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(v) 1*corrected

1)4 2)9 3)9 4)9 5)46)0 7)6 8)5 9)6

thanks

a)976812,527439,446162,200 613b)6754827,5728789,4586021,3192469c)5876189,3435913,3387517,3238391d)8531486,7625925,7525925 7325925e)1538775,1537775,1438775,1437775

I think E is correct answer

(a) 976812, 527439, 446162, 200613

(b) 6754827, 5728789, 4586021, 3192469

(c) 5876189, 3435913, 3387517, 3238391

(D) 8531486, 7625925, 7525925, 7325925

(e) 1538775, 1537775, 1438775, 1437775

The numbers having only two prime factors I.e 1 and the number itself are the prime numbers remaining all are composite numbers

1 is neither nor composite.

2 is the only even prime number

There are total of 25 prime numbers between 0 to 100. They are

2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89, 97

13105 and 13951

Let the two numbers be X and y hence X+y=27056and x-y=846add both 2x=27902X=13951hencey=27056-13951=13105

let the two number is= x and yso, x + y = 27056 ......equation 1x - y = 846 ...........equation 2according to the question, x + y + x - y = 27056+8462x = 27902x = 27902/2x = 13951hence, put the value of x in equation 1,x + y = 2705613951 + y = 27056y = 27056 - 13951 y = 13105

please like and when it is helpful please accept as best

x+y=27056 ------(1)x-y=846 ------(2)by adding (1) & (2)2x=27902x=13951 ------(3)put (3) in (1)13951+y=27056y=27056-13951y=13105

ax^3+bx^2+cx+d= 0

by comparison

a= 5 b= 3c= -8d= 2

Draw a line from the vertex to the point where the arcs intersect. Use a straightedge to ensure that the line is accurate. This line bisects the angle. For example, use a straightedge to draw a line connecting points F and A.

Given a quadrilateral ABCD with internal angle bisectors AF, BH, CHand DF of angles A, B, C and D respectively and the points E, F, Gand H form a quadrilateral EFGH.

To prove that EFGH is a cyclic quadrilateral.

∠HEF = ∠AEB [Vertically opposite angles] -------- (1)

Consider triangle AEB,

∠AEB + ½ ∠A + ½ ∠ B = 180°

∠AEB = 180° – ½ (∠A + ∠ B) -------- (2)

From (1) and (2),

∠HEF = 180° – ½ (∠A + ∠ B) --------- (3)

Similarly, ∠HGF = 180° – ½ (∠C + ∠ D) -------- (4)

From 3 and 4,

∠HEF + ∠HGF = 360° – ½ (∠A + ∠B + ∠C + ∠ D)= 360° – ½(360°) = 360° – 180° = 180°So, EFGH is a cyclic quadrilateral since the sum of the oppositeangles of the quadrilateral is 180°.]

8x + 5y = 11 3x - 4y = 10 24x + 15y - 24x + 32y = 33 - 80 47y = -47 y = -1 8x - 5 = 11 8x = 11 + 5 = 16 x = 2

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numeric no.is the correct answer of the given question

magic number is the correct answer

magic number is the correct answer

magic number is the correct answer

Magic number is the right answer.

magic number is the right answer.

Magic number is the your answer.

magic number is the answer

First find prime factors of 18, 81 and 108

18 = 2*3*381 = 3*3*3*3108 = 2*2*3*3*3

HCF is Highest common factor so HCF for given numbers = 9

LCM Least Common MultipleSo LCM = 2*2*3*3*3*3 = 324

First find prime factors of 12, 15 and 30

12 = 2*2*315 = 3*530 = 2*3*5

HCF is highest common factorSo HCF for given numbers is 3

LCM is least common multipleSo LCM = 2*2*3*5 = 60

a^3+ab+c is the correct answer of the given question

let 0.36 =xtherefore,10x=3.6---(i)100x=36.6-----(ii)subtracting (i) from (ii);

100x-10x=36.6-3.6therefore;90x=33x=33/90

0.36= 36/100 = 9/25

p/q form of 0.36= 9/25

36/100.9/25.by cancelling 4 table.

0,36=36/100=9/25p/q from of 0,36=9/25

x/4=36/9x/4=4x= 16

Let food last for more n daysNow total students =125 + 15 = 140

As per the rule of inverse variation 125*21 = 140*nn = 125*3/20 = 375/20 = 18.75

Food last for 18.75 days

write properly in simple

4x+3y-11=0; 9x+4y=0; 4(4x+3y-11); 3(9x+4y=0); 16x+12y+44=0; 27x+12y==0; /11x=44; x=44/11=4; 4(4)+3y-11=3y=11-16=5; y=5/3

fast plz

Answer:3).y=3Explanation:

Let the required ratio be k:1Then, 2 = 6k-4×1/k+1 ⇒ k = 3/2.Therefore the required ratio is 3/2: 1 i.e. 3:2

Also, y = 3×3+2×3/3+2 = 3.

tq

tq

x + 2 = 3 x = 3 - 2 = 1

one............. is the answer

1/3+4/3 = 1+4/3 = 5/3 = 1.66

1/3+4/3 = 1+4/3 =5/3

5/3 or 1.66 both are right

1.66 is a correct answer

1/3+4/3=1+4/3=5/3=1.66

1/3 + 4/3 = 5/3 = 1.66

5/3 or 1.66 he is right

1/3 + 4/3 = 1+4/3 = 5/3 = 1.66

4/4 is the answer of this

solution = 1/3+4/3 = 3 L. C. M. lene par =5/3 Hota hai

1/3+4/3=5/3=1.66

v

1.66666666666666666667

The answer is :- 1.66

first LCM to denominator and and add above and finally ans .is 5/3

1/3+4/3=1+4/3=5/3=1.6666.......

1/3 + 4/3 =1+4/3=5/3=0.16

Suppose, 1/3 + 4/3 = 5/3Then, 5/3 = 1.66

1/3+4/3 = 5/3 = 1.667

5/3 is the correct answer is this question

3+12=15÷9 =1.666666666

1/3 + 4/3 = 1+4/3 =5/3

1÷3 + 4÷3= 5÷3 answer is 5 by 3

the answer will be1/3+4/3=(1+4)/3=5/3

1\3+4\3=1+4\3=5\3=5\3

however it still uses the fact that (tgx)' = (secx)^2.

I = ∫(secx)^2 * dxlet u = tanx∴du = (secx)^2 * dxI = ∫du=u + c=tanx + c

hope it helps to you 😄😄😄😄😄😄😄😄😄😄😄

1/6 is the correct answer of the given question

8/15 is the correct answer of the given question

x = 2 y = 1 solutions so k = 2x + 3y = 2*2 + 3*1 = 4 + 3 = 7 2x + 3y = 7 two more solution x = 1/2 y = 2 x = -1 y = 3

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wrong answer

I=integ.of 1/(1-cosa.cosx). dx.

I=integ.of 1/[1-cosa(1-tan^2x/2)/(1+tan^2x/2)].dx

I=integ.of (1+tan^2x/2)/[1+tan^2x/2-cosa+cosa.tan^2x/2].dx

I=integ.of sec^2x/2.dx/[(1+cosa).tan^2x/2+(1-cosa)]

Let tan x/2=p

1/2.sec^2x/2.dx=dp. or. sec^2x/2.dx=2.dp

I=integ.of 2.dp/[(1-cosa)+(1+cosa).p^2]

I=integ.of 2.dp/[2.sin^a/2+2.cos^2a/2..p^2]

I=integ.of. dp/[sin^2a/2+ cos^2a/2.p^2]

I=integ.of. cosec^2a/2.dp/[1 + ( p.cot a/2)^2]

I=cosec^a/2.(1/cot a/2).tan^-1(p.cot a/2). + C

I=(1/sin^2a/2).(sin a/2/cos a/2).tan^-1(cot a/2.tan x/2). + C.

I=2.cosec a. tan^-1(cot a/2.tan x/2). + C.

well tried but it is incorrect

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(y^2+3/2)(y^2-3/2)=y^2(y^2-3/2)+3/2(y^2-3/2)=y^4-3/2y^2+3/2y^2-9/4=y^4-9/4

thanks

(4^-1)(1/4)=(1/4)(1/4)=1/16