months, 1 week ago

-3√2/√6+√3+4√3/√6+√2 =-3√2/√3*√2+√3+4√3/√3*√2+√2 =-3/√3+√3+4/√2+√2 =(-3√2+3√2+4√3+2√3) /√6 =6√3/√6 =6/√2 =3√2
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Govardhan C Chinthi 4 months, 2 weeks ago

2020√3

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