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Given,x = 3 is root of Quadratic eqn. x² - 2kx - 6

Thus,

(3)² - 2k× 3 - 6 = 0

⇒ 9 - 6k - 6 = 0

⇒ 3 - 6k = 0

⇒ 6k = 3

⇒ k = 3/6 = 1/2

1)Please like the solution 👍 ✔️

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thank u sis

(xsinx)/(1+cosx).

f'(x)g(x)-g'(x)f(x) all over g(x)^2. This is your quotient rule.

f'(x)g(x)+g'(x)f(x) is your prodcut rule.

(1+cosx)(sinx+xcosx)-[(xsinx)(-sinx)]

sinx+xcos^2x-[-xsin^2x]

[sinx+xcos^2x+xsin^2x]/(1-cosx)^2

[sinx+xcos^2x+xsin^2x]/(1-2cosx+cos^2x

not

zxx meansz(x^2)where z and x are variables

(3³)⁴ⁿ = (27)⁴ⁿ = (27⁴)ⁿ = (531441)ⁿ

since the last digit here 1 , so every time on putting any power on 1 , we will get 1 as the last digit.

and since the question is ((3)³)⁴ⁿ +1 so, we have to add +1 to the last digit

so, the last digit will become 1+1 = 2.

option a

(3³)⁴ⁿ = (27)⁴ⁿ = (27⁴)ⁿ = (531441)ⁿ

since the last digit here 1 , so every time on putting any power on 1 , we will get 1 as the last digit.

and since the question is ((3)³)⁴ⁿ +1 so, we have to add +1 to the last digit

so, the last digit will become 1+1 = 2.

option a

no ans is (d) but how I don't know

no answer should be 2 only.. check your answer.

alternate method.. assume n = any natural no. say 1

now ((3)³)⁴*¹)+1 = (27)⁴*¹ = (531441)¹+1 = 531441+1 = 531442 ...

laat Digit is 2 only....( my method is correct) you can check by putting different values of n , such as n=2,3 also.. you will get this result always..

remark : In exams if there comes question like these , always suppose some values on your one.. and calculate in your assumption , like i did taking n=1, you will get your answers easily.

This can be solved in a simple way.

We toss dice twice,

So total no. of outcomes :6*6=36

We need even numbers on both dice.

So possible number of arrangements are

{2,4,6}x{2,4,6}

=3x3

=9 possible combination of even numbers.

Therefore required probability is:9/36=1/4.

thank u

Simplifying ydx + x(1 + -3x2y2) * dy = 0 Reorder the terms for easier multiplication: dxy + x * dy(1 + -3x2y2) = 0 Multiply x * dy dxy + dxy(1 + -3x2y2) = 0 dxy + (1 * dxy + -3x2y2 * dxy) = 0 dxy + (1dxy + -3dx3y3) = 0 Combine like terms: dxy + 1dxy = 2dxy 2dxy + -3dx3y3 = 0 Solving 2dxy + -3dx3y3 = 0 Solving for variable 'd'. Move all terms containing d to the left, all other terms to the right. Factor out the Greatest Common Factor (GCF), 'dxy'. dxy(2 + -3x2y2) = 0

Subproblem 1

Set the factor 'dxy' equal to zero and attempt to solve: Simplifying dxy = 0 Solving dxy = 0 Move all terms containing d to the left, all other terms to the right. Simplifying dxy = 0 The solution to this equation could not be determined. This subproblem is being ignored because a solution could not be determined.

Subproblem 2

Set the factor '(2 + -3x2y2)' equal to zero and attempt to solve: Simplifying 2 + -3x2y2 = 0 Solving 2 + -3x2y2 = 0 Move all terms containing d to the left, all other terms to the right. Add '-2' to each side of the equation. 2 + -2 + -3x2y2 = 0 + -2 Combine like terms: 2 + -2 = 0 0 + -3x2y2 = 0 + -2 -3x2y2 = 0 + -2 Combine like terms: 0 + -2 = -2 -3x2y2 = -2 Add '3x2y2' to each side of the equation. -3x2y2 + 3x2y2 = -2 + 3x2y2 Combine like terms: -3x2y2 + 3x2y2 = 0 0 = -2 + 3x2y2 Simplifying 0 = -2 + 3x2y2 The solution to this equation could not be determined. This subproblem is being ignored because a solution could not be determined. The solution to this equation could not be determined.

Triangle ABC is right angled at C.Let BC = a, CA = b, AB = c.

(i) Area ofΔABC = 1/2 × Base × Height = 1/2 × BC × AC = 1/2ab Area ofΔABC = 1/2 × Base × Height = 1/2 × AB × CD = 1/2cp⇒1/2ab= 1/2cp⇒ab=cp Hence proved.

(ii) In right angled triangle ABC,AB^2= BC^2+AC^2c^2=a^2+b^2(ab/ p)^2=a^2+b^2a^2b^2/p^2=a^2+b^2-------- From proof (1)1/p^2= (a^2+b^2) /a^2b^21/p^2= (a^2/a^2b^2+b^2/a^2b^2)1/p^2= (1/b^2+ 1/a^2)1/p^2= (1/a^2+ 1/b^2)

Hence proved.

2sinx + tanx = 02sinx + sinx/cosx = 0(2sinx.cosx + sinx)/cosx = 02sinx.cosx + cosx = 0sinx(2cosx + 1) = 0

So, sinx = 0x = 2npi, pi + 2npi

And if 2cosx + 1 = 0cosx = - 1/2

x = 2pi/3 + 2pin, 4pi/3 + 2pin

Combine all solutionsx = 2pi/3 + 2pin, 4pi/3 + 2pin, 2pin, pi + 2npi

27 will be ur answer because 3×9 =27

24 is the right answer

27 will be right answer because the person is working for 9hours and the earning is Rs 3 per hour. So 3×9=27.

27 is the right answer.

sin 3x = sin(x + 2x)sin(x + 2x) = sinxcos2x + cosxsin2x

Subtituting the values of sin2x and cos2x we get

sin 3x = (sinx)(1−2sin^2x) +(cosx)(2sinxcosx)

sin 3x = 3sinx−4sin^2 3x

sin3x = 3sinx-4sin^3x

Its a 0/0 form so we will diffrentiate itnowd/dx(sin3x)/d/dx(sin5x)

3cos3x/5cos5xnow putting limits

3/5answer

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given, log (x+y)/2 =1/2 (logx+logy)log (x+y)/2=log (xy)^1/2(x+y)/2=(xy)^1/2(x+y)^2 /4=xy(x+y)^2 =4xy(x+y)^2 -4xy=0(x-y)^2=0(x-y)=0x=y

Given, x, y, z are three consecutive numbers

Let y = athen , x = a - 1 , z = a + 1

Now,xz = ( a + 1 ) ( a - 1 ) = a² - 1

log ( 1 + zx )

= log ( 1 + a² - 1)

= log (a²)

= log(y²)

= 2logy

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Note thatsin2x=1−cos2xsin2⁡x=1−cos2⁡x. Then:

∫sinx(1−cos²x)dx

=∫sinx−sinxcos²xdx=

=−cosx−∫sinxcos²xdx=

Letu=cosx⟺du=−sinxdx, so:

=−cosx+∫u²du=

=−cosx+1/3u³+C=1/3cos³x−cosx

find the value integration sin3xdx

41 is corect answee....

x+y=9xy=20so. x=9-y(9-y)y=209y -y^2=20solving eqn, we getx=5 and y=4,or x=4and y=5

value of x is =5 & value of y= 4 Hence 25+16=41.

x=5.y=4.or x=4.y=5 solving

x^2+y^2= 77 is the correct answer this question

x^2 +y^2 =41 is the correct answer

4 is the correct answer

x²+y²= 41

this is the correct answer

let assume x=5. y=4 then 5+4=9. ®5*4=20. ®5*5+4*4= 41answer

x=5,y=4then5+4=95×4=205×5+4×5=41

X=5,y=4

5+4=95✖4=205✖5➕4✖5=41

value of x is 5 or value of y is 4 and answer is 41

put value of x=5&y=4x+y=5+4=9xy=5×4=20 hence,x^2+y^2=5^2+4^2=25+16=41

41 is right answer hai

x 4 y 5corrects naswer

x+y=9.

xy=20

x2+y2 = 81-40=41

x+y=9 squaring on both sides

answer 45 is the right answer

right answer is 16+25=41

question answer is 41

according to the maths it's answer must be 41

🇹🇯🇹🇯🇹🇯🇹🇯🇹🇯🇹🇯📑25 +16=41✓✓✓✓✓✓✓✓✓✓✓✓✓✓✓✓✓

x=5 & y=4

Eaka answer fourty one hai

Given that, x+y=9xy=20(x+y)²=81Again , (X +y) ²=x²+Y²+2xy =81=x²+y²+40 =81-40=x²+y² = 41=x²+y²(x-y) ²=x²+y²-2xy(x-y) ²=41-40=1x-y=1x+y+x-y =9+1=10=2x=10=x=5Again, x+y-(x-y) =9-1 =2y=8 =y=4

×=5 & y=4 is right answer

(x+y=9,xy=20,) based on this (x×x+y×y= 41)

right the correct answer 41

41 is the correct answer .....

41 is the correct answer.

41

If xy=20,Find. x^+y^=?..... 1Then x =20/yPut the value of x =20/y in equation 1=(20/y) ^+( y) ^=400/y^+y^

x^2+y^2= 41 is correct answer

41is right answer 😋😋

Given, x+y=9x. y=20x²+y²=? Now, We know that, (x+y) ²= x²+y²+2xyx²+y² = (x+y)² -2xy = (9)² - 2×20 = 81-40 = 41

41 is the answer for this question

41 is correct answer

sin2x + sinx = 2sinxcosx + sinx= sinx(2cosx+1)

sin2x + sinx =0

Let one side of the triangle be x cm.then the other sides will be (x + 3) cm and (x + 6) cm.Perimeter of the triangle = (x + x + 3 + x + 6) = 3x + 9 cm

But , perimeter = 81 cmso, 3x + 9 = 813x = 72so,x = 24 cm.So, Length of all sides = 24 cm , 27 cm and 30 cm.

i want my question answer

Hi,

Here is the answer to your question.

-41x-125 is the correct answer