The work done by a force acting on a body is as shown in the graph. what is the total work done in covering an initial distance of 15 m?

If A and G are A.M and G.M between two positive numbers a and b are connected by the relation A+G=a-b then the numbers are in the ratio

We have to be careful while solving quadratic equation questions as there are chances of mistakes with the signs while finding the roots. Sometimes the students try to use factorisation methods to solve the quadratic equation, but in this type of questions, it is not recommended at all. Always try to use the quadratic formula for solving. You can also remember the statement to be proved, given in the question as a property for two positive numbers.

If A and G are A.M and G.M between two positive numbers a and b are connected by the relation A+G=a-b then the numbers are in the ratio

If $x equals sum subscript n equals 0 end subscript superscript infinity end superscript a to the power of n end exponent comma y equals sum subscript n equals 0 end subscript superscript infinity end superscript b to the power of n end exponent comma z equals sum subscript n equals 0 end subscript superscript infinity end superscript left parenthesis a b right parenthesis to the power of n end exponent$ where a<1, b<1 then

If l, m, n are the $p to the power of text th end text end exponent comma q to the power of text th end text end exponent comma r to the power of text th end text end exponent$ terms of a G.P which are +ve, then $open vertical bar table attributes columnalign left end attributes row cell l o g invisible function application l end cell p 1 row cell l o g invisible function application m end cell q 1 row cell l o g invisible function application n end cell r 1 end table close vertical bar equals$

The relationship between force and position is shown in the figure given (in one dimensional case) calculate the work done by the force in displacing a body from x=0 cm to x=5 cm

Fifth term of G.P is 2 The product of its first nine terms is

Here note that the fifth term is having fourth power of 2 and not fifth power. We need not to find all nine terms separately; only finding the product is enough because that product will then be written in the form of the term that is known. Terms in a G.P. are having a common ratio in between. That’s why the power of r is increasing as the terms are increasing.

Fifth term of G.P is 2 The product of its first nine terms is

If a, b, c are in A.P then $a open parentheses fraction numerator 1 over denominator b end fraction plus fraction numerator 1 over denominator c end fraction close parentheses$ , $b open parentheses 1 over c plus 1 over a close parentheses comma c open parentheses 1 over a plus 1 over b close parentheses$ are in

biology

The ultimate source of organic evolution is

biologyGeneral

If $a subscript 1 end subscript comma a subscript 2 end subscript comma a subscript 3 end subscript horizontal ellipsis horizontal ellipsis horizontal ellipsis a subscript n end subscript$ are in A.P with common difference $d$ then sind $open square brackets c o s e c invisible function application a subscript 1 end subscript c o s e c invisible function application a subscript 2 end subscript plus close$ $c o s e c invisible function application a subscript 2 end subscript c o s e c invisible function application a subscript 3 end subscript plus c o s e c invisible function application a subscript 3 end subscript c o s e c invisible function application a subscript 4 end subscript plus horizontal ellipsis horizontal ellipsis n$ terms] $equals$

The sum of all integers between 50 and 500 which are divisible by 7 is

$left parenthesis 666 horizontal ellipsis text n digits end text right parenthesis squared plus left parenthesis 888 horizontal ellipsis n$ digits )=

maths-

In a $increment blank P Q R$ as shown in figure given that $x colon y colon z equals 2 colon 3 colon 6$ , then the value of $angle Q P R$ is

maths-General

A bob of mass M is suspended by a massless string of length L. The horizontal velocity $v$ at position A is just sufficient to make it reach the point B. The angle $theta$ at which the speed of the bob is half of that at A, satisfies

A particle of mass $m$ attracted with a string of length $l$ is just revolving on the vertical circle without slacking of the string. If $v subscript A end subscript comma v subscript B end subscript$ and $v subscript D end subscript$ are speed at position $A comma B$ and $D$ then

A bob of mass $M$ is suspended by a massless string of length $L$ . The horizontal velocity $V$ at position $A$ is just sufficient to make it reach the point $B$ . The angle $theta$ at which the speed of the bob is half of that at $A$ , satisfies