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Room A with dimesnion 5m and 4mArea of given room A = 5×4 = 20m²

Tiles needed = Area of room/ area of a tile

500 = 20m² /Area of a tileArea of tile = 20m²/500 = 1m²/25 = 0.04m²

So number of tiles needed for hall with dimesnion 8m wide and 15 m long is= Area of Hall/ area of a tile= (8× 15)m² /(0.04m²)= 25× 8 × 15 = 3000 tiles

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Total length=11mTotal cut length=13/5+33/10=26+33/10=59/10=5.9mLength of remaining rope=11-5.9=5.1m

9/64

total length of rope =5mthen it is cutted into 2parts2/5 and 7/4the remaining length=x2/5+7/4+X=5

2/5+7/4 = 43/205-43/20 =57/20so, 57/20m is the right answer

First rope length = 40m 200cm = 40m + 2m = 42 m1m = 100 cm

Second rope length = 16m 370cm = 16m 3.7m = 19.7m

Total rope length = 42 + 19.7 = 61.7 m

33 × 42 = 1386 22 × 63 = 1386 33 × 42 = 22 × 63 So all 4 are in proportion.

In the given A.P ;a = 21d = 42-21 = 21l = 210n = ?

l = a +(n-1)d210 = 21+(n-1)21n-1 = 189/21n = 9+1n = 10th term

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When two numbers are co - prime then their HCF is 1.

The factors of 207 are:1, 3, 9, 23, 69, 207

The factors of 259 are:1, 7, 37, 259

HCF(259, 207) = 1.So, (3) option is correct

thanks bro for answer today is my test in tution

1

Growth rate = 10.7 - 3.2

= 7.5%

= 64000(1+7.5/100)³

= 7765.8

= 7766

population = 64000+7766

= 71766

2)In tri.ABCABD=ACD-------------(1)In tri. abd,BE is the medianAbe=bed-------------(2)now, ABD=BED=2(BED)-----------(3)ABC=ABD+ACDABC=2(ABD)-----using(1)ABC=2*2(BED)----using(3)ABC=4(BED)so, (BED)=1/4(ABC)

3)we know that diagonal of a parallelogram bisect each otherthere fore AO is equal to OCandDO is equal to OBin parallelogram ABCD AC is the diagonal and O is the median of triangle divide it into two eqaul parts of equal trianglesthis implies in triangle ABC, oc is median there fore ar(AOC)is equal to ar (BOC) (1)

similarly in triangle CBD, OB is median ar(COB) is equal to ar(BOD) (2)

in trianle BAD, OD is medianar(BOD) is equal to ar(AOD) (3)

NOW from 1,2 and 3 we get ar(AOC)=ar(BOC)=ar(BOD)=ar(AOD)hence, prooved

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GIVEN: ABCD is a ||gm, in which Pis a point on AB such that AP : PB = 3 : 2 and Qon CD such that CQ : QD= 4 : 1.TO PROVE:AR=3/7AC

PROOF : Since ABCD is a ||gm, then AB || CD and AD || BC.Since AB || CD and AC is a transversal, so∠PAR=∠QCR(Alternateinteriorangles).........(1)

Since ABCD is a ||gm, then AB = CD and AD = BC , as opposite sides of ||gmare equal.Let AB = CD =xso,AP=3x/5andPB=2x/5[as,AP:PB=3:2]andCQ=4x/5andQD=x/5[as,CQ:QD=4:1]

In△CQRand△APR,∠QCR=∠PAR[using(1)]∠QRC=∠PRA[verticallyoppositeangles]⇒△CQR~△APR[AAsimilarity]

⇒CR/AR=CQ/AP=QR/PR(correspondingsidesofsimilar△'sareproportional)

⇒CR/AR=CQ/AP⇒CR/AR=4x/5/3x/5⇒CR/AR=4/3⇒CR/AR+1=4/3+1⇒CR+AR/AR=7/3⇒AC/AR=7/3⇒7AR=3AC⇒AR=3/7AC

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[-86 + (-42)] + [-181 - (-63)]= (-128) + (-118)= - 246

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Let the speed of car from A is = Xthen car from B = X-8let distance of car which starts from A after 4 hours to point B is = Yand distance of car which starts from B to point A after 4 hrs = Zthen Y+Z+62= 350Y+z = 288kmDistance covered by B after 4 hrs= Z+62distance covered by A after 4 hrs = 350-Zspeed of car from A , X= (350-Z)/4 __(1speed of car from B , X-8 = (Z+62)/4 ___(2put value of X from 1 in eqn 2(350-Z)/4 - 8 = (Z+62)/4350-Z-32 = Z+62Z = 256/2 = 128 kmand X = (350-128)/4 = 55.5 km/hrSpeed of car from B = 55.5 - 8 =47.5 km/hr

Let the speed of car from A is = Xthen car from B = X-8let distance of car which starts from A after 4 hours to point B is = Yand distance of car which starts from B to point A after 4 hrs = Zthen Y+Z+62= 350Y+z = 288kmDistance covered by B after 4 hrs= Z+62distance covered by A after 4 hrs = 350-Zspeed of car from A , X= (350-Z)/4 __(1speed of car from B , X-8 = (Z+62)/4 ___(2put value of X from 1 in eqn 2(350-Z)/4 - 8 = (Z+62)/4350-Z-32 = Z+62Z = 256/2 = 128 kmand X = (350-128)/4 = 55.5 km/hrSpeed of car from B = 55.5 - 8 =47.5 km/hr

Let the speed of car from A is = Xthen car from B = X-8let distance of car which starts from A after 4 hours to point B is = Yand distance of car which starts from B to point A after 4 hrs = Zthen Y+Z+62= 350Y+z = 288kmDistance covered by B after 4 hrs= Z+62distance covered by A after 4 hrs = 350-Zspeed of car from A , X= (350-Z)/4 __(1speed of car from B , X-8 = (Z+62)/4 ___(2put value of X from 1 in eqn 2(350-Z)/4 - 8 = (Z+62)/4350-Z-32 = Z+62Z = 256/2 = 128 kmand X = (350-128)/4 = 55.5 km/hr*******Speed of car from B = 55.5 - 8 =47.5 km/hr******

Extra solution for proofhence car from A covers = 350 -128 =222 kmand car from B covers = 128+62 = 190 km

6 is the best answer for the question

6 is the best answer

Let S be the speed at which each worker builds the wall

Let N be the number of days required by 4 workers to build the wall

We assume S to be the same for all workers

S = 1 work divided by 3 workers then divided by 4 days = 1/12

S = 1 wall divided by 4 workers then divided by N days = 1/4N

1/12 = 1/4N

4N = 12

N = 12/4 = 3

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86×86=7348+ 48 =7396

7348 is a correct answer

86x86-7348+48=7396 is a right answer

Thanks

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30/42cut both by 215/21cut both by 35/7

answer is 30/42=15/21=5/7

(30/42)÷2=(15/21)÷3=5/7

30/42

Simplify both the numbers by 2,we get:-

15/21Again simplify both the numbers by 3,we get:-

5/7(simplest form)

5/7 is right answer for you question

Add 64 to 84008464 is obtained which is square of 92

find the latest number which must be added to 8400 to obtain a perfect square . find this perfect square and it's square root

add 64 to 84008464 is obtained which is square of 92

add the 64 to 8400 is obtained which is square of 92

thank you

1.1110your answer first question

40 men can cut 60 trees in 8 hours1 men can cut 60/40 trees in 8 hours40-8=32 men can cut 60×32/40 trees in 8 hours32 men can cut 60×32/(40×8) trees in 1 hours32 men can cut in 12 hours 60×32×12/(40×8)=72 trees.