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-16 is the following question answer.

-16 is the correct answer of the given question

-16 is the correct answer of the given question

answer of this question is;= -16.

answer of this question is -16

16 is the right answer

1-3+5-7+9-11+....................upto n terms,

1+5+9+................till n/2 terms

=n/4(2n-2)

=n(n-1)/2 --------- 1

3+7+11+15+......................till n/2 terms

=n/4(2n+4)

=n(n+2)/2 ------- 2

so,1-3+5-7+9-.................................n terms

=n(n-1)/2-n(n+1)/2 -------------------------------------from 1 and 2

=n/2(n-1-n-2)

=-3n/2

so,answer is (-3n/2)

ABC is a triangle right angled at B, and D and E are points of trisection of BC.Let BD = DE = EC= xThen BE = 2x and BC = 3xInΔ ABD,AD² = AB²+BD² AD² = AB²+x²InΔ ABE,AE² = AB² + BE²AE² = AB² + (2x)²AE² = AB² + 4x²InΔ ABC,AC² = AB² + BC² AC² =AB + (3x)² AC² =AB²+ 9x²Now,3AC²+5AD² = 3(AB² + 9x²) + 5(AB² + x²)8AB² + 32x²8(AB² + 4x²)= 8AE²⇒ 8AE² = 3AC² + 5AD²Hence proved.

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Given:ABCD is a parallelogram

ThusAB=CD(Opposite sides of a parallelogram)

1/2AB=1/2CD

AX=CY(X and Y are mid-points of AB and CD respectively)

Since AB is parallel to CD,

AX is parallel to CY.

Now in quadrilateral AXCY,opposite sides AX and CY are equal to and parallel to each other.

Therefore AXCY is a parallelogram.

Ans :- The basis of Euclidean divisionalgorithm isEuclid's divisionlemma. To calculate the Highest Common Factor (HCF) of two positive integers a and bweuseEuclid's division algorithm. HCF is the largest number which exactly divides two or more positive integers.

2x+2y=180°X+Y=180°Y=180°-35°Y=145° is the correct answer of the given question

y= 145 is the right answer

y+x+y+x=180degree=2y+2x=180degree2Y=180degree - 35 degree ×2 = 180degree - 70 degree=110degreeHence 2Y= 110 degree Y=110÷2 =55 degree

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(25)^1/2=5 hence 5^3=125(243)^1/5=3 hence 3^3=27(16)^1/4=2 hence 2^5=32(8)^1/3=2 hence 2^4=16hence125*27/32*16=3375/512

Using Euclid's division algorithm.

210 = (55*3) + 45

55 = (45*1) + 10

45 = (10*4) + 5

10 = (5*2) + 0

So, the remainder is 0 at the last stage, so HCF of 210 and 55 is 5.

∴ 5 = (210*5) + 55y

⇒ 5 = 1050 + 55y

⇒ - 55y = 1050 - 5

⇒ - 55y = 1045

⇒ y = - 1045/55

⇒ y = - 19

Hence, the value of x is 5 and y is - 19

This is true for n = 1, because 3^(4+2) + 5^(2+1) = 854 = 14 * 61.

Inductive Step:Assuming that this is true for n = k,3^(4(k+1) + 2) + 5^(2(k+1) + 1)= 3^(4k+6) + 5^(2k+3)= 3^4 * 3^(4k+2) + 5^2 * 5^(2k+1)= 81 * 3^(4k+2) + 25 * 5^(2k+1)= (25 + 56) * 3^(4k+2) + 25 * 5^(2k+1)= 25 [3^(4k+2) + 5^(2k+1)] + 56 * 3^(4k+2).

By inductive hypothesis, 3^(4k+2) + 5^(2k+1) is divisible by 14.Clearly, 56 * 3^(4k+2) = 14 * 4 * 3^(4k+2) is divisible by 14.

Therefore, 3^(4(k+1) + 2) + 5^(2(k+1) + 1) = 25 [3^(4k+2) + 5^(2k+1)] + 56 * 3^(4k+2)is divisible by 14, completing the inductive step.-------------------2) You meant to write x^n - 1 is divisible by x - 1 for x ≠ 1.

Base Case:This is true for n = 1, since x - 1 is clearly divisible by x - 1.

Inductive Step:Assuming that this is true for n = k,x^(k+1) - 1= (x^(k+1) - x) + (x - 1)= x(x^k - 1) + (x - 1).

By inductive hypothesis, x^k - 1 is divisible by x - 1.Clearly, x - 1 is divisible by x - 1.

So, x^(k+1) - 1 = x(x^k - 1) + (x - 1) is divisible by x -

thank you

AB=12cmandBC=8cmLetrbetheradiusofthehoopthen,OC=OA=rWeknowthatinrectangleJABCAB=JC=12cm&BC=AJ=8cm[oppositesidesofrectangleareequal]Now,OJ=OA−AJ=(r−8)cmInrightangled△OJCOC2=OJ2+JC2⇒r2=(r−8)2+(12)2⇒r2=r2+64−16r+144⇒r2=r2+208−16r⇒16r=208⇒r=20816=13cm

10

p = 500; a = ₹1000; r=5%I =1000 - 500 = ₹500t = 100×500/500×5 = 20 years

ANSWER IS 20 YEAR.. 20 is the correct answer

Let price of one book is xLet price of one pen is y

Then according to the given condition5x + 7y = 79.....(1)7x + 5y = 77......(2)

a1=√2,a2=√8= 2√2a3=√18=3√2a4=√32=4√2_________________________________common difference a2-a1= 2√2-√2=√2so d=√2hence Sn=n/2(2a+(n-1)d)=n/2(2*√2+(n-1)*√2))=n/2(2√2+√2n-√2)=n/2(√2+√2n)=n(n+1)/√2

Cos60° 1/2tan45= 1sin^2 + cos^2 = 1please like the solution 👍 ✔️👍

(x-5y)(x+10y) =x(x+10y)-5y(x+10y)=x^2+10xy-5xy-50y^2=x^2+5xy-50y^2

sukriya

cos x=(cosA-cosB)/(1-cosAcosB)cos x=[1-tan^2(x/2)]/[1+tan^2(x/2)]by componendo-dividendo on LHS andRHS[(-1)/tan^2(x/2)]=[(cosA-cosB+1-cosAcosB)/(cosA-cosB-1+c...factorising numerator and denominator of RHS[(-1)/tan^2(x/2)]=[{(1+cosA)(1-cosB)}/{(cosA-1)(1+cosB)}...[{cos^2(A/2)sin^2(B/2)}/{-sin^2(A/2)co...{-cot^2(a/2)tan^2(B/2)}

28/35 is the correct answer

sin30= 1/2 cos30= √3/2 sin30^2 - cos30^2 =1/4-3/4 =-2/4 =-1/2