General
Easy
Maths-

find the value of x in given figure.

Maths-General

  1. 34 to the power of ring operator end exponent    
  2. 24 to the power of ring operator end exponent    
  3. 54 to the power of ring operator end exponent    
  4. 44 to the power of ring operator end exponent    

    Answer:The correct answer is: 34 to the power of ring operator end exponent

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    find the value of x in given figure.blank l parallel to m and text PI|Q end text

    find the value of x in given figure.blank l parallel to m and text PI|Q end text

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    in triangle A B C,if A equals x plus 20 to the power of ring operator comma B with blank below equals 2 open parentheses x minus 10 to the power of ring operator close parentheses comma C C equals 3 over 2 x then triangle is ..... triangle

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    triangle A B C is an Isosceles Triangle, if A equals 92 to the power of ring operator end exponent then find ABD .

    triangle A B C is an Isosceles Triangle, if A equals 92 to the power of ring operator end exponent then find ABD .

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    In the given figure, angle 1 equals 8 x minus 6 to the power of ring operator end exponent comma angle 3 equals 3 x plus 4 to the power of ring operator end exponent and then find angle 4 equals 4 x plus 2 to the power of ring operator end exponent then find angle 1.

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    The value of not stretchy integral subscript 0 end subscript superscript pi end superscript open vertical bar sin to the power of 3 end exponent invisible function application rightwards arrow over short leftwards arrow theta close vertical bar d theta is

    I equals not stretchy integral subscript 0 end subscript superscript pi end superscript vertical line sin to the power of 3 end exponent invisible function application theta vertical line d theta
    Since sin invisible function application theta is positive in interval left parenthesis 0 comma pi right parenthesis
    therefore I equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin to the power of 3 end exponent invisible function application theta d theta equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin invisible function application theta left parenthesis 1 minus cos to the power of 2 end exponent invisible function application theta right parenthesis d theta
    equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin invisible function application theta d theta plus not stretchy integral subscript 0 end subscript superscript pi end superscript left parenthesis negative sin invisible function application theta right parenthesis cos to the power of 2 end exponent invisible function application theta d theta
    equals left square bracket negative cos invisible function application theta right square bracket subscript 0 end subscript superscript pi end superscript plus open parentheses fraction numerator cos to the power of 3 end exponent invisible function application theta over denominator 3 end fraction close parentheses subscript 0 end subscript superscript pi end superscript equals fraction numerator 4 over denominator 3 end fraction.

    The value of not stretchy integral subscript 0 end subscript superscript pi end superscript open vertical bar sin to the power of 3 end exponent invisible function application rightwards arrow over short leftwards arrow theta close vertical bar d theta is

    maths-General
    I equals not stretchy integral subscript 0 end subscript superscript pi end superscript vertical line sin to the power of 3 end exponent invisible function application theta vertical line d theta
    Since sin invisible function application theta is positive in interval left parenthesis 0 comma pi right parenthesis
    therefore I equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin to the power of 3 end exponent invisible function application theta d theta equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin invisible function application theta left parenthesis 1 minus cos to the power of 2 end exponent invisible function application theta right parenthesis d theta
    equals not stretchy integral subscript 0 end subscript superscript pi end superscript sin invisible function application theta d theta plus not stretchy integral subscript 0 end subscript superscript pi end superscript left parenthesis negative sin invisible function application theta right parenthesis cos to the power of 2 end exponent invisible function application theta d theta
    equals left square bracket negative cos invisible function application theta right square bracket subscript 0 end subscript superscript pi end superscript plus open parentheses fraction numerator cos to the power of 3 end exponent invisible function application theta over denominator 3 end fraction close parentheses subscript 0 end subscript superscript pi end superscript equals fraction numerator 4 over denominator 3 end fraction.
    General
    physics-

    The velocity-time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in 6 blank s e c are respectively

    Displacement = Summation of all the area with sign
    equals open parentheses A subscript 1 end subscript close parentheses plus open parentheses negative A subscript 2 end subscript close parentheses plus open parentheses A subscript 3 end subscript close parentheses equals open parentheses 2 cross times 4 close parentheses plus open parentheses negative 2 cross times 2 close parentheses plus open parentheses 2 cross times 2 close parentheses

    therefore D i s p l a c e m e n t equals 8 blank m
    Distance = Summation of all the areas without sign
    equals open vertical bar A subscript 1 end subscript close vertical bar plus open vertical bar negative A subscript 2 end subscript close vertical bar plus open vertical bar A subscript 3 end subscript close vertical bar equals open vertical bar 8 close vertical bar plus open vertical bar negative 4 close vertical bar plus open vertical bar 4 close vertical bar equals 8 plus 4 plus 4
    therefore D i s t a n c e equals 16 blank m

    The velocity-time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in 6 blank s e c are respectively

    physics-General
    Displacement = Summation of all the area with sign
    equals open parentheses A subscript 1 end subscript close parentheses plus open parentheses negative A subscript 2 end subscript close parentheses plus open parentheses A subscript 3 end subscript close parentheses equals open parentheses 2 cross times 4 close parentheses plus open parentheses negative 2 cross times 2 close parentheses plus open parentheses 2 cross times 2 close parentheses

    therefore D i s p l a c e m e n t equals 8 blank m
    Distance = Summation of all the areas without sign
    equals open vertical bar A subscript 1 end subscript close vertical bar plus open vertical bar negative A subscript 2 end subscript close vertical bar plus open vertical bar A subscript 3 end subscript close vertical bar equals open vertical bar 8 close vertical bar plus open vertical bar negative 4 close vertical bar plus open vertical bar 4 close vertical bar equals 8 plus 4 plus 4
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    physics-

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    therefore blank x equals acos omega t
    therefore v equals fraction numerator d x over denominator d t end fraction equals negative a omega sin invisible function application omega t
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    therefore blankspeedequals open vertical bar u close vertical bar equals open vertical bar negative a omega sin invisible function application omega t close vertical bar
    Hence, [c] is correct.

    The position of a particle at any instant t is given by x equals acos omega t. The speed-time graph of the particle is

    physics-General
    therefore blank x equals acos omega t
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    The instantaneous speed is given by modulus of instantaneous velocity.
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