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mahesh
Questions (11)
A freight train leaves a station traveling at 32 km/h. Two hours later, a passenger train leaves
the same station traveling in
2 answers
1,798 views
a train of length 100m runs at a speed of 120km/hr from north to south. another train of length 150m travels with a speed of 80
2 answers
2,772 views
the value of K+L/K-L , if K/L = 7 ?
1 answer
361 views
which of the following fraction is greater than 3/4 and less than 5/6. 1/2 , 2/3 ,4/5 , 9/10.
12 answers
3,510 views
Find the orthogonal trajectories of family of the circle x2+y2=a2 where a is a parameter
1 answer
1,880 views
Maths
If one third of the bucket can be filled in 3 min,how many min it take to fill whole bucket
1 answer
509 views
A man spend 1/3 of his income on food, of the rest 1/4 on bills and 1/5 on miscellaneous things. Finally he left with Rs.1760.
1 answer
1,833 views
A 40 kg slab rests on a friction less floor.a 10 kg block rests on top of the slab.the coefficient kinetic friction between the
6 answers
6,967 views
Prove that max & min values of asinx+bcosx are = +/- (a^2+b^2)^1/2
1 answer
2,336 views
Find max and min values of y=sinx sin2x
1 answer
623 views
. A boat is travelling on a bearing of 25° east of north at a speed of 5 knots (a
knot is 1.852 km/h). After travelling for 3
2 answers
1,320 views
Answers (37)
1. B. 3/7 = 3*6/7*6 = 18/42. 2. D.18/30 = 6/10 = 6*10/10*10 =60/100
Sorry. I wanted it to be (1-6x)^3-3x6x(1-6x)
1hr10min=1+10/60=1+1/6=7/6hr Distance=48x7/6=56miles
Distance covered in 4 hrs=4y Return distance=2x4y=8y and speed=5y Hence time of return trip=8y/5y=1.6hr Halt=1hr Total time=4+1+1.6=6.6hrs
Relative velocity=45+55=100mph Time taken to cover the total distance=150/100=1.5 hr i.e. 1hr 30min Meeting time is 2:15+1:30=3:45pm
Total money=x. Spent in I store=(x/2)+20 Remaining amt=x-[(x/2)+20]=(x/2)-20 Spent in II store=1/2[(x/2)-20]+20 =(x/4)+10 Total expenditure is x x=(x/2)+20+(x/4)+10 Or x=(3x/4)+30 Or x-3x/4=30 x/4=30 or x=120. Check: Spent in I store =60+20=80.
Please read second expression as 3(8x^2+14x+5)
24x^2+42x+15=3(8x^2+14+5) =3(8x^2+10x+4x+5) 3[2x(4x+5)+1(4x+5)] =3(4x+5)(2x+1)
(300)^1/2 /6^1/2=(300/6)^1/2 =50^1/2=7.07
(-16)^-2/3=- 1/(16)^2/3=-1/(16^2)^1/3 =-1/256^1/3=-1/6.35=-0.157
In 1.75 hrs it fills 100% In 1 hr it will fill 100/1.75=57.14%
Time for going upwqards=7/2=3.5 hrs Time for coming back=7/3.5=2 hrs Total 3.5+2=5.5 i.e. 5 1/2 hrs
-216x^3+1=1^3-(6x)^3 =(1-x)^3-3x(1-x)
Force is inversely proportional to square of distance between bodies, hence F/3.75x10^-5=18^2/52^2 F=(18^2x3.75x10^-5)/52^2 =0.46x10^-5 dynes
PS: s=11.47 which has appeared as 121.47 at one place in 4th line.
v=u-gt or 0=15-9.81t since final velocity is 0 and ball is going against gravity. Thus t=1.53 sec. s=ut-.5gt^2=15-.5x9.81x(1.53)^2 s=22.95-11.48=121.47 which is the height the ball will go. From there it falls freely for a height of 11.47+1=12.47 ft.
No. of $5=bills a, $10 bill=b and $20 bill=c Then a+b+c=35 5a+10b+20c=370, and b+c=a+7. Substituting value of b+c in first eq., a+(a+7)=35 or 2a=28 or a=14 and b+c=35-14=21. Substituting value of a in second eq., 14x5+10b+20c=370 or 10b+20c=370-70 Or
If house costs x and lot costs y then x+y=86000, and x=6y+2000 or x-6y=2000. Subtracting second eq. from first we get 6y+y=86000-2000 or 7y=84000 or y=12000. Thus lot costs $12000 House costs 86000-12000=$74000.
If cos^-1x=A then cosA=x and if (cosx)^-1=A then A=1/cosx
PS: There is slight typo. Please readt first line as 'Kath is' and third line as 2x+10.
If the number is x, then 3x-8 is=or greater than 2x. Agnus has x bones and Ignatz has x+5. Then x+x+5=27 or 2x=27-5=22 Or x=11. Ignatz has 11+5=16. Kathy mis x and John is x+10 Then x+x+10 is greater than 60 Or 2x+10is greater than 60 Or 2x 10 is greater
9.95x+11.5y=305.05 x+y=29 9.95x+9.95y=29*9.95 Or (11.50-9.95)y=305.05-288.55 1.55y=16.5 y=16.5/1.55=10.6, say 11 yellow shirts 29-11=18 white shirts 11*11.5+18*9.95=305.6 OK.
Lemon juice=x Oil=2x x+2x=3x=36
If one train has a velocity of V km/h, the other one will have velocity V+5 km/h approaching the first train. Therefore Relative velocity will be V+V+5=2V+5 km/h. Distance travelled by them in 2 hrs is 340-30=310km which is =2(2V+5) Thus 4V+10=310 or
v=u+ft or 30=10+2t t=(30-20)/2=10 secs.
Let it be x yrs ago, when Lisa was 15-x and her father was 40-x. Then- 40-x=6(15-x) or 40-x=90-6x Or 6x-x=90-40 or 5x=50 thus x=5. Let Bob be x yrs and Joel be 3x. Then (x+7)+(3x+7)=58 or 4x+14=85 Or 4x=44 hence x=11. Thus Bob is 11 and Joel is 33. After 7
Yes please. 10 cubed is 1000 and cuberoot of q^36=q^(36/3)=q^12
Let present age of Kristen be x and of Daniel be y.Then after 1 year, x+1=4(y+1)or x-4y=3.....Eq.1 After 10 yrs from then, ages will be (x+1)+10=2[(y+1)+10] Or x+11=2y+22 Or x-2y=11 .....Eq.2. Subtract eq. 1 from eq.2 and solve. We get y=4 and x=19. Check:
Let the integers be x and x/4.Then x/4+12=x-3 or x+48=4x-12 by multiplying both sides by 4. Or -3x=-20-48 or x=20 which is the larger integer. Smaller one is then 20/4=5. Check:5+12=17=20-3 OK.
v=u+at or 220/4=200/2+a*1 55=50+a or a=5m/s^2 Again, v=55+5*1=60m/sec at the start of 7th second.
Please see your third line: It should be 4y-3y=20-5 or y=15, by subtracting first equation from second. Now x+3*15=5 Or x=5-45=-40. Check:-40+4*15=20 by putting values in second equation. This -40=15-60=-40 OK.
Components of a force P due North-West acting at an angle A from North will be PcosA and PsinA in North and West directions respectively and should be equal in magnitude to the force acting on the object. Thus PcosA=83.4 and PsinA=47.2. Therefore
Heat released by 0.5 kg silver=heat gained by 1 kg water upto final temp T. Or 0.5(2.4x100)(150-T)=1(4.18x1000)(T-5) Or 1.2(150-T)=41.8(T-5) Or 180-1.2T=41.8T-209 Or 43T=389 Or T=9.05 deg Celsius
t^2=2x3.14x0.47/a Or, a=6.28x0.47/1.43^2=0.43 m/sec^2
Square has all sides equal hence 2x-5=x+8 or 2x-x=8+5 (by subtracting x and adding 5 on both sides) Thus x=13 inches. Check: 2x13-5=8+13 or 26-5=21.
Weight is mass m times acceleration due to gravity g, which is given by g=GMm/R^2 where M and R are mass & radius of Earth and G is Universal Gravitational Constant. For new planet, g`=G(0.1Mm/(R/2)^2 =0.1GMm/R^2/4 =GMm(0.1x4)/R^2=0.4GMm/R^2 Hence
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