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Let number of days required to complete work when 3 more men joined = x

Then,20*18 = 20*3 + 23*x

20*(18 - 3) = 23*x

20*15 = 23*x

x = 300/23 = 13

Therefore, 13 more days require to finish work

Actual cost is = 13 -20%(13) = 13-0.2(13) =13-2.6 = 10.4 Rs/kg

SI=4000*5*4/100=800amount=4000-8003200

area of the pavement = area of bigger rectangle -area of smaller rectangle=(12*18)-(15*9)=216-135 cm²=81 cm²

The relation R is both symmetric and reflexive.

Where is figure??? Please post figure also

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(ii) cd=be

Here,. 14 hours 20 minutes will be -. 20 minutes = 20/60 hours.

therefore, it will be (14 + 20/60) hours= (14+1/3) hours=43/3 hours. - i

5hours 20 minutes will be-

28 minutes = 28/60. hours

= 7/15 hours .

therefore, it will be,

= (5+7/15) hours

=82/15 hours. -ii

subtracting. ii. from i we get,

=. 43/3. - 82/15. hours

=215 - 82. /15. hours

= 133/15 hours= 8.8 hours.

= 82/15 hours. -

10% is answer of your question

it may be 90% bcoz 220-22=198 is the total day present nd the % is= total present ×100 by total day of school that is 198×100÷220=90%

90% attendance of the year

90% is the attendance percentage

90% is your question's answer

computer is a electronic devices which help us to perform complex tasks easily. Now they used in nearly every field. you can use computer for banking application, word processing, railway ticket reservation, etc,.

He was facing South direction.

Since the Sun rises in the East, the shadow must be towards West. West is right of Gopal if he faces South

1/10

x^2-3x-1= 0a= 1 b= -3 c= 1D= b^2-4ac9-4= 5roots will be3+√5/2 and 3-√5/2please like the solution

the measure of right angle is 90 degree

the measure of right angle is 90°

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the method is x=4y-3x=2y+15_________x-x=4y-2y-3-(+15)=2y-182y=18y=9

x=4y-3…........(1)x=2y+15.........(2)(1)=(2)4y-3=2y+15;2y=18y=9then x=4(9)-3=36-3=33

(a+b+c)² = a²+b²+c² + 2(ab+bc+ac)=> (a+b+c)² = 85+2(36) =157so, (a+b+c) = √157

distance walked = 6 km = 6000 mtime = 1 hr = 60 mindistance travelled in 20 min = (6000/60)20= 2000 m = 2 km ANS..

2km ans is a correct answer

Trapezoid looks like this

This region have most similar figure like trapezoid

Stanley speed in kilometres per hours is 12.

solution:-

20min-----4km60(1hr)----x

hence by crossmultiplication X=12km/hr

in 20 min Stanley runs 4kmin 1 min Stanley will run 4÷20in 60 min Stanley will run 4÷20×60 =12kmAND THE SPEED ROUND TO THE NEAREST TENTH IS 10 KM.

We have to find the LCM of 40 cm, 42 cm and 45 cm to get the required minimum distance. Because we are asked the minimum distanceNow,40 = 2 × 2 × 2 × 5,42 = 2 × 3 × 7 45 = 3 × 3 × 5 ∴ LCM (40, 42, 45) = 2 × 3 × 5 × 2 × 2 × 3 × 7 = 30 × 12 × 7 = 210 × 12 = 2520

Minimum distance each should walk 2520 cm. so that, each can cover the same distance in complete steps.

the answer of the following is 432cm

wrong anwer.... check it again🤣🤣

hi

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Let speed of A be Km/h and speed of B be km/h

And we know that distance =80km

80=(a-b) ×8 and 80=(a+b)×(4/3)

a-b=10 and a+b=60

a= 35. , b=25

To repair 1.6 km cost = 24000To repair 1 km cost = 24000/1.6 = 60000/4 = 15000To repair 1 km cost = Rs 15000

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the direction cosines of -2i+j-5k is

DCx = -2/√4+1+15 = -2/√30 DCy = 1/√30DCz = -5/√30

Direction ratios are =

a = 1, b = 1, c = -2

Magnitude of a vector =

√(1²+1²+(-2)²

= √6

The direction cosines are

= (a/|a|, b/|a|, c/|a|)

= (1/√6, 1/√6, -2/√6)

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Given,l=1/√3 ; m=1/√6; n=?we know thatl^2 + m^2 + n^2 =11/3 + 1/6 + n^2 =1n = 1/√2 or n=-1/√2since we need the angle with positive direction of x axiswe take n=1/√2

we know that cos gamma=n=1/√2gamma = π/4