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hey vandana I think b is right answer

18 km/h is its speedinkm/h was the answer

18km per hour is speed km per hour is answer

Ans: option(b) 10 km/hr

Volume of the water required for rising the water level by 4.5 metre= 7 × 4 × 4.5 = 126 m3

Time in which the tank needs to be filled = 6 hours and 18 minutes =6+18/60hr =6+3/10hr =63/10hr

Therefore, amount of water that needs to be supplied by the pipe in 1 hr126(63/10)=126×10/63=20m3/hr

Area of the pipe = 5×4 = 20 cm2 = 0.002 m2

Required speed of the water = 20/0.002

= 10000 m/hr = 10 km/hr

thank you

thanks

it is similar questions like your I hope it helps you

The speed of the stream is 1.5km per hour

70cm =0.7 mthen area =lb=1(0.7)=0.7sqm

cost =₹80(0.7)

525×16780 rupees is the total price of rice purchased

1 quital rice=rs 52516780 quintal rice=525*16780= rs 8809500

8809500 is the current answer for this question

1 quital rice =rs 52516780 quital rice 525*16780=rs 8809500

the total price of the rice is Rs 8809500

8809500.

let the rate be r

Interest= PxRxT/100340= 1700x r x 2/100340= 34rr=340/34r= 10%

formula interest=PXRXT/100; 340=1700×R×2/100= 340=34r; r=340/34=10°\°

formula interest=P×R×T/100;340=1700×R×2/100=R=340/34=10°\°

Option (D) is correct.

Let a = first term of the AP.andLet d = common difference of the AP

Nowa = A+(p-1).d.......(1)b = A+(q-1).d.......(2)c = A+(r-1).d........(3)

Subtracting 2nd from 1st, 3rd from 2nd and 1st from 3rd we get

a-b = (p-q).d......(4)b-c = (q-r).d........(5)c-a = (r-p).d.......(6)

multiply 4,5,6 by c,a,b respectively we have

c.(a-b) = c.(p-q).d......(4)a.(b-c) = a.(q-r).d........(5)b.(c-a) = b.(r-p).d.......(6)

a(q-r).d+b(r-p).d+c(p-q).d = 0(a(q-r)+b(r-p)+c(p-q)).d = 0

Now since d is common difference it should be non zero

Hencea(q-r)+b(r-p)+c(p-q)= 0

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1.When the solutions mix, theleadparticles andiodideparticles combine and create two new compounds, a yellow solid calledlead iodideand a white solid calledpotassium nitrate.2.Pb(NO3)2+2KI => PbI2 +2KNO33.Yes it is a double displacement reaction Like the solution if u find it useful

(i) ans=14 is the square root of 196

i) 14ii)3×4√2iii)3/3√9

i) 14ii)3×3√2iii)1/√3

i) √196=14ii) 3√18= 3×3√2

_/196 is correct answer

1) √196= 142) 3√18= 3×3√23)√9/27 = 1/√3

the answer would be ii

ans 1:4ans 2:3*3root2ans 3:3*3root6

the answer is 1. ✓ 197

i option is the correct answer

14 is your correct answer

√196 = 14 is the correct answer

¡) 14¡¡) 3×4√2¡¡¡)3/3√9

Question no 1.Answer is (i) √196

ans=14 is the square root of 196

option A is a correct answer

The correct ans is option d.

(i) is the correctness answer

1) 142) 3×4✓23) 1/√9

14 is a sqare root of 196.

a is the correct answer

answer is 14 is t

ans =14 is the square root of 196

If V(average) = 36km/h

Then initial speed, is V.

Final Speed is, V+15.

V(average) = 36 = (V + V + 15)/2

72 = 2V + 15

2V = 57

V = 28.5 km/h (initial speed)

V + 15 = 43.5 km/h (final speed)

If a : b = c : d then a, b , c ,d are in

proportion.

a × d = b × d

or

Product of extremes = product of means

Now ,

( 4x² - 3y²) : ( 2x² + 5y² ) = 12 : 19

( 4x² - 3y² ) 19 = ( 2x² + 5y² )12

76x² - 57y² = 24x² + 60y²

76x² - 24x² = 60y² +57y²

52x² = 117y²

x² / y² = 117 / 52

( x/ y )² = ( 13 × 9 ) / ( 13 × 4 )

= 9 / 4

( x / y )² = ( 3 / 2 )²

Therefore ,

x / y = 3 / 2

x : y = 3 : 2

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Let a be any positive integer and b = 6.

Then, by Euclids algorithm,

a = 6q + rfor some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 3q + 1 = 2k1+ 1, where k1is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2+ 1, where k2is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3+ 1, where k3is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3, or 6q + 5.

*From which book have you taken this question? Please tell us so that we can provide you faster answer.*

Sum of first n terms Sn=n/2[2a+(n-1)*d]n/2[2×8+(n-1)×20]=2n/2[2×-30+(2n-1)×8]16+20n-20=2(-60+16n-8)-4+20n=2(-68+16n)-2+10n=-68+16n66=6nn=11

Elder one = xYounger one = yx - y = 10......(1)x -15 = (y - 15)2x - 15= 2y - 30x -2y = -15......(2)equ.(1) - equ.(2)y = 25x = 10+25 x = 35Elder one =x=35yearsYounger one=y=25years

35 years and 25 years is the correct answer of the given question

35 years and 25 year is answer your question

35year and 25 year is your answer

72;£9-iecifudvsosowwuevxj

elder ka 35 yrs.and younger ka 25 yrs.

Step-by-step explanation:

Let Sandhya be x years old

Sandeep is x + 6 years old

5 years ago:

Sandhya = x - 5 years old

Sandeep = x + 6 - 5 = x + 1 years old

Solve x:

Five years ago, sandeep was three times as old as Sandhya

x + 1 = 3(x - 5)

x + 1 = 3x - 15

2x = 16

x = 8

Find their age:

Sandhya = x = 8 years old

Sandeep = x + 6 = 8 + 6 = 14 years old

Answer: Sandhya is 8 years old and Sandeep is 14 years old

thank you