Have a query? Ask a Tutor!!

Access to best educators and exclusive paper discussions

Offsite Campus Turito Championship

Can’t find what you’re looking for?

Our tutors are from top schools around the world, and they are waiting for you 24/7, anytime, anywhere.

Homework- One to One Solutions
Understand concepts to solve better
Get your doubts solved instantly

Maths-

General

Easy

Question

The vectors $stack x with ‾ on top$ and $stack y with ‾ on top$ satisfy the equation $p stack x with rightwards arrow on top plus q stack y with ‾ on top equals stack a with rightwards arrow on top$ (where $p comma q$ are scalar constants and $stack a with rightwards arrow on top$ is a known vector). It is given that $stack x with ‾ on top times stack y with rightwards arrow on top greater or equal than fraction numerator vertical line stack a with rightwards arrow on top vertical line to the power of 2 end exponent over denominator 4 p q end fraction$ , then $∣ x over y ∣$ is equal to $left parenthesis p q greater than 0 right parenthesis$

1
$\frac{p^{2}}{q^{2}}$
$\frac{p}{q}$
$\frac{q}{p}$

The correct answer is: $fraction numerator q over denominator p end fraction$

Related Questions to study

$L t subscript left parenthesis x rightwards arrow 1 right parenthesis l o g subscript e invisible function application left parenthesis 1 plus x cubed right parenthesis divided by s i n cubed invisible function application x$

$L t subscript left parenthesis x rightwards arrow 1 right parenthesis l o g subscript e invisible function application left parenthesis 1 plus x cubed right parenthesis divided by s i n cubed invisible function application x$

$L t subscript left parenthesis x rightwards arrow 1 right parenthesis invisible function application left parenthesis ∛ x minus 1 right parenthesis divided by left parenthesis x minus 1 right parenthesis$

$L t subscript left parenthesis x rightwards arrow 1 right parenthesis invisible function application left parenthesis ∛ x minus 1 right parenthesis divided by left parenthesis x minus 1 right parenthesis$

A: For a reaction at equilibrium, the Gibb’s free energy of reaction is minimum at constant temperature and pressure.

A: For a reaction at equilibrium, the Gibb’s free energy of reaction is minimum at constant temperature and pressure.

Chemistry-General

parallel

$L subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis 3 to the power of t minus 1 right parenthesis divided by left parenthesis √ left parenthesis 1 plus x right parenthesis minus 1 right parenthesis$

$L subscript left parenthesis x rightwards arrow 0 right parenthesis left parenthesis 3 to the power of t minus 1 right parenthesis divided by left parenthesis √ left parenthesis 1 plus x right parenthesis minus 1 right parenthesis$

$L t subscript left parenthesis x rightwards arrow 2 right parenthesis invisible function application left parenthesis left parenthesis 2 x squared minus 7 x minus 4 right parenthesis right parenthesis divided by left parenthesis left parenthesis 2 x minus 1 right parenthesis left parenthesis √ x minus 2 right parenthesis right parenthesis$

$L t subscript left parenthesis x rightwards arrow 2 right parenthesis invisible function application left parenthesis left parenthesis 2 x squared minus 7 x minus 4 right parenthesis right parenthesis divided by left parenthesis left parenthesis 2 x minus 1 right parenthesis left parenthesis √ x minus 2 right parenthesis right parenthesis$

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis invisible function application left parenthesis c o s invisible function application a x minus c o s invisible function application b x right parenthesis divided by x squared$

$L t subscript left parenthesis x rightwards arrow 0 right parenthesis invisible function application left parenthesis c o s invisible function application a x minus c o s invisible function application b x right parenthesis divided by x squared$

parallel

The value of $sin invisible function application open parentheses a over 2 close parentheses open curly brackets 1 half plus sum from k equals 1 to n of cos invisible function application left parenthesis k a right parenthesis close curly brackets$ is equal to

The value of $sin invisible function application open parentheses a over 2 close parentheses open curly brackets 1 half plus sum from k equals 1 to n of cos invisible function application left parenthesis k a right parenthesis close curly brackets$ is equal to

depending upon its numerical value 8. Solubility of a substance which dissolves with a decrease in volume and absorption of heat will be favoured by

depending upon its numerical value 8. Solubility of a substance which dissolves with a decrease in volume and absorption of heat will be favoured by

Chemistry-General

$L t subscript left parenthesis x rightwards arrow a right parenthesis left parenthesis x s i n invisible function application a minus a s i n invisible function application x right parenthesis divided by left parenthesis x minus a right parenthesis$

$L t subscript left parenthesis x rightwards arrow a right parenthesis left parenthesis x s i n invisible function application a minus a s i n invisible function application x right parenthesis divided by left parenthesis x minus a right parenthesis$

parallel

$L t subscript left parenthesis x rightwards arrow a right parenthesis left parenthesis t a n invisible function application left parenthesis x minus a right parenthesis right parenthesis divided by left parenthesis x squared minus a squared right parenthesis left parenthesis a not equal to 0 right parenthesis$

$L t subscript left parenthesis x rightwards arrow a right parenthesis left parenthesis t a n invisible function application left parenthesis x minus a right parenthesis right parenthesis divided by left parenthesis x squared minus a squared right parenthesis left parenthesis a not equal to 0 right parenthesis$

For the reaction $N subscript 2 end subscript O subscript 4 end subscript left parenthesis g right parenthesis 2 N O subscript 2 end subscript left parenthesis g right parenthesis$ , the degree of dissociation at equilibrium is 0.2 at 1 atm pressure. The equilibrium constant Kp will be

For the reaction $N subscript 2 end subscript O subscript 4 end subscript left parenthesis g right parenthesis 2 N O subscript 2 end subscript left parenthesis g right parenthesis$ , the degree of dissociation at equilibrium is 0.2 at 1 atm pressure. The equilibrium constant Kp will be

Chemistry-General

1 mole of A (g) is heated to 3000⁰ C in closed one liter vessel till the following equilibrium is reached A(g) B(g). The equilibrium constant for the reaction at 3000C is 4. What is the concentration of B (in mol. lit^–1at equilibrium?

1 mole of A (g) is heated to 3000⁰ C in closed one liter vessel till the following equilibrium is reached A(g) B(g). The equilibrium constant for the reaction at 3000C is 4. What is the concentration of B (in mol. lit^–1at equilibrium?

Chemistry-General

parallel

$L t subscript left parenthesis y rightwards arrow 0 right parenthesis invisible function application left parenthesis s i n invisible function application left parenthesis a plus b x right parenthesis minus s i n invisible function application left parenthesis a minus b x right parenthesis right parenthesis divided by x$

$L t subscript left parenthesis y rightwards arrow 0 right parenthesis invisible function application left parenthesis s i n invisible function application left parenthesis a plus b x right parenthesis minus s i n invisible function application left parenthesis a minus b x right parenthesis right parenthesis divided by x$

$L t subscript left parenthesis x plus x divided by 2 right parenthesis left parenthesis c o s x right parenthesis divided by left parenthesis x minus pi divided by 2 right parenthesis$

$L t subscript left parenthesis x plus x divided by 2 right parenthesis left parenthesis c o s x right parenthesis divided by left parenthesis x minus pi divided by 2 right parenthesis$

$L t left parenthesis x rightwards arrow 1 right parenthesis left parenthesis l o g subscript e superscript x right parenthesis divided by left parenthesis x minus 1 right parenthesis$

$L t left parenthesis x rightwards arrow 1 right parenthesis left parenthesis l o g subscript e superscript x right parenthesis divided by left parenthesis x minus 1 right parenthesis$

parallel

card img

With Turito Academy.

card img

With Turito Foundation.

card img

Get an Expert Advice From Turito.

Turito Academy

card img

With Turito Academy.

Test Prep

card img

With Turito Foundation.