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nth term = 9 - 5n 1st term = 9 - 5*1 = 9 - 5 = 4 2nd term = 9 - 5*2 = 9 - 10 = -1 3rd term = 9 - 5*3 = 9 - 15 = -6 Sum of first three terms = 4 - 1 - 6 = 3 - 6 = -3

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Let in place of blank value = nThen,395*n = 156025n = 156025/395n = 396

Therefore ans is 395

A parallelogram does have zero lines of symmetry.

sin^(-1)(3/5)=tan^(-1)(3/4)sin^(-1)(8/17)=tan^(-1)(8/15)tan^(-1)(3/4)+tan^(-1)(8/15)=tan^(-1)((3/4+8/15)/(1-3/4*8/15))=tan^(-1)((77/60)/(9/15))=tan^(-1)(77/36)=cos^(-1)(36/85)

8√x+x-5 x=88√8+8-516√2+3

After comparison,x + y = 6, ( i )5 + z = 5, ( ii )x * y = 8, ( III )From equation ( ii ),z = 0,From equation ( i ) and ( iii ),( x - y )^2 = ( x + y )^2 - 4 * x * y, = 36 - 32, = 4,( x - y ) = 2, ( iv )From equation ( i ) and ( iv ),x = 4,y = 2.

sin^-1(3/5) + sin^-1(8/17) = sin^-1{3/5√[1 - (8/17)^2] + 8/17√[1 - (3/5)^2}

=sin^-1[ 3/5 *15/17 + 8/17 * 4/5 ]

=sin^-1[ 45/85 + 32/85 ]

= sin^-1[ 77/85 ]

1- p, -p is the answer

2- 2(a,r,b,t) is the your answer

20.x²+xy+9x+9y=x(x+y)+9(x+y)=(x+y)(x+9)

b is the best answer

SI = (P×R×T)/10082.50 = (1650×R×2)/10082.50×100 = 1650×R×28250/1650×2 =RR =2.5A = 1650+82.50 = 1732.50

answer 1732.50 hoga right

P= 1650T= 2 years SI = 82.50R=?1) R=SI×100/P×T= 82.50×100/1650×2=2.5% p.c.p.a.2) Amount = P+SI= 1,650 + 82.50=1,732.50 rs.

thanks

We make infinity lines to only one point

the possibility of no of line in a point is : infinite

100 percentage price= 1900now it was sold to B on profit of 5 means 105 percentageso 19*105= 1995nowhe sent it to 10 percentage of 1995now 115 percentage of 1995= 2294.25

Consider A (0,0),B (1,0),C (0,1)

=>AB=root(X2-X1)^2+(Y2-Y1)^2

=>AB=root (1-0)^2+(1-0)^2

=>AB=1

similarly,

BC=root2

and

AC=1

Perimeter=AB+BC+AC

=1+root1+1=2+root2

Side=aPerimeter=4aP=48mm4a=48mma=48/4a=12mm

Given solution is not correct as given equation is quadratic equation so it should have 2 roots

We solve this equation as follows:2x^2 = 3x2x^2 - 3x = 0x(2x - 3) = 0x = 0, x = 3/2

So given equation have 2 roots 0 and 3/2

Sum of roots is -L/(L-m)=3r, product of roots is 1/(L-m)=2r^2 where r is one of the roots. So 1 / 2(L-m) = L^2 / 9(L-m)^2 9(L-m)=2L^2 2L^2 - 9L + 9m=0 For L to be real, the discriminant is non-negative. 81>=8X9m m<=9/8 The greatest value of m is therefore 9/8.

AB= FLAG POLE BC = TOWER

CD = BASE = 9M

TAN 60 = AC/CD

ROOT3 = AC/9

AC = 9 ROOT3

TAN 30 = BC/CD

1/ROOT3 = BC /9

BC = 9/ROOT3 (ROOT3/ROOT3)

BC = 3 ROOT3

AC = AB +BC

9 ROOT 3 = AB + 3 ROOT3

AB = 9 ROOT 3 - 3 TOOT 3

AB = 6 ROOT 3

Total possibilities = 4 + 3 = 7

Favourable possiblity = 4

So, Probablity = 4/7

Sorry we are currently not taking questions from accounts.

Median = 55.Mode = 56.Range = 63.Mean = (55+56+63)/3= 58

45 = (9)(5)= (3)(3)(5) is the prime factorization of 45 since 3 and 5 are both prime numbers, and therefore, each cannot be "broken down" or factored any further.Three (3) and 5 are prime numbers because each is a positive integer that is exactly divisible (a zero remainder) by exactly two positive integers: itself and 1, i.e., 3/3 = 1 and 3/1 = 3; Also, 5/5 = 1 and 5/1 = 5.45 = 9 x 5= 3×3×5 done. 3 and 5 are prime

7s+4p=28003s+2p=300

7s+4p-6s-4p=2800-600so s=22008800+4p=2800so p=-1500

5 dozens =5*12=60 and 2 scores=2*20=40ratio =60/40=3/2=3:2

Sahi h

Any odd positive integerncan be written in form of 4q+ 1 or 4q+ 3.

Ifn= 4q+ 1, Thenn^2- 1 = (4q+ 1)^2- 1 = 16q^2+ 8q+ 1 - 1 = 8q(2q+ 1)which is divisible by 8.

Ifn= 4q+ 3, Thenn^2- 1 = (4q+ 3)^2- 1 = 16q^2+ 24q+ 9 - 1 = 8(2q^2+ 3q+ 1) which is divisible by 8.

So, it is clear thatn^2- 1 is divisible by 8, if n is an odd positive integer.

{(√3+√2)³}²+{(√3-√2)³}2=2× {(√3)^6+(√2)^6}=2×{27+8}= 70

a²-b² = (a+b)(a-b)

= (√3)² - (√2)²

= 3-2

= 1

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Let a be any positive integer and b = 3

a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴ r = 0,1,2 .

Therefore, every number can be represented as these three forms. There are three cases.

Case 1: When a = 3q,

Where m is an integer such that m =

Case 2: When a = 3q + 1,

a = (3q +1) ³

a = 27q ³+ 27q ² + 9q + 1

a = 9(3q ³ + 3q ² + q) + 1

a = 9m + 1 [ Where m = 3q³ + 3q² + q ) .

Case 3: When a = 3q + 2,

a = (3q +2) ³

a = 27q³ + 54q² + 36q + 8

a = 9(3q³ + 6q² + 4q) + 8

a = 9m + 8

Where m is an integer such that m = (3q³ + 6q² + 4q)

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

Hence, it is proved .