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(AUB) ={1, 2,3,4}.N(AnB) ={ }.

n(P(A)) = { (1),(2), (3) (4) ( 1,2) ( 1,3 ) ( 1,4) ,{} }

Given,Radius = 3.5 cm

Perimeter of quarter of circle= 1/4*pi*r + 2r= 1/4*22/7*3.5 + 2*3.5= 11/2*.5 + 7= 11/4 + 7= (11 + 28)/4= 39/4 = 9.7

need all steps

Actual cost is = 13 -20%(13) = 13-0.2(13) =13-2.6 = 10.4 Rs/kg

27 days he stay in shimla.

3 days because 29 we cannot count because he has left to shimla and he come back on to Delhi on25th June we cannot come

Mr Gupta stay in may =3 days In June=24 days

Total days he stay in Shimla =3+24= 27 days.

So, correct Ans is = 27 days

27 days is the answer of the following

27 days is the correct answer of the given question

27 days is the right answer

27 days is the best answer

27 days is the correct answer of the following question

the correct answer is 27 days

29th may to 31st may = 3 days1st june to 24th june = 24 days= 3 days + 24 days = 27 daysso Mr Gupta stay 27 days in shimla.

27 days is the right answer.

27 days he stayed in shimla

Mr. Gupta stay in may=3 days, in june=24 days, total days in shimla=3+24=27 days.

27is correct answer

27 is correct answer

27 daysis correct answer

The diameter is equal to the shortest side of the rectangle.

So radius= 14/2 = 7cm.

Therefore,

Area of circle = πr2 = (22/7) * 49

= 154 cm2

1. -cos2x/2 2. sin3x/33. e^2x/24. (ax+b)³/3a5. -cos2x/2 -4e^3x/3

As the shadow increase by 10 mlet the total distance when the sun elevation was 45 is xwhereas when the sun elevation was 30 the distance increased by 10 m so x + 10

let h be the height of towerwhen the sun elevationtan 45 = h/x1 = h/xx = h .............(i)when the sun elevation was 30tan 30 = h / (x +10)1/√3 = h / (x + 10)(x + 10) /√3 = h..................(ii)

from (i) and (ii) we getx =(x + 10) /√3√3x = x + 10√3 x - x = 10x(√3 - 1) = 10x = 10 / (√3 - 1)x = 10 / (√3 - 1) * (√3 +1)/ (√3 +1) ................ rationalising the denominatorx = 10 ( √3 +1)/ (√3)² - (1)²x = 10 ( √3 +1) /3 -1x = 10 ( √3 +1) /2x = 5 ( √3 +1)from (i) we getx = h5 ( √3 +1) m = height of tower

thanks

we know that, when x→0 , sinx/x = 1

so, sin²ax/x² = (sinax/x) *(sinax/x) = a*a = a²

so, make it continuous a² = 1

=> a = ±1

thank you very much

it will be 0

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At x = 0, the value of the given function takes the form 0/0So we will diffrentiate itAcosax/bcosbxNow after putting limit It will be a/bPlease like the solution 👍 ✔️

yar dekh woh function yadh kar sinax/ax=1.then multiply ax/ax in numeretor and also bx/bx in denominator it will become in the form ax/bx cancel the x and result becomes a/b

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Thank you.

1/8!+1/(9*8!)=x/(10*9*8!)(9+1)/(9*8!)=x/(10*9*8!)10=x/10so x=100

1930's: N. Bourbaki used Z todenotetheset of integersand possibly popularized its usage. The letter stands for german word Zahlen which means numbers.

Z is the right answer

z is the right because integration is adding

option A ) N is the correct answer

z is the correct answer

z is correct ans.......it is option A

set of integers is denoted by (Z)

the answer could be option c

option (z) is the right answer

c) z is the right answer of the following

integers are denoted by z

How did you get 26-x

Ap=1/3acdp=1/2bpis the right answer......

In Triangle DAC ,

AD = AC ( GIVEN )

so, angle ADC = angle ACD (ANGLES OPPOSITE TO EQUAL SIDES ARE EQUAL )

now, angle ADC is an exterior angle for triangle ABD

so, angle ADC > angle ABD

or , angle ACD > angle ABD

or, angle ACB > angle ABC

so, AB > AC ( SIDE OPPOSITE TO THE LARGER ANGLE IS LONGEST )

therefor AB > AD (AD =AC)