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this theorem (lebinitz theorem) also called theorem for successive differentiation.

this rule is used to find nth derivative of multiplication of two functions

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1

Thanks a lot to you

Solution:-Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d)So, according to the question.a-3d + a - d + a + d + a + 3d = 324a = 32a = 32/4a = 8 ......(1)Now, (a - 3d)(a + 3d)/(a - d)(a + d) = 7/1515(a² - 9d²) = 7(a² - d²)15a² - 135d² = 7a² - 7d²15a² - 7a² = 135d² - 7d²8a² = 128d²Putting the value of a = 8 in above we get.8(8)² = 128d²128d² = 512d² = 512/128d² = 4d = 2So, the four consecutive numbers are 8 - (3*2)8 - 6 = 28 - 2 = 68 + 2 = 108 + (3*2)8 + 6 = 14Four consecutive numbers are 2, 6, 10 and 14

Area of the square=2500=>a²=√2500=50 m

∴Diagonal=a√2= 50√2 m

Escape velocity doesn't depend on the mass of the projectile .

For a spherically symmetric, massive body such as a star, or planet, the escape velocity for that body, at a given distance, is calculated by the formula

whereGis the universalgravitational constant(G≈6.67×10−11m3·kg−1·s−2),Mthe mass of the body to be escaped from, andrthe distance from thecenter of massof the body to the object.The relationship is independent of the mass of the object escaping the massive body.

nPr ÷ nCr = 24 n!/(n-r)! ÷ n!/(n-r)!×r! = 24 n!/(n-r)! × (n-r)!r!/n! = 24 r! = 24 = 4! r = 4

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1

the sum of 7 times natural no x what? please provide full context regarding question

If a and b are co prime then. either a or b , have to odd.

so, the square of odd is always odd. and for even . it is even.

therefore adding both squares we will get a²+b² odd+even = odd. option c is correct.

the answer of this question is -3a

1/4a is the correct answer

a^2+1=3a a^2+1÷a^2+1+4a=1÷4a here's the answer

a+1/a=3a^2+1=3aa^2+1/a^2+1+4q1/4a =answer

if a+1/a=3then (a^2+1)/a=3so 3a=a^2+1&a=(a^2+1)/3&a^2=3a-1so 3a/a^2+1+4a=(3a^2+3)/4a^2+9a+4

11655859896988487465875

SecA=x+1/4x∴, sec²A=(x+1/4x)²=x²+2.x.1/4x+1/16x²=x²+1/2+1/16x²Now, sec²A-tan²A=1or, tan²A=sec²A-1or, tan²A=x²+1/2+1/16x²-1or, tan²A=x²+1/16x²-1/2or, tan²A=x²-2.x.1/4x+1/16x²or, tan²A=(x-1/4x)²or, tanA=+-(x-1/4x)∴, either,secA+tanA=x+1/4x+x-1/4x [when tanA=x+1/4x]=2x or, secA+tanA=x+1/4x-x+1/4x [when tanA=-(x+1/4x)]=1/4x+1/4x=2/4x=1/2x (Proved)

n=12

Option a is the correct answer

u r wrong

+,-,÷,× is the answer of the following

The answer 10*15-1=149

Let a = 3x and b = 4x

Substituting a and b, we get

= (3*3x + 4*4x)/(4*3x - 4x)

= 9x+16x/12x-4x

= 25x/8x

= 25/8

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4y²-20y-2y+104y(y-5)-2(y-5) (4y-2) (y-5)

= 22*7*7*10/7*2*2

= 11*7*5

= 385

thanks

l think it is 1,if it is wrong I am sorry

-1 is the right answer.

-1 is the right answer of the following

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Thanks

+ is sign of addition

Right answer is + additiona.

By first principle if f[x]=tanx

F'[x]= f[x+h]-f[x]/x+h-x at Lth tends to 0

F'(x) = tan(x+h)-tanx/h

By simple trigonometry

tan(x+h)=tanx+tanh/1-tanxtanh

Thus,

f'(x) =

tanx+tanh-tanx+tan^2xtanh/(1-tanxtanh) h

=tanh(1+tan^2x)/(1-tanxtanh) h

But Lt (tanx)/x. =1, 1+tan^2x=sec^2x

h~0

Hence f'(x) =sec^2x(1)/(1-tanxtanh)

= sec^2x

What's the question not a proper answer

Answer is correct but plz explain

By first principle if f[x]=tanx

F'[x]= f[x+h]-f[x]/x+h-x at Lth tends to 0

F'(x) = tan(x+h)-tanx/h

By simple trigonometry

tan(x+h)=tanx+tanh/1-tanxtanh

Thus,

f'(x) =

tanx+tanh-tanx+tan^2xtanh/(1-tanxtanh) h

=tanh(1+tan^2x)/(1-tanxtanh) h

But Lt (tanx)/x. =1, 1+tan^2x=sec^2x

h~0

Hence f'(x) =sec^2x(1)/(1-tanxtanh)

= sec^2x

By first principle if f[x]=tanx

F'[x]= f[x+h]-f[x]/x+h-x at Lth tends to 0

F'(x) = tan(x+h)-tanx/h

By simple trigonometry

tan(x+h)=tanx+tanh/1-tanxtanh

Thus,

f'(x) =

tanx+tanh-tanx+tan^2xtanh/(1-tanxtanh) h

=tanh(1+tan^2x)/(1-tanxtanh) h

But Lt (tanx)/x. =1, 1+tan^2x=sec^2x

h~0

Hence f'(x) =sec^2x(1)/(1-tanxtanh)

= sec^2x

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Since x + z = 2y, z = 2y - x and: b^(z - x) = b^(2y - 2x) = (b^2)^(y - x)

But b^2 = ac, so that becomes:

b^(z - x) = a^(y - x) c^(y - x)

a^(y - z) b^(z - x) c^(x - y) = a^(y - z + y - x) c^(x - y + y - x) = a^(2y - z - x) c^0 = a^(2y - z - x)

But x + z = 2y also means 2y - z - x = 0, so the exponent on a is zero too a^(y - z) b^(z - x) c^(x - y) = a^(0) = 1

This, of course, requires a,b,c to be nonzero.