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Chemistry-

General

Easy

Question

The products (A) , (B) and (C) are:

The correct answer is:

Related Questions to study

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

$Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript plus straight X not stretchy ⟶ with straight H to the power of ⊖ on top Cr to the power of 3 plus end exponent plus straight H subscript 2 straight O plus$ 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

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

parallel

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

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

The products (A) and (B) are:

The products (A) and (B) are:

chemistry-General

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-

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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-

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–

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

parallel

The greatest possible difference between two of the roots if $theta element of$ [0, 2 $straight pi$ ] is

The greatest possible difference between two of the roots if $theta element of$ [0, 2 $straight pi$ ] is

Statement I : $open square brackets table row 5 0 0 row 0 3 0 row 0 0 2 end table close square brackets$ is a diagonal matrix.
Statement II : A square matrix A = (aij) is a diagonal matrix if aij = 0 . $for all straight i times not equal to straight j$

Statement I : $open square brackets table row 5 0 0 row 0 3 0 row 0 0 2 end table close square brackets$ is a diagonal matrix.
Statement II : A square matrix A = (aij) is a diagonal matrix if aij = 0 . $for all straight i times not equal to straight j$

Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity. Denote by tr(A), The sum of diagonal entries of A, Assume that A² = I.
Statement-I :If A $not equal to$ I and A $not equal to$ – I, then det A= – 1
Statement-II : If A $not equal to$ I and A $not equal to$ – I then tr(A) $not equal to$ 0.

Let A be a 2 × 2 matrix with real entries. Let I be the 2 × 2 identity. Denote by tr(A), The sum of diagonal entries of A, Assume that A² = I.
Statement-I :If A $not equal to$ I and A $not equal to$ – I, then det A= – 1
Statement-II : If A $not equal to$ I and A $not equal to$ – I then tr(A) $not equal to$ 0.

parallel

Suppose $A equals open square brackets table row 1 0 row 0 cell negative 1 end cell end table close square brackets$ , $B equals open square brackets table row 2 0 row 0 cell negative 2 end cell end table close square brackets$ let x be a 2×2 matrix such that $X to the power of straight prime$ AX = B
Statement-I : X is non singular & |x| = ±2
Statement-II : X is a singular matrix

Suppose $A equals open square brackets table row 1 0 row 0 cell negative 1 end cell end table close square brackets$ , $B equals open square brackets table row 2 0 row 0 cell negative 2 end cell end table close square brackets$ let x be a 2×2 matrix such that $X to the power of straight prime$ AX = B
Statement-I : X is non singular & |x| = ±2
Statement-II : X is a singular matrix

If $2 tan invisible function application A equals 3 tan invisible function application B comma$ then $fraction numerator sin invisible function application 2 B over denominator 5 minus cos invisible function application 2 B end fraction$ is equal to

If $2 tan invisible function application A equals 3 tan invisible function application B comma$ then $fraction numerator sin invisible function application 2 B over denominator 5 minus cos invisible function application 2 B end fraction$ is equal to

Statement-I : If A & B are two 3×3 matrices such that AB = 0, then A = 0 or B = 0
Statement-II : If A, B & X are three 3×3 matrices such that AX = B, |A| $not equal to$ 0, then X = A^–1B

Statement-I : If A & B are two 3×3 matrices such that AB = 0, then A = 0 or B = 0
Statement-II : If A, B & X are three 3×3 matrices such that AX = B, |A| $not equal to$ 0, then X = A^–1B

parallel

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