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(2-√x)^(1/3)=3Cube on both sides

2-√x=27-25=√xSquaring both sides625=x

A+ 13 d = 2(a+7d)

a + 13d= 2a+ 14d

​-a -d = 0

a+ d = 0 = a2 ---- --------> 1a + 5d = -8 -----> 22-1

a +5d = -8a +. d = 0------------------a = 2D = -2

a 20 = 2+ 19*-2 = -36

S20 = 10(2+(-36)) = -340

Squaring both sides3+√a=16√a=13Squaring in both sidesa=169

root(3 + root(a)) = 4

Squaring both sides 3 + root(a) = 4^2

root(a) = 16 - 3

root(a) = 13

Again, Squaring both sidesa = (13)^2 = 169

Therefore,Value of a = 169

169 is the correct answer of the given question

Let a be first term and d be common differenceA/q(8/2)[2a+(8-1)d] = 100⇒2a + 7d =25_____(1)and (19/2)[2a + (19-1)d] = 551⇒2a + 18d = 58____(2)subtracting equation (1) and (2)11d = 33⇒d = 3so a = (25 - 21)/2 = 2so, 2,5,8.11,14.......

a = 9- under root 21. under root = ur.ur.a -1/ur. a(ur.a *ur. a - 1)/ur. a

it becomesa - 1 = 0/ur.a

so,a - 1 = 0a = 1.

the answer is -4

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sir parallel sides are 40 cm and 60 cm

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4(sin⁴30°cos⁴60°)-2/3(sin²60°-cos²45°)+1/2tan²60°

=4{(1/2)⁴(1/2)⁴}-2/3{(√3/2)²-(1/√2)²}+1/2(√3)²

=4(1/16)*(1/16)-2/3(3/4-1/2)+3/2

=1/64-2/3(1/4)+3/2

=1/64+3/2-1/6

=97/64-1/6

=259/192

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if x=79 & y=97,so z=?

y=x^tanxTaking log on both sideshencelogy=tanxlogxDifferentiation on both sides1/y*dy/dx=tanx/X+logx*sec^2xhencedy/dx=x^tanx(tanx/x +logx*sec^2x)

it can verified that in (-252)÷ (-18) - sign will cancel out so only (252)÷ (18) will remain which same other given term so quotient will be same

Thank you

Let a be first term and d be common differenceA/q(8/2)[2a+(8-1)d] = 100⇒2a + 7d =25_____(1)and (19/2)[2a + (19-1)d] = 551⇒2a + 18d = 58____(2)subtracting equation (1) and (2)11d = 33⇒d = 3so a = (25 - 21)/2 = 2so AP = 2,5,8.11,14.......

Let a be first term and d be common difference

Sum of first 8 terms (8/2)[2a+(8-1)d] = 100⇒2a + 7d =25_____(1)

Sum of first 19 termsand (19/2)[2a + (19-1)d] = 551⇒2a + 18d = 58____(2)

subtracting eq (1) and eq (2)11d = 33⇒d = 3

so a = (25 - 21)/2 = 2

Therefore, AP = 2, 5, 8, 11, 14.......

a) 58÷ 13 let x x= 58 × 13 x= 754

58 + 85=143; 143/11=33/33=3;

b=11 answer of this question

AD=AB+CD+BCAD=2AB+8cmAD-8cm=2AB2AB=(12-8)cm2AB=4cmAB=2cm:- CD=2cm

3)8.....................

Let the number be n

According to the given conditionn + 60 = 75n = 75 - 60n = 15

Number is 15

For APnth term = a + (n-1)d

For given APa = p-1, d = (p+3 - p +1) =4

For 3rd term3p - 1= (p-1) + (3-1)43p - p = 2*42p = 8p = 4

Value of p = 4

Let's start arbitrarily with the term a(1).sum of first m terms = m/2 * (a(1) + a(m))= m/2 * [a(1) + a(1) + (m-1)*d]= m/2 * [2*a(1) + (m-1)*d] ...(1)

Sum of next n terms = n/2 * [a(m+1) + a(m+n)]= n/2 * [a(1) + m*d + a(1) + (m+n-1)*d]= n/2 * [2*a(1) + (2m+n-1)*d] ...(2)

Sum of next p terms = p/2 * [a(m+n+1) + a(m+n+p)]= p/2 * [a(1) + (m+n)*d + a(1) + (m+n+p-1)*d]= p/2 * [2*a(1) + (2m+2n+p-1)*d] ...(3)

All of these equal S. So now we need to eliminate S and a(1) and d.

S*(2/n - 2/m) = (m+n)*d ...(4)S*(2/p - 2/n) = (n+p)*d ... (5)

Divide the two to get(m+n)/(n+p) = (1/n - 1/m) / (1/p - 1/n)(m+n) * (1/n -1/p) = (n+p) * (1/m - 1/n)

60km/hr=60/60=1km/minhencein 45mindistance =speed*time=45km

Let radius of three concentric circles are r, 2r, 3r

Area of first region A1= pi*r*r

Area of second region A2= pi*2r*2r - pi*r*r= 3pi*r*r

Area of third region A3= pi*3r*3r - pi*2r*2r= 5pi*r*r

Ratio of area of three regions areA1:A2:A3 = pi*r*r:3pi*r*r:5pi*r*r= 1:3:5

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LHS = sin105° + cos105°

= sin(60 + 45 ) + cos ( 60 + 45 )

= ( sin 60 cos 45 + cos 60 sin 45 )

+ ( cos 60 cos 45 - sin60 din 45)

= { ( √3/2 × 1/√2 ) + ( 1/2 × 1 √2)} + { ( 1/2 × 1√2 ) - ( √3/2 × 1/√2)

= ( √3 /2√2 + 1 /2√2 + 1 2√2 + √3 /2√2 ) = 1/√2

The number multiplied by 5 gives 60 and the number is 12

Let the number be xGiven,5 × x =60x=60/5=12

this number is x 1212 multiply 5 =60