Chemistry-
General
Easy

Question

 The product (A) is:

The correct answer is:

Related Questions to study

General
maths-

If A plus B plus C equals pi comma then tan squared invisible function application A over 2 plus tan squared invisible function application B over 2 plus tan squared invisible function application C over 2 is always

If A plus B plus C equals pi comma then tan squared invisible function application A over 2 plus tan squared invisible function application B over 2 plus tan squared invisible function application C over 2 is always

maths-General
General
chemistry-

  The compound (A) is:

  The compound (A) is:

chemistry-General
General
chemistry-

The oxidation number of C in CH subscript 2 straight O is:

The oxidation number of C in CH subscript 2 straight O is:

chemistry-General
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General
Maths-

vertical line tan space x vertical line equals tan space x plus fraction numerator 1 over denominator cos begin display style space end style x end fraction left parenthesis 0 less or equal than x less or equal than 2 pi right parenthesis has

In this question, we have to find the number of solution. In each quadrant tan value is different. Make two different case, one is where tan is positive ,[ 0 , π/2  ] U [π , 3 π/2 ] . And second case , tan is negative at [π/2 , π] U [3 π/2 , 2 π ] .

vertical line tan space x vertical line equals tan space x plus fraction numerator 1 over denominator cos begin display style space end style x end fraction left parenthesis 0 less or equal than x less or equal than 2 pi right parenthesis has

Maths-General

In this question, we have to find the number of solution. In each quadrant tan value is different. Make two different case, one is where tan is positive ,[ 0 , π/2  ] U [π , 3 π/2 ] . And second case , tan is negative at [π/2 , π] U [3 π/2 , 2 π ] .

General
maths-

If A equals sin to the power of 2 end exponent invisible function application theta plus cos to the power of 4 end exponent invisible function application theta comma then for all real values of q

If A equals sin to the power of 2 end exponent invisible function application theta plus cos to the power of 4 end exponent invisible function application theta comma then for all real values of q

maths-General
General
chemistry-

 The products (A) , (B) and (C) are:

 The products (A) , (B) and (C) are:

chemistry-General
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General
maths-

In the interval left square bracket negative pi divided by 4 comma pi divided by 2 right square bracket, the equation, cos space 4 x plus fraction numerator 10 tan space x over denominator 1 plus tan squared space x end fraction equals 3 has

In the interval left square bracket negative pi divided by 4 comma pi divided by 2 right square bracket, the equation, cos space 4 x plus fraction numerator 10 tan space x over denominator 1 plus tan squared space x end fraction equals 3 has

maths-General
General
chemistry-

Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript plus straight X not stretchy ⟶ with straight H to the power of ⊖ on top Cr to the power of 3 plus end exponent plus straight H subscript 2 straight O plus oxidised product of , in the above reaction cannot be

Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript plus straight X not stretchy ⟶ with straight H to the power of ⊖ on top Cr to the power of 3 plus end exponent plus straight H subscript 2 straight O plus oxidised product of , in the above reaction cannot be

chemistry-General
General
maths-

Let a, b, c, d element of R. Then the cubic equation of the type a x cubed plus b x squared plus c x plus d equals 0 has either one root real or all three roots are real. But in case of trigonometric equations of the type sin cubed space x plus b sin squared space x plus c sin space x plus d equals 0 can possess several solutions depending upon the domain of x. To solve an equation of the type a cos space theta plus b sin space theta equals c. The equation can be written as cos space left parenthesis theta minus alpha right parenthesis equals c divided by square root of open parentheses a squared plus b squared close parentheses end root The solution is theta equals 2 straight n pi plus alpha plus-or-minus beta where tan space alpha = b divided by a comma cos space beta equals c divided by square root of open parentheses a squared plus b squared close parentheses end root

On the domain [–straight pistraight pi] the equation 4 sin cubed space x plus 2 sin squared space x minus 2 sin space x minus 1 equals 0 possess

Let a, b, c, d element of R. Then the cubic equation of the type a x cubed plus b x squared plus c x plus d equals 0 has either one root real or all three roots are real. But in case of trigonometric equations of the type sin cubed space x plus b sin squared space x plus c sin space x plus d equals 0 can possess several solutions depending upon the domain of x. To solve an equation of the type a cos space theta plus b sin space theta equals c. The equation can be written as cos space left parenthesis theta minus alpha right parenthesis equals c divided by square root of open parentheses a squared plus b squared close parentheses end root The solution is theta equals 2 straight n pi plus alpha plus-or-minus beta where tan space alpha = b divided by a comma cos space beta equals c divided by square root of open parentheses a squared plus b squared close parentheses end root

On the domain [–straight pistraight pi] the equation 4 sin cubed space x plus 2 sin squared space x minus 2 sin space x minus 1 equals 0 possess

maths-General
parallel
General
maths-

cos to the power of 4 invisible function application pi over 8 plus cos to the power of 4 invisible function application fraction numerator 3 pi over denominator 8 end fraction plus cos to the power of 4 invisible function application fraction numerator 5 pi over denominator 8 end fraction plus cos to the power of 4 invisible function application fraction numerator 7 pi over denominator 8 end fraction equals

cos to the power of 4 invisible function application pi over 8 plus cos to the power of 4 invisible function application fraction numerator 3 pi over denominator 8 end fraction plus cos to the power of 4 invisible function application fraction numerator 5 pi over denominator 8 end fraction plus cos to the power of 4 invisible function application fraction numerator 7 pi over denominator 8 end fraction equals

maths-General
General
chemistry-

 The products (A) and (B) are:

 The products (A) and (B) are:

chemistry-General
General
maths-

If w is a complex cube root of unity, then the matrix A = open square brackets table row 1 cell w to the power of 2 end exponent end cell w row cell w to the power of 2 end exponent end cell w 1 row w 1 cell w to the power of 2 end exponent end cell end table close square brackets is a-

If w is a complex cube root of unity, then the matrix A = open square brackets table row 1 cell w to the power of 2 end exponent end cell w row cell w to the power of 2 end exponent end cell w 1 row w 1 cell w to the power of 2 end exponent end cell end table close square brackets is a-

maths-General
parallel
General
maths-

Matrix [1 2] open parentheses open square brackets table row cell negative 2 end cell 5 row 3 2 end table close square brackets open square brackets table row 1 row 2 end table close square brackets close parentheses is equal to-

Matrix [1 2] open parentheses open square brackets table row cell negative 2 end cell 5 row 3 2 end table close square brackets open square brackets table row 1 row 2 end table close square brackets close parentheses is equal to-

maths-General
General
maths-

If A –2B = open square brackets table row 1 5 row 3 7 end table close square brackets and 2A – 3B = open square brackets table row cell negative 2 end cell 5 row 0 7 end table close square brackets, then matrix B is equal to–

If A –2B = open square brackets table row 1 5 row 3 7 end table close square brackets and 2A – 3B = open square brackets table row cell negative 2 end cell 5 row 0 7 end table close square brackets, then matrix B is equal to–

maths-General
General
maths-

The value of x for which the matrix A = open square brackets table row 2 0 7 row 0 1 0 row 1 cell negative 2 end cell 1 end table close square brackets is inverse of B = open square brackets table row cell negative x end cell cell 14 x end cell cell 7 x end cell row 0 1 0 row x cell negative 4 x end cell cell negative 2 x end cell end table close square brackets is

The value of x for which the matrix A = open square brackets table row 2 0 7 row 0 1 0 row 1 cell negative 2 end cell 1 end table close square brackets is inverse of B = open square brackets table row cell negative x end cell cell 14 x end cell cell 7 x end cell row 0 1 0 row x cell negative 4 x end cell cell negative 2 x end cell end table close square brackets is

maths-General
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