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answer is ( B) option

(+5)+(+7) = +5+7 = +12

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n( n+1)/2 = 55; n^2+ n = 55(2); n^2 + n =110, n^2+ n -110; n^2+11n -10n -110=0; n( n+11)-10( n+11); ( n + 11)( n-10); n = -11/10

x^2 - x - 30 = 0x^2 - 6x + 5x - 30 = 0x(x - 6) + 5(x - 6) = 0(x + 5)(x - 6) = 0x = - 5, 6

Therfore,Zeroes of given equation are - 5 and 6

Now,Sum of zeroes = - coefficient of x/ coefficient of x^2= - (-1) = 1

Product of zeroes = constant/ coefficient of x^2= - 30/1 = - 30

alpha and beta are zeroes of a quadratic polynomial x^2 + px + q,Then,alpha + beta = - palpha*beta = q

(alpha + beta)^2 = alpha^2 + beta^2 + 2alpha*beta

alpha^2 + beta^2 = p^2 - 2q

Therefore,Value of (alpha/beta + 2)(beta/alpha + 2)

= (alpha + 2 beta)(beta + 2 alpha)/alpha*beta

= (5alpha*beta + 2(alpha^2 + beta^2)) /alpha*beta

= 5q + 2(p^2 - 2q)/q

x = sec φ - tan φy = csc φ + cot φ

x = 1/cos φ - sin φ/cos φ= (1 - sin φ)/cos φy = 1/sin φ + cos φ/sin φ= (1 + cos φ)/sin φxy + x - y - 1=(1 - sin φ)/cos φ * (1 + cos φ)/sin φ + (1 - sin φ)/cos φ - (1 + cos φ)/sin φ + 1=[(1 - sin φ)(1 + cos φ)]/(cos φsin φ)+ (sin φ-sin^2 φ- cos φ-cos^2 φ)/(cos φsin φ)+1=(1+cosφ-sin φ-sinφcos φ)/(cos φsin φ) -(sin^2φ+cos^2φ+cos φ-sin φ)/(cos φsin φ)+1=(1+cosφ-sin φ-sinφcos φ - 1- cos φ+ sin φ)/(cos φsin φ) -1=(-sinφcos φ)/(cos φsin φ) +1= -1 + 1= 0

that, mth term=1/n and nth term=1/m.then ,let a and d be the first term and the common difference of the A.P.so a+(m-1)d=1/n...........(1) and a+(n-1)d=1/m...........(2).subtracting equation (1) by (2) we get,md-d-nd+d=1/n-1/m=>d(m-n)=m-n/mn=>d=1/mn. again if we put this value in equation (1) or (2) we get, a=1/mn.then, let A be the mnth term of the APa+(mn-1)d=1/mn+1+(-1/mn)=1 hence proved.

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Area of a square of side 3.4 cm long=(3.4 cm)²=11.56 cm²

Given that, mth term=1/n and nth term=1/m

then, let a and d be the first term and the common difference of the AP

So, a+(m-1)d=1/n...........(1) and a+(n-1)d=1/m...........(2)subtracting equation (1) by (2) we get,md-d-nd+d=1/n-1/m=>d(m-n)=m-n/mn=>d=1/mn

again if we put this value in equation (1) or (2) we get, a=1/mn

then, let A be the mnth term of the APA = a+(mn-1)d =1/mn+1+(-1/mn) =1

hence proved

thanka

Am=1/nAn=1/m

a+(m-1)d=1/n.........(1)a+(n-1)d=1/m.........(2)

eqⁿ(2) - eqⁿ(1)

(n-m)d=1/m - 1/nd=1/mn

put in eqⁿ 1a+(m-1)(1/mn)=1/na+1/n -1/mn=1/na-1/mn=0a=1/mn

Amn=a+(mn-1)dAmn=1/mn +(mn-1)1/mnAmn=1/mn+1-1/mnAmn=1

L.H.S.=(1+1/cos2x)/(sin2x/cos2x) =(1+cos2x)(sin2x) =2cos^2x/2sinxcosx =cosx/sinx =cotx =R.H.S.Hence Proved

please explain the third step, Kadambari

mean = (1+2+3+...+10)/10 = 10(10+1)/(2×10) = 5(11)/10 = 55/10 = 5.5

we know that sum of zeroes = -(a+1) = 2-3 = -1,a+ 1 = 1a = 0product of zeroes = 2×(-3) = bb = -6

Cosec(90°-48)+tan (90°-88°)=Sec 42°+cot 2°

x^2-7x+10=0; x^2-5x-2x+10=0; x(x-5)-2( x-5)=0; ( x-5)(x-2); x=2, 5

X²+7x+10=0

x²+2x+5x+10=0

x(x+2)+5(x+2)=0

(x+2)(x+5) = 0

x+2 = 0 ; x = -2

x+5 = 0 ; x = -5

Relationship between the zeroes and coefficients :-

Sum of zeroes = -2+(-5) = -2-5 = -7/1 = -x coefficient /x² coefficient

Product of zeroes = (-2)(-5) = 10/1 = constant/x² coefficient

Hope it helps

x^2+7x+10=0; x^2+2x+5x+10=0; x( x+2)+5( x+2)=0 ( x+5)( x+2); x=2, 5 ( or) -2, -5

let alpha=p bita= qx^2-7x+10sum of zeros = -b/ap+q= 7product of zeros= c/apq = 10p^2+q^2= (p+q)^2-2pq = (7)^2-2×10 = 49-20= 29

x=2,5

correct answer is 2,5

We know(a+b)² + (a-b)²2 (a² + b²) So,(7m-8n)² + (7m + 8n)²2( (7m)² + (8n)²)2( 49m² + 64n²)98m² + 128n²

For us to get this area, we need to draw the diagram and analyze the shapes that are within the diagram.

In this case, the diagram consists of a rectangle of 150m by 21m and two semicircles of diameter 84m.

Area of rectangle =21×150=3150m2

Area of semicircles=3.142 × 42 ×42=5538.96m2

Total area =5538.96+3150=8688.96m2

wrong answer 10458m2

use appropriate answer

nCr = nC(n-r)

=> r = 8 and n-r = 2 => n = r+2 = 8+2 = 10

now, nC2 = 10C2 = 10*9/2*1 = 45

so thanks

Area = Length××BreadthSo, Breadth = Area ÷ Length=440÷22=20m=440÷22=20m

Now, perimeter = 2(length + breadth)=2(22+20)=2×42=84m

Α and β are the zeros of 3x²-4x +1 polynomial,

first of all we factorise 3x²-4x+13x² -4x + 1

=3x² -3x -x +1

=3x( x -1) -1(x -1)

=(3x -1)(x -1)

hence. (3x -1) and (x -1) are the factors of given polynomial .

so, x = 1/3 and 1 are the zeros of that polynomial.

hence, α = 1/3. and β = 1or α = 1 and β. = 1/3you can choose any one in both

I choose α = 1. and β = 1/3

now,let any unknown. polynomial. whose zeros are α²/β and β²/α

α²/β = (1)²/(1/3) = 3

β²/α = (1/3)²/1 = 1/9

now, equation of unknown polynomial.

x²- ( sum of roots)x + product of roots

= x²- ( α²/β + β²/α)x +(α²/β)(β²/α)

put α²/β = 3 and β²/α = 1/9

= x²- ( 3 +1/9)x + 3 × 1/9

= x² -28x/9 + 3/9

={ 9x² -28x + 3 }1/9

hence, 9x² -28x + 3 is answer

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To factorize :4u^2+8u4u^2+8u =04u(u+2)=0u=-2,0

if the angle between unit and vector a"and bak

25 xsquare + 49 ysquare

74xy2 is the correct answer

25xsqure+49ysqure+70xy

= 5x square+7y square-70xy is the right answer of this question

(a+b)²=a²+2ab+b²: (5x7y)²=(5x)²+2(5x)(7y)+(7y)²=25x²+49y²=25x²+70xy+49y²

not clean photo and not see

1

thank you

Given∠ ABX = ∠PQY so,180° - ∠ ABX = 180° - ∠PQY ∠ ABC = ∠PQR ......(1)In ∆ ABC and ∆ PQR,AB = PQ (given)∠ ABC = ∠PQR ( proved in (1))BC = QR ( given)by SAS rule ∆ABC and ∆PQR ar congurent

thanks