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JAC OB254
Questions and answers by
JAC OB254
Questions (8)
Prove that one half the sum of any two real numbers is between the two real numbers.
1 answer
95 views
1. Prove that there is no upper bound for the set E=(2,4,6,8,....)
2. Prove that there is no lower bound for the set I of
1 answer
80 views
A rod 2.5m long is supported at A and B. The rod is carrying a load of 5kN at a distance of 1m from A. What are the reactions at
1 answer
77 views
A body of weight 500N rests on a plane inclined at 20 degrees to the horizontal. The coefficient of friction is 0.4, determine a
1 answer
128 views
Derive the formula for calculating the arithmetic series : Sn=n/2(2a+(n-1)d)
2 answers
96 views
Find one rational number and one irrational number between 0.3333.... and 0.4444....
2 answers
61 views
Express -2/3, 2/9, - 2/19, 2/33,... in sequence notation.
1 answer
55 views
If f(x) =1/(x² - 8)²,find the expression for inverse function g(f) =f^(-1) (x)
2 answers
69 views
Answers (80)
A number is divisible by 9 if the sum of its digits is 9 or a multiple of 9 3+7+1=11 The next multiple of 9 nearer to 11 is 18 thus, 18-11=7
The height of the tree above the boy is 12-1.4=10.6 Using tan#= opposite /adjacent where # is the angle of elevation tan#=10.6/10 hence #= 46.67 degrees
Potential energy is lost while kinetic energy is gained.
They are not the same
11/8 /(6) 11/48
A=P(1+r/100)^n =650(1+4/100)^3 =731.16 I=A-P =731.16-650 =81.16
I don't have a lot of money, but if I had a lot of money, I would have travelled.
Find the cube root of 512 =8 inches
(-2,3)+(2,3) =(0,6)
a
2Li(s)+2H2O(l)->2LiOH(aq)+H2(g) RAM of Li=7 moles of Li=3/7 moles mole ratio of Li to H2 is 2:1 moles of H2 =1/2×3/7=3/14 moles mass=moles×molar mass =3/14×2 =0.429g 1mole at STP= 22.4dm^3 3/14moles=?? =3/14×22.4 =4.8dm^3
2KClO3(s)>2KCl(s)+3O2(g) RFM of KClO3 =39+35.5+3×16 =122.5 moles of KClO3 =29.4/122.5 =0.33moles mole ratio of KClO3 to O2 is 2:3 moles of O2 =3/2×0.33= 0.495moles mass=moles×molar mass =0.495×16 =7.92g 1 mole at STP=22.4dm^3 0.495 moles=??
93×7=651 points
the expression is in the form of ax^2+bx+c, where a=9, b=t and c=16 thus, b^2=4ac t^2=4×9×16 t^2=576 t=24
Longitude difference=45-15 =30° 1°=4 min 30°=?? =30×4= 120 min 120/60= 2 hours since Nigeria is to the west of Ethiopia, subtract 2 hours from 5:00p.m 5:00-2:00 =3:00p.m.
(a) 250cm^3=1.4g 1000cm^3=?? =1000/250×1.4 =5.6g/dm^3 molarity=concentration(g/dm^3)/ RFM RFM of KOH=39+16+1=56 Molarity=5.6/56 =0.1M=0.1mol/dm^3 (b) H2SO4(aq)+ 2KOH(aq)> K2SO4(aq)+ 2H2O(l) moles of 25cm^3 of KOH=25/1000×0.1 =0.0025 moles mole ratio of
2/5×55=22km 22km=1 litre 55km=?? =55/22 =2.5 litres
No of moles= molarity×volume =0.85×0.4 =0.34moles RFM of CuSO4= 64+32+4×16= 160 mass=no of moles×RFM =0.34×160 =54.4g
1hr>30sec from mon 0445hr to fri 1845hr is 4 days 14 hours 4×24+14=110 hours 1hr=30sec 110hrs= 110×30 =3300sec 3300/3600=11/12hr 11/12×60=55 min 1845hr-55min =1750hrs =5:50p.m
Where the car maintains a constant speed, that is the maximum speed. v=u+at but u=0m/s since the car starts from rest. v=0+(4×5) v=20m/s maximum speed= 20m/s
-7-5=-12
120 what has tempted you to ask such a question you could use a calculator or mind?
Transmission of impulses
They neither gain nor lose electrons and so their outermost energy level is full.
2SO2(g)+O2(g)>2SO3(g) RFM of SO2= 32+2×16=64 no of mole of SO2 =96/64=1.5moles mole ratio of SO2 to SO3 is 2:2, thus no of moles of SO3 is 1.5moles 1 mole at STP=22.4Litres 1.5 moles=?? =1.5×22.4 =33.6 Litres
14-(-3)^2 14-9 =5
Sn=n/2[2a+(n-1)d] =11/2[2×2+(11-1)3] =11/2[4+30] =11/2×34 =187
0.2
(a) Let the number of pencils bought be x Ruler=x+7 cost of pencils=0.7x cost of rulers= 0.4(x+7)=0.4x+2.8 0.4x+2.8=0.7x-1.1 2.8+1.1=0.7x-0.4x 3.9=0.3x x=13 number of pencils bought= 13 (b) number of rulers bought=13+7=20 cost=20(0.40) =$8.00
Let the money Michael had first=x Bryant=60/100x= 3/5x let the bill paid for lunch=y Bryant=25/100y= 1/4y Michael=75/100y= 3/4y 3/5x-1/4y=56 12x-5y=1120 ....eqn(i) x-3/4y=25/100x x-3/4y=1/4x 4x-3y=x 3x-3y=0....eqn(ii) solve by elimination method by
Let the cost of one pen=x 2 pens=2x file=2(2x) =4x 4x:x =4:1
Let the number of boys be x girls=x+40 girls left the hall= 2/5(x+40)= 2/5x+16 boys left the hall= 1/3x girls remained in the hall= x+40-(2/5x+16) =x+40-2/5x-16 =3/5x+24 boys remained in the hall= x-1/3x=2/3x 3/5x+24=2/3x+18 24-18=2/3x-3/5x 6=1/15x x=6×15
80000/800=100
(13-4)/(3-0) 9/3=3
F=ma a=F/m a=8/2 a=4m/s^2 F=(2+1)4 F=12N
Molecular mass of gas x=7×2=14
nth term=a+(n-1)d 10=a+(4-1)d 10=a+3d a+3d=10....eqn(i) 16=a+(7-1)d 16=a+6d a+6d=16....eqn(ii) a+3d=10 a+6d=16 solve using elimination method -3d=-6 d=2 a+3×2=10 a+6=10 a=4 Sn=n/2[2a+(n-1)d] =9/2[2×4+(9-1)2] =9/2[8+16] =9/2×24 =108
Six hundred and ninety one and four tenths
volume of the tank =66×35×45 =103,950cm^3 rate= 103950cm^3/18 min =5775cm^3/min
P=2(l+w) =2(1750+1500) =2(3250) =6500m 1000m=1km 6500m=?? =6500/1000 =6.5km
sin 30= x/36 x=36sin 30 x=18km 18-7=11km =11km east from the starting point
(a) v^2=u^2-2gs but v=0 0=20^2-(2×10×s) 0=400-20s 20s=400 s=20m (b) v=u-gt but v=0 0=20-(10×t) 0=20-10t 10t=20 t=2 seconds (c) v=u+gt when the stone reaches the maximum height, it returns and experiences free fall and u becomes zero (u=0m/s) v=0+10×2
Let the original number of bags be b new number of bags after the discount= 160/100b =1.6b total price for b bags before discount= 40b 100%=40b 144%=?? 144/100×40b =57.6b this is the price after the discount price of each bag during sale= 57.6b/1.6b =$36
LCM of 9, 15 & 15 is 45 1/9×45=5 2/15×45=6 3/15×45=9 1/9, 2/15, 3/15
Are waves that require a material medium for transmission.
Relative density of a substance= density of that substance/ density of water 7.8=density of steel/ 1000 density of steel=7.8×1000 =7800kg/m^3 volume=10×10×10 =1000cm^3 1000/1000000= 0.001m^3 mass=density×volume =7800×0.001 =7.8kg
Let Amina's age=x Ado=4x x+4x=20 5x=20 x=4 thus, Amina is 4 years old and Ado is 16 years old 16-4=12 years
a) A=#r^2 =22/7×3.5×3.5 =38.5cm^2 b) curved surface area=2#rh =2×22/7×3.5×20 =440cm^2 c)area of the hemispherical part= 2#r^2 =2×22/7×3.5×3.5 =77cm^2 total surface area= 38.5+440+77 =555.5cm^2
Time=distance/speed =6/3×10^8 =2×10^-9seconds
16-2(15-1) 16-2(14) 16-28 =-12
