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105×106=11130 is the answer

11130 is the answer of this question

Let O be the centre of the circle.

Given that,

OQ = 25cm and PQ = 24 cm

As the radius is perpendicular to the tangent at the point of contact,

Therefore, OP ⊥ PQ

Applying Pythagoras theorem in ΔOPQ, we obtain

OP^2 + PQ^2 = OQ^2

OP^2 + 24^2 = 25^2

OP^2 = 625 − 576

OP^2 = 49

OP = 7

Therefore, the radius of the circle is 7 cm.

Please specify the question

To find the number of poles=50/2=25 poles

thanks for helping me

thanks for helping me

Let three consecutive terms of an AP be ( A - D ) , A and ( A + D ).

According to the question

Sum of three consecutive terms = 48

A - D + A + A + D = 48

3A = 48

A = (48/3) = 16.

First term ( A ) = 16.

Again,

Product of first term and last term = 252

( A - D ) × ( A + D ) = 252

( A)² - ( D)² = 252

(16)² - ( D )² = 252

(D)² = 256 - 252

D² = 4

D = ✓4 = 2

Hence,

Common difference ( D ) = 2.

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bhai maine check kiya tha ki ye aap sahi hai ki nhi

Here given,x+y = 8 cm ....(1) y+z = 12 cm .....(2)x+z = 10 cm .....(3)

KUMAR

Adding (1), (2) and (3), we get 2(x+y+z) = 30x+y+z = 30/2x+y+z = 15 .....(4)(4) - (2), we getx+12 = 15x = 15 - 12x =3(4) - (3), we get10+y = 15y = 15 - 10y = 5(4) - (1), we get 8+z = 15z = 15 - 8z = 7So,AD = 3 cmBE = 5 cmCF = 7 cm Answer.

x+y=8, y+z=12, x+z=10, adding (x+y-8+y+z-12+x+z-10)=(2x+ 2y+ 2z=-30); 2(x+y+z)=30; x+y+z=30/2=15;; we get x+12=15; x=3; y=15-10=5; z=15-8=7

See it will be 5:30 + 4:10 = 9:40 mins.

first term,a=p^2 +1and n=2p+1d=2 (given is the series of consecutive integers)and S=(n/2)(2a+(n-1)d)S=((2p+1)/2)(2 (p^2 +1)+((2p+1)-1)×2)S=((2p+1)/2)(2p^2 +2 +4p)S=((2p+1)/2)(2(p^2 +2p+1))S=(2p+1)(p^2 +2p +1)S=(2p+1)(p+1)^2

wrong answer

how u select Nth term

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- 2 1/9 - 6= - 19/9 - 6= (-19 - 54)/9= (-73/9)= - 8 1/9

(a) is correct option

a is correct because if u do the math the right way then u see what i mean

a is correct because if u do the math the right way then u see what i mean

S = ar + ar^2 + ar^3 + ... + ar^nSr = ar^2 + ar^3 + ar^4 + ... + ar^(n + 1)Sr - S = ar^(n + 1) - arS * (r - 1) = ar * (r^n - 1)S = ar * (1 - r^n) / (1 - r)

S is the sumar is the initial termr is the common ration is the number of terms

S = 2110ar = 160r = 2/3n = ?

Let's see if this will even ever sum to 2110

S = 160 * (1 - (2/3)^inf) / (1 - 2/3)S = 160 * 3 * (1 - 0)S = 480

This will never sum to 2110. Are you sure you copied this correctly

Question you have submitted is incomplete. Please post a complete question

it's a complete question

a/9 +a/3 =36 a+3a/9=364a/9 =36

4a =324 a=324/4 81answer

We applied Euclid Division algorithm on n and 3.a = bq +r on putting a = n and b = 3n = 3q +r , 0<r<3i.e n = 3q -------- (1),n = 3q +1 --------- (2), n = 3q +2 -----------(3)n = 3q is divisible by 3or n +2 = 3q +1+2 = 3q +3 also divisible by 3or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3Hence n, n+2 , n+4 are divisible by 3.

The highest common factor of k,2k,3k,4k,5 k is k

Volume of rectangle block = l x b x h= 6cm x 12cm x 5cm= 3 x 2 x 3 x 2 x 2 x 5= 2³ x 3² x 5.At least one cube can be formed of side 2cm.

Ans : a) 1

A trade man lose a discount of 25% on the MP how much above the CP must he mark his articles to make a profit of 17%

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(a) 1Volume of rectangle blockL×B×H6cm×12cm×5cm3×2×3×2×2×52^3×3^2×5One cube can be formed

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volume of block= l×b×h=6×12×5=3×2×3×2×2×5=2cube×3sqaure×5

possible cube of side 2cm

answer=a)1

option no. 1 is correct

vbsi73+827+sii+882shs+#-88ndn(£(! £8929999

the answer is night be 1

volume of rect angle block=l×b×h6cm× 12cm ×5cm3×2×3×2×2×5

ans:a)1

=l×b×h=6cm×12cm×5cm=3×2×3×2×2×5

ans=a)1

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volume of the rectangle block is 1.

(a) 1 is correct answer

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= l × b × h= 6cm × 12cm × 5cm= 3 × 2 × 3 × 2 × 2 × 5= 1

vol. of rectangular block = l*b*h = 6*12*5=3*2*3*2*2*5=2^3 *3^2 *5at least one cube can formed of side 2 cmso, option a is the right answer

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12cm me know if you have any questions about the

volume of rectangle=l×b× h= 6cm× 12cm×5cm=3×2×3×2×2×5

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volume of rectangle block=6cm × 12cm × 5cm=3 × 2 × 3 × 2 × 2 × 52³ × 3²× 5at least one cube can be formed of side 2cm ans A

a rectangle 6 2 5 cm so it is 1

at least one cube can be formed of side 2cmAnd):a)1

the correct answer is one (1)

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volume of rectangle block is 6cm x 12cm x 5cm3x2x3x2x2x52^3x3^2x5at least one cube can formed of side 2cmso answer is option 1

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volume of rectangle block is =lxbxh ATQ there sides are =6cm x 12cm x 5cmconvert them into fraction 6cm =2x312cm=2x2x35cm=1x5=in cube we have take pair of three numbers =write them into a line 2x2x2x3x3x5we have only one pair of three numbers that is 2so we can sa that only 1 cube can make from that rectangle that is a perfact cube

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At least one cube can be formed

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this question answer is (a) 1

volume of rectangle block. =l×b×h. =6cm×12cm×5cm. 3×2×3×2×2×5. =2cube×3square×5. At least one cube can be formed of side 2cm. Ans:a)1

option a is correct answer ans will be 1

The least possible number 1

1. =. Is the Right options

Volume of rectangle = l×b×h= (6×12×5)cm= 3×2×3×2×2×5= (3)2×(2)3×5So.Ans:- (a)1

(1+1+1)/3=3/3=1, LCM of 3,3,3 is 3

(1/3)+(1/3)+(1/3)=(1+1+1)/3=3/3=1

2k + 1 = a3k + 3 = b5k - 1 = c

To form an AP,a + c = 2b2k + 1 + 5k - 1 = 2( 3k + 3)7k = 6k + 6k = 6

So, k = 6

6-24x+14+35x=5320+11x=5311x=53-2011x=33X=3

Let,k + 9 = a2k - 1 = b2k + 7 = c

To be in AP,

a + c = 2b(k + 9) + (2k + 7) = 2(2k - 1)k + 9 + 2k + 7 = 4k - 23k + 16 = 4k - 23k - 4k = - 2 - 16- k = - 18k = 18

For k = 18, the terms k+9 , 2k - 1 , 2k + 7 are in AP

Hope This helps you!

If a,b,c are in AP then, 2b=a+c

2(2k-1)=k+9+2k+74k-2=3k+16k=18