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c that is X axis as 4 is.

Y-axis is the answer of yoyr question

12/2+29/4+30/6=72+87+70/12=229/12

229/12 is the correct answer of the given question

229/12 is the answer of the

229/12 is correct answer.

229/12 is the right answer.

229/12 is the correct answer

12/2+29/4+30/6=229/12

229/12 is the right answer

Let A/2= 2B/5=c/3= kA= 2kB= 5k/2c= 3kA:B:C= 2k:5k/2:3k= 4:5:6

Let A/2= 2B/5=c/3= kA= 2kB= 5k/2c= 3kA:B:C= 2k:5k/2:3k= 4:5:6

Option (c) is the right ans

Answer will be 0because we divide two same quantities

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7+42+958

58

according to BODMAS7+42+949+958

6(X+(-4))=60-24x-4=36/6=6x=6+4=10

Given

px^2-14x+8=0let a and b be the roots of the equationb=6asum of roots = -b/aproduct of roots = c/ahere a=p , b=-14 , c=8a+b=14/pab=8/p

a+6a=14/p7a=14/pa=2/p

a*6a=8/p6a^2=8/p3a^2=4/p

3*(2/p)^2=4/p

3*4/p^2=4/p

p=3

because All connected to their vetices

445-4=441572-5= 567699-6= 693now HCF of 441 , 567, 693441)693(1 441 --------- 252)441(1 252 ---------+ 189)252(1 189 ---------- 63)189(4 18963)567(81 504 -------- 63 63 --------+ the greatest number is 63

Apointon the y-axis willalways have anx-coordinateofzero.

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given that, area of circle is 1256cm²πr² =1256put π=3.14 now, r²=1256/3.14r²=400 : r=20now, diagonal of rhombus : 2r : 2(20)=40cmarea of rhombus : 1/2*d1*d21/2*(40)(40)800 cm²hope this helps you

the sum of first 25 terms are 2000

An AP hasd = 6a18 = 110

an = a + (n-1)dWhen n = 18a18 = a + (18-1)6110 = a + 17 × 6a + 102 = 110a = 110 - 102a = 8

Sn = n/2 [ 2a + (n-1)d ]When n = 25S25 = 25/2 [ 2×8 + (25-1)6 ]S25 = 25/2 × 2[ 8 + 24×3 ]S25 = 25 ( 8 + 72 )S25 = 25 × 80S25 = 2000

5 person can save 15 litres of waterthen one person can save = 15/5= 3 litrenow 25 persons can save = 25*3= 75 litres of waterPlease like the solution 👍 ✔️

3is the right answer

3 is the correct answer

3 is the correct answer of the given question

66/7tell me grab cv

3 is the right answer

3 is the right answer...

Common difference of APd = - 6

nth term = a + (n-1)d

Then,a16 = a + 15da12 = a + 11d

a16 - a12 = (a + 15d) - (a + 11d) = 15d - 11d = 4d = 4*(-6) = - 24

look we have No.of boys =12No. of girls=8 Now we choose 1 boy's from the total No. of boy'sSo1.12P1 npr=n!/(n-r)!=12!/11!=12×11!/11!=12Ans

Two points are,

(22,20)&(0,16)

Lets suppose that the coordinates of midpoint of the segment joining (22,20) and (0,16) is, (x,y)

We know that,If coordinates of two points are (x1,y1) and (x2,y2), then the coordinates of midpoint (x,y) joinng the segment of those points are -(x,y) = {(x1+x2)/2 , (y1+y2)/2}

lets go back to the question

(x,y) = {(x1+x2)/2 , (y1+y2)/2}

=> x = (x1+x2)/2=> x = (22+0)/2=> x = 11

&

=> y = (y1+y2)/2=> y = (20+16)/2=> y = 18

So the coordinates of midpoint of segnent joining (22,20) and (0,16) are

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P = 3125T = 3 yearsR1 % = 4% , R2% = 5%, R3% = 6%A = P (1 + R/100)^TA = 3125 ( 1 + 4 / 100) ( 1+ 5/100) (1+6/100)= 3125 × 104/100 × 105/100 × 106/100=3,617.25CI = A - P=3,617.25 - 3125=492.25

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answer is 25 plz like as a best answer

25 is the best answer

(d) is the correct answer

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real root doesn't exist

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We take a point (at^2,2at) on the parabola

the normal is

y-2at=-t(x-at^2)

y-2at=-xt+at^3

on x axis y=0,(x-2a)t=at^3

x=a(2+t^3)

let the locus be (h,k)

now find midpoint get he relation between the points.

( i ) 343( ii ) 1,331( iii ) 4,096( iv ) 12,167

(1) 343(2) 1331(3) 4096(4) 12167

1) 7=3432)11=1,3313)16=4,0964)23=12,167

(1) 343(2) 1,331(3) 4,096(4) 12,167

i) 343ii) 1331iii) 4096iv) 12167

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(1) 343(2) 1334(3) 4096(4) 12167

1- 3432-13313- 40964- 12167

1 3432 133134096 4 12167

(I) 49(ii)121(iii)256(iv)529

1. 3432. 1,3313. 4,0964. 12,167

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(I) 7^3 = 343(ii) 11^3 = 1331(III) 16^3 = 4096(iv) 23^3 = 12167

(1)343(2)1331(3)4069(4)12167

343,1,331,4096,12,167

(1)=343(2)=1,331(3)=4,096(4)=12,167

7 34311 133116 409623 12167

1 - 3432 - 1,3313 - 4,8964 - 12,167

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1) 3432) 13313) 40964) 12167

7^3 = 34311^3 = 133116^3 = 409623^3 = 12167

(¡) 7=343(¡¡) 11=1331(¡¡¡) 16=4096(iv)23=12167

i) 343ii) 1331iii) 4096iv) 12167

(1)343(2)1331(3)4096(4)12167

7- 34311-133116-409623-12167

1)7=3432)11=1,3313)16=4,0964)23=12,167

7=34311=133116=409623=12167

1-3432-13313-0964-12167

1)3432)13313)40964)12167

34313314096,12167

7*7*7=34311*11*11=133116*16*16=409623*23*23=12167

Age of mother=30yearsAge of daughter=4yearsage aftee x years is 30+x and 4+xsum of their ages aftee x years=30+x+4+x==34+2x

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(-1+V-3 -1-V3/2)^17= (1)^17=1

1 is the correct answer of the given question

Thank you!

how many lakh make a crore

We know, 4*4*4=64Cube root of 64 is 4.

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Let present age of son be X years.Age of mother = ( X + 25 ) years.After 8 years age of son= ( X + 8 ) years.After 8 years age of mother = ( X + 25 + 8 ) = ( X + 33 ) years.According to the question,X+ 8 / X + 33 = 4/99 ( X + 8 ) = 4 ( X + 33 )9X + 72 = 4X + 1329X - 4X = 132 - 725X = 60X = 60/5X = 12 years.Present age of son = X = 12 Years.And,Present age of mother = X + 25 = 12 + 25 = 37 years.

this is theTriangular NumberSequence: 1, 3, 6, 10, 15, 21, 28, 36, 45, ... This sequence comes from a pattern of dots that form atriangle: By adding another row of dots and counting all the dots we can.