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yaar very simple

slope at (4,3)is dy/dx=-4/3.

the tangent will meet the axis at x=25/4.

length of tangent is 3root17/4.

slope of normal is 3/4,normal will meet axis at x=0.

length of normal is 5.

area is 1/2(3root17/4)(5)=15root17/8.

yaar very simple take 3^9(1+3^3+3^6+3^{n-9))=x^3
3^9(1+27+729+3^{n-9})=x^3
3^9is a cube 1+27+729+3^{n-9}) should be a perfect cube
757+3^{n-9})=x^3(x=10,11,12,13,14,................)
for x=10 n=14 the entire equatin becomes a perfect cube
for any other x there is no n such thatl equation is a perfect cube

-------------------------------soory wrong answer-----------------------------------------------------------

yaaaaaaar iam very weak in integration can give all integration equation plzzzzzzzzzzzzzzzzzzz rates assured for full answer.

Yaar its very simple a metal is one which prefers to loose electron.Since loosing of 7 electrons is absurd (or) it requires a lot of energy. but its very easy to accept 1electron

flourine is more electronegetive than oxygen hence the partial charge formed on the atoms will be more and so, more polarisation so more stronger hydrogen bonding

$tan^{-1}(\frac{1}{\sqrt{x^2-1}})=2tan^{-1}(\frac{1}{7})+tan^{-1}(\frac{4}{3})$

$tan^{-1}(\frac{2/7}{48/49})+tan^{-1}(4/3)$

$=tan^{-1}(\frac{7/24+4/3}{1-(7/24)(4/3)}$

$=tan^{-1}(117/44)$

$\frac{1}{\sqrt{x^2-1}}=117/44$

simplfy, x=125/117

In simple way take S=n+(n-1)x+(n-2)x^2+.........................+x^n --------------------------------(1)

Sx=nx+(n-1)x^2+(n-2)x^3+.......................+x^(n+1)-------------------------------------(2)

subtract (2)-(1)

S(x-1)={x^(n+1)+x^n+............................+x+1}+(n-1)

$S(x-1)=\frac{x^{n+1}-1}{x-1}+n-1=\frac{x^{n+1}-1+(x-1)n-x+1}{x-1}$

$S=\frac{x^{n-1}+(n-1)x-n}{{(x-1)}^2}$

For a triangle of given perimeter the area is maximum for a equilateral triangle.

In the triangle perimeter is (c+5+3)=c+8

For the equilateral triangle side is (c+8)/3

so, area of given triangle is $<\frac{\sqrt{3}}{4}a^2$ $\Rightarrow$ $\frac{\sqrt{3}}{4}(\frac{c+8}{3})^2=\frac{c^2+16c+64}{12\sqrt{3}}$

2) For a triangle of given perimeter the area is maximum for a equilateral triangle.

In the triangle perimeter is (c+5+3)=c+8

For the equilateral triangle side is (c+8)/3

so, area of given triangle is $<\frac{\sqrt{3}}{4}a^2$ $\Rightarrow$ $\frac{\sqrt{3}}{4}(\frac{c+8}{3})^2=\frac{c^2+16c+64}{12\sqrt{3}}$

But the equality can never be holded as 2 sides are givenas 3,5

I think 2nd one is wrong equality doesnot hold.

Its simple in translatory motion all the particles of the body has velocity vector parallel or all the particles move in same direction.
While in rotatory motion all particles have velocity tangential to the body.

I think 3hydrogen bonds 1is intermolecular hydrogen bonding & other 2 are covalent bonds

Inorganic O.P.Tandon verrrrry good book

the equation can also be writen as

$\frac{3}{{log}_{4}x}+\frac{2}{{log}_{4}4x}+\frac{3}{{log}_{4}16x}=0$

Take ${log}_{4}x$ =y,the equation reduces to

$3(\frac{1}{y}+\frac{1}{2+y})+\frac{2}{1+y}$ =0

Now simplify the the equation.

you will get y=-1/2,-3/2

so, x=1/2 or 1/8

cheersssss.............